INSERT INTO `exam_simulation2` (`id`, `examYear`, `questions`, `parent_que`, `child_que`, `gen_child_que_desc`, `imageDir`, `explanation`, `parent_que_expl`, `child_que_exp`, `gen_child_que_expl`, `package_id`, `subject_id`, `status`) VALUES
(572, '2016', '\n\nQuestion\n\nThe cost of maintaining a school is partly constant and partly varies as the number of pupils. With 50 pupils, the cost is$15,705.00 and with 40 pupils, it is$13,305.00.\n(a) Find the cost when there are 44 pupils.\n(b) If the fee per pupil is$360.00, what is the least number of pupils for which the school can run without a loss?\n\n', '', 'The cost of maintaining a school is partly constant and partly varies as the number of pupils. With 50 pupils, the cost is$15,705.00 and with 40 pupils, it is$13,305.00.\n(a) Find the cost when there are 44 pupils.\n(b) If the fee per pupil is$360.00, what is the least number of pupils for which the school can run without a loss?', 'The cost of maintaining a school is partly constant and partly varies as the number of pupils. With 50 pupils, the cost is$15,705.00 and with 40 pupils, it is$13,305.00.\n(a) Find the cost when there are 44 pupils.~(b) If the fee per pupil is$360.00, what is the least number of pupils for which the school can run without a loss?', NULL, '\n\nExplanation\nLet P represent the cost ; n, the number of pupils and k and c,constant terms.\n\\(\\therefore P = k + cn\\)\n\\(15705 = k + 50c ... (1)\\)\n\\(13305 = k + 40c .... (2)\\)\n(1) - (2) :\n\\(2400 = 10c \\implies c = \\frac{2400}{10} = 240\\)\nPut c = 240 in (1), so that\n\\(15705 = k + (50)(240)\\)\n\\(15705 = k + 12000 \\implies k = 3705\\)\n\\(P = 3705 + 240n\\) ------ formula connecting price and number of pupils.\n(a) When number of pupils, n = 44\n\\(P = 3705 + 240(44) = 3705 + 10560 = $14265\\)\n(b) For school not to be at loss:\n\\(360 \\geq k + cn\\)\n\\(360n \\geq 3705 + 240n\\)\n\\(360n - 240n \\geq 3705 \\implies 120n \\geq 3705\\)\n\\(n \\geq \\frac{3705}{120}\\)\n\\(n \\geq 30.875 \\)\n\\(n \\geq 31\\)\nThe pupils should be 31 at least for the school not to run at loss.\n\n', '', 'Explanation\nLet P represent the cost ; n, the number of pupils and k and c,constant terms.\n\\(\\therefore P = k + cn\\)\n\\(15705 = k + 50c ... (1)\\)\n\\(13305 = k + 40c .... (2)\\)\n(1) - (2) :\n\\(2400 = 10c \\implies c = \\frac{2400}{10} = 240\\)\nPut c = 240 in (1), so that\n\\(15705 = k + (50)(240)\\)\n\\(15705 = k + 12000 \\implies k = 3705\\)\n\\(P = 3705 + 240n\\) ------ formula connecting price and number of pupils.\n(a) When number of pupils, n = 44\n\\(P = 3705 + 240(44) = 3705 + 10560 = $14265\\)\n(b) For school not to be at loss:\n\\(360 \\geq k + cn\\)\n\\(360n \\geq 3705 + 240n\\)\n\\(360n - 240n \\geq 3705 \\implies 120n \\geq 3705\\)\n\\(n \\geq \\frac{3705}{120}\\)\n\\(n \\geq 30.875 \\)\n\\(n \\geq 31\\)\nThe pupils should be 31 at least for the school not to run at loss.\n\n', '(a) When number of pupils, n = 44\n\\(P = 3705 + 240(44) = 3705 + 10560 = $14265\\)~(b) For school not to be at loss:\n\\(360 \\geq k + cn\\)\n\\(360n \\geq 3705 + 240n\\)\n\\(360n - 240n \\geq 3705 \\implies 120n \\geq 3705\\)\n\\(n \\geq \\frac{3705}{120}\\)\n\\(n \\geq 30.875 \\)\n\\(n \\geq 31\\)\nThe pupils should be 31 at least for the school not to run at loss.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(573, '2016', '\n\nQuestion\n\nOn a graph sheet, using a scale of 2cm to 2 units on both axes,\n(a) Draw the straight line joining points P(-5, 3) and Q(2, 3);\n(b) construct the locus L of points equidistant from P and Q;\n(c) by construction, locate points R and S on L, such that PRQS forms a rhombus of sides 5cm;\n(d) find : (i) coordinates of R and S; (ii) area of the rhombus in cm\\(^{2}\\).\n\n', '', 'On a graph sheet, using a scale of 2cm to 2 units on both axes,\n(a) Draw the straight line joining points P(-5, 3) and Q(2, 3);\n(b) construct the locus L of points equidistant from P and Q;\n(c) by construction, locate points R and S on L, such that PRQS forms a rhombus of sides 5cm;\n(d) find : (i) coordinates of R and S; (ii) area of the rhombus in cm\\(^{2}\\).', 'On a graph sheet, using a scale of 2cm to 2 units on both axes,\n(a) Draw the straight line joining points P(-5, 3) and Q(2, 3);~(b) construct the locus L of points equidistant from P and Q;\n(c) by construction, locate points R and S on L, such that PRQS forms a rhombus of sides 5cm;\n(d) find : (i) coordinates of R and S; (ii) area of the rhombus in cm\\(^{2}\\).~(c) by construction, locate points R and S on L, such that PRQS forms a rhombus of sides 5cm;\n(d) find : (i) coordinates of R and S; (ii) area of the rhombus in cm\\(^{2}\\).~(d) find : (i) coordinates of R and S; (ii) area of the rhombus in cm\\(^{2}\\).', NULL, '\n\nExplanation\nCoordinate of R and S : R(-1.6, 6.5) and S(-1.6, 0.6).\nArea of rhombus PRQS : \\(\\frac{(5 + 2) \\times (6.5 + 0.6)}{2} = \\frac{49.7}{2}\\)\n= \\(24.85 cm^{2}\\)\n\n', 'Coordinate of R and S : R(-1.6, 6.5) and S(-1.6, 0.6).\nArea of rhombus PRQS : \\(\\frac{(5 + 2) \\times (6.5 + 0.6)}{2} = \\frac{49.7}{2}\\)\n= \\(24.85 cm^{2}\\)\n\n', '', '', 'WAEC-PK-1', 'WAEC-MATH', 1),
(574, '2016', '\n\nQuestion\n\n(a) A shop owner marked a shirt at a price to enable him to make a gain of 20%. During a special sales period, the shirt was sold at 10% reduction to a customer at N864.00. What was the original cost to the shop owner?\n(b) A rectangular lawn of length (x + 5) metres is (x - 2) metres wide. If the diagonal is (x + 6) metres, find ;\n(i) the value of x ; (ii) the area of lawn.\n\n', '', '(a) A shop owner marked a shirt at a price to enable him to make a gain of 20%. During a special sales period, the shirt was sold at 10% reduction to a customer at N864.00. What was the original cost to the shop owner?\n(b) A rectangular lawn of length (x + 5) metres is (x - 2) metres wide. If the diagonal is (x + 6) metres, find ;\n(i) the value of x ; (ii) the area of lawn.', '(a) A shop owner marked a shirt at a price to enable him to make a gain of 20%. During a special sales period, the shirt was sold at 10% reduction to a customer at N864.00. What was the original cost to the shop owner?~(b) A rectangular lawn of length (x + 5) metres is (x - 2) metres wide. If the diagonal is (x + 6) metres, find ;\n(i) the value of x ; (ii) the area of lawn.', NULL, '\n\nExplanation\n(a) Price with 20% gain = 100% = x\nSelling price = 100% - 10% = 90%\ni.e. 90% of x = 864\n\\(\\therefore x = 864 \\times \\frac{100}{90} = N960\\)\nLet cost price = y = 100%\nx = 20% of y + y = 120% of y.\n\\(y = \\frac{100}{120} x = \\frac{100}{120} \\times N960 = N800\\)\n(b)\n(i) \\((x + 6)^{2} = (x + 5)^{2} + (x - 2)^{2}\\)\n\\(x^{2} + 12x + 36 = x^{2} + 10x + 25 + x^{2} - 4x + 4\\)\n\\(x^{2} + 12x + 36 = 2x^{2} + 6x + 29\\)\n\\(2x^{2} - x^{2} + 6x - 12x + 29 - 36 = 0\\)\n\\(x^{2} - 6x - 7 = 0\\)\n\\(x^{2} - 7x + x - 7 = 0 \\implies x(x - 7) + 1(x - 7) = 0\\)\n\\((x - 7)(x + 1) = 0 \\implies \\text{x = 7 or -1}\\)\nSince measurements cannot be negative, then x = 7 is the suitable answer.\n(ii) Length of the lawn = (x + 5) metres = (7 + 5) = 12 metres.\nWidth of the lawn = (x - 2) metres = (7 - 2) = 5 metres\n\\(\\therefore \\text{The area of the lawn} = 12 \\times 5 = 60 m^{2}\\)\n\n', '', 'Explanation\n(a) Price with 20% gain = 100% = x\nSelling price = 100% - 10% = 90%\ni.e. 90% of x = 864\n\\(\\therefore x = 864 \\times \\frac{100}{90} = N960\\)\nLet cost price = y = 100%\nx = 20% of y + y = 120% of y.\n\\(y = \\frac{100}{120} x = \\frac{100}{120} \\times N960 = N800\\)\n(b)\n(i) \\((x + 6)^{2} = (x + 5)^{2} + (x - 2)^{2}\\)\n\\(x^{2} + 12x + 36 = x^{2} + 10x + 25 + x^{2} - 4x + 4\\)\n\\(x^{2} + 12x + 36 = 2x^{2} + 6x + 29\\)\n\\(2x^{2} - x^{2} + 6x - 12x + 29 - 36 = 0\\)\n\\(x^{2} - 6x - 7 = 0\\)\n\\(x^{2} - 7x + x - 7 = 0 \\implies x(x - 7) + 1(x - 7) = 0\\)\n\\((x - 7)(x + 1) = 0 \\implies \\text{x = 7 or -1}\\)\nSince measurements cannot be negative, then x = 7 is the suitable answer.\n(ii) Length of the lawn = (x + 5) metres = (7 + 5) = 12 metres.\nWidth of the lawn = (x - 2) metres = (7 - 2) = 5 metres\n\\(\\therefore \\text{The area of the lawn} = 12 \\times 5 = 60 m^{2}\\)\n\n', '(a) Price with 20% gain = 100% = x\nSelling price = 100% - 10% = 90%\ni.e. 90% of x = 864\n\\(\\therefore x = 864 \\times \\frac{100}{90} = N960\\)\nLet cost price = y = 100%\nx = 20% of y + y = 120% of y.\n\\(y = \\frac{100}{120} x = \\frac{100}{120} \\times N960 = N800\\)~(b)\n(i) \\((x + 6)^{2} = (x + 5)^{2} + (x - 2)^{2}\\)\n\\(x^{2} + 12x + 36 = x^{2} + 10x + 25 + x^{2} - 4x + 4\\)\n\\(x^{2} + 12x + 36 = 2x^{2} + 6x + 29\\)\n\\(2x^{2} - x^{2} + 6x - 12x + 29 - 36 = 0\\)\n\\(x^{2} - 6x - 7 = 0\\)\n\\(x^{2} - 7x + x - 7 = 0 \\implies x(x - 7) + 1(x - 7) = 0\\)\n\\((x - 7)(x + 1) = 0 \\implies \\text{x = 7 or -1}\\)\nSince measurements cannot be negative, then x = 7 is the suitable answer.\n(ii) Length of the lawn = (x + 5) metres = (7 + 5) = 12 metres.\nWidth of the lawn = (x - 2) metres = (7 - 2) = 5 metres\n\\(\\therefore \\text{The area of the lawn} = 12 \\times 5 = 60 m^{2}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(575, '2016', '\n\nQuestion\n\n(a) Copy and complete the following table of values for \\(y = 9 \\cos x + 5 \\sin x\\) to one decimal place.\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n\n\ny\n\n10.3\n\n\n-0.2\n-5.3\n\n-10.3\n\n\n\n(b) Using a scale of 2cm to 30° on the x- axis and 2 cm to 1 unit on the y- axis, draw the graph of \\(y = 9 \\cos x + 5 \\sin x\\) for \\(0° \\leq x \\leq 210°\\).\n(c) Use your graph to solve the equation: (i) \\(9\\cos x + 5\\sin x = 0\\); (ii) \\(9\\cos x+ 5\\sin x = 3.5\\), correct to the nearest degree.\n(d) Find the maximum value of y correct to one decimal place.\n\n\n\n(a) Copy and complete the following table of values for \\(y = 9 \\cos x + 5 \\sin x\\) to one decimal place.\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n\n\ny\n\n10.3\n\n\n-0.2\n-5.3\n\n-10.3\n\n\n', '', '(a) Copy and complete the following table of values for \\(y = 9 \\cos x + 5 \\sin x\\) to one decimal place.\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n\n\ny\n\n10.3\n\n\n-0.2\n-5.3\n\n-10.3\n\n\n\n(b) Using a scale of 2cm to 30° on the x- axis and 2 cm to 1 unit on the y- axis, draw the graph of \\(y = 9 \\cos x + 5 \\sin x\\) for \\(0° \\leq x \\leq 210°\\).\n(c) Use your graph to solve the equation: (i) \\(9\\cos x + 5\\sin x = 0\\); (ii) \\(9\\cos x+ 5\\sin x = 3.5\\), correct to the nearest degree.\n(d) Find the maximum value of y correct to one decimal place.\n\n\n\n(a) Copy and complete the following table of values for \\(y = 9 \\cos x + 5 \\sin x\\) to one decimal place.\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n\n\ny\n\n10.3\n\n\n-0.2\n-5.3\n\n-10.3\n', '(a) Copy and complete the following table of values for \\(y = 9 \\cos x + 5 \\sin x\\) to one decimal place.\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n\n\ny\n\n10.3\n\n\n-0.2\n-5.3\n\n-10.3\n\n\n~(b) Using a scale of 2cm to 30° on the x- axis and 2 cm to 1 unit on the y- axis, draw the graph of \\(y = 9 \\cos x + 5 \\sin x\\) for \\(0° \\leq x \\leq 210°\\).\n(c) Use your graph to solve the equation: (i) \\(9\\cos x + 5\\sin x = 0\\); (ii) \\(9\\cos x+ 5\\sin x = 3.5\\), correct to the nearest degree.\n(d) Find the maximum value of y correct to one decimal place.\n\n\n\n(a) Copy and complete the following table of values for \\(y = 9 \\cos x + 5 \\sin x\\) to one decimal place.\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n\n\ny\n\n10.3\n\n\n-0.2\n-5.3\n\n-10.3\n~(c) Use your graph to solve the equation: (i) \\(9\\cos x + 5\\sin x = 0\\); (ii) \\(9\\cos x+ 5\\sin x = 3.5\\), correct to the nearest degree.\n(d) Find the maximum value of y correct to one decimal place.\n\n\n\n(a) Copy and complete the following table of values for \\(y = 9 \\cos x + 5 \\sin x\\) to one decimal place.\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n\n\ny\n\n10.3\n\n\n-0.2\n-5.3\n\n-10.3\n~(d) Find the maximum value of y correct to one decimal place.\n\n\n\n(a) Copy and complete the following table of values for \\(y = 9 \\cos x + 5 \\sin x\\) to one decimal place.\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n\n\ny\n\n10.3\n\n\n-0.2\n-5.3\n\n-10.3\n', NULL, '\n\nExplanation\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n\n\ny\n9\n10.3\n8.8\n5\n-0.2\n-5.3\n-9\n-10.3\n\n\n\n(b)\n(c)(i) From the graph, \\(9\\cos x + 5\\sin x = 0 \\implies x = 119°\\)\n(d) Maximum value of y = 10.3.\n\n', '', 'Explanation\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n\n\ny\n9\n10.3\n8.8\n5\n-0.2\n-5.3\n-9\n-10.3\n\n\n\n(b)\n(c)(i) From the graph, \\(9\\cos x + 5\\sin x = 0 \\implies x = 119°\\)\n(d) Maximum value of y = 10.3.\n\n', '~(b)\n(c)(i) From the graph, \\(9\\cos x + 5\\sin x = 0 \\implies x = 119°\\)\n(d) Maximum value of y = 10.3.\n~(c)(i) From the graph, \\(9\\cos x + 5\\sin x = 0 \\implies x = 119°\\)\n(d) Maximum value of y = 10.3.\n~(d) Maximum value of y = 10.3.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(576, '2016', '\n\nQuestion\n\nA, B and C are subsets of the universal set U such that : \\(U = {0, 1, 2, 3,..., 12}; A = {x : 0 \\leq x \\leq 7}; B = {4, 6, 8, 10, 12}; C = {1 < y < 8}\\), where y is a prime number.\n(a) Draw a venn diagram to illustrate the information given above;\n(b) Find: (i) \\((B \\cup C)\'\\); (ii) \\(A\'\\cap B \\cap C\\).\n\n', '', 'A, B and C are subsets of the universal set U such that : \\(U = {0, 1, 2, 3,..., 12}; A = {x : 0 \\leq x \\leq 7}; B = {4, 6, 8, 10, 12}; C = {1 < y < 8}\\), where y is a prime number.\n(a) Draw a venn diagram to illustrate the information given above;\n(b) Find: (i) \\((B \\cup C)\'\\); (ii) \\(A\'\\cap B \\cap C\\).', 'A, B and C are subsets of the universal set U such that : \\(U = {0, 1, 2, 3,..., 12}; A = {x : 0 \\leq x \\leq 7}; B = {4, 6, 8, 10, 12}; C = {1 < y < 8}\\), where y is a prime number.\n(a) Draw a venn diagram to illustrate the information given above;~(b) Find: (i) \\((B \\cup C)\'\\); (ii) \\(A\'\\cap B \\cap C\\).', NULL, '\n\nExplanation\n(a)\n(b) A = {0, 1, 2, 3, 4, 5, 6, 7}\nB = {4, 6, 8, 10, 12}\nC = {2, 3, 5, 7}\nU = {0,1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}\n(i) \\(B \\cup C = {2, 3, 4, 5, 6, 7, 8, 10, 12}\\)\n \\((B \\cup C)\' = {0, 1, 9, 11}\\)\n(ii) \\(A\' = {8, 9, 10, 11, 12}\\)\n\\(A\' \\cap B \\cap C = { }\\)\n\n', '', 'Explanation\n(a)\n(b) A = {0, 1, 2, 3, 4, 5, 6, 7}\nB = {4, 6, 8, 10, 12}\nC = {2, 3, 5, 7}\nU = {0,1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}\n(i) \\(B \\cup C = {2, 3, 4, 5, 6, 7, 8, 10, 12}\\)\n \\((B \\cup C)\' = {0, 1, 9, 11}\\)\n(ii) \\(A\' = {8, 9, 10, 11, 12}\\)\n\\(A\' \\cap B \\cap C = { }\\)\n\n', '~(b) A = {0, 1, 2, 3, 4, 5, 6, 7}\nB = {4, 6, 8, 10, 12}\nC = {2, 3, 5, 7}\nU = {0,1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}\n(i) \\(B \\cup C = {2, 3, 4, 5, 6, 7, 8, 10, 12}\\)\n \\((B \\cup C)\' = {0, 1, 9, 11}\\)\n(ii) \\(A\' = {8, 9, 10, 11, 12}\\)\n\\(A\' \\cap B \\cap C = { }\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(577, '2016', '\n\nQuestion\n\n(a) Given that \\(\\log_{10} 2 = 0.3010, \\log_{10} 7 = 0.8451\\) and \\(\\log_{10} 5 = 0.6990\\), evaluate without using logarithm tables:\n(i) \\(\\log_{10} 35\\); (ii) \\(\\log_{10} 2.8\\).\n(b) Given that \\(N^{0.8942} = 2.8\\), use your result in (a)(ii) to find the value of N.\n\n', '', '(a) Given that \\(\\log_{10} 2 = 0.3010, \\log_{10} 7 = 0.8451\\) and \\(\\log_{10} 5 = 0.6990\\), evaluate without using logarithm tables:\n(i) \\(\\log_{10} 35\\); (ii) \\(\\log_{10} 2.8\\).\n(b) Given that \\(N^{0.8942} = 2.8\\), use your result in (a)(ii) to find the value of N.', '(a) Given that \\(\\log_{10} 2 = 0.3010, \\log_{10} 7 = 0.8451\\) and \\(\\log_{10} 5 = 0.6990\\), evaluate without using logarithm tables:\n(i) \\(\\log_{10} 35\\); (ii) \\(\\log_{10} 2.8\\).~(b) Given that \\(N^{0.8942} = 2.8\\), use your result in (a)(ii) to find the value of N.', NULL, '\n\nExplanation\n(a)(i) \\(\\log_{10} 35 = \\log_{10} (7 \\times 5) = \\log_{10} 7 + \\log_{10} 5\\)\n= \\(0.8451 + 0.6990 = 1.5441\\)\n(ii) \\(\\log_{10} 2.8 = \\log_{10} (\\frac{4 \\times 7}{10} = \\log_{10} 4 + \\log_{10} 7 - \\log_{10} 10\\)\n= \\(\\log_{10} 2^{2} + \\log_{10} 7 - 1\\)\n= \\(2\\log_{10} 2 + \\log_{10} 7 - 1 = 2(0.3010) + 0.8451- 1\\)\n= \\(0.4471\\)\n(b) \\(N^{0.8942} = 2.8\\)\n\\(0.8942 \\log N = \\log 2.8\\)\n\\(\\log N = \\frac{\\log 2.8}{0.8942} = \\frac{0.4471}{0.8942}\\)\n\\(\\log N = 0.5\\)\n\n', '', 'Explanation\n(a)(i) \\(\\log_{10} 35 = \\log_{10} (7 \\times 5) = \\log_{10} 7 + \\log_{10} 5\\)\n= \\(0.8451 + 0.6990 = 1.5441\\)\n(ii) \\(\\log_{10} 2.8 = \\log_{10} (\\frac{4 \\times 7}{10} = \\log_{10} 4 + \\log_{10} 7 - \\log_{10} 10\\)\n= \\(\\log_{10} 2^{2} + \\log_{10} 7 - 1\\)\n= \\(2\\log_{10} 2 + \\log_{10} 7 - 1 = 2(0.3010) + 0.8451- 1\\)\n= \\(0.4471\\)\n(b) \\(N^{0.8942} = 2.8\\)\n\\(0.8942 \\log N = \\log 2.8\\)\n\\(\\log N = \\frac{\\log 2.8}{0.8942} = \\frac{0.4471}{0.8942}\\)\n\\(\\log N = 0.5\\)\n\n', '(a)(i) \\(\\log_{10} 35 = \\log_{10} (7 \\times 5) = \\log_{10} 7 + \\log_{10} 5\\)\n= \\(0.8451 + 0.6990 = 1.5441\\)\n(ii) \\(\\log_{10} 2.8 = \\log_{10} (\\frac{4 \\times 7}{10} = \\log_{10} 4 + \\log_{10} 7 - \\log_{10} 10\\)\n= \\(\\log_{10} 2^{2} + \\log_{10} 7 - 1\\)\n= \\(2\\log_{10} 2 + \\log_{10} 7 - 1 = 2(0.3010) + 0.8451- 1\\)\n= \\(0.4471\\)~(b) \\(N^{0.8942} = 2.8\\)\n\\(0.8942 \\log N = \\log 2.8\\)\n\\(\\log N = \\frac{\\log 2.8}{0.8942} = \\frac{0.4471}{0.8942}\\)\n\\(\\log N = 0.5\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(578, '2016', '\n\nQuestion\n\n(a) The value of the expression \\(2Ax - Kx^{2}\\) is 7 when x = 1 and 4 when x = 2. Find the values of the constants A and K.\n(b) Solve the equation \\(x^{2} - 3x - 1 = 0\\), giving your answers correct to 1 decimal place.\n\n', '', '(a) The value of the expression \\(2Ax - Kx^{2}\\) is 7 when x = 1 and 4 when x = 2. Find the values of the constants A and K.\n(b) Solve the equation \\(x^{2} - 3x - 1 = 0\\), giving your answers correct to 1 decimal place.', '(a) The value of the expression \\(2Ax - Kx^{2}\\) is 7 when x = 1 and 4 when x = 2. Find the values of the constants A and K.~(b) Solve the equation \\(x^{2} - 3x - 1 = 0\\), giving your answers correct to 1 decimal place.', NULL, '\n\nExplanation\n(a) \\(2Ax - Kx^{2}\\)\nWhen x = 1, \\(2A(1) - K(1^{2}) = 7\\)\n\\(2A - K = 7 ... (1)\\)\nWhen x = 2, \\(2A(2) - K(2^{2}) = 4\\)\n\\(4A - 4K = 4 \\implies A - K = 1 ... (2)\\)\nFrom (2), \\(A = 1 + K\\)\n(1) becomes : \\(2(1 + K) - K = 7\\)\n\\(2 + 2K - K = 7 \\implies 2 + K = 7\\)\n\\(\\therefore K = 5\\)\n\\(A = 1 + K = 1 + 5 = 6\\)\n\n\\(A, K = 6, 5\\)\n(b) \\(x^{2} - 3x - 1 = 0\\)\n\\(a = 1, b = -3, c = -1\\)\nUsing the quadratic formula,\n\\(x = \\frac{-b \\pm \\sqrt{b^{2} - 4ac}}{2a}\\)\n\\(x = \\frac{-(-3) \\pm \\sqrt{(-3)^{2} - 4(1)(-3)}}{2(1)}\\)\n\\(x = \\frac{3 \\pm \\sqrt{9 + 12}}{2}\\)\n\\(x = \\frac{3 \\pm \\sqrt{21}}{2} = \\frac{3 \\pm 4.583}{2}\\)\n\\(x = \\frac{3 + 4.583}{2} ; x = \\frac{3 - 4.583}{2}\\)\n\\(x = \\frac{7.583}{2} = 3.7915 \\approxeq 3.8\\) (to 1 d.p)\n\\(x = \\frac{-1.583}{2} = -0.7915 \\approxeq -0.8\\) (to 1 d.p)\n\n', '', 'Explanation\n(a) \\(2Ax - Kx^{2}\\)\nWhen x = 1, \\(2A(1) - K(1^{2}) = 7\\)\n\\(2A - K = 7 ... (1)\\)\nWhen x = 2, \\(2A(2) - K(2^{2}) = 4\\)\n\\(4A - 4K = 4 \\implies A - K = 1 ... (2)\\)\nFrom (2), \\(A = 1 + K\\)\n(1) becomes : \\(2(1 + K) - K = 7\\)\n\\(2 + 2K - K = 7 \\implies 2 + K = 7\\)\n\\(\\therefore K = 5\\)\n\\(A = 1 + K = 1 + 5 = 6\\)\n\n\\(A, K = 6, 5\\)\n(b) \\(x^{2} - 3x - 1 = 0\\)\n\\(a = 1, b = -3, c = -1\\)\nUsing the quadratic formula,\n\\(x = \\frac{-b \\pm \\sqrt{b^{2} - 4ac}}{2a}\\)\n\\(x = \\frac{-(-3) \\pm \\sqrt{(-3)^{2} - 4(1)(-3)}}{2(1)}\\)\n\\(x = \\frac{3 \\pm \\sqrt{9 + 12}}{2}\\)\n\\(x = \\frac{3 \\pm \\sqrt{21}}{2} = \\frac{3 \\pm 4.583}{2}\\)\n\\(x = \\frac{3 + 4.583}{2} ; x = \\frac{3 - 4.583}{2}\\)\n\\(x = \\frac{7.583}{2} = 3.7915 \\approxeq 3.8\\) (to 1 d.p)\n\\(x = \\frac{-1.583}{2} = -0.7915 \\approxeq -0.8\\) (to 1 d.p)\n\n', '(a) \\(2Ax - Kx^{2}\\)\nWhen x = 1, \\(2A(1) - K(1^{2}) = 7\\)\n\\(2A - K = 7 ... (1)\\)\nWhen x = 2, \\(2A(2) - K(2^{2}) = 4\\)\n\\(4A - 4K = 4 \\implies A - K = 1 ... (2)\\)\nFrom (2), \\(A = 1 + K\\)\n(1) becomes : \\(2(1 + K) - K = 7\\)\n\\(2 + 2K - K = 7 \\implies 2 + K = 7\\)\n\\(\\therefore K = 5\\)\n\\(A = 1 + K = 1 + 5 = 6\\)\n\n\\(A, K = 6, 5\\)~(b) \\(x^{2} - 3x - 1 = 0\\)\n\\(a = 1, b = -3, c = -1\\)\nUsing the quadratic formula,\n\\(x = \\frac{-b \\pm \\sqrt{b^{2} - 4ac}}{2a}\\)\n\\(x = \\frac{-(-3) \\pm \\sqrt{(-3)^{2} - 4(1)(-3)}}{2(1)}\\)\n\\(x = \\frac{3 \\pm \\sqrt{9 + 12}}{2}\\)\n\\(x = \\frac{3 \\pm \\sqrt{21}}{2} = \\frac{3 \\pm 4.583}{2}\\)\n\\(x = \\frac{3 + 4.583}{2} ; x = \\frac{3 - 4.583}{2}\\)\n\\(x = \\frac{7.583}{2} = 3.7915 \\approxeq 3.8\\) (to 1 d.p)\n\\(x = \\frac{-1.583}{2} = -0.7915 \\approxeq -0.8\\) (to 1 d.p)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(579, '2016', '\n\nQuestion\n\nThe area of a rectangular floor is 13.5m\\(^{2}\\). One side is 1.5m longer than the other.\n(a) Calculate the dimensions of the floor ;\n(b) If it costs N250.00 per square metre to carpet the floor and only N2,000.00 is available, what area of the floor can be covered with carpet?\n\n', '', 'The area of a rectangular floor is 13.5m\\(^{2}\\). One side is 1.5m longer than the other.\n(a) Calculate the dimensions of the floor ;\n(b) If it costs N250.00 per square metre to carpet the floor and only N2,000.00 is available, what area of the floor can be covered with carpet?', 'The area of a rectangular floor is 13.5m\\(^{2}\\). One side is 1.5m longer than the other.\n(a) Calculate the dimensions of the floor ;~(b) If it costs N250.00 per square metre to carpet the floor and only N2,000.00 is available, what area of the floor can be covered with carpet?', NULL, '\n\nExplanation\nLet one of the sides be x metres.\nThe other is (x + 1.5) metres.\n(a) \\(Area = Length \\times breadth\\)\n\\(x(x + 1.5) = 13.5\\)\n\\(x^{2} + 1.5x = 13.5 \\implies x^{2} + 1.5x - 13.5 = 0\\)\n\\(x^{2} + 4.5x - 3x - 13.5 = 0 \\implies x(x + 4.5) - 3(x + 4.5) = 0\\)\n\\((x - 3)(x + 4.5) = 0 \\implies x = 3 ; x = -4.5\\)\nSince x cannot be negative, therefore x = 3m.\nThe longer side = (3 + 1.5) = 4.5m\n(b) \\(N250 = 1m^{2}\\)\n\\(N2000 = \\frac{2000}{250} = 8m^{2}\\)\n\\(\\therefore \\text{The area of the floor that will be covered by the carpet} = 8m^{2}\\)\n\n', '', 'Explanation\nLet one of the sides be x metres.\nThe other is (x + 1.5) metres.\n(a) \\(Area = Length \\times breadth\\)\n\\(x(x + 1.5) = 13.5\\)\n\\(x^{2} + 1.5x = 13.5 \\implies x^{2} + 1.5x - 13.5 = 0\\)\n\\(x^{2} + 4.5x - 3x - 13.5 = 0 \\implies x(x + 4.5) - 3(x + 4.5) = 0\\)\n\\((x - 3)(x + 4.5) = 0 \\implies x = 3 ; x = -4.5\\)\nSince x cannot be negative, therefore x = 3m.\nThe longer side = (3 + 1.5) = 4.5m\n(b) \\(N250 = 1m^{2}\\)\n\\(N2000 = \\frac{2000}{250} = 8m^{2}\\)\n\\(\\therefore \\text{The area of the floor that will be covered by the carpet} = 8m^{2}\\)\n\n', '(a) \\(Area = Length \\times breadth\\)\n\\(x(x + 1.5) = 13.5\\)\n\\(x^{2} + 1.5x = 13.5 \\implies x^{2} + 1.5x - 13.5 = 0\\)\n\\(x^{2} + 4.5x - 3x - 13.5 = 0 \\implies x(x + 4.5) - 3(x + 4.5) = 0\\)\n\\((x - 3)(x + 4.5) = 0 \\implies x = 3 ; x = -4.5\\)\nSince x cannot be negative, therefore x = 3m.\nThe longer side = (3 + 1.5) = 4.5m~(b) \\(N250 = 1m^{2}\\)\n\\(N2000 = \\frac{2000}{250} = 8m^{2}\\)\n\\(\\therefore \\text{The area of the floor that will be covered by the carpet} = 8m^{2}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(580, '2016', '\n\nQuestion\n\n(a) A number is selected at random from each of the sets {2, 3, 4} and {1, 3, 5}. What is the probability that the sum of the two numbers will be less than 7 but greater than 3?\n(b)\nIn the diagram, ABCD is a circle. DAE, CBE, ABF and DCF are straight lines. If y + m = 90°, find the value of x.\n\n', '', '(a) A number is selected at random from each of the sets {2, 3, 4} and {1, 3, 5}. What is the probability that the sum of the two numbers will be less than 7 but greater than 3?\n(b)\nIn the diagram, ABCD is a circle. DAE, CBE, ABF and DCF are straight lines. If y + m = 90°, find the value of x.', '(a) A number is selected at random from each of the sets {2, 3, 4} and {1, 3, 5}. What is the probability that the sum of the two numbers will be less than 7 but greater than 3?~(b)\nIn the diagram, ABCD is a circle. DAE, CBE, ABF and DCF are straight lines. If y + m = 90°, find the value of x.', NULL, '\n\nExplanation\n(a)\n\n\n\n+\n1\n3\n5\n\n\n2\n3\n5\n7\n\n\n3\n4\n6\n8\n\n\n4\n5\n7\n9\n\n\n\nSample space = 9\nSum greater than 3 but less than 7 = {5, 4, 6, 5} = 4\nP(sum greater than 3 but less than 7) = \\(\\frac{4}{9}\\).\n(b) Given that y + m = 90°\n< ABE = x° (vertically opp. angles)\n< BAD = (x + m)° (exterior angles)\n< DCB = (x + y)° (exterior angles)\n(x + m)° + (x + y)° = 180° (opp. angles of a cyclic quadrilateral)\n(2x + y + m)° = 180°\n(2x + 90) = 180\n2x = 180 - 90 = 90°\nx = 45°\n\n\n', '', 'Explanation\n(a)\n\n\n\n+\n1\n3\n5\n\n\n2\n3\n5\n7\n\n\n3\n4\n6\n8\n\n\n4\n5\n7\n9\n\n\n\nSample space = 9\nSum greater than 3 but less than 7 = {5, 4, 6, 5} = 4\nP(sum greater than 3 but less than 7) = \\(\\frac{4}{9}\\).\n(b) Given that y + m = 90°\n< ABE = x° (vertically opp. angles)\n< BAD = (x + m)° (exterior angles)\n< DCB = (x + y)° (exterior angles)\n(x + m)° + (x + y)° = 180° (opp. angles of a cyclic quadrilateral)\n(2x + y + m)° = 180°\n(2x + 90) = 180\n2x = 180 - 90 = 90°\nx = 45°\n\n\n', '(a)\n\n\n\n+\n1\n3\n5\n\n\n2\n3\n5\n7\n\n\n3\n4\n6\n8\n\n\n4\n5\n7\n9\n\n\n\nSample space = 9\nSum greater than 3 but less than 7 = {5, 4, 6, 5} = 4\nP(sum greater than 3 but less than 7) = \\(\\frac{4}{9}\\).~(b) Given that y + m = 90°\n< ABE = x° (vertically opp. angles)\n< BAD = (x + m)° (exterior angles)\n< DCB = (x + y)° (exterior angles)\n(x + m)° + (x + y)° = 180° (opp. angles of a cyclic quadrilateral)\n(2x + y + m)° = 180°\n(2x + 90) = 180\n2x = 180 - 90 = 90°\nx = 45°\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(581, '2016', '\n\nQuestion\n\nThe third term of a Geometric Progression (G.P) is 360 and the sixth term is 1215. Find the\n(a) common ratio;\n(b) first term ;\n(c) sum of the first four terms.\n\n', '', 'The third term of a Geometric Progression (G.P) is 360 and the sixth term is 1215. Find the\n(a) common ratio;\n(b) first term ;\n(c) sum of the first four terms.', 'The third term of a Geometric Progression (G.P) is 360 and the sixth term is 1215. Find the\n(a) common ratio;~(b) first term ;\n(c) sum of the first four terms.~(c) sum of the first four terms.', NULL, '\n\nExplanation\n\\(T_{n} = ar^{n - 1}\\) (terms of a geometric progression)\n\\(T_{3} = ar^{3 - 1} = ar^{2} = 360 ... (1)\\)\n\\(T_{6} = ar^{6 - 1} = ar^{5} = 1215 ... (2)\\)\nDivide (2) by (1).\n\\(\\frac{ar^{5}}{ar^{2}} = \\frac{1215}{360}\\)\n\\(r^{3} = \\frac{27}{8}\\)\n\\(\\therefore r = \\sqrt[3]{\\frac{27}{8}} = \\frac{3}{2}\\)\n(b) Putting r in (1), we have\n\\(ar^{2} = 360\\)\n\\(a \\times (\\frac{3}{2})^{2} = 360 \\implies \\frac{9a}{4} = 360\\)\n\\(a = \\frac{360 \\times 4}{9} = 160\\)\n(c) \\(S_{n} = \\frac{a(r^{n} - 1)}{r - 1}\\) (sum of terms of a G.P)\n\\(S_{4} = \\frac{160((\\frac{3}{2})^{4} - 1)}{\\frac{3}{2} - 1}\\)\n= \\(\\frac{160(\\frac{81}{16} - 1)}{\\frac{1}{2}}\\)\n= \\(\\frac{160 \\times \\frac{65}{16}}{\\frac{1}{2}}\\)\n= \\(1300\\)\n\n', '', 'Explanation\n\\(T_{n} = ar^{n - 1}\\) (terms of a geometric progression)\n\\(T_{3} = ar^{3 - 1} = ar^{2} = 360 ... (1)\\)\n\\(T_{6} = ar^{6 - 1} = ar^{5} = 1215 ... (2)\\)\nDivide (2) by (1).\n\\(\\frac{ar^{5}}{ar^{2}} = \\frac{1215}{360}\\)\n\\(r^{3} = \\frac{27}{8}\\)\n\\(\\therefore r = \\sqrt[3]{\\frac{27}{8}} = \\frac{3}{2}\\)\n(b) Putting r in (1), we have\n\\(ar^{2} = 360\\)\n\\(a \\times (\\frac{3}{2})^{2} = 360 \\implies \\frac{9a}{4} = 360\\)\n\\(a = \\frac{360 \\times 4}{9} = 160\\)\n(c) \\(S_{n} = \\frac{a(r^{n} - 1)}{r - 1}\\) (sum of terms of a G.P)\n\\(S_{4} = \\frac{160((\\frac{3}{2})^{4} - 1)}{\\frac{3}{2} - 1}\\)\n= \\(\\frac{160(\\frac{81}{16} - 1)}{\\frac{1}{2}}\\)\n= \\(\\frac{160 \\times \\frac{65}{16}}{\\frac{1}{2}}\\)\n= \\(1300\\)\n\n', '~(b) Putting r in (1), we have\n\\(ar^{2} = 360\\)\n\\(a \\times (\\frac{3}{2})^{2} = 360 \\implies \\frac{9a}{4} = 360\\)\n\\(a = \\frac{360 \\times 4}{9} = 160\\)\n(c) \\(S_{n} = \\frac{a(r^{n} - 1)}{r - 1}\\) (sum of terms of a G.P)\n\\(S_{4} = \\frac{160((\\frac{3}{2})^{4} - 1)}{\\frac{3}{2} - 1}\\)\n= \\(\\frac{160(\\frac{81}{16} - 1)}{\\frac{1}{2}}\\)\n= \\(\\frac{160 \\times \\frac{65}{16}}{\\frac{1}{2}}\\)\n= \\(1300\\)\n~(c) \\(S_{n} = \\frac{a(r^{n} - 1)}{r - 1}\\) (sum of terms of a G.P)\n\\(S_{4} = \\frac{160((\\frac{3}{2})^{4} - 1)}{\\frac{3}{2} - 1}\\)\n= \\(\\frac{160(\\frac{81}{16} - 1)}{\\frac{1}{2}}\\)\n= \\(\\frac{160 \\times \\frac{65}{16}}{\\frac{1}{2}}\\)\n= \\(1300\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(582, '2016', '\n\nQuestion\n\n(a) A radio which a dealer bought for N6,000.00 and marked to give a profit of 30% was reduced in a sales by 10%. Find : (i) the final sales price ; (ii) the percentage profit.\n(b) Solve the equation : \\(2^{(2x + 1)} - 9(2^{x}) + 4 = 0\\).\n\n', '', '(a) A radio which a dealer bought for N6,000.00 and marked to give a profit of 30% was reduced in a sales by 10%. Find : (i) the final sales price ; (ii) the percentage profit.\n(b) Solve the equation : \\(2^{(2x + 1)} - 9(2^{x}) + 4 = 0\\).', '(a) A radio which a dealer bought for N6,000.00 and marked to give a profit of 30% was reduced in a sales by 10%. Find : (i) the final sales price ; (ii) the percentage profit.~(b) Solve the equation : \\(2^{(2x + 1)} - 9(2^{x}) + 4 = 0\\).', NULL, '\n\nExplanation\n(a) Cost price of a radio = N6,000.00\n\\(\\therefore\\) 30% profit = \\(\\N6000 \\times \\frac{30}{100} = N1800\\)\n\\(\\therefore\\) Marked price = N(6000 + 1800) = N7800\nFinal Selling price = \\(\\frac{100 - 10}{100} \\times N7800 = N7020\\)\nProfit = N(7020 - 6000) = N1020.\n\\(\\therefore\\) % profit = \\(\\frac{1020}{6000} \\times 100% = 17%\\)\n(b) \\(2^{(2x + 1)} - 9(2^{x}) + 4 = 0\\)\n\\((2^{x})^{2} \\times 2 - 9(2^{x}) + 4 = 0\\)\nLet \\(2^{x} = y\\)\n\\(2y^{2} - 9y + 4 = 0\\)\n\\(2y^{2} - 8y - y + 4 = 0\\)\n\\(2y(y - 4) - 1(y - 4) = 0\\)\n\\((2y - 1)(y - 4) = 0\\)\n\\(2y = 1 \\implies y = \\frac{1}{2}; y = 4\\)\n\\(y = \\frac{1}{2} \\implies 2^{x} = \\frac{1}{2} = 2^{-1} \\)\n\\(x = -1\\)\n\\(y = 4 \\implies 2^{x} = 4 = 2^{2}\\)\n\\(x = 2\\)\n\\(\\therefore x = -1, 2\\)\n\n', '', 'Explanation\n(a) Cost price of a radio = N6,000.00\n\\(\\therefore\\) 30% profit = \\(\\N6000 \\times \\frac{30}{100} = N1800\\)\n\\(\\therefore\\) Marked price = N(6000 + 1800) = N7800\nFinal Selling price = \\(\\frac{100 - 10}{100} \\times N7800 = N7020\\)\nProfit = N(7020 - 6000) = N1020.\n\\(\\therefore\\) % profit = \\(\\frac{1020}{6000} \\times 100% = 17%\\)\n(b) \\(2^{(2x + 1)} - 9(2^{x}) + 4 = 0\\)\n\\((2^{x})^{2} \\times 2 - 9(2^{x}) + 4 = 0\\)\nLet \\(2^{x} = y\\)\n\\(2y^{2} - 9y + 4 = 0\\)\n\\(2y^{2} - 8y - y + 4 = 0\\)\n\\(2y(y - 4) - 1(y - 4) = 0\\)\n\\((2y - 1)(y - 4) = 0\\)\n\\(2y = 1 \\implies y = \\frac{1}{2}; y = 4\\)\n\\(y = \\frac{1}{2} \\implies 2^{x} = \\frac{1}{2} = 2^{-1} \\)\n\\(x = -1\\)\n\\(y = 4 \\implies 2^{x} = 4 = 2^{2}\\)\n\\(x = 2\\)\n\\(\\therefore x = -1, 2\\)\n\n', '(a) Cost price of a radio = N6,000.00\n\\(\\therefore\\) 30% profit = \\(\\N6000 \\times \\frac{30}{100} = N1800\\)\n\\(\\therefore\\) Marked price = N(6000 + 1800) = N7800\nFinal Selling price = \\(\\frac{100 - 10}{100} \\times N7800 = N7020\\)\nProfit = N(7020 - 6000) = N1020.\n\\(\\therefore\\) % profit = \\(\\frac{1020}{6000} \\times 100% = 17%\\)~(b) \\(2^{(2x + 1)} - 9(2^{x}) + 4 = 0\\)\n\\((2^{x})^{2} \\times 2 - 9(2^{x}) + 4 = 0\\)\nLet \\(2^{x} = y\\)\n\\(2y^{2} - 9y + 4 = 0\\)\n\\(2y^{2} - 8y - y + 4 = 0\\)\n\\(2y(y - 4) - 1(y - 4) = 0\\)\n\\((2y - 1)(y - 4) = 0\\)\n\\(2y = 1 \\implies y = \\frac{1}{2}; y = 4\\)\n\\(y = \\frac{1}{2} \\implies 2^{x} = \\frac{1}{2} = 2^{-1} \\)\n\\(x = -1\\)\n\\(y = 4 \\implies 2^{x} = 4 = 2^{2}\\)\n\\(x = 2\\)\n\\(\\therefore x = -1, 2\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(583, '2016', '\n\nQuestion\n\n(a) Copy and complete the table of values for the relation \\(y = 5 - 7x - 6x^{2}\\) for \\(-3 \\leq x \\leq 2\\).\n\n\n\nx\n-3\n-2\n-1\n-0.5\n0\n1\n2\n\n\ny\n-28\n\n6\n\n5\n\n\n\n\n\n(b) Using a scale of 2cm to 1 unit on the x- axis and 2cm to 5 units on the y- axis, draw the :\n(i) graph of \\(y = 5 - 7x - 6x^{2}\\) ; (ii) line \\(y = 3\\) on the same axis.\n(c) Use your graph to find the : (i) roots of the equation \\(2 - 7x - 6x^{2} = 0\\) ; (ii) maximum value of \\(y = 5 - 7x - 6x^{2}\\).\n\n\n(a) Copy and complete the table of values for the relation \\(y = 5 - 7x - 6x^{2}\\) for \\(-3 \\leq x \\leq 2\\).\n\n\n\nx\n-3\n-2\n-1\n-0.5\n0\n1\n2\n\n\ny\n-28\n\n6\n\n5\n\n\n\n\n', '', '(a) Copy and complete the table of values for the relation \\(y = 5 - 7x - 6x^{2}\\) for \\(-3 \\leq x \\leq 2\\).\n\n\n\nx\n-3\n-2\n-1\n-0.5\n0\n1\n2\n\n\ny\n-28\n\n6\n\n5\n\n\n\n\n\n(b) Using a scale of 2cm to 1 unit on the x- axis and 2cm to 5 units on the y- axis, draw the :\n(i) graph of \\(y = 5 - 7x - 6x^{2}\\) ; (ii) line \\(y = 3\\) on the same axis.\n(c) Use your graph to find the : (i) roots of the equation \\(2 - 7x - 6x^{2} = 0\\) ; (ii) maximum value of \\(y = 5 - 7x - 6x^{2}\\).\n\n\n(a) Copy and complete the table of values for the relation \\(y = 5 - 7x - 6x^{2}\\) for \\(-3 \\leq x \\leq 2\\).\n\n\n\nx\n-3\n-2\n-1\n-0.5\n0\n1\n2\n\n\ny\n-28\n\n6\n\n5\n\n\n', '(a) Copy and complete the table of values for the relation \\(y = 5 - 7x - 6x^{2}\\) for \\(-3 \\leq x \\leq 2\\).\n\n\n\nx\n-3\n-2\n-1\n-0.5\n0\n1\n2\n\n\ny\n-28\n\n6\n\n5\n\n\n\n\n~(b) Using a scale of 2cm to 1 unit on the x- axis and 2cm to 5 units on the y- axis, draw the :\n(i) graph of \\(y = 5 - 7x - 6x^{2}\\) ; (ii) line \\(y = 3\\) on the same axis.\n(c) Use your graph to find the : (i) roots of the equation \\(2 - 7x - 6x^{2} = 0\\) ; (ii) maximum value of \\(y = 5 - 7x - 6x^{2}\\).\n\n\n(a) Copy and complete the table of values for the relation \\(y = 5 - 7x - 6x^{2}\\) for \\(-3 \\leq x \\leq 2\\).\n\n\n\nx\n-3\n-2\n-1\n-0.5\n0\n1\n2\n\n\ny\n-28\n\n6\n\n5\n\n\n~(c) Use your graph to find the : (i) roots of the equation \\(2 - 7x - 6x^{2} = 0\\) ; (ii) maximum value of \\(y = 5 - 7x - 6x^{2}\\).\n\n\n(a) Copy and complete the table of values for the relation \\(y = 5 - 7x - 6x^{2}\\) for \\(-3 \\leq x \\leq 2\\).\n\n\n\nx\n-3\n-2\n-1\n-0.5\n0\n1\n2\n\n\ny\n-28\n\n6\n\n5\n\n\n', NULL, '\n\nExplanation\nx\n-3\n-2\n-1\n-0.5\n0\n1\n2\n\n\ny\n-28\n-5\n6\n7\n5\n-8\n-33\n\n\n\n(b)(i)\n(c)(i) \\(2 - 7x - 6x^{2} = 0\\)\n\\(2 + 3 - 7x - 6x^{2} = 3\\)\n\\(5- 7x - 6x^{2} = 3\\)\n\\(\\therefore \\text{the roots at y = 3 are } x = -1.4; 0.2\\)\n(ii) Maximum value of \\(y = 5 - 7x - 6x^{2}\\) is y = 7.\n\n', '', 'Explanation\nx\n-3\n-2\n-1\n-0.5\n0\n1\n2\n\n\ny\n-28\n-5\n6\n7\n5\n-8\n-33\n\n\n\n(b)(i)\n(c)(i) \\(2 - 7x - 6x^{2} = 0\\)\n\\(2 + 3 - 7x - 6x^{2} = 3\\)\n\\(5- 7x - 6x^{2} = 3\\)\n\\(\\therefore \\text{the roots at y = 3 are } x = -1.4; 0.2\\)\n(ii) Maximum value of \\(y = 5 - 7x - 6x^{2}\\) is y = 7.\n\n', '~(b)(i)\n(c)(i) \\(2 - 7x - 6x^{2} = 0\\)\n\\(2 + 3 - 7x - 6x^{2} = 3\\)\n\\(5- 7x - 6x^{2} = 3\\)\n\\(\\therefore \\text{the roots at y = 3 are } x = -1.4; 0.2\\)\n(ii) Maximum value of \\(y = 5 - 7x - 6x^{2}\\) is y = 7.\n~(c)(i) \\(2 - 7x - 6x^{2} = 0\\)\n\\(2 + 3 - 7x - 6x^{2} = 3\\)\n\\(5- 7x - 6x^{2} = 3\\)\n\\(\\therefore \\text{the roots at y = 3 are } x = -1.4; 0.2\\)\n(ii) Maximum value of \\(y = 5 - 7x - 6x^{2}\\) is y = 7.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(584, '2016', '\n\nQuestion\n\nUsing a ruler and a pair of compasses only, construct (a) triangle QRT with |QR| = 8cm, |RT| = 6cm and |QT| = 4.5cm.\n(b) a quadrilateral QRSP which has a common base QR with \\(\\Delta\\)QRT such that QTP is a straight line, PQ || SR, |QP| = 9cm and |RS| = 4.5cm.\n(i) Measure |PS| ; (ii) Find the perpendicular distance between RS and PQ ; (iii) What is QRSP?\n\n', '', 'Using a ruler and a pair of compasses only, construct (a) triangle QRT with |QR| = 8cm, |RT| = 6cm and |QT| = 4.5cm.\n(b) a quadrilateral QRSP which has a common base QR with \\(\\Delta\\)QRT such that QTP is a straight line, PQ || SR, |QP| = 9cm and |RS| = 4.5cm.\n(i) Measure |PS| ; (ii) Find the perpendicular distance between RS and PQ ; (iii) What is QRSP?', 'Using a ruler and a pair of compasses only, construct (a) triangle QRT with |QR| = 8cm, |RT| = 6cm and |QT| = 4.5cm.~(b) a quadrilateral QRSP which has a common base QR with \\(\\Delta\\)QRT such that QTP is a straight line, PQ || SR, |QP| = 9cm and |RS| = 4.5cm.\n(i) Measure |PS| ; (ii) Find the perpendicular distance between RS and PQ ; (iii) What is QRSP?', NULL, '\n\nExplanation\n(a)\n(b)(i) PS = 6.0cm\n(ii) Perpendicular disstance between RS and PQ i.e. RK = 5.9cm\n(iii) QRSP is a trapezium.\n\n', '', 'Explanation\n(a)\n(b)(i) PS = 6.0cm\n(ii) Perpendicular disstance between RS and PQ i.e. RK = 5.9cm\n(iii) QRSP is a trapezium.\n\n', '~(b)(i) PS = 6.0cm\n(ii) Perpendicular disstance between RS and PQ i.e. RK = 5.9cm\n(iii) QRSP is a trapezium.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(585, '2016', '\n\nQuestion\n\nA surveyor standing at a point X sights a pole Y due east of him and a tower Z of a building on a bearing of 046°. After walking to a point W, a distance of 180m in the South- East direction, he observes the bearing of Z and Y to be 337° and 050° respectively.\n(a) Calculate, correct to the nearest metre : (i) |XY| ; (ii) |ZW|\n(b) If N is on XY such that XZ = ZN, find the bearing of Z from N.\n\n', '', 'A surveyor standing at a point X sights a pole Y due east of him and a tower Z of a building on a bearing of 046°. After walking to a point W, a distance of 180m in the South- East direction, he observes the bearing of Z and Y to be 337° and 050° respectively.\n(a) Calculate, correct to the nearest metre : (i) |XY| ; (ii) |ZW|\n(b) If N is on XY such that XZ = ZN, find the bearing of Z from N.', 'A surveyor standing at a point X sights a pole Y due east of him and a tower Z of a building on a bearing of 046°. After walking to a point W, a distance of 180m in the South- East direction, he observes the bearing of Z and Y to be 337° and 050° respectively.\n(a) Calculate, correct to the nearest metre : (i) |XY| ; (ii) |ZW|~(b) If N is on XY such that XZ = ZN, find the bearing of Z from N.', NULL, '\n\nExplanation\n\nXWZ = 90° - (45° + 23°) = 22°\nWYX = 180° - (45° + 95°) = 40°\n< XZW = 180° - (89° + 22°) = 69°\n(a)(i) \\(\\frac{XY}{\\sin 95} = \\frac{180}{\\sin 40°}\\)\n\\(XY = \\frac{180 \\sin 95}{\\sin 40} = 278.96m\\)\n\\(\\approxeq 279m\\)\n(ii) \\(\\frac{ZW}{\\sin 89°} = \\frac{180}{\\sin 69°}\\)\n\\(ZW = \\frac{180 \\sin 89°}{\\sin 69°} = 192.76m\\)\n\\(\\approxeq 193m\\)\n(b) < ZNX = <ZXN (base angles of an isosceles triangle)\n= 44°^.\n\\(\\therefore\\) Bearing of Z from N = 270° + 44° = 314°.\n\n', '', 'Explanation\n\nXWZ = 90° - (45° + 23°) = 22°\nWYX = 180° - (45° + 95°) = 40°\n< XZW = 180° - (89° + 22°) = 69°\n(a)(i) \\(\\frac{XY}{\\sin 95} = \\frac{180}{\\sin 40°}\\)\n\\(XY = \\frac{180 \\sin 95}{\\sin 40} = 278.96m\\)\n\\(\\approxeq 279m\\)\n(ii) \\(\\frac{ZW}{\\sin 89°} = \\frac{180}{\\sin 69°}\\)\n\\(ZW = \\frac{180 \\sin 89°}{\\sin 69°} = 192.76m\\)\n\\(\\approxeq 193m\\)\n(b) < ZNX = <ZXN (base angles of an isosceles triangle)\n= 44°^.\n\\(\\therefore\\) Bearing of Z from N = 270° + 44° = 314°.\n\n', '(a)(i) \\(\\frac{XY}{\\sin 95} = \\frac{180}{\\sin 40°}\\)\n\\(XY = \\frac{180 \\sin 95}{\\sin 40} = 278.96m\\)\n\\(\\approxeq 279m\\)\n(ii) \\(\\frac{ZW}{\\sin 89°} = \\frac{180}{\\sin 69°}\\)\n\\(ZW = \\frac{180 \\sin 89°}{\\sin 69°} = 192.76m\\)\n\\(\\approxeq 193m\\)~(b) < ZNX = <ZXN (base angles of an isosceles triangle)\n= 44°^.\n\\(\\therefore\\) Bearing of Z from N = 270° + 44° = 314°.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(586, '2016', '\n\nQuestion\n\nIn the diagram, ASRTB represents a piece of string passing over a pulley of radius 10cm in a vertical plane. O is the centre of the pulley and AMB is a horizontal straight line touching the pulley at M. Angle SAB = 90° and angle TBA = 60°.\n(a) Calculate (i) the obtuse angle SOT ; (ii) arc SRT ; (iii) |BT|\n(b) Find, correct to the nearest cm, the length of the string. (Take \\(\\pi = \\frac{22}{7}\\)).\n\n\n', '', 'In the diagram, ASRTB represents a piece of string passing over a pulley of radius 10cm in a vertical plane. O is the centre of the pulley and AMB is a horizontal straight line touching the pulley at M. Angle SAB = 90° and angle TBA = 60°.\n(a) Calculate (i) the obtuse angle SOT ; (ii) arc SRT ; (iii) |BT|\n(b) Find, correct to the nearest cm, the length of the string. (Take \\(\\pi = \\frac{22}{7}\\)).\n', 'In the diagram, ASRTB represents a piece of string passing over a pulley of radius 10cm in a vertical plane. O is the centre of the pulley and AMB is a horizontal straight line touching the pulley at M. Angle SAB = 90° and angle TBA = 60°.\n(a) Calculate (i) the obtuse angle SOT ; (ii) arc SRT ; (iii) |BT|~(b) Find, correct to the nearest cm, the length of the string. (Take \\(\\pi = \\frac{22}{7}\\)).\n', NULL, '\n\nExplanation\n(a)(i) < ASO = < AMO = 90° (radius to a tangent)\n< SOM = 360° - (3 x 90°) (sum of angles of a rectangle)\n= 90°.\n\\(\\therefore\\) obtuse < SOT = 360° - (90° + 60° + 60°) = 150°.\n(ii) Arc SRT = \\(\\frac{150°}{360°} \\times 2\\pi r\\)\n= \\(\\frac{150}{360} \\times 2 \\times \\frac{22}{7} \\times 10 = \\frac{550}{21}\\)\n= \\(26.19cm\\)\n(iii) \\(\\tan 60° = \\frac{TB}{10}\\)\n\\(TB = 10\\tan 60° = 17.32cm\\)\n(b) Length of the string ASRTB = AS + arc SRT + TB\n= 10cm + 26.19cm + 17.32cm\n= 53.51cm \\(\\approxeq\\) 54cm.\n\n', '', 'Explanation\n(a)(i) < ASO = < AMO = 90° (radius to a tangent)\n< SOM = 360° - (3 x 90°) (sum of angles of a rectangle)\n= 90°.\n\\(\\therefore\\) obtuse < SOT = 360° - (90° + 60° + 60°) = 150°.\n(ii) Arc SRT = \\(\\frac{150°}{360°} \\times 2\\pi r\\)\n= \\(\\frac{150}{360} \\times 2 \\times \\frac{22}{7} \\times 10 = \\frac{550}{21}\\)\n= \\(26.19cm\\)\n(iii) \\(\\tan 60° = \\frac{TB}{10}\\)\n\\(TB = 10\\tan 60° = 17.32cm\\)\n(b) Length of the string ASRTB = AS + arc SRT + TB\n= 10cm + 26.19cm + 17.32cm\n= 53.51cm \\(\\approxeq\\) 54cm.\n\n', '(a)(i) < ASO = < AMO = 90° (radius to a tangent)\n< SOM = 360° - (3 x 90°) (sum of angles of a rectangle)\n= 90°.\n\\(\\therefore\\) obtuse < SOT = 360° - (90° + 60° + 60°) = 150°.\n(ii) Arc SRT = \\(\\frac{150°}{360°} \\times 2\\pi r\\)\n= \\(\\frac{150}{360} \\times 2 \\times \\frac{22}{7} \\times 10 = \\frac{550}{21}\\)\n= \\(26.19cm\\)\n(iii) \\(\\tan 60° = \\frac{TB}{10}\\)\n\\(TB = 10\\tan 60° = 17.32cm\\)~(b) Length of the string ASRTB = AS + arc SRT + TB\n= 10cm + 26.19cm + 17.32cm\n= 53.51cm \\(\\approxeq\\) 54cm.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(587, '2016', '\n\nQuestion\n\nThe table below shows the number of eggs laid by the chickens in a man\'s farm in a year.\n\n\n\nNo of eggs per year\nNo of chickens\n\n\n45 - 49\n10\n\n\n50 - 54\n36\n\n\n55 - 59\n64\n\n\n60 - 64\n52\n\n\n65 - 69\n28\n\n\n70 - 74\n10\n\n\n\n(a) Draw a cumulative frequency curve for the distribution.\n(b) Use your graph to find the interquartile range.\n(c) If a woman buys a chicken from the farm, what is the probability that the chicken lays at least 60 eggs in a year?\n\n\n\nThe table below shows the number of eggs laid by the chickens in a man\'s farm in a year.\n\n\n\nNo of eggs per year\nNo of chickens\n\n\n45 - 49\n10\n\n\n50 - 54\n36\n\n\n55 - 59\n64\n\n\n60 - 64\n52\n\n\n65 - 69\n28\n\n\n70 - 74\n10\n\n\n', '', 'The table below shows the number of eggs laid by the chickens in a man\'s farm in a year.\n\n\n\nNo of eggs per year\nNo of chickens\n\n\n45 - 49\n10\n\n\n50 - 54\n36\n\n\n55 - 59\n64\n\n\n60 - 64\n52\n\n\n65 - 69\n28\n\n\n70 - 74\n10\n\n\n\n(a) Draw a cumulative frequency curve for the distribution.\n(b) Use your graph to find the interquartile range.\n(c) If a woman buys a chicken from the farm, what is the probability that the chicken lays at least 60 eggs in a year?\n\n\n\nThe table below shows the number of eggs laid by the chickens in a man\'s farm in a year.\n\n\n\nNo of eggs per year\nNo of chickens\n\n\n45 - 49\n10\n\n\n50 - 54\n36\n\n\n55 - 59\n64\n\n\n60 - 64\n52\n\n\n65 - 69\n28\n\n\n70 - 74\n10\n', 'The table below shows the number of eggs laid by the chickens in a man\'s farm in a year.\n\n\n\nNo of eggs per year\nNo of chickens\n\n\n45 - 49\n10\n\n\n50 - 54\n36\n\n\n55 - 59\n64\n\n\n60 - 64\n52\n\n\n65 - 69\n28\n\n\n70 - 74\n10\n\n\n\n(a) Draw a cumulative frequency curve for the distribution.~(b) Use your graph to find the interquartile range.\n(c) If a woman buys a chicken from the farm, what is the probability that the chicken lays at least 60 eggs in a year?\n\n\n\nThe table below shows the number of eggs laid by the chickens in a man\'s farm in a year.\n\n\n\nNo of eggs per year\nNo of chickens\n\n\n45 - 49\n10\n\n\n50 - 54\n36\n\n\n55 - 59\n64\n\n\n60 - 64\n52\n\n\n65 - 69\n28\n\n\n70 - 74\n10\n~(c) If a woman buys a chicken from the farm, what is the probability that the chicken lays at least 60 eggs in a year?\n\n\n\nThe table below shows the number of eggs laid by the chickens in a man\'s farm in a year.\n\n\n\nNo of eggs per year\nNo of chickens\n\n\n45 - 49\n10\n\n\n50 - 54\n36\n\n\n55 - 59\n64\n\n\n60 - 64\n52\n\n\n65 - 69\n28\n\n\n70 - 74\n10\n', NULL, '\n\nExplanation\n(a)\n\n\n\n\nNo of eggs\nper year\n\nClass boundaries\nFreq\nCum Freq\n\n\n45 - 49\n44.5 - 49.5\n10\n10\n\n\n50 - 54\n49.5 - 54.5\n36\n46\n\n\n55 - 59\n54.5 - 59.5\n64\n110\n\n\n60 - 64\n59.5 - 64.5\n52\n162\n\n\n65 - 69\n64.5 - 69.5\n28\n190\n\n\n70 - 74\n69.5 - 74.5\n10\n200\n\n\n\n\n\n(b) From the graph, the interquartile range = \\(Q_{3} - Q_{1} = 63 - 54 = 9\\)\n(c) Probability of a chicken laying at least 60 eggs in a year = \\(\\frac{52 + 28 + 10}{200}\\)\n= \\(\\frac{90}{200} = \\frac{9}{20}\\)\n\n\n', '', 'Explanation\n(a)\n\n\n\n\nNo of eggs\nper year\n\nClass boundaries\nFreq\nCum Freq\n\n\n45 - 49\n44.5 - 49.5\n10\n10\n\n\n50 - 54\n49.5 - 54.5\n36\n46\n\n\n55 - 59\n54.5 - 59.5\n64\n110\n\n\n60 - 64\n59.5 - 64.5\n52\n162\n\n\n65 - 69\n64.5 - 69.5\n28\n190\n\n\n70 - 74\n69.5 - 74.5\n10\n200\n\n\n\n\n\n(b) From the graph, the interquartile range = \\(Q_{3} - Q_{1} = 63 - 54 = 9\\)\n(c) Probability of a chicken laying at least 60 eggs in a year = \\(\\frac{52 + 28 + 10}{200}\\)\n= \\(\\frac{90}{200} = \\frac{9}{20}\\)\n\n\n', '(a)\n\n\n\n\nNo of eggs\nper year\n\nClass boundaries\nFreq\nCum Freq\n\n\n45 - 49\n44.5 - 49.5\n10\n10\n\n\n50 - 54\n49.5 - 54.5\n36\n46\n\n\n55 - 59\n54.5 - 59.5\n64\n110\n\n\n60 - 64\n59.5 - 64.5\n52\n162\n\n\n65 - 69\n64.5 - 69.5\n28\n190\n\n\n70 - 74\n69.5 - 74.5\n10\n200\n\n\n\n\n~(b) From the graph, the interquartile range = \\(Q_{3} - Q_{1} = 63 - 54 = 9\\)\n(c) Probability of a chicken laying at least 60 eggs in a year = \\(\\frac{52 + 28 + 10}{200}\\)\n= \\(\\frac{90}{200} = \\frac{9}{20}\\)\n\n~(c) Probability of a chicken laying at least 60 eggs in a year = \\(\\frac{52 + 28 + 10}{200}\\)\n= \\(\\frac{90}{200} = \\frac{9}{20}\\)\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1);
INSERT INTO `exam_simulation2` (`id`, `examYear`, `questions`, `parent_que`, `child_que`, `gen_child_que_desc`, `imageDir`, `explanation`, `parent_que_expl`, `child_que_exp`, `gen_child_que_expl`, `package_id`, `subject_id`, `status`) VALUES
(588, '2016', '\n\nQuestion\n\n(a) Copy and complete the binary multiplication table:\n\n\n\nx\n10\n11\n100\n101\n\n\n10\n100\n\n1000\n\n\n\n11\n110\n\n1100\n\n\n\n100\n\n\n10000\n10100\n\n\n\n(b) Convert \\(11.011_{two}\\) to a number in base ten.\n(c) Simplify \\(\\frac{9.6 \\times 10^{18}}{0.24 \\times 10^{5}}\\) and express your answer in the form \\(P \\times 10^{m}\\) where 1 < P < 10 and m is an integer.\n\n\n(a) Copy and complete the binary multiplication table:\n\n\n\nx\n10\n11\n100\n101\n\n\n10\n100\n\n1000\n\n\n\n11\n110\n\n1100\n\n\n\n100\n\n\n10000\n10100\n\n\n', '', '(a) Copy and complete the binary multiplication table:\n\n\n\nx\n10\n11\n100\n101\n\n\n10\n100\n\n1000\n\n\n\n11\n110\n\n1100\n\n\n\n100\n\n\n10000\n10100\n\n\n\n(b) Convert \\(11.011_{two}\\) to a number in base ten.\n(c) Simplify \\(\\frac{9.6 \\times 10^{18}}{0.24 \\times 10^{5}}\\) and express your answer in the form \\(P \\times 10^{m}\\) where 1 < P < 10 and m is an integer.\n\n\n(a) Copy and complete the binary multiplication table:\n\n\n\nx\n10\n11\n100\n101\n\n\n10\n100\n\n1000\n\n\n\n11\n110\n\n1100\n\n\n\n100\n\n\n10000\n10100\n', '(a) Copy and complete the binary multiplication table:\n\n\n\nx\n10\n11\n100\n101\n\n\n10\n100\n\n1000\n\n\n\n11\n110\n\n1100\n\n\n\n100\n\n\n10000\n10100\n\n\n~(b) Convert \\(11.011_{two}\\) to a number in base ten.\n(c) Simplify \\(\\frac{9.6 \\times 10^{18}}{0.24 \\times 10^{5}}\\) and express your answer in the form \\(P \\times 10^{m}\\) where 1 < P < 10 and m is an integer.\n\n\n(a) Copy and complete the binary multiplication table:\n\n\n\nx\n10\n11\n100\n101\n\n\n10\n100\n\n1000\n\n\n\n11\n110\n\n1100\n\n\n\n100\n\n\n10000\n10100\n~(c) Simplify \\(\\frac{9.6 \\times 10^{18}}{0.24 \\times 10^{5}}\\) and express your answer in the form \\(P \\times 10^{m}\\) where 1 < P < 10 and m is an integer.\n\n\n(a) Copy and complete the binary multiplication table:\n\n\n\nx\n10\n11\n100\n101\n\n\n10\n100\n\n1000\n\n\n\n11\n110\n\n1100\n\n\n\n100\n\n\n10000\n10100\n', NULL, '\n\nExplanation\n(a)\n\n\n\nx\n10\n11\n100\n101\n\n\n10\n100\n110\n1000\n1010\n\n\n11\n110\n1001\n1100\n1111\n\n\n100\n1000\n1100\n10000\n10100\n\n\n\n(b) \\(11.011_{two} = 1 \\times 2^{1} + 1 \\times 2^{0} + 0 \\times 2^{-1} + 1 \\times 2^{-2} + 1 \\times 2^{-3}\\)\n= \\(2 + 1 + 0 + \\frac{1}{4} + \\frac{1}{8}\\)\n= \\(3\\frac{3}{8}\\)\n= \\(3.375\\)\n(c) \\(\\frac{9.6 \\times 10^{18}}{0.24 \\times 10^{5}} \\equiv \\frac{9.6 \\times 10^{18}}{2.4 \\times 10^{4}}\\)\n= \\(\\frac{9.6}{2.4} \\times 10^{18 - 4}\\)\n= \\(4 \\times 10^{14}\\)\n\n', '', 'Explanation\n(a)\n\n\n\nx\n10\n11\n100\n101\n\n\n10\n100\n110\n1000\n1010\n\n\n11\n110\n1001\n1100\n1111\n\n\n100\n1000\n1100\n10000\n10100\n\n\n\n(b) \\(11.011_{two} = 1 \\times 2^{1} + 1 \\times 2^{0} + 0 \\times 2^{-1} + 1 \\times 2^{-2} + 1 \\times 2^{-3}\\)\n= \\(2 + 1 + 0 + \\frac{1}{4} + \\frac{1}{8}\\)\n= \\(3\\frac{3}{8}\\)\n= \\(3.375\\)\n(c) \\(\\frac{9.6 \\times 10^{18}}{0.24 \\times 10^{5}} \\equiv \\frac{9.6 \\times 10^{18}}{2.4 \\times 10^{4}}\\)\n= \\(\\frac{9.6}{2.4} \\times 10^{18 - 4}\\)\n= \\(4 \\times 10^{14}\\)\n\n', '(a)\n\n\n\nx\n10\n11\n100\n101\n\n\n10\n100\n110\n1000\n1010\n\n\n11\n110\n1001\n1100\n1111\n\n\n100\n1000\n1100\n10000\n10100\n\n\n~(b) \\(11.011_{two} = 1 \\times 2^{1} + 1 \\times 2^{0} + 0 \\times 2^{-1} + 1 \\times 2^{-2} + 1 \\times 2^{-3}\\)\n= \\(2 + 1 + 0 + \\frac{1}{4} + \\frac{1}{8}\\)\n= \\(3\\frac{3}{8}\\)\n= \\(3.375\\)\n(c) \\(\\frac{9.6 \\times 10^{18}}{0.24 \\times 10^{5}} \\equiv \\frac{9.6 \\times 10^{18}}{2.4 \\times 10^{4}}\\)\n= \\(\\frac{9.6}{2.4} \\times 10^{18 - 4}\\)\n= \\(4 \\times 10^{14}\\)\n~(c) \\(\\frac{9.6 \\times 10^{18}}{0.24 \\times 10^{5}} \\equiv \\frac{9.6 \\times 10^{18}}{2.4 \\times 10^{4}}\\)\n= \\(\\frac{9.6}{2.4} \\times 10^{18 - 4}\\)\n= \\(4 \\times 10^{14}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(589, '2016', '\n\nQuestion\n\n(a) The 6th term of an A.P is 35 and the 13th term is 77. Find the 20th term.\n(b)\nThe Venn diagram represents three subsets P, Q and R of the universal set U. Copy the Venn diagram. Shade and indicate the regions represented by (i) \\(P \\cap Q\' \\cap R\\) ; (ii) \\(P\'\\cap Q \\cap R\'\\).\n\n', '', '(a) The 6th term of an A.P is 35 and the 13th term is 77. Find the 20th term.\n(b)\nThe Venn diagram represents three subsets P, Q and R of the universal set U. Copy the Venn diagram. Shade and indicate the regions represented by (i) \\(P \\cap Q\' \\cap R\\) ; (ii) \\(P\'\\cap Q \\cap R\'\\).', '(a) The 6th term of an A.P is 35 and the 13th term is 77. Find the 20th term.~(b)\nThe Venn diagram represents three subsets P, Q and R of the universal set U. Copy the Venn diagram. Shade and indicate the regions represented by (i) \\(P \\cap Q\' \\cap R\\) ; (ii) \\(P\'\\cap Q \\cap R\'\\).', NULL, '\n\nExplanation\n(a) \\(T_{n} = a + (n - 1)d\\) (terms of an A.P)\n\\(T_{6} = a + 5d = 35 ... (1)\\)\n\\(T_{13} = a + 12d = 77 ... (2)\\)\n(2) - (1) : \\(7d = 42 \\implies d = 6\\)\n\\(a + 5(6) = 35 \\implies a + 30 = 35\\)\n\\(a = 5\\)\n\\(\\therefore T_{20} = a + 19d = 5 + 19(6)\\)\n= \\(5 + 114 = 119\\)\n(b)\n\n', '', 'Explanation\n(a) \\(T_{n} = a + (n - 1)d\\) (terms of an A.P)\n\\(T_{6} = a + 5d = 35 ... (1)\\)\n\\(T_{13} = a + 12d = 77 ... (2)\\)\n(2) - (1) : \\(7d = 42 \\implies d = 6\\)\n\\(a + 5(6) = 35 \\implies a + 30 = 35\\)\n\\(a = 5\\)\n\\(\\therefore T_{20} = a + 19d = 5 + 19(6)\\)\n= \\(5 + 114 = 119\\)\n(b)\n\n', '(a) \\(T_{n} = a + (n - 1)d\\) (terms of an A.P)\n\\(T_{6} = a + 5d = 35 ... (1)\\)\n\\(T_{13} = a + 12d = 77 ... (2)\\)\n(2) - (1) : \\(7d = 42 \\implies d = 6\\)\n\\(a + 5(6) = 35 \\implies a + 30 = 35\\)\n\\(a = 5\\)\n\\(\\therefore T_{20} = a + 19d = 5 + 19(6)\\)\n= \\(5 + 114 = 119\\)~(b)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(590, '2016', '\n\nQuestion\n\n(a) Given that \\(\\sin x = \\frac{5}{13}, 0° \\leq x \\leq 90°\\), find \\(\\frac{\\cos x - 2 \\sin x }{2\\tan x}\\).\n(b)\nThe diagram represents the vertical cross-section of a mountain with height NQ standing on a horizontal ground PRN. If the angles of elevation of the top of the mountain from P and R are 30° and 70° respectively and PR = 500m, calculate, correct to 3 significant figures :\n(i) |QP| ; (ii) the height of the mountain.\n\n', '', '(a) Given that \\(\\sin x = \\frac{5}{13}, 0° \\leq x \\leq 90°\\), find \\(\\frac{\\cos x - 2 \\sin x }{2\\tan x}\\).\n(b)\nThe diagram represents the vertical cross-section of a mountain with height NQ standing on a horizontal ground PRN. If the angles of elevation of the top of the mountain from P and R are 30° and 70° respectively and PR = 500m, calculate, correct to 3 significant figures :\n(i) |QP| ; (ii) the height of the mountain.', '(a) Given that \\(\\sin x = \\frac{5}{13}, 0° \\leq x \\leq 90°\\), find \\(\\frac{\\cos x - 2 \\sin x }{2\\tan x}\\).~(b)\nThe diagram represents the vertical cross-section of a mountain with height NQ standing on a horizontal ground PRN. If the angles of elevation of the top of the mountain from P and R are 30° and 70° respectively and PR = 500m, calculate, correct to 3 significant figures :\n(i) |QP| ; (ii) the height of the mountain.', NULL, '\n\nExplanation\n(a)\n\\(\\sin x = \\frac{5}{13}\\)\n\\(AB = \\sqrt{13^{2} - 5^{2}} = 12\\)\n\\(\\cos x = \\frac{12}{13}\\)\n\\(\\tan x = \\frac{5}{12}\\)\n\\(\\frac{\\cos x - 2 \\sin x}{2 \\tan x} = \\frac{\\frac{12}{15} - 2(\\frac{5}{13})}{2(\\frac{5}{12})}\\)\n= \\(\\frac{\\frac{12}{13} - \\frac{10}{13}}{\\frac{5}{6}}\\)\n= \\(\\frac{2}{13} \\times \\frac{6}{5} = \\frac{12}{65}\\)\n(b) (i)\n< QRP = 180° - 70° = 110° (angles on a straight line)\n< RQP = 180° - (110° + 30°) = 40°\n\\(\\frac{PR}{\\sin 40°} = \\frac{PQ}{\\sin 110°}\\)\n\\(\\frac{500}{\\sin 40°} = \\frac{PQ}{\\sin 110°}\\)\n\\(PQ = \\frac{500 \\sin 110°}{\\sin 40°} = 730.94m\\)\n\\(\\approxeq 731 m\\)\n(ii) \\(\\sin 30° = \\frac{QN}{PQ} = \\frac{QN}{730.94}\\)\n\\(QN = 730.94 \\times 0.5 = 365.47 m\\)\n\\(\\approxeq 365 m\\)\n\n', '', 'Explanation\n(a)\n\\(\\sin x = \\frac{5}{13}\\)\n\\(AB = \\sqrt{13^{2} - 5^{2}} = 12\\)\n\\(\\cos x = \\frac{12}{13}\\)\n\\(\\tan x = \\frac{5}{12}\\)\n\\(\\frac{\\cos x - 2 \\sin x}{2 \\tan x} = \\frac{\\frac{12}{15} - 2(\\frac{5}{13})}{2(\\frac{5}{12})}\\)\n= \\(\\frac{\\frac{12}{13} - \\frac{10}{13}}{\\frac{5}{6}}\\)\n= \\(\\frac{2}{13} \\times \\frac{6}{5} = \\frac{12}{65}\\)\n(b) (i)\n< QRP = 180° - 70° = 110° (angles on a straight line)\n< RQP = 180° - (110° + 30°) = 40°\n\\(\\frac{PR}{\\sin 40°} = \\frac{PQ}{\\sin 110°}\\)\n\\(\\frac{500}{\\sin 40°} = \\frac{PQ}{\\sin 110°}\\)\n\\(PQ = \\frac{500 \\sin 110°}{\\sin 40°} = 730.94m\\)\n\\(\\approxeq 731 m\\)\n(ii) \\(\\sin 30° = \\frac{QN}{PQ} = \\frac{QN}{730.94}\\)\n\\(QN = 730.94 \\times 0.5 = 365.47 m\\)\n\\(\\approxeq 365 m\\)\n\n', '(a)\n\\(\\sin x = \\frac{5}{13}\\)\n\\(AB = \\sqrt{13^{2} - 5^{2}} = 12\\)\n\\(\\cos x = \\frac{12}{13}\\)\n\\(\\tan x = \\frac{5}{12}\\)\n\\(\\frac{\\cos x - 2 \\sin x}{2 \\tan x} = \\frac{\\frac{12}{15} - 2(\\frac{5}{13})}{2(\\frac{5}{12})}\\)\n= \\(\\frac{\\frac{12}{13} - \\frac{10}{13}}{\\frac{5}{6}}\\)\n= \\(\\frac{2}{13} \\times \\frac{6}{5} = \\frac{12}{65}\\)~(b) (i)\n< QRP = 180° - 70° = 110° (angles on a straight line)\n< RQP = 180° - (110° + 30°) = 40°\n\\(\\frac{PR}{\\sin 40°} = \\frac{PQ}{\\sin 110°}\\)\n\\(\\frac{500}{\\sin 40°} = \\frac{PQ}{\\sin 110°}\\)\n\\(PQ = \\frac{500 \\sin 110°}{\\sin 40°} = 730.94m\\)\n\\(\\approxeq 731 m\\)\n(ii) \\(\\sin 30° = \\frac{QN}{PQ} = \\frac{QN}{730.94}\\)\n\\(QN = 730.94 \\times 0.5 = 365.47 m\\)\n\\(\\approxeq 365 m\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(591, '2016', '\n\nQuestion\n\nThe table below shows how a company\'s sales manager spent his 1995 annual salary.\n\n\n\nFood\n30%\n\n\nRent\n18%\n\n\nCar Maintenance\n25%\n\n\nSavings\n12%\n\n\nTaxes\n5%\n\n\nOthers\n10%\n\n\n\n(a) Represent this information on a pie chart.\n(b) Find his savings at the end of the year if his annual salary was N60,000.00.\n\n\n\nThe table below shows how a company\'s sales manager spent his 1995 annual salary.\n\n\n\nFood\n30%\n\n\nRent\n18%\n\n\nCar Maintenance\n25%\n\n\nSavings\n12%\n\n\nTaxes\n5%\n\n\nOthers\n10%\n\n\n', '', 'The table below shows how a company\'s sales manager spent his 1995 annual salary.\n\n\n\nFood\n30%\n\n\nRent\n18%\n\n\nCar Maintenance\n25%\n\n\nSavings\n12%\n\n\nTaxes\n5%\n\n\nOthers\n10%\n\n\n\n(a) Represent this information on a pie chart.\n(b) Find his savings at the end of the year if his annual salary was N60,000.00.\n\n\n\nThe table below shows how a company\'s sales manager spent his 1995 annual salary.\n\n\n\nFood\n30%\n\n\nRent\n18%\n\n\nCar Maintenance\n25%\n\n\nSavings\n12%\n\n\nTaxes\n5%\n\n\nOthers\n10%\n', 'The table below shows how a company\'s sales manager spent his 1995 annual salary.\n\n\n\nFood\n30%\n\n\nRent\n18%\n\n\nCar Maintenance\n25%\n\n\nSavings\n12%\n\n\nTaxes\n5%\n\n\nOthers\n10%\n\n\n\n(a) Represent this information on a pie chart.~(b) Find his savings at the end of the year if his annual salary was N60,000.00.\n\n\n\nThe table below shows how a company\'s sales manager spent his 1995 annual salary.\n\n\n\nFood\n30%\n\n\nRent\n18%\n\n\nCar Maintenance\n25%\n\n\nSavings\n12%\n\n\nTaxes\n5%\n\n\nOthers\n10%\n', NULL, '\n\nExplanation\nAngle of sector:\n\\(Food = \\frac{30}{100} \\times 360° = 108°\\)\n\\(Rent = \\frac{18}{100} \\times 360° = 64.8°\\)\n\\(\\text{Car Maintenance} = \\frac{25}{100} \\times 360° = 90°\\)\n\\(Savings = \\frac{12}{100} \\times 360° = 43.2°\\)\n\\(Taxes = \\frac{5}{100} \\times 360° = 18°\\)\n\\(Others = \\frac{10}{100} \\times 360° = 36°\\)\n(a)\n(b) Annual Salary = N60,000.00\n\\(\\therefore Savings = N(\\frac{12}{100} \\times 60,000) = N7,200\\)\n\n', '', 'Explanation\nAngle of sector:\n\\(Food = \\frac{30}{100} \\times 360° = 108°\\)\n\\(Rent = \\frac{18}{100} \\times 360° = 64.8°\\)\n\\(\\text{Car Maintenance} = \\frac{25}{100} \\times 360° = 90°\\)\n\\(Savings = \\frac{12}{100} \\times 360° = 43.2°\\)\n\\(Taxes = \\frac{5}{100} \\times 360° = 18°\\)\n\\(Others = \\frac{10}{100} \\times 360° = 36°\\)\n(a)\n(b) Annual Salary = N60,000.00\n\\(\\therefore Savings = N(\\frac{12}{100} \\times 60,000) = N7,200\\)\n\n', '~(b) Annual Salary = N60,000.00\n\\(\\therefore Savings = N(\\frac{12}{100} \\times 60,000) = N7,200\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(592, '2016', '\n\nQuestion\n\n(a) Given that \\(\\frac{5y - x}{8y + 3x} = \\frac{1}{5}\\), find the value of \\(\\frac{x}{y}\\) to two decimal places.\n(b) If 3 is a root of the quadratic equation \\(x^{2} + bx - 15 = 0\\), determine the value of b.Find the other root.\n\n', '', '(a) Given that \\(\\frac{5y - x}{8y + 3x} = \\frac{1}{5}\\), find the value of \\(\\frac{x}{y}\\) to two decimal places.\n(b) If 3 is a root of the quadratic equation \\(x^{2} + bx - 15 = 0\\), determine the value of b.Find the other root.', '(a) Given that \\(\\frac{5y - x}{8y + 3x} = \\frac{1}{5}\\), find the value of \\(\\frac{x}{y}\\) to two decimal places.~(b) If 3 is a root of the quadratic equation \\(x^{2} + bx - 15 = 0\\), determine the value of b.Find the other root.', NULL, '\n\nExplanation\n(a) \\(\\frac{5y - x}{8y + 3x} = \\frac{1}{5}\\)\n\\(5(5y - x) = 8y + 3x \\implies 25y - 5x = 8y + 3x\\)\n\\(25y - 8y = 3x + 5x \\implies 17y = 8x\\)\n\\(\\therefore \\frac{x}{y} = \\frac{17}{8} = 2.125 \\approxeq 2.13\\)\n(b) \\(x^{2} + bx - 15 = 0\\)\nSince x = 3 is a root of the equation, f(3) = 0.\n\\(3^{2} + 3b - 15 = 0 \\implies 3b = 6\\)\n\\(b = 2\\)\n\\(\\therefore\\) The equation is \\(x^{2} + 2x - 15 = 0\\)\n\\(x^{2} - 3x + 5x - 15 = 0 \\implies x(x - 3) + 5(x - 3) = 0\\)\n\\((x - 3)(x + 5) = 0\\)\nThe second root of the equation is x = -5.\n\n', '', 'Explanation\n(a) \\(\\frac{5y - x}{8y + 3x} = \\frac{1}{5}\\)\n\\(5(5y - x) = 8y + 3x \\implies 25y - 5x = 8y + 3x\\)\n\\(25y - 8y = 3x + 5x \\implies 17y = 8x\\)\n\\(\\therefore \\frac{x}{y} = \\frac{17}{8} = 2.125 \\approxeq 2.13\\)\n(b) \\(x^{2} + bx - 15 = 0\\)\nSince x = 3 is a root of the equation, f(3) = 0.\n\\(3^{2} + 3b - 15 = 0 \\implies 3b = 6\\)\n\\(b = 2\\)\n\\(\\therefore\\) The equation is \\(x^{2} + 2x - 15 = 0\\)\n\\(x^{2} - 3x + 5x - 15 = 0 \\implies x(x - 3) + 5(x - 3) = 0\\)\n\\((x - 3)(x + 5) = 0\\)\nThe second root of the equation is x = -5.\n\n', '(a) \\(\\frac{5y - x}{8y + 3x} = \\frac{1}{5}\\)\n\\(5(5y - x) = 8y + 3x \\implies 25y - 5x = 8y + 3x\\)\n\\(25y - 8y = 3x + 5x \\implies 17y = 8x\\)\n\\(\\therefore \\frac{x}{y} = \\frac{17}{8} = 2.125 \\approxeq 2.13\\)~(b) \\(x^{2} + bx - 15 = 0\\)\nSince x = 3 is a root of the equation, f(3) = 0.\n\\(3^{2} + 3b - 15 = 0 \\implies 3b = 6\\)\n\\(b = 2\\)\n\\(\\therefore\\) The equation is \\(x^{2} + 2x - 15 = 0\\)\n\\(x^{2} - 3x + 5x - 15 = 0 \\implies x(x - 3) + 5(x - 3) = 0\\)\n\\((x - 3)(x + 5) = 0\\)\nThe second root of the equation is x = -5.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(593, '2016', '\n\nQuestion\n\n(a) Use logarithm tables to evaluate \\(\\frac{15.05 \\times \\sqrt{0.00695}}{6.95 \\times 10^{2}}\\).\n(b) The first 5 students to arrive in a school on a Monday morning were 2 boys and 3 girls. Of these, two were chosen at random for an assignment. Find the probability that :\n(i) both were boys ; (ii) the two were of different sexes.\n\n', '', '(a) Use logarithm tables to evaluate \\(\\frac{15.05 \\times \\sqrt{0.00695}}{6.95 \\times 10^{2}}\\).\n(b) The first 5 students to arrive in a school on a Monday morning were 2 boys and 3 girls. Of these, two were chosen at random for an assignment. Find the probability that :\n(i) both were boys ; (ii) the two were of different sexes.', '(a) Use logarithm tables to evaluate \\(\\frac{15.05 \\times \\sqrt{0.00695}}{6.95 \\times 10^{2}}\\).~(b) The first 5 students to arrive in a school on a Monday morning were 2 boys and 3 girls. Of these, two were chosen at random for an assignment. Find the probability that :\n(i) both were boys ; (ii) the two were of different sexes.', NULL, '\n\nExplanation\n(a) \\(\\frac{15.05 \\times \\sqrt{0.00695}}{6.95 \\times 10^{2}}\\)\n\n\n\nNo\nLog\n\n\n15.05\n\\(1.1775 = 1.1775\\)\n\n\n\\(\\sqrt{0.00695}\\)\n\\(\\bar{3}.8420 \\div 2 = \\bar{2}.9210\\)\n\n\n\n           = 0.0985\n\n\n\\(6.95 \\times 10^{2}\\)\n           - 2.8420\n\n\nAntilog = 0.001805\n        = \\(\\bar{3}.2565\\)\n\n\n\n\\(\\therefore \\frac{15.05 \\times \\sqrt{0.00695}}{6.95 \\times 10^{2}} \\approxeq 0.00181\\) (3 sig. figs)\n(b) No of boys = 2, No of girls = 3\n\\(\\therefore\\) Total students = 5\n(i) P(both are boys) = \\(\\frac{2}{5} \\times \\frac{1}{4} = \\frac{1}{10}\\)\n(ii) P(both are of different sexes) = P(first a boy, then a girl) or P(first a girl, then a boy)\n= \\(\\frac{2}{5} \\times \\frac{3}{4} + \\frac{3}{5} \\times \\frac{2}{4} \\)\n= \\(\\frac{6}{20} + \\frac{6}{20}\\)\n= \\(\\frac{12}{20} = \\frac{3}{5}\\)\n\n\n', '', 'Explanation\n(a) \\(\\frac{15.05 \\times \\sqrt{0.00695}}{6.95 \\times 10^{2}}\\)\n\n\n\nNo\nLog\n\n\n15.05\n\\(1.1775 = 1.1775\\)\n\n\n\\(\\sqrt{0.00695}\\)\n\\(\\bar{3}.8420 \\div 2 = \\bar{2}.9210\\)\n\n\n\n           = 0.0985\n\n\n\\(6.95 \\times 10^{2}\\)\n           - 2.8420\n\n\nAntilog = 0.001805\n        = \\(\\bar{3}.2565\\)\n\n\n\n\\(\\therefore \\frac{15.05 \\times \\sqrt{0.00695}}{6.95 \\times 10^{2}} \\approxeq 0.00181\\) (3 sig. figs)\n(b) No of boys = 2, No of girls = 3\n\\(\\therefore\\) Total students = 5\n(i) P(both are boys) = \\(\\frac{2}{5} \\times \\frac{1}{4} = \\frac{1}{10}\\)\n(ii) P(both are of different sexes) = P(first a boy, then a girl) or P(first a girl, then a boy)\n= \\(\\frac{2}{5} \\times \\frac{3}{4} + \\frac{3}{5} \\times \\frac{2}{4} \\)\n= \\(\\frac{6}{20} + \\frac{6}{20}\\)\n= \\(\\frac{12}{20} = \\frac{3}{5}\\)\n\n\n', '(a) \\(\\frac{15.05 \\times \\sqrt{0.00695}}{6.95 \\times 10^{2}}\\)\n\n\n\nNo\nLog\n\n\n15.05\n\\(1.1775 = 1.1775\\)\n\n\n\\(\\sqrt{0.00695}\\)\n\\(\\bar{3}.8420 \\div 2 = \\bar{2}.9210\\)\n\n\n\n           = 0.0985\n\n\n\\(6.95 \\times 10^{2}\\)\n           - 2.8420\n\n\nAntilog = 0.001805\n        = \\(\\bar{3}.2565\\)\n\n\n\n\\(\\therefore \\frac{15.05 \\times \\sqrt{0.00695}}{6.95 \\times 10^{2}} \\approxeq 0.00181\\) (3 sig. figs)~(b) No of boys = 2, No of girls = 3\n\\(\\therefore\\) Total students = 5\n(i) P(both are boys) = \\(\\frac{2}{5} \\times \\frac{1}{4} = \\frac{1}{10}\\)\n(ii) P(both are of different sexes) = P(first a boy, then a girl) or P(first a girl, then a boy)\n= \\(\\frac{2}{5} \\times \\frac{3}{4} + \\frac{3}{5} \\times \\frac{2}{4} \\)\n= \\(\\frac{6}{20} + \\frac{6}{20}\\)\n= \\(\\frac{12}{20} = \\frac{3}{5}\\)\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(594, '2016', '\n\nQuestion\n\n(a) Using a ruler and a pair of compasses only, construst \\(\\Delta\\) ABC in which |AB| = 7cm, |BC| = 5cm and < ABC = 75°. Measure |AC|.\n(b) In (a) above, locate by construction, a point D such that CD is parallel to AB and D is equidistant from points A and C. Measure < BAD.\n\n', '', '(a) Using a ruler and a pair of compasses only, construst \\(\\Delta\\) ABC in which |AB| = 7cm, |BC| = 5cm and < ABC = 75°. Measure |AC|.\n(b) In (a) above, locate by construction, a point D such that CD is parallel to AB and D is equidistant from points A and C. Measure < BAD.', '(a) Using a ruler and a pair of compasses only, construst \\(\\Delta\\) ABC in which |AB| = 7cm, |BC| = 5cm and < ABC = 75°. Measure |AC|.~(b) In (a) above, locate by construction, a point D such that CD is parallel to AB and D is equidistant from points A and C. Measure < BAD.', NULL, '\n\nExplanation\n|AC| = 8.35 cm, < BAD = 70°.\n\n', '|AC| = 8.35 cm, < BAD = 70°.\n\n', '', '', 'WAEC-PK-1', 'WAEC-MATH', 1),
(595, '2016', '\n\nQuestion\n\n(a) Solve the simultaneous equation : \\(\\log_{10} x + \\log_{10} y = 4\\)\n                              \\(\\log_{10} x + 2\\log_{10} y = 3\\)\n(b) The time, t,taken to buy fuel at a petrol station varies directly as the number of vehicles V on queue and jointly varies inversely as the number of pumps P available in the station. In a station with 5 pumps, it took 10 minutes to fuel 20 vehicles. Find :\n(i) the relationship between t, P and V ; (ii) the time it will take to fuel 50 vehicles in the station with 2 pumps ; (iii) the number of pumps required to fuel 40 vehicles in 20 minutes.\n\n', '', '(a) Solve the simultaneous equation : \\(\\log_{10} x + \\log_{10} y = 4\\)\n                              \\(\\log_{10} x + 2\\log_{10} y = 3\\)\n(b) The time, t,taken to buy fuel at a petrol station varies directly as the number of vehicles V on queue and jointly varies inversely as the number of pumps P available in the station. In a station with 5 pumps, it took 10 minutes to fuel 20 vehicles. Find :\n(i) the relationship between t, P and V ; (ii) the time it will take to fuel 50 vehicles in the station with 2 pumps ; (iii) the number of pumps required to fuel 40 vehicles in 20 minutes.', '(a) Solve the simultaneous equation : \\(\\log_{10} x + \\log_{10} y = 4\\)\n                              \\(\\log_{10} x + 2\\log_{10} y = 3\\)~(b) The time, t,taken to buy fuel at a petrol station varies directly as the number of vehicles V on queue and jointly varies inversely as the number of pumps P available in the station. In a station with 5 pumps, it took 10 minutes to fuel 20 vehicles. Find :\n(i) the relationship between t, P and V ; (ii) the time it will take to fuel 50 vehicles in the station with 2 pumps ; (iii) the number of pumps required to fuel 40 vehicles in 20 minutes.', NULL, '\n\nExplanation\n(a) \\(\\log_{10} x + \\log_{10} y = 4 ... (1)\\)\n\\(\\log_{10} x + 2 \\log_{10} y = 3 ...(2)\\)\n\\((1) \\implies \\log_{10} (xy) = 4 \\therefore xy = 10^{4} ... (3)\\)\n\\((2) \\implies \\log_{10} (xy^{2}) = 3 \\therefore xy^{2} = 10^{3} ... (4)\\)\nDivide (4) by (3) :\n\\(y = 10^{-1} = 0.1\\)\nPut y in (3), we have\n\\(\\frac{x}{10} = 10^{4} \\implies x = 10^{5}\\)\n\\((x, y) = (10^{5}, 10^{-1})\\)\n(b) (i) \\(t \\propto \\frac{V}{P}\\)\n\\(t = \\frac{kV}{P}\\)\nwhen t = 10, V= 20 and P = 5.\n\\(10 = \\frac{20k}{5} \\implies 50 = 20k\\)\n\\(k = \\frac{5}{2}\\)\n\\(\\therefore t = \\frac{5V}{2P}\\)\n(ii) When V = 50, P = 2, we have\n\\(t = \\frac{5(50)}{2(2)} = \\frac{250}{4}\\)\n= \\(62.5 minutes\\)\n(iii) when V = 40, t = 20 minutes\n\\(20 = \\frac{5(40)}{2P} \\implies 40P = 200\\)\n\\(P = \\frac{200}{40} = 5 pumps\\)\n\n', '', 'Explanation\n(a) \\(\\log_{10} x + \\log_{10} y = 4 ... (1)\\)\n\\(\\log_{10} x + 2 \\log_{10} y = 3 ...(2)\\)\n\\((1) \\implies \\log_{10} (xy) = 4 \\therefore xy = 10^{4} ... (3)\\)\n\\((2) \\implies \\log_{10} (xy^{2}) = 3 \\therefore xy^{2} = 10^{3} ... (4)\\)\nDivide (4) by (3) :\n\\(y = 10^{-1} = 0.1\\)\nPut y in (3), we have\n\\(\\frac{x}{10} = 10^{4} \\implies x = 10^{5}\\)\n\\((x, y) = (10^{5}, 10^{-1})\\)\n(b) (i) \\(t \\propto \\frac{V}{P}\\)\n\\(t = \\frac{kV}{P}\\)\nwhen t = 10, V= 20 and P = 5.\n\\(10 = \\frac{20k}{5} \\implies 50 = 20k\\)\n\\(k = \\frac{5}{2}\\)\n\\(\\therefore t = \\frac{5V}{2P}\\)\n(ii) When V = 50, P = 2, we have\n\\(t = \\frac{5(50)}{2(2)} = \\frac{250}{4}\\)\n= \\(62.5 minutes\\)\n(iii) when V = 40, t = 20 minutes\n\\(20 = \\frac{5(40)}{2P} \\implies 40P = 200\\)\n\\(P = \\frac{200}{40} = 5 pumps\\)\n\n', '(a) \\(\\log_{10} x + \\log_{10} y = 4 ... (1)\\)\n\\(\\log_{10} x + 2 \\log_{10} y = 3 ...(2)\\)\n\\((1) \\implies \\log_{10} (xy) = 4 \\therefore xy = 10^{4} ... (3)\\)\n\\((2) \\implies \\log_{10} (xy^{2}) = 3 \\therefore xy^{2} = 10^{3} ... (4)\\)\nDivide (4) by (3) :\n\\(y = 10^{-1} = 0.1\\)\nPut y in (3), we have\n\\(\\frac{x}{10} = 10^{4} \\implies x = 10^{5}\\)\n\\((x, y) = (10^{5}, 10^{-1})\\)~(b) (i) \\(t \\propto \\frac{V}{P}\\)\n\\(t = \\frac{kV}{P}\\)\nwhen t = 10, V= 20 and P = 5.\n\\(10 = \\frac{20k}{5} \\implies 50 = 20k\\)\n\\(k = \\frac{5}{2}\\)\n\\(\\therefore t = \\frac{5V}{2P}\\)\n(ii) When V = 50, P = 2, we have\n\\(t = \\frac{5(50)}{2(2)} = \\frac{250}{4}\\)\n= \\(62.5 minutes\\)\n(iii) when V = 40, t = 20 minutes\n\\(20 = \\frac{5(40)}{2P} \\implies 40P = 200\\)\n\\(P = \\frac{200}{40} = 5 pumps\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(596, '2016', '\n\nQuestion\n\n The solid is a cylinder surmounted by a hemispherical bowl. Calculate its\n(a) total surface area ;\n(b) volume (Take \\(\\pi = \\frac{22}{7}\\))\n\n', '', ' The solid is a cylinder surmounted by a hemispherical bowl. Calculate its\n(a) total surface area ;\n(b) volume (Take \\(\\pi = \\frac{22}{7}\\))', ' The solid is a cylinder surmounted by a hemispherical bowl. Calculate its\n(a) total surface area ;~(b) volume (Take \\(\\pi = \\frac{22}{7}\\))', NULL, '\n\nExplanation\n(a) Area of curved surface of cylinder = \\(2\\pi r h\\)\n= \\(2 \\times \\frac{22}{7} \\times 7 \\times 10cm^{2} = 440 cm^{2}\\)\nArea of the base of cylinder = \\(\\pi r^{2}\\)\n= \\(\\frac{22}{7} \\times 7 \\times 7 = 154 cm^{2}\\)\nSurface area of hemisphere = \\(\\frac{4\\pi r^{2}}{2} = 2 \\pi r^{2}\\)\n= \\(2 \\times \\frac{22}{7} \\times 7 \\times 7 = 308 cm^{2}\\)\n\\(\\therefore\\) Total surface area = 440 + 154 + 308 = 902cm\\(^{2}\\)\n(b) Volume of cylinder = \\(\\pi r^{2} h\\)\n= \\(\\frac{22}{7} \\times 7 \\times 7 \\times 10 = 1540 cm^{3}\\)\nVolume of hemisphere = \\(\\frac{1}{2}(\\frac{4\\pi r^{3}}{3})\\)\n= \\(\\frac{4 \\times 22 \\times 7 \\times 7 \\times 7}{2 \\times 7 \\times 3}\\)\n= \\(718.67 cm^{3}\\)\nTotal volume = 1540 + 718.67 = 2258.67 cm\\(^{3}\\).\n\n', '', 'Explanation\n(a) Area of curved surface of cylinder = \\(2\\pi r h\\)\n= \\(2 \\times \\frac{22}{7} \\times 7 \\times 10cm^{2} = 440 cm^{2}\\)\nArea of the base of cylinder = \\(\\pi r^{2}\\)\n= \\(\\frac{22}{7} \\times 7 \\times 7 = 154 cm^{2}\\)\nSurface area of hemisphere = \\(\\frac{4\\pi r^{2}}{2} = 2 \\pi r^{2}\\)\n= \\(2 \\times \\frac{22}{7} \\times 7 \\times 7 = 308 cm^{2}\\)\n\\(\\therefore\\) Total surface area = 440 + 154 + 308 = 902cm\\(^{2}\\)\n(b) Volume of cylinder = \\(\\pi r^{2} h\\)\n= \\(\\frac{22}{7} \\times 7 \\times 7 \\times 10 = 1540 cm^{3}\\)\nVolume of hemisphere = \\(\\frac{1}{2}(\\frac{4\\pi r^{3}}{3})\\)\n= \\(\\frac{4 \\times 22 \\times 7 \\times 7 \\times 7}{2 \\times 7 \\times 3}\\)\n= \\(718.67 cm^{3}\\)\nTotal volume = 1540 + 718.67 = 2258.67 cm\\(^{3}\\).\n\n', '(a) Area of curved surface of cylinder = \\(2\\pi r h\\)\n= \\(2 \\times \\frac{22}{7} \\times 7 \\times 10cm^{2} = 440 cm^{2}\\)\nArea of the base of cylinder = \\(\\pi r^{2}\\)\n= \\(\\frac{22}{7} \\times 7 \\times 7 = 154 cm^{2}\\)\nSurface area of hemisphere = \\(\\frac{4\\pi r^{2}}{2} = 2 \\pi r^{2}\\)\n= \\(2 \\times \\frac{22}{7} \\times 7 \\times 7 = 308 cm^{2}\\)\n\\(\\therefore\\) Total surface area = 440 + 154 + 308 = 902cm\\(^{2}\\)~(b) Volume of cylinder = \\(\\pi r^{2} h\\)\n= \\(\\frac{22}{7} \\times 7 \\times 7 \\times 10 = 1540 cm^{3}\\)\nVolume of hemisphere = \\(\\frac{1}{2}(\\frac{4\\pi r^{3}}{3})\\)\n= \\(\\frac{4 \\times 22 \\times 7 \\times 7 \\times 7}{2 \\times 7 \\times 3}\\)\n= \\(718.67 cm^{3}\\)\nTotal volume = 1540 + 718.67 = 2258.67 cm\\(^{3}\\).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(597, '2016', '\n\nQuestion\n\nAbove is the graph of the quadratic function \\(y = ax^{2} + bx + c\\) where a, b and c are constants. Using the graph, find :\n(a)(i) the scales on both axes ; (ii) the equation of the line of symmetry of the curve ; (iii) the roots of the quadratic equation \\(ax^{2} + bx + c = 0\\)\n(b) Use the coordinates of D, E and G to find the values of the constants a, b and c hence write down the quadratic function illustrated in the graph.\n(c) Find the greatest value of y within the range \\(-3 \\leq x \\leq 5\\).\n\n\n', '', 'Above is the graph of the quadratic function \\(y = ax^{2} + bx + c\\) where a, b and c are constants. Using the graph, find :\n(a)(i) the scales on both axes ; (ii) the equation of the line of symmetry of the curve ; (iii) the roots of the quadratic equation \\(ax^{2} + bx + c = 0\\)\n(b) Use the coordinates of D, E and G to find the values of the constants a, b and c hence write down the quadratic function illustrated in the graph.\n(c) Find the greatest value of y within the range \\(-3 \\leq x \\leq 5\\).\n', 'Above is the graph of the quadratic function \\(y = ax^{2} + bx + c\\) where a, b and c are constants. Using the graph, find :\n(a)(i) the scales on both axes ; (ii) the equation of the line of symmetry of the curve ; (iii) the roots of the quadratic equation \\(ax^{2} + bx + c = 0\\)~(b) Use the coordinates of D, E and G to find the values of the constants a, b and c hence write down the quadratic function illustrated in the graph.\n(c) Find the greatest value of y within the range \\(-3 \\leq x \\leq 5\\).\n~(c) Find the greatest value of y within the range \\(-3 \\leq x \\leq 5\\).\n', NULL, '\n\nExplanation\n(a)(i) Scale : On x- axis, 2cm = 1 unit\n          On y- axis, 2cm = 5 units.\n(ii) Equation of line of symmetry is x = 1.25.\n(iii) Roots of the equation \\(ax^{2} + bx + c = 0\\) are x = 0.25 and x = 2.25.\n(b) Coordinates are D(0, 1), E(1, -2) and G(3, 4).\nSubstituting for y and x in \\(ax^{2} + bx + c = y\\)\nD(0, 1) : \\(1 = a(0^{2}) + b(0) + c \\implies c = 1\\)\nE(1, -2) : \\(-2 = a(1^{2}) + b(1) + c \\implies -2 = a + b + c\\)\n\\(a + b = -2 - 1 = -3 ... (1)\\)\nG(3, 4) : \\(4 = a(3^{2}) + b(3) + c \\implies 4 = 9a + 3b + c\\)\n\\(9a + 3b = 4 - 1 = 3 ... (2)\\)\n\\(\\implies 3a + b = 1 ... (2a)\\)\n(2a) - (1) : \\(2a = 4 \\implies a = 2\\)\n\\(a + b = -3 \\implies 2 + b = -3\\)\n\\(b = -3 - 2 = -5\\)\n\\(\\therefore\\) The equation is \\(y = 2x^{2} - 5x + 1\\)\n(c) The greatest value of y = 33.5.\n\n', '', 'Explanation\n(a)(i) Scale : On x- axis, 2cm = 1 unit\n          On y- axis, 2cm = 5 units.\n(ii) Equation of line of symmetry is x = 1.25.\n(iii) Roots of the equation \\(ax^{2} + bx + c = 0\\) are x = 0.25 and x = 2.25.\n(b) Coordinates are D(0, 1), E(1, -2) and G(3, 4).\nSubstituting for y and x in \\(ax^{2} + bx + c = y\\)\nD(0, 1) : \\(1 = a(0^{2}) + b(0) + c \\implies c = 1\\)\nE(1, -2) : \\(-2 = a(1^{2}) + b(1) + c \\implies -2 = a + b + c\\)\n\\(a + b = -2 - 1 = -3 ... (1)\\)\nG(3, 4) : \\(4 = a(3^{2}) + b(3) + c \\implies 4 = 9a + 3b + c\\)\n\\(9a + 3b = 4 - 1 = 3 ... (2)\\)\n\\(\\implies 3a + b = 1 ... (2a)\\)\n(2a) - (1) : \\(2a = 4 \\implies a = 2\\)\n\\(a + b = -3 \\implies 2 + b = -3\\)\n\\(b = -3 - 2 = -5\\)\n\\(\\therefore\\) The equation is \\(y = 2x^{2} - 5x + 1\\)\n(c) The greatest value of y = 33.5.\n\n', '(a)(i) Scale : On x- axis, 2cm = 1 unit\n          On y- axis, 2cm = 5 units.\n(ii) Equation of line of symmetry is x = 1.25.\n(iii) Roots of the equation \\(ax^{2} + bx + c = 0\\) are x = 0.25 and x = 2.25.~(b) Coordinates are D(0, 1), E(1, -2) and G(3, 4).\nSubstituting for y and x in \\(ax^{2} + bx + c = y\\)\nD(0, 1) : \\(1 = a(0^{2}) + b(0) + c \\implies c = 1\\)\nE(1, -2) : \\(-2 = a(1^{2}) + b(1) + c \\implies -2 = a + b + c\\)\n\\(a + b = -2 - 1 = -3 ... (1)\\)\nG(3, 4) : \\(4 = a(3^{2}) + b(3) + c \\implies 4 = 9a + 3b + c\\)\n\\(9a + 3b = 4 - 1 = 3 ... (2)\\)\n\\(\\implies 3a + b = 1 ... (2a)\\)\n(2a) - (1) : \\(2a = 4 \\implies a = 2\\)\n\\(a + b = -3 \\implies 2 + b = -3\\)\n\\(b = -3 - 2 = -5\\)\n\\(\\therefore\\) The equation is \\(y = 2x^{2} - 5x + 1\\)\n(c) The greatest value of y = 33.5.\n~(c) The greatest value of y = 33.5.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(598, '2016', '\n\nQuestion\n\n(a)PQRST is a circle with centre C. PCS is a straight line, RS // QT, |QR| = |RS| and < QTS = 56°. Find (i) SQT (ii) PQT.\n(b) In the diagram, points B and C are on a horizontal plane and |BC| = 30cm. A and D are points vertically above B and C respectively. |DC| = 40 cm and |AB| = 26 cm. Calculate the angles of depression of : (i) B from D ; (ii) A from D ; correct to the nearest degree.\n\n', '', '(a)PQRST is a circle with centre C. PCS is a straight line, RS // QT, |QR| = |RS| and < QTS = 56°. Find (i) SQT (ii) PQT.\n(b) In the diagram, points B and C are on a horizontal plane and |BC| = 30cm. A and D are points vertically above B and C respectively. |DC| = 40 cm and |AB| = 26 cm. Calculate the angles of depression of : (i) B from D ; (ii) A from D ; correct to the nearest degree.', '(a)PQRST is a circle with centre C. PCS is a straight line, RS // QT, |QR| = |RS| and < QTS = 56°. Find (i) SQT (ii) PQT.~(b) In the diagram, points B and C are on a horizontal plane and |BC| = 30cm. A and D are points vertically above B and C respectively. |DC| = 40 cm and |AB| = 26 cm. Calculate the angles of depression of : (i) B from D ; (ii) A from D ; correct to the nearest degree.', NULL, '\n\nExplanation\n(a) From the figure < QRS = 180° - 56° = 124° (opposite angles of a parallelogram)\n\\(\\therefore\\) < RQS = < PQS = \\(\\frac{180° - 124°}{2} = 28°\\) (base angles of an isosceles triangle)\n\\(\\therefore\\) < SQT = < RSQ = 28° (alternate angles PS // QT)\n(ii) < PQT = 90° - 28° = 62° (< PQS is in a semi-circle)\n(b)\n\\(\\tan < BDC = \\frac{30}{40} = 0.75\\)\n\\(< BDC = \\tan^{-1} (0.75) = 36.87°\\)\n\\(\\therefore < BDF = 90° - 36.87° = 53.13° \\approxeq 53°\\)\n\\(\\tan < ADF = \\frac{14}{30} = 0.4667\\)\n\\(< ADF = \\tan^{-1} (0.4667) = 25.02° \\approxeq 25°\\)\n\n', '', 'Explanation\n(a) From the figure < QRS = 180° - 56° = 124° (opposite angles of a parallelogram)\n\\(\\therefore\\) < RQS = < PQS = \\(\\frac{180° - 124°}{2} = 28°\\) (base angles of an isosceles triangle)\n\\(\\therefore\\) < SQT = < RSQ = 28° (alternate angles PS // QT)\n(ii) < PQT = 90° - 28° = 62° (< PQS is in a semi-circle)\n(b)\n\\(\\tan < BDC = \\frac{30}{40} = 0.75\\)\n\\(< BDC = \\tan^{-1} (0.75) = 36.87°\\)\n\\(\\therefore < BDF = 90° - 36.87° = 53.13° \\approxeq 53°\\)\n\\(\\tan < ADF = \\frac{14}{30} = 0.4667\\)\n\\(< ADF = \\tan^{-1} (0.4667) = 25.02° \\approxeq 25°\\)\n\n', '(a) From the figure < QRS = 180° - 56° = 124° (opposite angles of a parallelogram)\n\\(\\therefore\\) < RQS = < PQS = \\(\\frac{180° - 124°}{2} = 28°\\) (base angles of an isosceles triangle)\n\\(\\therefore\\) < SQT = < RSQ = 28° (alternate angles PS // QT)\n(ii) < PQT = 90° - 28° = 62° (< PQS is in a semi-circle)~(b)\n\\(\\tan < BDC = \\frac{30}{40} = 0.75\\)\n\\(< BDC = \\tan^{-1} (0.75) = 36.87°\\)\n\\(\\therefore < BDF = 90° - 36.87° = 53.13° \\approxeq 53°\\)\n\\(\\tan < ADF = \\frac{14}{30} = 0.4667\\)\n\\(< ADF = \\tan^{-1} (0.4667) = 25.02° \\approxeq 25°\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(599, '2016', '\n\nQuestion\n\nThe table below shows the mark distribution of candidates in an aptitude test for selection into the public service.\n\n\n\nMarks (in %)\nFreq\n\n\n44 - 46\n2\n\n\n47 - 49\n5\n\n\n50 - 52\n11\n\n\n53 - 55\n20\n\n\n56 - 61\n42\n\n\n62 - 64\n46\n\n\n65 - 67\n36\n\n\n68 - 70\n9\n\n\n71 - 73\n3\n\n\n\n(a) Make a cumulative frequency for the distribution\n(b) Draw the cumulative frequency curve.\n(c) From your graph, estimate the median mark.\n(d) The cut-off mark was 63%. What percentage of the candidates was selected?\n\n\nThe table below shows the mark distribution of candidates in an aptitude test for selection into the public service.\n\n\n\nMarks (in %)\nFreq\n\n\n44 - 46\n2\n\n\n47 - 49\n5\n\n\n50 - 52\n11\n\n\n53 - 55\n20\n\n\n56 - 61\n42\n\n\n62 - 64\n46\n\n\n65 - 67\n36\n\n\n68 - 70\n9\n\n\n71 - 73\n3\n\n\n', '', 'The table below shows the mark distribution of candidates in an aptitude test for selection into the public service.\n\n\n\nMarks (in %)\nFreq\n\n\n44 - 46\n2\n\n\n47 - 49\n5\n\n\n50 - 52\n11\n\n\n53 - 55\n20\n\n\n56 - 61\n42\n\n\n62 - 64\n46\n\n\n65 - 67\n36\n\n\n68 - 70\n9\n\n\n71 - 73\n3\n\n\n\n(a) Make a cumulative frequency for the distribution\n(b) Draw the cumulative frequency curve.\n(c) From your graph, estimate the median mark.\n(d) The cut-off mark was 63%. What percentage of the candidates was selected?\n\n\nThe table below shows the mark distribution of candidates in an aptitude test for selection into the public service.\n\n\n\nMarks (in %)\nFreq\n\n\n44 - 46\n2\n\n\n47 - 49\n5\n\n\n50 - 52\n11\n\n\n53 - 55\n20\n\n\n56 - 61\n42\n\n\n62 - 64\n46\n\n\n65 - 67\n36\n\n\n68 - 70\n9\n\n\n71 - 73\n3\n', 'The table below shows the mark distribution of candidates in an aptitude test for selection into the public service.\n\n\n\nMarks (in %)\nFreq\n\n\n44 - 46\n2\n\n\n47 - 49\n5\n\n\n50 - 52\n11\n\n\n53 - 55\n20\n\n\n56 - 61\n42\n\n\n62 - 64\n46\n\n\n65 - 67\n36\n\n\n68 - 70\n9\n\n\n71 - 73\n3\n\n\n\n(a) Make a cumulative frequency for the distribution~(b) Draw the cumulative frequency curve.\n(c) From your graph, estimate the median mark.\n(d) The cut-off mark was 63%. What percentage of the candidates was selected?\n\n\nThe table below shows the mark distribution of candidates in an aptitude test for selection into the public service.\n\n\n\nMarks (in %)\nFreq\n\n\n44 - 46\n2\n\n\n47 - 49\n5\n\n\n50 - 52\n11\n\n\n53 - 55\n20\n\n\n56 - 61\n42\n\n\n62 - 64\n46\n\n\n65 - 67\n36\n\n\n68 - 70\n9\n\n\n71 - 73\n3\n~(c) From your graph, estimate the median mark.\n(d) The cut-off mark was 63%. What percentage of the candidates was selected?\n\n\nThe table below shows the mark distribution of candidates in an aptitude test for selection into the public service.\n\n\n\nMarks (in %)\nFreq\n\n\n44 - 46\n2\n\n\n47 - 49\n5\n\n\n50 - 52\n11\n\n\n53 - 55\n20\n\n\n56 - 61\n42\n\n\n62 - 64\n46\n\n\n65 - 67\n36\n\n\n68 - 70\n9\n\n\n71 - 73\n3\n~(d) The cut-off mark was 63%. What percentage of the candidates was selected?\n\n\nThe table below shows the mark distribution of candidates in an aptitude test for selection into the public service.\n\n\n\nMarks (in %)\nFreq\n\n\n44 - 46\n2\n\n\n47 - 49\n5\n\n\n50 - 52\n11\n\n\n53 - 55\n20\n\n\n56 - 61\n42\n\n\n62 - 64\n46\n\n\n65 - 67\n36\n\n\n68 - 70\n9\n\n\n71 - 73\n3\n', NULL, '\n\nExplanation\n(a)\n\n\n\nMarks (in %)\nFreq\nClass boundaries\nCum Freq\n\n\n44 - 46\n2\n43.5 - 46.5\n2\n\n\n47 - 49\n5\n46.5 - 49.5\n7\n\n\n50 - 52\n11\n49.5 - 52.5\n18\n\n\n53 - 55\n20\n52.5 - 55.5\n38\n\n\n56 - 61\n42\n55.5 - 61.5\n80\n\n\n62 - 64\n46\n61.5 - 64.5\n126\n\n\n65 - 67\n36\n64.5 - 67.5\n162\n\n\n68 - 70\n9\n67.5 - 70.5\n171\n\n\n71 - 73\n3\n70.5 - 73.5\n174\n\n\n\n\n\n\n', '(a)\n\n\n\nMarks (in %)\nFreq\nClass boundaries\nCum Freq\n\n\n44 - 46\n2\n43.5 - 46.5\n2\n\n\n47 - 49\n5\n46.5 - 49.5\n7\n\n\n50 - 52\n11\n49.5 - 52.5\n18\n\n\n53 - 55\n20\n52.5 - 55.5\n38\n\n\n56 - 61\n42\n55.5 - 61.5\n80\n\n\n62 - 64\n46\n61.5 - 64.5\n126\n\n\n65 - 67\n36\n64.5 - 67.5\n162\n\n\n68 - 70\n9\n67.5 - 70.5\n171\n\n\n71 - 73\n3\n70.5 - 73.5\n174\n\n\n\n\n\n\n', '', '', 'WAEC-PK-1', 'WAEC-MATH', 1),
(600, '2016', '\n\nQuestion\n\nA shopkeeper buys 40 kg of fruits for N120.00. He sells 20 kg at N5.00 per kg, 10 kg at N3.00 per kg, 5 kg at N2.00 per kg and the remaining 5 kg at 50k per kg. Calculate the :\n(a) amount he realises from the sales ;\n(b) total profit / loss ;\n(c) percentage profit/ loss on his outlay of N120.00.\n\n', '', 'A shopkeeper buys 40 kg of fruits for N120.00. He sells 20 kg at N5.00 per kg, 10 kg at N3.00 per kg, 5 kg at N2.00 per kg and the remaining 5 kg at 50k per kg. Calculate the :\n(a) amount he realises from the sales ;\n(b) total profit / loss ;\n(c) percentage profit/ loss on his outlay of N120.00.', 'A shopkeeper buys 40 kg of fruits for N120.00. He sells 20 kg at N5.00 per kg, 10 kg at N3.00 per kg, 5 kg at N2.00 per kg and the remaining 5 kg at 50k per kg. Calculate the :\n(a) amount he realises from the sales ;~(b) total profit / loss ;\n(c) percentage profit/ loss on his outlay of N120.00.~(c) percentage profit/ loss on his outlay of N120.00.', NULL, '\n\nExplanation\n(a) 20 kg at N5.00 each = N100.00\n 10 kg at N3.00 each = N30.00\n 5 kg at N2.00 each = N10.00\n 5 kg at 50k each = N2.50\n Total = N 142.50.\n(b) Selling price = N142.50\nCost price = N120.00\n\\(\\therefore\\) Profit = N(142.50 - 120.00) = N22.50.\n(c) \\(% profit = \\frac{22.5}{120} \\times 100%\\)\n= \\(\\frac{45}{240} \\times 100% = 18.75%\\)\n\n', '', 'Explanation\n(a) 20 kg at N5.00 each = N100.00\n 10 kg at N3.00 each = N30.00\n 5 kg at N2.00 each = N10.00\n 5 kg at 50k each = N2.50\n Total = N 142.50.\n(b) Selling price = N142.50\nCost price = N120.00\n\\(\\therefore\\) Profit = N(142.50 - 120.00) = N22.50.\n(c) \\(% profit = \\frac{22.5}{120} \\times 100%\\)\n= \\(\\frac{45}{240} \\times 100% = 18.75%\\)\n\n', '(a) 20 kg at N5.00 each = N100.00\n 10 kg at N3.00 each = N30.00\n 5 kg at N2.00 each = N10.00\n 5 kg at 50k each = N2.50\n Total = N 142.50.~(b) Selling price = N142.50\nCost price = N120.00\n\\(\\therefore\\) Profit = N(142.50 - 120.00) = N22.50.\n(c) \\(% profit = \\frac{22.5}{120} \\times 100%\\)\n= \\(\\frac{45}{240} \\times 100% = 18.75%\\)\n~(c) \\(% profit = \\frac{22.5}{120} \\times 100%\\)\n= \\(\\frac{45}{240} \\times 100% = 18.75%\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(601, '2016', '\n\nQuestion\n\n(a) Solve the simultaneous equation : \\(6y + 5x = 12 ; 4y - 3x = 11\\).\n(b)\nIn the diagram, ADC is a straight line. /CD/ = 48 cm, /BD/ = 36 cm and /AD/ = y cm. Find the value of y.\n\n', '', '(a) Solve the simultaneous equation : \\(6y + 5x = 12 ; 4y - 3x = 11\\).\n(b)\nIn the diagram, ADC is a straight line. /CD/ = 48 cm, /BD/ = 36 cm and /AD/ = y cm. Find the value of y.', '(a) Solve the simultaneous equation : \\(6y + 5x = 12 ; 4y - 3x = 11\\).~(b)\nIn the diagram, ADC is a straight line. /CD/ = 48 cm, /BD/ = 36 cm and /AD/ = y cm. Find the value of y.', NULL, '\n\nExplanation\n(a) \\(6y + 5x = -12 ... (1)\\)\n\\(4y - 3x = 11 ... (2)\\)\nMultiply (1) by 2 and (2) by 3,\n\\(12y + 10x = -24 ... (1a)\\)\n\\(12y - 9x = 33... (2a)\\)\n\\((1a) - (2a) : 19x = -57 \\implies x = -3\\)\n\\(4y - 3(-3) = 11 \\implies 4y + 9 = 11\\)\n\\(4y = 2 \\implies y = \\frac{1}{2}\\)\n(b)\n\\(/BC/^{2} = 36^{2} + 48^{2}\\)\n\\(/BC/^{2} = 1296 + 2304 = 3600\\)\n\\(/BC/ = \\sqrt{3600} = 60 cm\\)\n\\(/AC/^{2} = /BC/^{2} + /AB/^{2}\\)\n\\((y + 48)^{2} = 60^{2} + 36^{2} + y^{2}\\)\n\\(y^{2} + 96y + 2304 = 3600 + 1296 + y^{2}\\)\n\\(96y = 3600 + 1296 - 2304 = 2592\\)\n\\(96y = 2592 \\implies y = \\frac{2592}{96} = 27 cm\\)\n\n', '', 'Explanation\n(a) \\(6y + 5x = -12 ... (1)\\)\n\\(4y - 3x = 11 ... (2)\\)\nMultiply (1) by 2 and (2) by 3,\n\\(12y + 10x = -24 ... (1a)\\)\n\\(12y - 9x = 33... (2a)\\)\n\\((1a) - (2a) : 19x = -57 \\implies x = -3\\)\n\\(4y - 3(-3) = 11 \\implies 4y + 9 = 11\\)\n\\(4y = 2 \\implies y = \\frac{1}{2}\\)\n(b)\n\\(/BC/^{2} = 36^{2} + 48^{2}\\)\n\\(/BC/^{2} = 1296 + 2304 = 3600\\)\n\\(/BC/ = \\sqrt{3600} = 60 cm\\)\n\\(/AC/^{2} = /BC/^{2} + /AB/^{2}\\)\n\\((y + 48)^{2} = 60^{2} + 36^{2} + y^{2}\\)\n\\(y^{2} + 96y + 2304 = 3600 + 1296 + y^{2}\\)\n\\(96y = 3600 + 1296 - 2304 = 2592\\)\n\\(96y = 2592 \\implies y = \\frac{2592}{96} = 27 cm\\)\n\n', '(a) \\(6y + 5x = -12 ... (1)\\)\n\\(4y - 3x = 11 ... (2)\\)\nMultiply (1) by 2 and (2) by 3,\n\\(12y + 10x = -24 ... (1a)\\)\n\\(12y - 9x = 33... (2a)\\)\n\\((1a) - (2a) : 19x = -57 \\implies x = -3\\)\n\\(4y - 3(-3) = 11 \\implies 4y + 9 = 11\\)\n\\(4y = 2 \\implies y = \\frac{1}{2}\\)~(b)\n\\(/BC/^{2} = 36^{2} + 48^{2}\\)\n\\(/BC/^{2} = 1296 + 2304 = 3600\\)\n\\(/BC/ = \\sqrt{3600} = 60 cm\\)\n\\(/AC/^{2} = /BC/^{2} + /AB/^{2}\\)\n\\((y + 48)^{2} = 60^{2} + 36^{2} + y^{2}\\)\n\\(y^{2} + 96y + 2304 = 3600 + 1296 + y^{2}\\)\n\\(96y = 3600 + 1296 - 2304 = 2592\\)\n\\(96y = 2592 \\implies y = \\frac{2592}{96} = 27 cm\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(602, '2016', '\n\nQuestion\n\nThe table below shows the marks obtained by forty pupils in a Mathematics test.\n\n\n\nMarks\n0 - 9\n10 - 19\n20 - 29\n30 - 39\n40 - 49\n50 - 59\n\n\nNo of pupils\n4\n5\n6\n12\n8\n5\n\n\n\n(a) Draw a histogram for the mark distribution ;\n(b) Use your histogram to estimate the mode ;\n(c) Calculate the median of the distribution.\n\n\n\nThe table below shows the marks obtained by forty pupils in a Mathematics test.\n\n\n\nMarks\n0 - 9\n10 - 19\n20 - 29\n30 - 39\n40 - 49\n50 - 59\n\n\nNo of pupils\n4\n5\n6\n12\n8\n5\n\n\n', '', 'The table below shows the marks obtained by forty pupils in a Mathematics test.\n\n\n\nMarks\n0 - 9\n10 - 19\n20 - 29\n30 - 39\n40 - 49\n50 - 59\n\n\nNo of pupils\n4\n5\n6\n12\n8\n5\n\n\n\n(a) Draw a histogram for the mark distribution ;\n(b) Use your histogram to estimate the mode ;\n(c) Calculate the median of the distribution.\n\n\n\nThe table below shows the marks obtained by forty pupils in a Mathematics test.\n\n\n\nMarks\n0 - 9\n10 - 19\n20 - 29\n30 - 39\n40 - 49\n50 - 59\n\n\nNo of pupils\n4\n5\n6\n12\n8\n5\n', 'The table below shows the marks obtained by forty pupils in a Mathematics test.\n\n\n\nMarks\n0 - 9\n10 - 19\n20 - 29\n30 - 39\n40 - 49\n50 - 59\n\n\nNo of pupils\n4\n5\n6\n12\n8\n5\n\n\n\n(a) Draw a histogram for the mark distribution ;~(b) Use your histogram to estimate the mode ;\n(c) Calculate the median of the distribution.\n\n\n\nThe table below shows the marks obtained by forty pupils in a Mathematics test.\n\n\n\nMarks\n0 - 9\n10 - 19\n20 - 29\n30 - 39\n40 - 49\n50 - 59\n\n\nNo of pupils\n4\n5\n6\n12\n8\n5\n~(c) Calculate the median of the distribution.\n\n\n\nThe table below shows the marks obtained by forty pupils in a Mathematics test.\n\n\n\nMarks\n0 - 9\n10 - 19\n20 - 29\n30 - 39\n40 - 49\n50 - 59\n\n\nNo of pupils\n4\n5\n6\n12\n8\n5\n', NULL, '\n\nExplanation\n(a)\n\n\n\nMarks\nClass boundary\nFreq\nCum freq\n\n\n0 - 9\n0 - 9.5\n4\n4\n\n\n10 - 19\n9.5 - 19.5\n5\n9\n\n\n20 - 29\n19.5 - 29.5\n6\n15\n\n\n30 - 39\n29.5 - 39.5\n12\n27\n\n\n40 - 49\n39.5 - 49.5\n8\n35\n\n\n50 - 59\n49.5 - 59.5\n5\n40\n\n\n\n\n(b) Mode = 35.5\n(c) Median = \\(L_{1} + \\frac{(\\frac{N}{2} - \\sum f_{p}) \\times c}{f_{m}}\\)\nwhere \\(L_{1}\\) = lower class boundary of median class = 29.5\n\\(N = \\sum f = 40\\) ; \\(\\sum f_{p}\\) = cumulative frequency before median class = 15.\n\\(f_{m}\\) = frequency of median class = 12, c = class interval = 10.\nMedian = \\(29.5 + \\frac{(\\frac{40}{2} - 15) \\times 10}{12}\\)\n= \\(29.5 + \\frac{(20 - 15) \\times 10}{12}\\)\n= \\(29.5 + 4.17 = 33.67\\)\n\n', '', 'Explanation\n(a)\n\n\n\nMarks\nClass boundary\nFreq\nCum freq\n\n\n0 - 9\n0 - 9.5\n4\n4\n\n\n10 - 19\n9.5 - 19.5\n5\n9\n\n\n20 - 29\n19.5 - 29.5\n6\n15\n\n\n30 - 39\n29.5 - 39.5\n12\n27\n\n\n40 - 49\n39.5 - 49.5\n8\n35\n\n\n50 - 59\n49.5 - 59.5\n5\n40\n\n\n\n\n(b) Mode = 35.5\n(c) Median = \\(L_{1} + \\frac{(\\frac{N}{2} - \\sum f_{p}) \\times c}{f_{m}}\\)\nwhere \\(L_{1}\\) = lower class boundary of median class = 29.5\n\\(N = \\sum f = 40\\) ; \\(\\sum f_{p}\\) = cumulative frequency before median class = 15.\n\\(f_{m}\\) = frequency of median class = 12, c = class interval = 10.\nMedian = \\(29.5 + \\frac{(\\frac{40}{2} - 15) \\times 10}{12}\\)\n= \\(29.5 + \\frac{(20 - 15) \\times 10}{12}\\)\n= \\(29.5 + 4.17 = 33.67\\)\n\n', '(a)\n\n\n\nMarks\nClass boundary\nFreq\nCum freq\n\n\n0 - 9\n0 - 9.5\n4\n4\n\n\n10 - 19\n9.5 - 19.5\n5\n9\n\n\n20 - 29\n19.5 - 29.5\n6\n15\n\n\n30 - 39\n29.5 - 39.5\n12\n27\n\n\n40 - 49\n39.5 - 49.5\n8\n35\n\n\n50 - 59\n49.5 - 59.5\n5\n40\n\n\n\n~(b) Mode = 35.5\n(c) Median = \\(L_{1} + \\frac{(\\frac{N}{2} - \\sum f_{p}) \\times c}{f_{m}}\\)\nwhere \\(L_{1}\\) = lower class boundary of median class = 29.5\n\\(N = \\sum f = 40\\) ; \\(\\sum f_{p}\\) = cumulative frequency before median class = 15.\n\\(f_{m}\\) = frequency of median class = 12, c = class interval = 10.\nMedian = \\(29.5 + \\frac{(\\frac{40}{2} - 15) \\times 10}{12}\\)\n= \\(29.5 + \\frac{(20 - 15) \\times 10}{12}\\)\n= \\(29.5 + 4.17 = 33.67\\)\n~(c) Median = \\(L_{1} + \\frac{(\\frac{N}{2} - \\sum f_{p}) \\times c}{f_{m}}\\)\nwhere \\(L_{1}\\) = lower class boundary of median class = 29.5\n\\(N = \\sum f = 40\\) ; \\(\\sum f_{p}\\) = cumulative frequency before median class = 15.\n\\(f_{m}\\) = frequency of median class = 12, c = class interval = 10.\nMedian = \\(29.5 + \\frac{(\\frac{40}{2} - 15) \\times 10}{12}\\)\n= \\(29.5 + \\frac{(20 - 15) \\times 10}{12}\\)\n= \\(29.5 + 4.17 = 33.67\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1);
INSERT INTO `exam_simulation2` (`id`, `examYear`, `questions`, `parent_que`, `child_que`, `gen_child_que_desc`, `imageDir`, `explanation`, `parent_que_expl`, `child_que_exp`, `gen_child_que_expl`, `package_id`, `subject_id`, `status`) VALUES
(603, '2016', '\n\nQuestion\n\n(a) If \\(\\log_{10} 2 = 0.3010\\) and \\(\\log_{10} 3 = 0.4771\\), calculate without using tables, the value of \\(\\log_{10} 0.72\\).\n(b) A hawk on top of a tree, 20 metres high views a chick on the ground at an angle of depression of 39°. Find, correct to 2 significant figures, the distance of the chick from the bottom of the tree.\n\n', '', '(a) If \\(\\log_{10} 2 = 0.3010\\) and \\(\\log_{10} 3 = 0.4771\\), calculate without using tables, the value of \\(\\log_{10} 0.72\\).\n(b) A hawk on top of a tree, 20 metres high views a chick on the ground at an angle of depression of 39°. Find, correct to 2 significant figures, the distance of the chick from the bottom of the tree.', '(a) If \\(\\log_{10} 2 = 0.3010\\) and \\(\\log_{10} 3 = 0.4771\\), calculate without using tables, the value of \\(\\log_{10} 0.72\\).~(b) A hawk on top of a tree, 20 metres high views a chick on the ground at an angle of depression of 39°. Find, correct to 2 significant figures, the distance of the chick from the bottom of the tree.', NULL, '\n\nExplanation\n(a) \\(\\log_{10} 0.72 =\\log_{10} (\\frac{72}{100})\\)\n= \\(\\log_{10} 72 - \\log_{10} 100\\)\n= \\(72 = 2^{3} \\times 3^{2}\\)\n\\(\\log_{10} 72 = \\log_{10} 2^{3} + \\log_{10} 3^{2} = 3\\log_{10} 2 + 2 \\log_{10} 3\\)\n= \\(3(0.3010) + 2(0.4771) = 0.9030 + 0.9542 = 1.8572\\)\n\\(\\therefore \\log_{10} 0.72 \\equiv \\log_{10} 72 - \\log_{10} 100\\)\n= \\(1.8572 - 2 = -0.1428\\)\n(b) \\(\\tan 39° = \\frac{20}{AB} \\)\n= \\(\\frac{20}{\\tan 39°}\\)\n\\(AB = 24.698 m \\approxeq 25m\\)\n\n', '', 'Explanation\n(a) \\(\\log_{10} 0.72 =\\log_{10} (\\frac{72}{100})\\)\n= \\(\\log_{10} 72 - \\log_{10} 100\\)\n= \\(72 = 2^{3} \\times 3^{2}\\)\n\\(\\log_{10} 72 = \\log_{10} 2^{3} + \\log_{10} 3^{2} = 3\\log_{10} 2 + 2 \\log_{10} 3\\)\n= \\(3(0.3010) + 2(0.4771) = 0.9030 + 0.9542 = 1.8572\\)\n\\(\\therefore \\log_{10} 0.72 \\equiv \\log_{10} 72 - \\log_{10} 100\\)\n= \\(1.8572 - 2 = -0.1428\\)\n(b) \\(\\tan 39° = \\frac{20}{AB} \\)\n= \\(\\frac{20}{\\tan 39°}\\)\n\\(AB = 24.698 m \\approxeq 25m\\)\n\n', '(a) \\(\\log_{10} 0.72 =\\log_{10} (\\frac{72}{100})\\)\n= \\(\\log_{10} 72 - \\log_{10} 100\\)\n= \\(72 = 2^{3} \\times 3^{2}\\)\n\\(\\log_{10} 72 = \\log_{10} 2^{3} + \\log_{10} 3^{2} = 3\\log_{10} 2 + 2 \\log_{10} 3\\)\n= \\(3(0.3010) + 2(0.4771) = 0.9030 + 0.9542 = 1.8572\\)\n\\(\\therefore \\log_{10} 0.72 \\equiv \\log_{10} 72 - \\log_{10} 100\\)\n= \\(1.8572 - 2 = -0.1428\\)~(b) \\(\\tan 39° = \\frac{20}{AB} \\)\n= \\(\\frac{20}{\\tan 39°}\\)\n\\(AB = 24.698 m \\approxeq 25m\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(604, '2016', '\n\nQuestion\n\n(a) ABCD is a trapezium with AB parallel to DC and /AD/ = /AB/. If < BAD = 106°, find < BDC.\n(b) The table below shows the distribution of 20 cards labelled A - E.\n\n\n\nCard\nA\nB\nC\nD\nE\n\n\nFrequency\n3\n4\n7\n5\n1\n\n\n\n(i) If a card is selected at random from the pack, what is the probability that the card is E? (ii) If two cards are selected at random one after the other without replacement from the pack, what is the probability that one of the two cards is B?\n\n\n\n(a) ABCD is a trapezium with AB parallel to DC and /AD/ = /AB/. If < BAD = 106°, find < BDC.\n(b) The table below shows the distribution of 20 cards labelled A - E.\n\n\n\nCard\nA\nB\nC\nD\nE\n\n\nFrequency\n3\n4\n7\n5\n1\n\n\n', '', '(a) ABCD is a trapezium with AB parallel to DC and /AD/ = /AB/. If < BAD = 106°, find < BDC.\n(b) The table below shows the distribution of 20 cards labelled A - E.\n\n\n\nCard\nA\nB\nC\nD\nE\n\n\nFrequency\n3\n4\n7\n5\n1\n\n\n\n(i) If a card is selected at random from the pack, what is the probability that the card is E? (ii) If two cards are selected at random one after the other without replacement from the pack, what is the probability that one of the two cards is B?\n\n\n\n(a) ABCD is a trapezium with AB parallel to DC and /AD/ = /AB/. If < BAD = 106°, find < BDC.\n(b) The table below shows the distribution of 20 cards labelled A - E.\n\n\n\nCard\nA\nB\nC\nD\nE\n\n\nFrequency\n3\n4\n7\n5\n1\n', '(a) ABCD is a trapezium with AB parallel to DC and /AD/ = /AB/. If < BAD = 106°, find < BDC.\n(b) The table below shows the distribution of 20 cards labelled A - E.\n\n\n\nCard\nA\nB\nC\nD\nE\n\n\nFrequency\n3\n4\n7\n5\n1\n\n\n\n(i) If a card is selected at random from the pack, what is the probability that the card is E? (ii) If two cards are selected at random one after the other without replacement from the pack, what is the probability that one of the two cards is B?\n\n\n\n(a) ABCD is a trapezium with AB parallel to DC and /AD/ = /AB/. If < BAD = 106°, find < BDC.~(b) The table below shows the distribution of 20 cards labelled A - E.\n\n\n\nCard\nA\nB\nC\nD\nE\n\n\nFrequency\n3\n4\n7\n5\n1\n\n\n\n(i) If a card is selected at random from the pack, what is the probability that the card is E? (ii) If two cards are selected at random one after the other without replacement from the pack, what is the probability that one of the two cards is B?\n\n\n\n(a) ABCD is a trapezium with AB parallel to DC and /AD/ = /AB/. If < BAD = 106°, find < BDC.\n(b) The table below shows the distribution of 20 cards labelled A - E.\n\n\n\nCard\nA\nB\nC\nD\nE\n\n\nFrequency\n3\n4\n7\n5\n1\n', NULL, '\n\nExplanation\n(a)\n\\(< BAD = 106°\\) (Given)\n\\(/AD/ = /AB/\\) (given)\n\\(\\therefore < ADC = \\frac{180° - 106°}{2} = 37°\\) (base angles of an isosceles triangle)\n(b) Total card = 20\n(i) P(card picked is E) = \\(\\frac{1}{20}\\)\n(ii) P(B and A or C or D or E) or P(A or C or D or E and B)\n= \\(\\frac{4}{20} \\times \\frac{3 + 7 + 5 + 1}{19} + \\frac{3 + 7 + 5 + 1}{20} \\times \\frac{4}{19}\\)\n= \\(\\frac{64}{380} + \\frac{64}{380} = \\frac{32}{95}\\)\n\n', '', 'Explanation\n(a)\n\\(< BAD = 106°\\) (Given)\n\\(/AD/ = /AB/\\) (given)\n\\(\\therefore < ADC = \\frac{180° - 106°}{2} = 37°\\) (base angles of an isosceles triangle)\n(b) Total card = 20\n(i) P(card picked is E) = \\(\\frac{1}{20}\\)\n(ii) P(B and A or C or D or E) or P(A or C or D or E and B)\n= \\(\\frac{4}{20} \\times \\frac{3 + 7 + 5 + 1}{19} + \\frac{3 + 7 + 5 + 1}{20} \\times \\frac{4}{19}\\)\n= \\(\\frac{64}{380} + \\frac{64}{380} = \\frac{32}{95}\\)\n\n', '(a)\n\\(< BAD = 106°\\) (Given)\n\\(/AD/ = /AB/\\) (given)\n\\(\\therefore < ADC = \\frac{180° - 106°}{2} = 37°\\) (base angles of an isosceles triangle)~(b) Total card = 20\n(i) P(card picked is E) = \\(\\frac{1}{20}\\)\n(ii) P(B and A or C or D or E) or P(A or C or D or E and B)\n= \\(\\frac{4}{20} \\times \\frac{3 + 7 + 5 + 1}{19} + \\frac{3 + 7 + 5 + 1}{20} \\times \\frac{4}{19}\\)\n= \\(\\frac{64}{380} + \\frac{64}{380} = \\frac{32}{95}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(605, '2016', '\n\nQuestion\n\n(a) Without using tables, find the value of \\(\\frac{0.45 \\times 0.91}{0.0117}\\)\n(b) Find the number which is exactly halfway between \\(1\\frac{6}{7}\\) and \\(2\\frac{11}{28}\\).\n(c) If each interior angle of a regular polygon is five times the exterior angle, how many sides has the polygon?\n(d) Calculate the volume of the material used in making a pipe 20cm long, with an internal diameter 6cm and external diameter 8cm. [Take \\(pi = \\frac{22}{7}\\)].\n\n', '', '(a) Without using tables, find the value of \\(\\frac{0.45 \\times 0.91}{0.0117}\\)\n(b) Find the number which is exactly halfway between \\(1\\frac{6}{7}\\) and \\(2\\frac{11}{28}\\).\n(c) If each interior angle of a regular polygon is five times the exterior angle, how many sides has the polygon?\n(d) Calculate the volume of the material used in making a pipe 20cm long, with an internal diameter 6cm and external diameter 8cm. [Take \\(pi = \\frac{22}{7}\\)].', '(a) Without using tables, find the value of \\(\\frac{0.45 \\times 0.91}{0.0117}\\)~(b) Find the number which is exactly halfway between \\(1\\frac{6}{7}\\) and \\(2\\frac{11}{28}\\).\n(c) If each interior angle of a regular polygon is five times the exterior angle, how many sides has the polygon?\n(d) Calculate the volume of the material used in making a pipe 20cm long, with an internal diameter 6cm and external diameter 8cm. [Take \\(pi = \\frac{22}{7}\\)].~(c) If each interior angle of a regular polygon is five times the exterior angle, how many sides has the polygon?\n(d) Calculate the volume of the material used in making a pipe 20cm long, with an internal diameter 6cm and external diameter 8cm. [Take \\(pi = \\frac{22}{7}\\)].~(d) Calculate the volume of the material used in making a pipe 20cm long, with an internal diameter 6cm and external diameter 8cm. [Take \\(pi = \\frac{22}{7}\\)].', NULL, '\n\nExplanation\n(a) \\(\\frac{0.45 \\times 0.91}{0.0117} = \\frac{45 \\times 91}{117}\\)\n= \\(5 \\times 7 = 35\\)\n(b) Halfway between \\(1\\frac{6}{7}\\) and \\(2\\frac{11}{28}\\)\n= \\(\\frac{1\\frac{6}{7} + 2\\frac{11}{28}}{2}\\)\n= \\((\\frac{13}{7} + \\frac{67}{28}) \\times \\frac{1}{2}\\)\n= \\(\\frac{119}{28} \\times \\frac{1}{2} \\)\n= \\(\\frac{119}{56} = 2\\frac{7}{56} = 2\\frac{1}{8}\\)\n(c) Let the interior angle = 5x° and the exterior angle = x°.\n\\(\\therefore 5x° + x° = 180°\\)\n\\(6x° = 180° \\implies x = 30°\\)\nExterior angle = \\(\\frac{360°}{n} \\implies 30° = \\frac{360°}{n}\\)\n\\(n = \\frac{360°}{30°} = 12\\)\nThe polygon has 12 sides.\n(d) Length of pipe = 20 cm\nInternal diameter = 6 cm ; radius = 3cm\nExternal diameter = 8 cm ; radius = 4cm\nLet the external radius = R and internal radius = r\n\\(\\therefore\\) Volume of material used = \\(\\pi R^{2} l - \\pi r^{2} l\\)\nWhere l = length.\n\\(Volume = \\pi l (R^{2} - r^{2}) cm^{3}\\)\n= \\(\\frac{22}{7} \\times 20 (4^{2} - 3^{2}) = \\frac{22}{7} \\times 20 \\times 7 = 440 cm^{3}\\)\n\n\n', '', 'Explanation\n(a) \\(\\frac{0.45 \\times 0.91}{0.0117} = \\frac{45 \\times 91}{117}\\)\n= \\(5 \\times 7 = 35\\)\n(b) Halfway between \\(1\\frac{6}{7}\\) and \\(2\\frac{11}{28}\\)\n= \\(\\frac{1\\frac{6}{7} + 2\\frac{11}{28}}{2}\\)\n= \\((\\frac{13}{7} + \\frac{67}{28}) \\times \\frac{1}{2}\\)\n= \\(\\frac{119}{28} \\times \\frac{1}{2} \\)\n= \\(\\frac{119}{56} = 2\\frac{7}{56} = 2\\frac{1}{8}\\)\n(c) Let the interior angle = 5x° and the exterior angle = x°.\n\\(\\therefore 5x° + x° = 180°\\)\n\\(6x° = 180° \\implies x = 30°\\)\nExterior angle = \\(\\frac{360°}{n} \\implies 30° = \\frac{360°}{n}\\)\n\\(n = \\frac{360°}{30°} = 12\\)\nThe polygon has 12 sides.\n(d) Length of pipe = 20 cm\nInternal diameter = 6 cm ; radius = 3cm\nExternal diameter = 8 cm ; radius = 4cm\nLet the external radius = R and internal radius = r\n\\(\\therefore\\) Volume of material used = \\(\\pi R^{2} l - \\pi r^{2} l\\)\nWhere l = length.\n\\(Volume = \\pi l (R^{2} - r^{2}) cm^{3}\\)\n= \\(\\frac{22}{7} \\times 20 (4^{2} - 3^{2}) = \\frac{22}{7} \\times 20 \\times 7 = 440 cm^{3}\\)\n\n\n', '(a) \\(\\frac{0.45 \\times 0.91}{0.0117} = \\frac{45 \\times 91}{117}\\)\n= \\(5 \\times 7 = 35\\)~(b) Halfway between \\(1\\frac{6}{7}\\) and \\(2\\frac{11}{28}\\)\n= \\(\\frac{1\\frac{6}{7} + 2\\frac{11}{28}}{2}\\)\n= \\((\\frac{13}{7} + \\frac{67}{28}) \\times \\frac{1}{2}\\)\n= \\(\\frac{119}{28} \\times \\frac{1}{2} \\)\n= \\(\\frac{119}{56} = 2\\frac{7}{56} = 2\\frac{1}{8}\\)\n(c) Let the interior angle = 5x° and the exterior angle = x°.\n\\(\\therefore 5x° + x° = 180°\\)\n\\(6x° = 180° \\implies x = 30°\\)\nExterior angle = \\(\\frac{360°}{n} \\implies 30° = \\frac{360°}{n}\\)\n\\(n = \\frac{360°}{30°} = 12\\)\nThe polygon has 12 sides.\n(d) Length of pipe = 20 cm\nInternal diameter = 6 cm ; radius = 3cm\nExternal diameter = 8 cm ; radius = 4cm\nLet the external radius = R and internal radius = r\n\\(\\therefore\\) Volume of material used = \\(\\pi R^{2} l - \\pi r^{2} l\\)\nWhere l = length.\n\\(Volume = \\pi l (R^{2} - r^{2}) cm^{3}\\)\n= \\(\\frac{22}{7} \\times 20 (4^{2} - 3^{2}) = \\frac{22}{7} \\times 20 \\times 7 = 440 cm^{3}\\)\n\n~(c) Let the interior angle = 5x° and the exterior angle = x°.\n\\(\\therefore 5x° + x° = 180°\\)\n\\(6x° = 180° \\implies x = 30°\\)\nExterior angle = \\(\\frac{360°}{n} \\implies 30° = \\frac{360°}{n}\\)\n\\(n = \\frac{360°}{30°} = 12\\)\nThe polygon has 12 sides.\n(d) Length of pipe = 20 cm\nInternal diameter = 6 cm ; radius = 3cm\nExternal diameter = 8 cm ; radius = 4cm\nLet the external radius = R and internal radius = r\n\\(\\therefore\\) Volume of material used = \\(\\pi R^{2} l - \\pi r^{2} l\\)\nWhere l = length.\n\\(Volume = \\pi l (R^{2} - r^{2}) cm^{3}\\)\n= \\(\\frac{22}{7} \\times 20 (4^{2} - 3^{2}) = \\frac{22}{7} \\times 20 \\times 7 = 440 cm^{3}\\)\n\n~(d) Length of pipe = 20 cm\nInternal diameter = 6 cm ; radius = 3cm\nExternal diameter = 8 cm ; radius = 4cm\nLet the external radius = R and internal radius = r\n\\(\\therefore\\) Volume of material used = \\(\\pi R^{2} l - \\pi r^{2} l\\)\nWhere l = length.\n\\(Volume = \\pi l (R^{2} - r^{2}) cm^{3}\\)\n= \\(\\frac{22}{7} \\times 20 (4^{2} - 3^{2}) = \\frac{22}{7} \\times 20 \\times 7 = 440 cm^{3}\\)\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(606, '2016', '\n\nQuestion\n\n(a) The universal set U is the set of integers, P, Q and R are subsets of U defined as follows:\n\\(P = x :x \\leq 2 \\) ; \\(Q =x : -7 < x < 15\\) ; \\(R = x : -2 \\leq x < 19\\).\nFind (i) \\(P \\cap Q\\) ; (ii) \\(P \\cap (Q \\cup R\')\\), where R\' is the complement of R with respect to U.\n(b) The following data shows the marks of 40 students in a History examination.\n41 52 37 56 63 48 65 46 54 32 51 66 74 23 35 61 58 44 49 53 45 57 56 38 59 28 50 49 67 56 36 45 79 68 43 56 26 47 55 71.\n(i) Form a grouped frequency table with the class intervals 20 - 29, 30 - 39, 40 - 49 etc; (ii) Find the mean of the distribution.\n\n', '', '(a) The universal set U is the set of integers, P, Q and R are subsets of U defined as follows:\n\\(P = x :x \\leq 2 \\) ; \\(Q =x : -7 < x < 15\\) ; \\(R = x : -2 \\leq x < 19\\).\nFind (i) \\(P \\cap Q\\) ; (ii) \\(P \\cap (Q \\cup R\')\\), where R\' is the complement of R with respect to U.\n(b) The following data shows the marks of 40 students in a History examination.\n41 52 37 56 63 48 65 46 54 32 51 66 74 23 35 61 58 44 49 53 45 57 56 38 59 28 50 49 67 56 36 45 79 68 43 56 26 47 55 71.\n(i) Form a grouped frequency table with the class intervals 20 - 29, 30 - 39, 40 - 49 etc; (ii) Find the mean of the distribution.', '(a) The universal set U is the set of integers, P, Q and R are subsets of U defined as follows:\n\\(P = x :x \\leq 2 \\) ; \\(Q =x : -7 < x < 15\\) ; \\(R = x : -2 \\leq x < 19\\).\nFind (i) \\(P \\cap Q\\) ; (ii) \\(P \\cap (Q \\cup R\')\\), where R\' is the complement of R with respect to U.~(b) The following data shows the marks of 40 students in a History examination.\n41 52 37 56 63 48 65 46 54 32 51 66 74 23 35 61 58 44 49 53 45 57 56 38 59 28 50 49 67 56 36 45 79 68 43 56 26 47 55 71.\n(i) Form a grouped frequency table with the class intervals 20 - 29, 30 - 39, 40 - 49 etc; (ii) Find the mean of the distribution.', NULL, '\n\nExplanation\n(a) \\(P = x : x \\leq 2 = {..., -1, 0, 1, 2}\\)\n\\(Q = x : -7 < x < 15 = {-6, -5, -4, -3, -2, -1, 0, ..., 13, 14}\\)\n\\(R = x : -2 \\leq x < 19 = {-2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18}\\)\n(i) \\(P \\cap Q = {-6, -5, -4, -3, -2, -1, 0, 1, 2}\\)\n(ii) \\(R\' = {..., -5, -4, -3, 19, 20, ...}\\)\n\\(Q \\cup R\' = {..., -1, 0, 1, 2, 3, ...}\\)\n\\(\\therefore P \\cap (Q \\cup R\') = {..., -1, 0, 1, 2}\\)\n(b)(i)\n\n\n\nClass intervals\nMidpoint (x)\nFreq (f)\n\\(fx\\)\n\n\n20 - 29\n24.5\n3\n73.5\n\n\n30 - 39\n34.5\n5\n172.5\n\n\n40 - 49\n44.5\n10\n445\n\n\n50 - 59\n54.5\n13\n708.5\n\n\n60 - 69\n64.5\n6\n387\n\n\n70 - 79\n74.5\n3\n223.5\n\n\n\\(\\sum \\)\n\n40\n2010\n\n\n\n(ii) \\(Mean = \\frac{\\sum fx}{\\sum f} = \\frac{2010}{40}\\)\n= \\(50.25\\)\n\n\n', '', 'Explanation\n(a) \\(P = x : x \\leq 2 = {..., -1, 0, 1, 2}\\)\n\\(Q = x : -7 < x < 15 = {-6, -5, -4, -3, -2, -1, 0, ..., 13, 14}\\)\n\\(R = x : -2 \\leq x < 19 = {-2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18}\\)\n(i) \\(P \\cap Q = {-6, -5, -4, -3, -2, -1, 0, 1, 2}\\)\n(ii) \\(R\' = {..., -5, -4, -3, 19, 20, ...}\\)\n\\(Q \\cup R\' = {..., -1, 0, 1, 2, 3, ...}\\)\n\\(\\therefore P \\cap (Q \\cup R\') = {..., -1, 0, 1, 2}\\)\n(b)(i)\n\n\n\nClass intervals\nMidpoint (x)\nFreq (f)\n\\(fx\\)\n\n\n20 - 29\n24.5\n3\n73.5\n\n\n30 - 39\n34.5\n5\n172.5\n\n\n40 - 49\n44.5\n10\n445\n\n\n50 - 59\n54.5\n13\n708.5\n\n\n60 - 69\n64.5\n6\n387\n\n\n70 - 79\n74.5\n3\n223.5\n\n\n\\(\\sum \\)\n\n40\n2010\n\n\n\n(ii) \\(Mean = \\frac{\\sum fx}{\\sum f} = \\frac{2010}{40}\\)\n= \\(50.25\\)\n\n\n', '(a) \\(P = x : x \\leq 2 = {..., -1, 0, 1, 2}\\)\n\\(Q = x : -7 < x < 15 = {-6, -5, -4, -3, -2, -1, 0, ..., 13, 14}\\)\n\\(R = x : -2 \\leq x < 19 = {-2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18}\\)\n(i) \\(P \\cap Q = {-6, -5, -4, -3, -2, -1, 0, 1, 2}\\)\n(ii) \\(R\' = {..., -5, -4, -3, 19, 20, ...}\\)\n\\(Q \\cup R\' = {..., -1, 0, 1, 2, 3, ...}\\)\n\\(\\therefore P \\cap (Q \\cup R\') = {..., -1, 0, 1, 2}\\)~(b)(i)\n\n\n\nClass intervals\nMidpoint (x)\nFreq (f)\n\\(fx\\)\n\n\n20 - 29\n24.5\n3\n73.5\n\n\n30 - 39\n34.5\n5\n172.5\n\n\n40 - 49\n44.5\n10\n445\n\n\n50 - 59\n54.5\n13\n708.5\n\n\n60 - 69\n64.5\n6\n387\n\n\n70 - 79\n74.5\n3\n223.5\n\n\n\\(\\sum \\)\n\n40\n2010\n\n\n\n(ii) \\(Mean = \\frac{\\sum fx}{\\sum f} = \\frac{2010}{40}\\)\n= \\(50.25\\)\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(607, '2016', '\n\nQuestion\n\n(a) Find the number N such that when \\(\\frac{1}{3}\\) of it is added to 8, the result is the same as when \\(\\frac{1}{2}\\) of it is subtracted from 18.\n(b) Using a ruler and a pair of compasses only, construct a trapezium ABCD, in which the parallel sides AB and DC are 4 cm apart. < DAB = 60°, /AB/ = 8 cm and /BC/ = 5 cm. Measure /DC/.\n\n', '', '(a) Find the number N such that when \\(\\frac{1}{3}\\) of it is added to 8, the result is the same as when \\(\\frac{1}{2}\\) of it is subtracted from 18.\n(b) Using a ruler and a pair of compasses only, construct a trapezium ABCD, in which the parallel sides AB and DC are 4 cm apart. < DAB = 60°, /AB/ = 8 cm and /BC/ = 5 cm. Measure /DC/.', '(a) Find the number N such that when \\(\\frac{1}{3}\\) of it is added to 8, the result is the same as when \\(\\frac{1}{2}\\) of it is subtracted from 18.~(b) Using a ruler and a pair of compasses only, construct a trapezium ABCD, in which the parallel sides AB and DC are 4 cm apart. < DAB = 60°, /AB/ = 8 cm and /BC/ = 5 cm. Measure /DC/.', NULL, '\n\nExplanation\n(a) Let the number be N.\n\\(\\frac{N}{3} + 8 = 18 - \\frac{N}{2}\\)\n\\(\\frac{N}{3} + \\frac{N}{2} = 18 - 8 = 10\\)\n\\(\\frac{5N}{6} = 10 \\implies 5N = 60\\)\n\\(N = 12\\)\n(b)\n\n', '', 'Explanation\n(a) Let the number be N.\n\\(\\frac{N}{3} + 8 = 18 - \\frac{N}{2}\\)\n\\(\\frac{N}{3} + \\frac{N}{2} = 18 - 8 = 10\\)\n\\(\\frac{5N}{6} = 10 \\implies 5N = 60\\)\n\\(N = 12\\)\n(b)\n\n', '(a) Let the number be N.\n\\(\\frac{N}{3} + 8 = 18 - \\frac{N}{2}\\)\n\\(\\frac{N}{3} + \\frac{N}{2} = 18 - 8 = 10\\)\n\\(\\frac{5N}{6} = 10 \\implies 5N = 60\\)\n\\(N = 12\\)~(b)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(608, '2016', '\n\nQuestion\n\n(a) Copy and complete the table for \\(y = 3x^{2} - 5x - 7\\)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\n\\(y = 3x^{2} - 5x - 7\\)\n35\n\n\n-7\n-9\n\n5\n\n\n\n\n(b) Using a scale of 2cm = 1 unit along the x- axis and 2cm = 5 units along the y- axis, draw the graph of \\(y = 3x^{2} - 5x - 7\\).\n(c) On the same axis, draw the graph of \\(y + 3x + 2 = 0\\).\n(d) From your graph, find the : (i) range of values of x for which \\(3x^{2} - 5x - 7 < 0\\) ; (ii) roots of the equation \\(3x^{2} - 2x - 5 = 0\\).\n\n\n\n(a) Copy and complete the table for \\(y = 3x^{2} - 5x - 7\\)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\n\\(y = 3x^{2} - 5x - 7\\)\n35\n\n\n-7\n-9\n\n5\n\n\n\n', '', '(a) Copy and complete the table for \\(y = 3x^{2} - 5x - 7\\)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\n\\(y = 3x^{2} - 5x - 7\\)\n35\n\n\n-7\n-9\n\n5\n\n\n\n\n(b) Using a scale of 2cm = 1 unit along the x- axis and 2cm = 5 units along the y- axis, draw the graph of \\(y = 3x^{2} - 5x - 7\\).\n(c) On the same axis, draw the graph of \\(y + 3x + 2 = 0\\).\n(d) From your graph, find the : (i) range of values of x for which \\(3x^{2} - 5x - 7 < 0\\) ; (ii) roots of the equation \\(3x^{2} - 2x - 5 = 0\\).\n\n\n\n(a) Copy and complete the table for \\(y = 3x^{2} - 5x - 7\\)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\n\\(y = 3x^{2} - 5x - 7\\)\n35\n\n\n-7\n-9\n\n5\n\n', '(a) Copy and complete the table for \\(y = 3x^{2} - 5x - 7\\)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\n\\(y = 3x^{2} - 5x - 7\\)\n35\n\n\n-7\n-9\n\n5\n\n\n\n~(b) Using a scale of 2cm = 1 unit along the x- axis and 2cm = 5 units along the y- axis, draw the graph of \\(y = 3x^{2} - 5x - 7\\).\n(c) On the same axis, draw the graph of \\(y + 3x + 2 = 0\\).\n(d) From your graph, find the : (i) range of values of x for which \\(3x^{2} - 5x - 7 < 0\\) ; (ii) roots of the equation \\(3x^{2} - 2x - 5 = 0\\).\n\n\n\n(a) Copy and complete the table for \\(y = 3x^{2} - 5x - 7\\)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\n\\(y = 3x^{2} - 5x - 7\\)\n35\n\n\n-7\n-9\n\n5\n\n~(c) On the same axis, draw the graph of \\(y + 3x + 2 = 0\\).\n(d) From your graph, find the : (i) range of values of x for which \\(3x^{2} - 5x - 7 < 0\\) ; (ii) roots of the equation \\(3x^{2} - 2x - 5 = 0\\).\n\n\n\n(a) Copy and complete the table for \\(y = 3x^{2} - 5x - 7\\)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\n\\(y = 3x^{2} - 5x - 7\\)\n35\n\n\n-7\n-9\n\n5\n\n~(d) From your graph, find the : (i) range of values of x for which \\(3x^{2} - 5x - 7 < 0\\) ; (ii) roots of the equation \\(3x^{2} - 2x - 5 = 0\\).\n\n\n\n(a) Copy and complete the table for \\(y = 3x^{2} - 5x - 7\\)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\n\\(y = 3x^{2} - 5x - 7\\)\n35\n\n\n-7\n-9\n\n5\n\n', NULL, '\n\nExplanation\n(a)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\n\\(y = 3x^{2} - 5x - 7\\)\n35\n15\n1\n-7\n-9\n-5\n5\n21\n\n\n\n(b)\n(c) \\(y + 3x + 2 = 0 \\implies y = -3x - 2\\)\nwhen x = -2, y = 6 - 2 = 4.\nwhen x = 3, y = -9 - 2 = -11.\n(d) \\(3x^{2} - 2x - 5 = 0 = 3x^{2} - 5x - 7 = -3x - 2\\)\nThe roots of the equation \\(3x^{2} - 2x - 5 = 0\\) is when the line \\(y = -3x - 2\\) cuts the equation i.e. x = 1.0 or 1.7.\n\n', '', 'Explanation\n(a)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\n\\(y = 3x^{2} - 5x - 7\\)\n35\n15\n1\n-7\n-9\n-5\n5\n21\n\n\n\n(b)\n(c) \\(y + 3x + 2 = 0 \\implies y = -3x - 2\\)\nwhen x = -2, y = 6 - 2 = 4.\nwhen x = 3, y = -9 - 2 = -11.\n(d) \\(3x^{2} - 2x - 5 = 0 = 3x^{2} - 5x - 7 = -3x - 2\\)\nThe roots of the equation \\(3x^{2} - 2x - 5 = 0\\) is when the line \\(y = -3x - 2\\) cuts the equation i.e. x = 1.0 or 1.7.\n\n', '(a)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\n\\(y = 3x^{2} - 5x - 7\\)\n35\n15\n1\n-7\n-9\n-5\n5\n21\n\n\n~(b)\n(c) \\(y + 3x + 2 = 0 \\implies y = -3x - 2\\)\nwhen x = -2, y = 6 - 2 = 4.\nwhen x = 3, y = -9 - 2 = -11.\n(d) \\(3x^{2} - 2x - 5 = 0 = 3x^{2} - 5x - 7 = -3x - 2\\)\nThe roots of the equation \\(3x^{2} - 2x - 5 = 0\\) is when the line \\(y = -3x - 2\\) cuts the equation i.e. x = 1.0 or 1.7.\n~(c) \\(y + 3x + 2 = 0 \\implies y = -3x - 2\\)\nwhen x = -2, y = 6 - 2 = 4.\nwhen x = 3, y = -9 - 2 = -11.\n(d) \\(3x^{2} - 2x - 5 = 0 = 3x^{2} - 5x - 7 = -3x - 2\\)\nThe roots of the equation \\(3x^{2} - 2x - 5 = 0\\) is when the line \\(y = -3x - 2\\) cuts the equation i.e. x = 1.0 or 1.7.\n~(d) \\(3x^{2} - 2x - 5 = 0 = 3x^{2} - 5x - 7 = -3x - 2\\)\nThe roots of the equation \\(3x^{2} - 2x - 5 = 0\\) is when the line \\(y = -3x - 2\\) cuts the equation i.e. x = 1.0 or 1.7.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(609, '2016', '\n\nQuestion\n\n(a) Make d the subject of the formula \\(S = \\frac{n}{2}[2a + (n - 1) d]\\).\n(b) (i)\nIn the diagram, O is the centre of the circle, A, B and P are points on the circumference. Prove that < AOB = 2 < APB.\n(ii)\nFind the angles x, v and z in the diagram.\n\n', '', '(a) Make d the subject of the formula \\(S = \\frac{n}{2}[2a + (n - 1) d]\\).\n(b) (i)\nIn the diagram, O is the centre of the circle, A, B and P are points on the circumference. Prove that < AOB = 2 < APB.\n(ii)\nFind the angles x, v and z in the diagram.', '(a) Make d the subject of the formula \\(S = \\frac{n}{2}[2a + (n - 1) d]\\).~(b) (i)\nIn the diagram, O is the centre of the circle, A, B and P are points on the circumference. Prove that < AOB = 2 < APB.\n(ii)\nFind the angles x, v and z in the diagram.', NULL, '\n\nExplanation\n(a) \\(S = \\frac{n}{2} [2a + (n - 1)d]\\)\n\n\\(2S = n[2a + (n - 1) d]\\)\n\\(\\frac{2S}{n} = 2a + (n - 1)d\\)\n\\(\\frac{2S}{n} - 2a = (n - 1) d\\)\n\\(d = \\frac{\\frac{2S}{n} - 2a}{(n - 1)}\\)\n= \\(\\frac{2S - 2an}{n(n - 1)}\\)\n(b)\n(i) Given: circle ABP\nTo prove < AOB = 2 < APB\nConstruction : Join PO produced to Q.\nProof : /OA/ = /OB/ (radii)\n\\(\\therefore x_{1} = x_{2} \\) (base angles of isosceles triangle)\n\\(\\therefore < AOQ = x_{1} + x_{2}\\) (exterior angle of triangle AOP)\n\\(\\therefore < AOQ = 2x_{2} (x_{1} = x_{2})\\)\nSimilarly, < AOB = < BOQ - < AOQ.\n= \\(2y_{2} - 2x_{2} = 2(y_{2} - x_{2}) = 2 \\times < APB\\)\n\\(\\therefore < AOB = 2 < APB.\\)\n(ii) \\(z = 2x \\) (angle subtended at the centre)\n\\(z = 2(126°) = 252°\\)\n\\(\\therefore x = 360° - 252° = 108°\\)\n\\(\\therefore y = \\frac{1}{2} \\times 108° = 54°\\)\n\\(x = 108° ; y = 54° ; z = 252°\\)\n\n', '', 'Explanation\n(a) \\(S = \\frac{n}{2} [2a + (n - 1)d]\\)\n\n\\(2S = n[2a + (n - 1) d]\\)\n\\(\\frac{2S}{n} = 2a + (n - 1)d\\)\n\\(\\frac{2S}{n} - 2a = (n - 1) d\\)\n\\(d = \\frac{\\frac{2S}{n} - 2a}{(n - 1)}\\)\n= \\(\\frac{2S - 2an}{n(n - 1)}\\)\n(b)\n(i) Given: circle ABP\nTo prove < AOB = 2 < APB\nConstruction : Join PO produced to Q.\nProof : /OA/ = /OB/ (radii)\n\\(\\therefore x_{1} = x_{2} \\) (base angles of isosceles triangle)\n\\(\\therefore < AOQ = x_{1} + x_{2}\\) (exterior angle of triangle AOP)\n\\(\\therefore < AOQ = 2x_{2} (x_{1} = x_{2})\\)\nSimilarly, < AOB = < BOQ - < AOQ.\n= \\(2y_{2} - 2x_{2} = 2(y_{2} - x_{2}) = 2 \\times < APB\\)\n\\(\\therefore < AOB = 2 < APB.\\)\n(ii) \\(z = 2x \\) (angle subtended at the centre)\n\\(z = 2(126°) = 252°\\)\n\\(\\therefore x = 360° - 252° = 108°\\)\n\\(\\therefore y = \\frac{1}{2} \\times 108° = 54°\\)\n\\(x = 108° ; y = 54° ; z = 252°\\)\n\n', '(a) \\(S = \\frac{n}{2} [2a + (n - 1)d]\\)\n\n\\(2S = n[2a + (n - 1) d]\\)\n\\(\\frac{2S}{n} = 2a + (n - 1)d\\)\n\\(\\frac{2S}{n} - 2a = (n - 1) d\\)\n\\(d = \\frac{\\frac{2S}{n} - 2a}{(n - 1)}\\)\n= \\(\\frac{2S - 2an}{n(n - 1)}\\)~(b)\n(i) Given: circle ABP\nTo prove < AOB = 2 < APB\nConstruction : Join PO produced to Q.\nProof : /OA/ = /OB/ (radii)\n\\(\\therefore x_{1} = x_{2} \\) (base angles of isosceles triangle)\n\\(\\therefore < AOQ = x_{1} + x_{2}\\) (exterior angle of triangle AOP)\n\\(\\therefore < AOQ = 2x_{2} (x_{1} = x_{2})\\)\nSimilarly, < AOB = < BOQ - < AOQ.\n= \\(2y_{2} - 2x_{2} = 2(y_{2} - x_{2}) = 2 \\times < APB\\)\n\\(\\therefore < AOB = 2 < APB.\\)\n(ii) \\(z = 2x \\) (angle subtended at the centre)\n\\(z = 2(126°) = 252°\\)\n\\(\\therefore x = 360° - 252° = 108°\\)\n\\(\\therefore y = \\frac{1}{2} \\times 108° = 54°\\)\n\\(x = 108° ; y = 54° ; z = 252°\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(610, '2016', '\n\nQuestion\n\n(a) P and Q are points on the parallel of latitude 68.7°S, their longitudes being 124°W and 56°E respectively. What is their distance apart measured along the parallel of latitude? [Take R = 6400km, \\(\\pi = 3.142\\)]. (Give your answers to 3 significant figures).\n(b) A bag contains four red, three white and five green balls. (i) If one ball is picked at random, what is the probability that it is not green? (ii) if two balls are picked at random without replacement, what is the probability that one is red and the other white?\n\n', '', '(a) P and Q are points on the parallel of latitude 68.7°S, their longitudes being 124°W and 56°E respectively. What is their distance apart measured along the parallel of latitude? [Take R = 6400km, \\(\\pi = 3.142\\)]. (Give your answers to 3 significant figures).\n(b) A bag contains four red, three white and five green balls. (i) If one ball is picked at random, what is the probability that it is not green? (ii) if two balls are picked at random without replacement, what is the probability that one is red and the other white?', '(a) P and Q are points on the parallel of latitude 68.7°S, their longitudes being 124°W and 56°E respectively. What is their distance apart measured along the parallel of latitude? [Take R = 6400km, \\(\\pi = 3.142\\)]. (Give your answers to 3 significant figures).~(b) A bag contains four red, three white and five green balls. (i) If one ball is picked at random, what is the probability that it is not green? (ii) if two balls are picked at random without replacement, what is the probability that one is red and the other white?', NULL, '\n\nExplanation\n(a)\nDifference in longitudes = 124° + 56° = 180°\nDistance PQ = \\(\\frac{180°}{360°} \\times 2\\pi r\\)\nwhere \\(r = R \\cos \\theta\\)\n\\(/PQ/ = \\frac{180}{360} \\times 2 \\times 3.142 \\times R \\cos 68.7°\\)\n\\(/PQ/ = 3.142 \\times 6400 \\times 0.3633\\)\n= \\(7305.53 km\\)\n\\(\\approxeq 7300 km\\) (3 sig. figs)\n(b) Red balls = 4, white = 3 ; green = 5\nTotal = 4 + 3 + 5 = 12 balls.\n(i) P(green) = \\(\\frac{5}{12}\\)\nP(not green) = \\(1 - \\frac{5}{12} = \\frac{7}{12}\\)\n(ii) P(picking red and white) = P( red then white) or P(white then red)\n= \\(\\frac{4}{12} \\times \\frac{3}{11} + \\frac{3}{12} \\times \\frac{4}{11}\\)\n= \\(\\frac{2}{11}\\)\n\n', '', 'Explanation\n(a)\nDifference in longitudes = 124° + 56° = 180°\nDistance PQ = \\(\\frac{180°}{360°} \\times 2\\pi r\\)\nwhere \\(r = R \\cos \\theta\\)\n\\(/PQ/ = \\frac{180}{360} \\times 2 \\times 3.142 \\times R \\cos 68.7°\\)\n\\(/PQ/ = 3.142 \\times 6400 \\times 0.3633\\)\n= \\(7305.53 km\\)\n\\(\\approxeq 7300 km\\) (3 sig. figs)\n(b) Red balls = 4, white = 3 ; green = 5\nTotal = 4 + 3 + 5 = 12 balls.\n(i) P(green) = \\(\\frac{5}{12}\\)\nP(not green) = \\(1 - \\frac{5}{12} = \\frac{7}{12}\\)\n(ii) P(picking red and white) = P( red then white) or P(white then red)\n= \\(\\frac{4}{12} \\times \\frac{3}{11} + \\frac{3}{12} \\times \\frac{4}{11}\\)\n= \\(\\frac{2}{11}\\)\n\n', '(a)\nDifference in longitudes = 124° + 56° = 180°\nDistance PQ = \\(\\frac{180°}{360°} \\times 2\\pi r\\)\nwhere \\(r = R \\cos \\theta\\)\n\\(/PQ/ = \\frac{180}{360} \\times 2 \\times 3.142 \\times R \\cos 68.7°\\)\n\\(/PQ/ = 3.142 \\times 6400 \\times 0.3633\\)\n= \\(7305.53 km\\)\n\\(\\approxeq 7300 km\\) (3 sig. figs)~(b) Red balls = 4, white = 3 ; green = 5\nTotal = 4 + 3 + 5 = 12 balls.\n(i) P(green) = \\(\\frac{5}{12}\\)\nP(not green) = \\(1 - \\frac{5}{12} = \\frac{7}{12}\\)\n(ii) P(picking red and white) = P( red then white) or P(white then red)\n= \\(\\frac{4}{12} \\times \\frac{3}{11} + \\frac{3}{12} \\times \\frac{4}{11}\\)\n= \\(\\frac{2}{11}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(611, '2016', '\n\nQuestion\n\nTwo men P and Q set off from a base camp R, prospecting for oil. P moves 20km on a bearing of 205° and Q moves 15km on a bearing of 060°. Calculate the:\n(a) distance of Q from P ;\n(b) bearing of Q from P.\n(Give your answer in each case to the nearest whole number)\n\n', '', 'Two men P and Q set off from a base camp R, prospecting for oil. P moves 20km on a bearing of 205° and Q moves 15km on a bearing of 060°. Calculate the:\n(a) distance of Q from P ;\n(b) bearing of Q from P.\n(Give your answer in each case to the nearest whole number)', 'Two men P and Q set off from a base camp R, prospecting for oil. P moves 20km on a bearing of 205° and Q moves 15km on a bearing of 060°. Calculate the:\n(a) distance of Q from P ;~(b) bearing of Q from P.\n(Give your answer in each case to the nearest whole number)', NULL, '\n\nExplanation\n(a) By cosine rule, \\(/PQ/^{2} = 20^{2} + 15^{2} - 2 \\times 20 \\times 15 \\times \\cos 145°\\)\n= \\(400 + 225 - 600 \\cos 145°\\)\n= \\(625 + 600 \\times 0.8192\\)\n\n= \\(491.52 + 625\\)\n\\(/PQ/^{2} = 1116.52 \\implies /PQ/ = \\sqrt{1116.52} = 33.41\\)\n\\(\\approxeq 33km\\) ( to the nearest whole number)\n(b) By sine rule,\n\\(\\frac{\\sin < QPR}{15} = \\frac{\\sin 145}{33.41}\\)\n\\(\\sin < QPR = \\frac{15 \\sin 145}{33.41}\\)\n\\(\\sin < QPR = \\frac{15 \\times 0.5736}{33.41} = 0.2575\\)\n\\(< QPR = 14.92°\\)\n\\(\\therefore\\) Bearing of Q from P = 25° + 14.92° = 39.92°.\n\\(\\approxeq 40°\\) (to nearest whole number)\n\n', '', 'Explanation\n(a) By cosine rule, \\(/PQ/^{2} = 20^{2} + 15^{2} - 2 \\times 20 \\times 15 \\times \\cos 145°\\)\n= \\(400 + 225 - 600 \\cos 145°\\)\n= \\(625 + 600 \\times 0.8192\\)\n\n= \\(491.52 + 625\\)\n\\(/PQ/^{2} = 1116.52 \\implies /PQ/ = \\sqrt{1116.52} = 33.41\\)\n\\(\\approxeq 33km\\) ( to the nearest whole number)\n(b) By sine rule,\n\\(\\frac{\\sin < QPR}{15} = \\frac{\\sin 145}{33.41}\\)\n\\(\\sin < QPR = \\frac{15 \\sin 145}{33.41}\\)\n\\(\\sin < QPR = \\frac{15 \\times 0.5736}{33.41} = 0.2575\\)\n\\(< QPR = 14.92°\\)\n\\(\\therefore\\) Bearing of Q from P = 25° + 14.92° = 39.92°.\n\\(\\approxeq 40°\\) (to nearest whole number)\n\n', '(a) By cosine rule, \\(/PQ/^{2} = 20^{2} + 15^{2} - 2 \\times 20 \\times 15 \\times \\cos 145°\\)\n= \\(400 + 225 - 600 \\cos 145°\\)\n= \\(625 + 600 \\times 0.8192\\)\n\n= \\(491.52 + 625\\)\n\\(/PQ/^{2} = 1116.52 \\implies /PQ/ = \\sqrt{1116.52} = 33.41\\)\n\\(\\approxeq 33km\\) ( to the nearest whole number)~(b) By sine rule,\n\\(\\frac{\\sin < QPR}{15} = \\frac{\\sin 145}{33.41}\\)\n\\(\\sin < QPR = \\frac{15 \\sin 145}{33.41}\\)\n\\(\\sin < QPR = \\frac{15 \\times 0.5736}{33.41} = 0.2575\\)\n\\(< QPR = 14.92°\\)\n\\(\\therefore\\) Bearing of Q from P = 25° + 14.92° = 39.92°.\n\\(\\approxeq 40°\\) (to nearest whole number)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(612, '2016', '\n\nQuestion\n\n(a) Factorise : \\(px - 2px - 4qy + 2py\\)\n(b) Given that the universal set U = {1, 2, 3, 4,5, 6, 7, 8, 9, 10}, P = {1, 2, 4, 6, 10} and Q = {2, 3, 6, 9}; show that \\((P \\cup Q)\' = P\' \\cap Q\'\\)\n\n', '', '(a) Factorise : \\(px - 2px - 4qy + 2py\\)\n(b) Given that the universal set U = {1, 2, 3, 4,5, 6, 7, 8, 9, 10}, P = {1, 2, 4, 6, 10} and Q = {2, 3, 6, 9}; show that \\((P \\cup Q)\' = P\' \\cap Q\'\\)', '(a) Factorise : \\(px - 2px - 4qy + 2py\\)~(b) Given that the universal set U = {1, 2, 3, 4,5, 6, 7, 8, 9, 10}, P = {1, 2, 4, 6, 10} and Q = {2, 3, 6, 9}; show that \\((P \\cup Q)\' = P\' \\cap Q\'\\)', NULL, '\n\nExplanation\n(a) \\(px - 2qx - 4qy + 2py\\)\n= \\(x(p - 2q) +2y(p - 2q)\\)\n= \\((x + 2y)(p - 2q)\\)\n(b) \\(U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}\\)\n\\(P = {1, 2, 4, 6, 10} ; Q = {2, 3, 6, 9}\\)\n\\((P \\cup Q) = {1, 2, 3, 4, 6, 9, 10}\\)\n\\((P \\cup Q)\' = {5, 7, 8}\\)\n\\(P\' \\cap Q\' = {3, 5, 7, 8, 9} \\cap {1, 4, 5, 7, 8, 10}\\)\n= \\({5, 7, 8}\\)\nHence, \\((P \\cup Q)\' = P\' \\cap Q\'\\).\n\n', '', 'Explanation\n(a) \\(px - 2qx - 4qy + 2py\\)\n= \\(x(p - 2q) +2y(p - 2q)\\)\n= \\((x + 2y)(p - 2q)\\)\n(b) \\(U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}\\)\n\\(P = {1, 2, 4, 6, 10} ; Q = {2, 3, 6, 9}\\)\n\\((P \\cup Q) = {1, 2, 3, 4, 6, 9, 10}\\)\n\\((P \\cup Q)\' = {5, 7, 8}\\)\n\\(P\' \\cap Q\' = {3, 5, 7, 8, 9} \\cap {1, 4, 5, 7, 8, 10}\\)\n= \\({5, 7, 8}\\)\nHence, \\((P \\cup Q)\' = P\' \\cap Q\'\\).\n\n', '(a) \\(px - 2qx - 4qy + 2py\\)\n= \\(x(p - 2q) +2y(p - 2q)\\)\n= \\((x + 2y)(p - 2q)\\)~(b) \\(U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}\\)\n\\(P = {1, 2, 4, 6, 10} ; Q = {2, 3, 6, 9}\\)\n\\((P \\cup Q) = {1, 2, 3, 4, 6, 9, 10}\\)\n\\((P \\cup Q)\' = {5, 7, 8}\\)\n\\(P\' \\cap Q\' = {3, 5, 7, 8, 9} \\cap {1, 4, 5, 7, 8, 10}\\)\n= \\({5, 7, 8}\\)\nHence, \\((P \\cup Q)\' = P\' \\cap Q\'\\).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(613, '2016', '\n\nQuestion\n\nThe quantity y is partly constant and partly varies inversely as the square of x.\n(a) Write down the relationship between x and y.\n(b) When x = 1, y = 11 and when x = 2, y = 5, find the value of y when x = 4.\n\n', '', 'The quantity y is partly constant and partly varies inversely as the square of x.\n(a) Write down the relationship between x and y.\n(b) When x = 1, y = 11 and when x = 2, y = 5, find the value of y when x = 4.', 'The quantity y is partly constant and partly varies inversely as the square of x.\n(a) Write down the relationship between x and y.~(b) When x = 1, y = 11 and when x = 2, y = 5, find the value of y when x = 4.', NULL, '\n\nExplanation\n(a) \\(y = k + \\frac{c}{x^{2}}\\)\nwhere c and k are constants.\n(b) When x = 1, y = 11\n\\(11 = k + \\frac{c}{1^{2}} \\implies 11 = k + c ... (1)\\)\nWhen x = 2, y = 5\n\\(5 = k + \\frac{c}{2^{2}} \\implies 5 = k + \\frac{c}{4}\\)\n\\(\\equiv 20 = 4k + c ... (2)\\)\n(2) - (1) :\n\\(4k - k = 20 - 11 \\implies 3k = 9\\)\n\\(k = 3\\)\nPut k = 3 in (1), we have\n\\(11 = 3 + c \\implies c = 11 - 3 = 8\\)\n\\(\\therefore y = 3 + \\frac{8}{x^{2}}\\)\nWhen x = 4,\n\\(y = 3 + \\frac{8}{4^{2}} = 3 + \\frac{1}{2}\\)\n= \\(3\\frac{1}{2}\\)\n\n', '', 'Explanation\n(a) \\(y = k + \\frac{c}{x^{2}}\\)\nwhere c and k are constants.\n(b) When x = 1, y = 11\n\\(11 = k + \\frac{c}{1^{2}} \\implies 11 = k + c ... (1)\\)\nWhen x = 2, y = 5\n\\(5 = k + \\frac{c}{2^{2}} \\implies 5 = k + \\frac{c}{4}\\)\n\\(\\equiv 20 = 4k + c ... (2)\\)\n(2) - (1) :\n\\(4k - k = 20 - 11 \\implies 3k = 9\\)\n\\(k = 3\\)\nPut k = 3 in (1), we have\n\\(11 = 3 + c \\implies c = 11 - 3 = 8\\)\n\\(\\therefore y = 3 + \\frac{8}{x^{2}}\\)\nWhen x = 4,\n\\(y = 3 + \\frac{8}{4^{2}} = 3 + \\frac{1}{2}\\)\n= \\(3\\frac{1}{2}\\)\n\n', '(a) \\(y = k + \\frac{c}{x^{2}}\\)\nwhere c and k are constants.~(b) When x = 1, y = 11\n\\(11 = k + \\frac{c}{1^{2}} \\implies 11 = k + c ... (1)\\)\nWhen x = 2, y = 5\n\\(5 = k + \\frac{c}{2^{2}} \\implies 5 = k + \\frac{c}{4}\\)\n\\(\\equiv 20 = 4k + c ... (2)\\)\n(2) - (1) :\n\\(4k - k = 20 - 11 \\implies 3k = 9\\)\n\\(k = 3\\)\nPut k = 3 in (1), we have\n\\(11 = 3 + c \\implies c = 11 - 3 = 8\\)\n\\(\\therefore y = 3 + \\frac{8}{x^{2}}\\)\nWhen x = 4,\n\\(y = 3 + \\frac{8}{4^{2}} = 3 + \\frac{1}{2}\\)\n= \\(3\\frac{1}{2}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(614, '2016', '\n\nQuestion\n\n(a) In the diagram, PQSR and SRYZ are parallelograms and PQYZ is a straight line. If /QY/ = 2cm and /RS/ = 3cm, find /PZ/.\n(b) P and Q are two towns on the earth\'s surface on latitude 56°N. Thei longitudes are 25°E and 95°E respectively. Find the distance PQ along their parallel of latitude, correct to the nearest km. [Take radius of the earth as 6400km and \\(\\pi = \\frac{22}{7}\\)]\n\n\n', '', '(a) In the diagram, PQSR and SRYZ are parallelograms and PQYZ is a straight line. If /QY/ = 2cm and /RS/ = 3cm, find /PZ/.\n(b) P and Q are two towns on the earth\'s surface on latitude 56°N. Thei longitudes are 25°E and 95°E respectively. Find the distance PQ along their parallel of latitude, correct to the nearest km. [Take radius of the earth as 6400km and \\(\\pi = \\frac{22}{7}\\)]\n', '(a) In the diagram, PQSR and SRYZ are parallelograms and PQYZ is a straight line. If /QY/ = 2cm and /RS/ = 3cm, find /PZ/.~(b) P and Q are two towns on the earth\'s surface on latitude 56°N. Thei longitudes are 25°E and 95°E respectively. Find the distance PQ along their parallel of latitude, correct to the nearest km. [Take radius of the earth as 6400km and \\(\\pi = \\frac{22}{7}\\)]\n', NULL, '\n\nExplanation\n(a) \\(/QY/ = 2cm; /RS/ = 3cm\\)\n\\(\\therefore /YZ/ = /RS/ = /PQ/ = 3cm\\) (opp sides of a parallelogram)\nBut \\(/PZ/ = /PQ/ + /QY/ + /YZ/\\)\n\\(\\therefore /PZ/ = 3cm + 2cm + 3cm = 8cm\\)\n(b) Difference in longitude = 95° - 25° = 70°\n\nDistance PQ = \\(\\frac{70}{360} \\times 2\\pi R \\cos \\theta\\)\n= \\(\\frac{70}{360} \\times 2 \\times \\frac{22}{7} \\times 6400 \\cos 56\\)\n= \\(\\frac{7}{36} \\times 2 \\times \\frac{22}{7} \\times 6400 \\times 0.5592\\)\n= \\(4374.19 km \\approxeq 4374 km\\) (to the nearest km)\n\n', '', 'Explanation\n(a) \\(/QY/ = 2cm; /RS/ = 3cm\\)\n\\(\\therefore /YZ/ = /RS/ = /PQ/ = 3cm\\) (opp sides of a parallelogram)\nBut \\(/PZ/ = /PQ/ + /QY/ + /YZ/\\)\n\\(\\therefore /PZ/ = 3cm + 2cm + 3cm = 8cm\\)\n(b) Difference in longitude = 95° - 25° = 70°\n\nDistance PQ = \\(\\frac{70}{360} \\times 2\\pi R \\cos \\theta\\)\n= \\(\\frac{70}{360} \\times 2 \\times \\frac{22}{7} \\times 6400 \\cos 56\\)\n= \\(\\frac{7}{36} \\times 2 \\times \\frac{22}{7} \\times 6400 \\times 0.5592\\)\n= \\(4374.19 km \\approxeq 4374 km\\) (to the nearest km)\n\n', '(a) \\(/QY/ = 2cm; /RS/ = 3cm\\)\n\\(\\therefore /YZ/ = /RS/ = /PQ/ = 3cm\\) (opp sides of a parallelogram)\nBut \\(/PZ/ = /PQ/ + /QY/ + /YZ/\\)\n\\(\\therefore /PZ/ = 3cm + 2cm + 3cm = 8cm\\)~(b) Difference in longitude = 95° - 25° = 70°\n\nDistance PQ = \\(\\frac{70}{360} \\times 2\\pi R \\cos \\theta\\)\n= \\(\\frac{70}{360} \\times 2 \\times \\frac{22}{7} \\times 6400 \\cos 56\\)\n= \\(\\frac{7}{36} \\times 2 \\times \\frac{22}{7} \\times 6400 \\times 0.5592\\)\n= \\(4374.19 km \\approxeq 4374 km\\) (to the nearest km)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(615, '2016', '\n\nQuestion\n\n(a) A pack of 52 playing cards is shuffled and a card is drawn at random. Calculate the probability that it is either a five or a red nine.\n[Hint : There are 4 fives and 2 red nines in a pack of 52 cards]\n(b) P, Q and R are points in the same horizontal plane. The bearing of Q from P is 150° and the bearing of R from Q is 060°. If /PQ/ = 5m and /QR/ = 3m, find the bearing of R from P, correct to the nearest degree.\n\n', '', '(a) A pack of 52 playing cards is shuffled and a card is drawn at random. Calculate the probability that it is either a five or a red nine.\n[Hint : There are 4 fives and 2 red nines in a pack of 52 cards]\n(b) P, Q and R are points in the same horizontal plane. The bearing of Q from P is 150° and the bearing of R from Q is 060°. If /PQ/ = 5m and /QR/ = 3m, find the bearing of R from P, correct to the nearest degree.', '(a) A pack of 52 playing cards is shuffled and a card is drawn at random. Calculate the probability that it is either a five or a red nine.\n[Hint : There are 4 fives and 2 red nines in a pack of 52 cards]~(b) P, Q and R are points in the same horizontal plane. The bearing of Q from P is 150° and the bearing of R from Q is 060°. If /PQ/ = 5m and /QR/ = 3m, find the bearing of R from P, correct to the nearest degree.', NULL, '\n\nExplanation\n(a) Total no of cards = 52; No of fives = 4; No of red nines = 2.\nProb(of 5) = \\(\\frac{4}{52} = \\frac{1}{13}\\)\nProb(red nine) = \\(\\frac{2}{52} = \\frac{1}{26}\\)\n\\(\\therefore\\) Prob (of 5 or red nine) = \\(\\frac{1}{13} + \\frac{1}{26} = \\frac{3}{26}\\)\n(b) \\(\\tan < RPQ = \\frac{3}{5} = 0.6\\)\n\n\\(< RPQ = 30.96°\\)\n\\(\\therefore\\) Bearing of R from P = 150° - 30.96° = 119.04°\n\\(\\approxeq\\) 119° (to the nearest degree).\n\n', '', 'Explanation\n(a) Total no of cards = 52; No of fives = 4; No of red nines = 2.\nProb(of 5) = \\(\\frac{4}{52} = \\frac{1}{13}\\)\nProb(red nine) = \\(\\frac{2}{52} = \\frac{1}{26}\\)\n\\(\\therefore\\) Prob (of 5 or red nine) = \\(\\frac{1}{13} + \\frac{1}{26} = \\frac{3}{26}\\)\n(b) \\(\\tan < RPQ = \\frac{3}{5} = 0.6\\)\n\n\\(< RPQ = 30.96°\\)\n\\(\\therefore\\) Bearing of R from P = 150° - 30.96° = 119.04°\n\\(\\approxeq\\) 119° (to the nearest degree).\n\n', '(a) Total no of cards = 52; No of fives = 4; No of red nines = 2.\nProb(of 5) = \\(\\frac{4}{52} = \\frac{1}{13}\\)\nProb(red nine) = \\(\\frac{2}{52} = \\frac{1}{26}\\)\n\\(\\therefore\\) Prob (of 5 or red nine) = \\(\\frac{1}{13} + \\frac{1}{26} = \\frac{3}{26}\\)~(b) \\(\\tan < RPQ = \\frac{3}{5} = 0.6\\)\n\n\\(< RPQ = 30.96°\\)\n\\(\\therefore\\) Bearing of R from P = 150° - 30.96° = 119.04°\n\\(\\approxeq\\) 119° (to the nearest degree).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(616, '2016', '\n\nQuestion\n\nThe frequency table shows the marks scored by 32 students in a test.\n\n\n\nMarks scored\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n\n\nNo of students\n2\n3\n4\n4\n4\n4\n5\n3\n2\n1\n\n\n\nFind the :\n(a)(i) mean ; (ii) median ; (iii) mode of the marks;\n(b) percentage of the students who scored at least 8 marks.\n\n\n\nThe frequency table shows the marks scored by 32 students in a test.\n\n\n\nMarks scored\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n\n\nNo of students\n2\n3\n4\n4\n4\n4\n5\n3\n2\n1\n\n\n', '', 'The frequency table shows the marks scored by 32 students in a test.\n\n\n\nMarks scored\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n\n\nNo of students\n2\n3\n4\n4\n4\n4\n5\n3\n2\n1\n\n\n\nFind the :\n(a)(i) mean ; (ii) median ; (iii) mode of the marks;\n(b) percentage of the students who scored at least 8 marks.\n\n\n\nThe frequency table shows the marks scored by 32 students in a test.\n\n\n\nMarks scored\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n\n\nNo of students\n2\n3\n4\n4\n4\n4\n5\n3\n2\n1\n', 'The frequency table shows the marks scored by 32 students in a test.\n\n\n\nMarks scored\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n\n\nNo of students\n2\n3\n4\n4\n4\n4\n5\n3\n2\n1\n\n\n\nFind the :\n(a)(i) mean ; (ii) median ; (iii) mode of the marks;~(b) percentage of the students who scored at least 8 marks.\n\n\n\nThe frequency table shows the marks scored by 32 students in a test.\n\n\n\nMarks scored\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n\n\nNo of students\n2\n3\n4\n4\n4\n4\n5\n3\n2\n1\n', NULL, '\n\nExplanation\nx\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\nTotal\n\n\nf\n2\n3\n4\n4\n4\n4\n5\n3\n2\n1\n32\n\n\nfx\n2\n6\n12\n16\n20\n24\n35\n24\n18\n10\n167\n\n\n\n(a)(i) Mean = \\(\\frac{\\sum fx}{\\sum f} = \\frac{167}{32}\\)\n= \\(5.21875 \\approxeq 5.22\\)\n(ii) By inspection, the average of the numbers in the 16th and 17th position give the median\n= \\(\\frac{5 + 5}{2} = 5\\)\n(iii) Mode = 7 marks having the largest frequency.\n(b) Students who scored at least 8 marks = Those who scored 8, 9, 10.\nTotal number of students that scored at least 8 = 3 + 2 + 1 = 6.\n\\(%age = \\frac{6}{32} \\times 100% = 18.75%\\)\n\n', '', 'Explanation\nx\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\nTotal\n\n\nf\n2\n3\n4\n4\n4\n4\n5\n3\n2\n1\n32\n\n\nfx\n2\n6\n12\n16\n20\n24\n35\n24\n18\n10\n167\n\n\n\n(a)(i) Mean = \\(\\frac{\\sum fx}{\\sum f} = \\frac{167}{32}\\)\n= \\(5.21875 \\approxeq 5.22\\)\n(ii) By inspection, the average of the numbers in the 16th and 17th position give the median\n= \\(\\frac{5 + 5}{2} = 5\\)\n(iii) Mode = 7 marks having the largest frequency.\n(b) Students who scored at least 8 marks = Those who scored 8, 9, 10.\nTotal number of students that scored at least 8 = 3 + 2 + 1 = 6.\n\\(%age = \\frac{6}{32} \\times 100% = 18.75%\\)\n\n', '(a)(i) Mean = \\(\\frac{\\sum fx}{\\sum f} = \\frac{167}{32}\\)\n= \\(5.21875 \\approxeq 5.22\\)\n(ii) By inspection, the average of the numbers in the 16th and 17th position give the median\n= \\(\\frac{5 + 5}{2} = 5\\)\n(iii) Mode = 7 marks having the largest frequency.~(b) Students who scored at least 8 marks = Those who scored 8, 9, 10.\nTotal number of students that scored at least 8 = 3 + 2 + 1 = 6.\n\\(%age = \\frac{6}{32} \\times 100% = 18.75%\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1);
INSERT INTO `exam_simulation2` (`id`, `examYear`, `questions`, `parent_que`, `child_que`, `gen_child_que_desc`, `imageDir`, `explanation`, `parent_que_expl`, `child_que_exp`, `gen_child_que_expl`, `package_id`, `subject_id`, `status`) VALUES
(617, '2016', '\n\nQuestion\n\n(a) Using mathematical tables, find ; (i) \\(2 \\sin 63.35°\\) ; (ii) \\(\\log \\cos 44.74°\\);\n(b) Find the value of K given that \\(\\log K - \\log (K - 2) = \\log 5\\);\n(c) Use logarithm tables to evaluate \\(\\frac{(3.68)^{2} \\times 6.705}{\\sqrt{0.3581}}\\)\n\n', '', '(a) Using mathematical tables, find ; (i) \\(2 \\sin 63.35°\\) ; (ii) \\(\\log \\cos 44.74°\\);\n(b) Find the value of K given that \\(\\log K - \\log (K - 2) = \\log 5\\);\n(c) Use logarithm tables to evaluate \\(\\frac{(3.68)^{2} \\times 6.705}{\\sqrt{0.3581}}\\)', '(a) Using mathematical tables, find ; (i) \\(2 \\sin 63.35°\\) ; (ii) \\(\\log \\cos 44.74°\\);~(b) Find the value of K given that \\(\\log K - \\log (K - 2) = \\log 5\\);\n(c) Use logarithm tables to evaluate \\(\\frac{(3.68)^{2} \\times 6.705}{\\sqrt{0.3581}}\\)~(c) Use logarithm tables to evaluate \\(\\frac{(3.68)^{2} \\times 6.705}{\\sqrt{0.3581}}\\)', NULL, '\n\nExplanation\n(a)(i) \\(\\sin 63.35° = 0.8938\\)\n\\(2 \\sin 63.35° = 2 \\times 0.8938 = 1.7876\\)\n(ii) \\(\\log \\cos 44.74° = \\bar{1}.8514\\)\n(b) \\(\\log K - \\log (K - 2) = \\log 5\\)\n\\(\\log (\\frac{K}{K - 2}) = \\log 5\\)\n\\(\\therefore \\frac{K}{K - 2} = 5 \\implies K = 5(K - 2)\\)\n\\(K = 5K - 10 \\implies 10 = 5K - K = 4K\\)\n\\(K = 2\\frac{1}{2}\\)\n(c) \\(\\frac{(3.68)^{2} \\times 6.705}{\\sqrt{0.3581}}\\)\n\n\n\nNo\nLog\n\n\n\\((3.68)^{2}\\)\n\\(0.5658 \\times 2 = 1.316\\)\n\n\n\\(6.705\\)\n\\(0.8264 = 0.8264\\)\n\n\n\n\n= \\(1.9580\\)\n\n\n\n\\(\\sqrt{0.3581}\\)\n\\(\\bar{1}.5540 \\div 2 = \\bar{1}.7770\\)\n\n\n\\(151.7\\)\n\\(\\gets 2.1810\\)\n\n\n\n\\(\\therefore \\frac{(3.68)^{2} \\times 6.705}{\\sqrt{0.3581}} = 151.7\\)\n\n', '', 'Explanation\n(a)(i) \\(\\sin 63.35° = 0.8938\\)\n\\(2 \\sin 63.35° = 2 \\times 0.8938 = 1.7876\\)\n(ii) \\(\\log \\cos 44.74° = \\bar{1}.8514\\)\n(b) \\(\\log K - \\log (K - 2) = \\log 5\\)\n\\(\\log (\\frac{K}{K - 2}) = \\log 5\\)\n\\(\\therefore \\frac{K}{K - 2} = 5 \\implies K = 5(K - 2)\\)\n\\(K = 5K - 10 \\implies 10 = 5K - K = 4K\\)\n\\(K = 2\\frac{1}{2}\\)\n(c) \\(\\frac{(3.68)^{2} \\times 6.705}{\\sqrt{0.3581}}\\)\n\n\n\nNo\nLog\n\n\n\\((3.68)^{2}\\)\n\\(0.5658 \\times 2 = 1.316\\)\n\n\n\\(6.705\\)\n\\(0.8264 = 0.8264\\)\n\n\n\n\n= \\(1.9580\\)\n\n\n\n\\(\\sqrt{0.3581}\\)\n\\(\\bar{1}.5540 \\div 2 = \\bar{1}.7770\\)\n\n\n\\(151.7\\)\n\\(\\gets 2.1810\\)\n\n\n\n\\(\\therefore \\frac{(3.68)^{2} \\times 6.705}{\\sqrt{0.3581}} = 151.7\\)\n\n', '(a)(i) \\(\\sin 63.35° = 0.8938\\)\n\\(2 \\sin 63.35° = 2 \\times 0.8938 = 1.7876\\)\n(ii) \\(\\log \\cos 44.74° = \\bar{1}.8514\\)~(b) \\(\\log K - \\log (K - 2) = \\log 5\\)\n\\(\\log (\\frac{K}{K - 2}) = \\log 5\\)\n\\(\\therefore \\frac{K}{K - 2} = 5 \\implies K = 5(K - 2)\\)\n\\(K = 5K - 10 \\implies 10 = 5K - K = 4K\\)\n\\(K = 2\\frac{1}{2}\\)\n(c) \\(\\frac{(3.68)^{2} \\times 6.705}{\\sqrt{0.3581}}\\)\n\n\n\nNo\nLog\n\n\n\\((3.68)^{2}\\)\n\\(0.5658 \\times 2 = 1.316\\)\n\n\n\\(6.705\\)\n\\(0.8264 = 0.8264\\)\n\n\n\n\n= \\(1.9580\\)\n\n\n\n\\(\\sqrt{0.3581}\\)\n\\(\\bar{1}.5540 \\div 2 = \\bar{1}.7770\\)\n\n\n\\(151.7\\)\n\\(\\gets 2.1810\\)\n\n\n\n\\(\\therefore \\frac{(3.68)^{2} \\times 6.705}{\\sqrt{0.3581}} = 151.7\\)\n~(c) \\(\\frac{(3.68)^{2} \\times 6.705}{\\sqrt{0.3581}}\\)\n\n\n\nNo\nLog\n\n\n\\((3.68)^{2}\\)\n\\(0.5658 \\times 2 = 1.316\\)\n\n\n\\(6.705\\)\n\\(0.8264 = 0.8264\\)\n\n\n\n\n= \\(1.9580\\)\n\n\n\n\\(\\sqrt{0.3581}\\)\n\\(\\bar{1}.5540 \\div 2 = \\bar{1}.7770\\)\n\n\n\\(151.7\\)\n\\(\\gets 2.1810\\)\n\n\n\n\\(\\therefore \\frac{(3.68)^{2} \\times 6.705}{\\sqrt{0.3581}} = 151.7\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(618, '2017', '\n\nQuestion\n\n(a) Given that \\(p = x + ym^{3}\\), find m in terms of p, x and y.\n(b) Using the method of completing the square, find the roots of the equation \\(x^{2} - 6x + 7 = 0\\), correct to 1 decimal place.\n(c) The product of two consecutive positive odd numbers is 195. By constructing a quadratic equation and solving it, find the two numbers.\n\n', '', '(a) Given that \\(p = x + ym^{3}\\), find m in terms of p, x and y.\n(b) Using the method of completing the square, find the roots of the equation \\(x^{2} - 6x + 7 = 0\\), correct to 1 decimal place.\n(c) The product of two consecutive positive odd numbers is 195. By constructing a quadratic equation and solving it, find the two numbers.', '(a) Given that \\(p = x + ym^{3}\\), find m in terms of p, x and y.~(b) Using the method of completing the square, find the roots of the equation \\(x^{2} - 6x + 7 = 0\\), correct to 1 decimal place.\n(c) The product of two consecutive positive odd numbers is 195. By constructing a quadratic equation and solving it, find the two numbers.~(c) The product of two consecutive positive odd numbers is 195. By constructing a quadratic equation and solving it, find the two numbers.', NULL, '\n\nExplanation\n(a) \\(p = x + ym^{3}\\)\n\\(ym^{3} = p - x\\)\n\\(m^{3} = \\frac{p - x}{y}\\)\n\\(m = \\sqrt[3]{\\frac{p - x}{y}}\\)\n(b) \\(x^{2} - 6x + 7 = 0\\)\n\\(\\implies x^{2} - 6x = -7\\)\n\\(x^{2} - 6x + (-3)^{2} = -7 + (-3)^{2}\\)\n\n\\(x^{2} - 6x + 9 = 2\\)\n\\((x - 3)^{2} = 2 \\implies x - 3 = \\pm \\sqrt{2}\\)\n\\(x = 3 \\pm \\sqrt{2} = 3 \\pm 1.414\\)\n\\(x = 4.414 ; 1.586\\)\n\\(x \\approxeq 4.4 ; 1.6\\)\n(c) Let the first odd number be z.\nThe next positive odd number = z + 2\n\\(\\therefore z(z + 2) = 195\\)\n\\(z^{2} + 2z = 195\\)\n\\(z^{2} + 2z - 195 = 0 \\implies z^{2} - 13z + 15z - 195 = 0\\)\n\\(z(z - 13) + 15(z - 13) = 0 \\implies \\text{z = 13 or -15}\\)\nSince we are told the odd numbers are positive, then z = 13 is the suitable answer.\nHence the second odd number = 13 + 2 = 15.\nThe two odd numbers are 13 and 15.\n\n', '', 'Explanation\n(a) \\(p = x + ym^{3}\\)\n\\(ym^{3} = p - x\\)\n\\(m^{3} = \\frac{p - x}{y}\\)\n\\(m = \\sqrt[3]{\\frac{p - x}{y}}\\)\n(b) \\(x^{2} - 6x + 7 = 0\\)\n\\(\\implies x^{2} - 6x = -7\\)\n\\(x^{2} - 6x + (-3)^{2} = -7 + (-3)^{2}\\)\n\n\\(x^{2} - 6x + 9 = 2\\)\n\\((x - 3)^{2} = 2 \\implies x - 3 = \\pm \\sqrt{2}\\)\n\\(x = 3 \\pm \\sqrt{2} = 3 \\pm 1.414\\)\n\\(x = 4.414 ; 1.586\\)\n\\(x \\approxeq 4.4 ; 1.6\\)\n(c) Let the first odd number be z.\nThe next positive odd number = z + 2\n\\(\\therefore z(z + 2) = 195\\)\n\\(z^{2} + 2z = 195\\)\n\\(z^{2} + 2z - 195 = 0 \\implies z^{2} - 13z + 15z - 195 = 0\\)\n\\(z(z - 13) + 15(z - 13) = 0 \\implies \\text{z = 13 or -15}\\)\nSince we are told the odd numbers are positive, then z = 13 is the suitable answer.\nHence the second odd number = 13 + 2 = 15.\nThe two odd numbers are 13 and 15.\n\n', '(a) \\(p = x + ym^{3}\\)\n\\(ym^{3} = p - x\\)\n\\(m^{3} = \\frac{p - x}{y}\\)\n\\(m = \\sqrt[3]{\\frac{p - x}{y}}\\)~(b) \\(x^{2} - 6x + 7 = 0\\)\n\\(\\implies x^{2} - 6x = -7\\)\n\\(x^{2} - 6x + (-3)^{2} = -7 + (-3)^{2}\\)\n\n\\(x^{2} - 6x + 9 = 2\\)\n\\((x - 3)^{2} = 2 \\implies x - 3 = \\pm \\sqrt{2}\\)\n\\(x = 3 \\pm \\sqrt{2} = 3 \\pm 1.414\\)\n\\(x = 4.414 ; 1.586\\)\n\\(x \\approxeq 4.4 ; 1.6\\)\n(c) Let the first odd number be z.\nThe next positive odd number = z + 2\n\\(\\therefore z(z + 2) = 195\\)\n\\(z^{2} + 2z = 195\\)\n\\(z^{2} + 2z - 195 = 0 \\implies z^{2} - 13z + 15z - 195 = 0\\)\n\\(z(z - 13) + 15(z - 13) = 0 \\implies \\text{z = 13 or -15}\\)\nSince we are told the odd numbers are positive, then z = 13 is the suitable answer.\nHence the second odd number = 13 + 2 = 15.\nThe two odd numbers are 13 and 15.\n~(c) Let the first odd number be z.\nThe next positive odd number = z + 2\n\\(\\therefore z(z + 2) = 195\\)\n\\(z^{2} + 2z = 195\\)\n\\(z^{2} + 2z - 195 = 0 \\implies z^{2} - 13z + 15z - 195 = 0\\)\n\\(z(z - 13) + 15(z - 13) = 0 \\implies \\text{z = 13 or -15}\\)\nSince we are told the odd numbers are positive, then z = 13 is the suitable answer.\nHence the second odd number = 13 + 2 = 15.\nThe two odd numbers are 13 and 15.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(619, '2017', '\n\nQuestion\n\n(a) Copy and complete the table for the relation \\(y = 2 \\cos 2x - 1\\).\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n\n\n\\(y = 2\\cos 2x - 1\\)\n1.0\n0\n\n\n\n\n1.0\n\n\n\n(b) Using a scale of 2cm = 30° on the x- axis and 2cm = 1 unit on the y- axis, draw the graph of \\(y = 2 \\cos 2x - 1\\) for \\(0° \\leq x \\leq 180°\\).\n(c) On the same axis, draw the graph of \\(y = \\frac{1}{180} (x - 360)\\)\n(d) Use your graphs to find the : (i) values of x for which \\(2 \\cos 2x + \\frac{1}{2} = 0\\); (ii) roots of the equation \\(2 \\cos 2x - \\frac{x}{180} + 1 = 0\\).\n\n\n(a) Copy and complete the table for the relation \\(y = 2 \\cos 2x - 1\\).\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n\n\n\\(y = 2\\cos 2x - 1\\)\n1.0\n0\n\n\n\n\n1.0\n\n\n', '', '(a) Copy and complete the table for the relation \\(y = 2 \\cos 2x - 1\\).\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n\n\n\\(y = 2\\cos 2x - 1\\)\n1.0\n0\n\n\n\n\n1.0\n\n\n\n(b) Using a scale of 2cm = 30° on the x- axis and 2cm = 1 unit on the y- axis, draw the graph of \\(y = 2 \\cos 2x - 1\\) for \\(0° \\leq x \\leq 180°\\).\n(c) On the same axis, draw the graph of \\(y = \\frac{1}{180} (x - 360)\\)\n(d) Use your graphs to find the : (i) values of x for which \\(2 \\cos 2x + \\frac{1}{2} = 0\\); (ii) roots of the equation \\(2 \\cos 2x - \\frac{x}{180} + 1 = 0\\).\n\n\n(a) Copy and complete the table for the relation \\(y = 2 \\cos 2x - 1\\).\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n\n\n\\(y = 2\\cos 2x - 1\\)\n1.0\n0\n\n\n\n\n1.0\n', '(a) Copy and complete the table for the relation \\(y = 2 \\cos 2x - 1\\).\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n\n\n\\(y = 2\\cos 2x - 1\\)\n1.0\n0\n\n\n\n\n1.0\n\n\n~(b) Using a scale of 2cm = 30° on the x- axis and 2cm = 1 unit on the y- axis, draw the graph of \\(y = 2 \\cos 2x - 1\\) for \\(0° \\leq x \\leq 180°\\).\n(c) On the same axis, draw the graph of \\(y = \\frac{1}{180} (x - 360)\\)\n(d) Use your graphs to find the : (i) values of x for which \\(2 \\cos 2x + \\frac{1}{2} = 0\\); (ii) roots of the equation \\(2 \\cos 2x - \\frac{x}{180} + 1 = 0\\).\n\n\n(a) Copy and complete the table for the relation \\(y = 2 \\cos 2x - 1\\).\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n\n\n\\(y = 2\\cos 2x - 1\\)\n1.0\n0\n\n\n\n\n1.0\n~(c) On the same axis, draw the graph of \\(y = \\frac{1}{180} (x - 360)\\)\n(d) Use your graphs to find the : (i) values of x for which \\(2 \\cos 2x + \\frac{1}{2} = 0\\); (ii) roots of the equation \\(2 \\cos 2x - \\frac{x}{180} + 1 = 0\\).\n\n\n(a) Copy and complete the table for the relation \\(y = 2 \\cos 2x - 1\\).\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n\n\n\\(y = 2\\cos 2x - 1\\)\n1.0\n0\n\n\n\n\n1.0\n~(d) Use your graphs to find the : (i) values of x for which \\(2 \\cos 2x + \\frac{1}{2} = 0\\); (ii) roots of the equation \\(2 \\cos 2x - \\frac{x}{180} + 1 = 0\\).\n\n\n(a) Copy and complete the table for the relation \\(y = 2 \\cos 2x - 1\\).\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n\n\n\\(y = 2\\cos 2x - 1\\)\n1.0\n0\n\n\n\n\n1.0\n', NULL, '\n\nExplanation\n(a)\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n\n\n\\(y = 2\\cos 2x - 1\\)\n1.0\n0\n-2\n-3\n-2\n0\n1.0\n\n\n\n(b)\n(c)\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n\n\n\\(y = \\frac{1}{180} (x - 360)\\)\n-2.0\n-1.83\n-1.67\n-1.5\n-1.33\n-1.17\n-1.0\n\n\n\n(d)(i) \\(2 \\cos 2x + \\frac{1}{2} = 0\\)\n\\(2 \\cos 2x + \\frac{1}{2} - 1\\frac{1}{2} = -1\\frac{1}{2}\\)\n\\(2 \\cos 2x - 1 = -1\\frac{1}{2}\\)\nThe values of x at \\(2 \\cos 2x + \\frac{1}{2} = 0\\) are x = 51° and 129°.\n(ii) \\(2 \\cos 2x - \\frac{x}{180} + 1 = 0\\)\n\\(2 \\cos 2x -1 = \\frac{1}{180} (x - 360)\\)\nThe values of x is where the two graphs intersect; which are at x = 54° and 132°.\n\n', '', 'Explanation\n(a)\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n\n\n\\(y = 2\\cos 2x - 1\\)\n1.0\n0\n-2\n-3\n-2\n0\n1.0\n\n\n\n(b)\n(c)\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n\n\n\\(y = \\frac{1}{180} (x - 360)\\)\n-2.0\n-1.83\n-1.67\n-1.5\n-1.33\n-1.17\n-1.0\n\n\n\n(d)(i) \\(2 \\cos 2x + \\frac{1}{2} = 0\\)\n\\(2 \\cos 2x + \\frac{1}{2} - 1\\frac{1}{2} = -1\\frac{1}{2}\\)\n\\(2 \\cos 2x - 1 = -1\\frac{1}{2}\\)\nThe values of x at \\(2 \\cos 2x + \\frac{1}{2} = 0\\) are x = 51° and 129°.\n(ii) \\(2 \\cos 2x - \\frac{x}{180} + 1 = 0\\)\n\\(2 \\cos 2x -1 = \\frac{1}{180} (x - 360)\\)\nThe values of x is where the two graphs intersect; which are at x = 54° and 132°.\n\n', '(a)\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n\n\n\\(y = 2\\cos 2x - 1\\)\n1.0\n0\n-2\n-3\n-2\n0\n1.0\n\n\n~(b)\n(c)\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n\n\n\\(y = \\frac{1}{180} (x - 360)\\)\n-2.0\n-1.83\n-1.67\n-1.5\n-1.33\n-1.17\n-1.0\n\n\n\n(d)(i) \\(2 \\cos 2x + \\frac{1}{2} = 0\\)\n\\(2 \\cos 2x + \\frac{1}{2} - 1\\frac{1}{2} = -1\\frac{1}{2}\\)\n\\(2 \\cos 2x - 1 = -1\\frac{1}{2}\\)\nThe values of x at \\(2 \\cos 2x + \\frac{1}{2} = 0\\) are x = 51° and 129°.\n(ii) \\(2 \\cos 2x - \\frac{x}{180} + 1 = 0\\)\n\\(2 \\cos 2x -1 = \\frac{1}{180} (x - 360)\\)\nThe values of x is where the two graphs intersect; which are at x = 54° and 132°.\n~(c)\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n\n\n\\(y = \\frac{1}{180} (x - 360)\\)\n-2.0\n-1.83\n-1.67\n-1.5\n-1.33\n-1.17\n-1.0\n\n\n\n(d)(i) \\(2 \\cos 2x + \\frac{1}{2} = 0\\)\n\\(2 \\cos 2x + \\frac{1}{2} - 1\\frac{1}{2} = -1\\frac{1}{2}\\)\n\\(2 \\cos 2x - 1 = -1\\frac{1}{2}\\)\nThe values of x at \\(2 \\cos 2x + \\frac{1}{2} = 0\\) are x = 51° and 129°.\n(ii) \\(2 \\cos 2x - \\frac{x}{180} + 1 = 0\\)\n\\(2 \\cos 2x -1 = \\frac{1}{180} (x - 360)\\)\nThe values of x is where the two graphs intersect; which are at x = 54° and 132°.\n~(d)(i) \\(2 \\cos 2x + \\frac{1}{2} = 0\\)\n\\(2 \\cos 2x + \\frac{1}{2} - 1\\frac{1}{2} = -1\\frac{1}{2}\\)\n\\(2 \\cos 2x - 1 = -1\\frac{1}{2}\\)\nThe values of x at \\(2 \\cos 2x + \\frac{1}{2} = 0\\) are x = 51° and 129°.\n(ii) \\(2 \\cos 2x - \\frac{x}{180} + 1 = 0\\)\n\\(2 \\cos 2x -1 = \\frac{1}{180} (x - 360)\\)\nThe values of x is where the two graphs intersect; which are at x = 54° and 132°.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(620, '2017', '\n\nQuestion\n\n(a) Using a ruler and a pair of compasses only, construct triangle ABC with /AB/ = 7.5cm, /BC/ = 8.1cm and < ABC = 105°.\n(b) Locate a point D on BC such that /BD/ : /DC/ is 3 : 2.\n(c) Through D, construct a line I perpendicular to BC.\n(d) If the line I meets AC at P, measure /BP/.\n\n', '', '(a) Using a ruler and a pair of compasses only, construct triangle ABC with /AB/ = 7.5cm, /BC/ = 8.1cm and < ABC = 105°.\n(b) Locate a point D on BC such that /BD/ : /DC/ is 3 : 2.\n(c) Through D, construct a line I perpendicular to BC.\n(d) If the line I meets AC at P, measure /BP/.', '(a) Using a ruler and a pair of compasses only, construct triangle ABC with /AB/ = 7.5cm, /BC/ = 8.1cm and < ABC = 105°.~(b) Locate a point D on BC such that /BD/ : /DC/ is 3 : 2.\n(c) Through D, construct a line I perpendicular to BC.\n(d) If the line I meets AC at P, measure /BP/.~(c) Through D, construct a line I perpendicular to BC.\n(d) If the line I meets AC at P, measure /BP/.~(d) If the line I meets AC at P, measure /BP/.', NULL, '\n\nExplanation\n(a - d)\n\n', '(a - d)\n\n', '', '', 'WAEC-PK-1', 'WAEC-MATH', 1),
(621, '2017', '\n\nQuestion\n\n(a) A man travels from a village X on a bearing of 060° to a village Y which is 20km away. From Y, he travels to a village Z, on a bearing of 195°. If Z is directly east of X, calculate, correct to three significant figures, the distance of :\n(i) Y from Z ; (ii) Z from X .\n(b) An aircraft flies due South from an airfield on latitude 36°N, longitude 138°E to an airfield on latitude 36°S, longitude 138°E.\n(i) Calculate the distance travelled, correct to three significant figures ; (ii) if the speed of the aircraft is 800km per hour, calculate the time taken, correct to the nearest hour.\n[Take \\(\\pi = \\frac{22}{7}\\), R = 6400km].\n\n', '', '(a) A man travels from a village X on a bearing of 060° to a village Y which is 20km away. From Y, he travels to a village Z, on a bearing of 195°. If Z is directly east of X, calculate, correct to three significant figures, the distance of :\n(i) Y from Z ; (ii) Z from X .\n(b) An aircraft flies due South from an airfield on latitude 36°N, longitude 138°E to an airfield on latitude 36°S, longitude 138°E.\n(i) Calculate the distance travelled, correct to three significant figures ; (ii) if the speed of the aircraft is 800km per hour, calculate the time taken, correct to the nearest hour.\n[Take \\(\\pi = \\frac{22}{7}\\), R = 6400km].', '(a) A man travels from a village X on a bearing of 060° to a village Y which is 20km away. From Y, he travels to a village Z, on a bearing of 195°. If Z is directly east of X, calculate, correct to three significant figures, the distance of :\n(i) Y from Z ; (ii) Z from X .~(b) An aircraft flies due South from an airfield on latitude 36°N, longitude 138°E to an airfield on latitude 36°S, longitude 138°E.\n(i) Calculate the distance travelled, correct to three significant figures ; (ii) if the speed of the aircraft is 800km per hour, calculate the time taken, correct to the nearest hour.\n[Take \\(\\pi = \\frac{22}{7}\\), R = 6400km].', NULL, '\n\nExplanation\n(a) < XZY = 180° - (30° + 45°) = 105°\n\n\\(\\frac{20}{\\sin 105°} = \\frac{YZ}{\\sin 30°}\\)\n\\(YZ = \\frac{20 \\sin 30°}{\\sin 105°} = 10.35km\\)\n\\(\\approxeq 10.4km \\) (to 3 sig. figs)\n(ii) \\(\\frac{20}{\\sin 75} = \\frac{XZ}{\\sin 45}\\)\n\\(XZ = \\frac{20 \\sin 45}{\\sin 75} = 14.64km\\)\n\\(\\approxeq 14.6km\\) ( to 3 sig. figs)\n(b) Let P and Q be two airfields.\n\nDifference in latitude = 36° + 36° = 72°\n(i) Distance between P and Q = \\(\\frac{72}{360} \\times 2 \\times \\frac{22}{7} \\times 6400\\)\n= \\(\\frac{22}{7} \\times 2560\\)\n= \\(8,045.71km \\approxeq 8,050 km\\) (3 sig. figs)\n(ii) Time taken = \\(\\frac{8,045.71}{800} = 10.06 hours\\)\n\\(\\approxeq 10hours\\) (to the nearest hour)\n\n', '', 'Explanation\n(a) < XZY = 180° - (30° + 45°) = 105°\n\n\\(\\frac{20}{\\sin 105°} = \\frac{YZ}{\\sin 30°}\\)\n\\(YZ = \\frac{20 \\sin 30°}{\\sin 105°} = 10.35km\\)\n\\(\\approxeq 10.4km \\) (to 3 sig. figs)\n(ii) \\(\\frac{20}{\\sin 75} = \\frac{XZ}{\\sin 45}\\)\n\\(XZ = \\frac{20 \\sin 45}{\\sin 75} = 14.64km\\)\n\\(\\approxeq 14.6km\\) ( to 3 sig. figs)\n(b) Let P and Q be two airfields.\n\nDifference in latitude = 36° + 36° = 72°\n(i) Distance between P and Q = \\(\\frac{72}{360} \\times 2 \\times \\frac{22}{7} \\times 6400\\)\n= \\(\\frac{22}{7} \\times 2560\\)\n= \\(8,045.71km \\approxeq 8,050 km\\) (3 sig. figs)\n(ii) Time taken = \\(\\frac{8,045.71}{800} = 10.06 hours\\)\n\\(\\approxeq 10hours\\) (to the nearest hour)\n\n', '(a) < XZY = 180° - (30° + 45°) = 105°\n\n\\(\\frac{20}{\\sin 105°} = \\frac{YZ}{\\sin 30°}\\)\n\\(YZ = \\frac{20 \\sin 30°}{\\sin 105°} = 10.35km\\)\n\\(\\approxeq 10.4km \\) (to 3 sig. figs)\n(ii) \\(\\frac{20}{\\sin 75} = \\frac{XZ}{\\sin 45}\\)\n\\(XZ = \\frac{20 \\sin 45}{\\sin 75} = 14.64km\\)\n\\(\\approxeq 14.6km\\) ( to 3 sig. figs)~(b) Let P and Q be two airfields.\n\nDifference in latitude = 36° + 36° = 72°\n(i) Distance between P and Q = \\(\\frac{72}{360} \\times 2 \\times \\frac{22}{7} \\times 6400\\)\n= \\(\\frac{22}{7} \\times 2560\\)\n= \\(8,045.71km \\approxeq 8,050 km\\) (3 sig. figs)\n(ii) Time taken = \\(\\frac{8,045.71}{800} = 10.06 hours\\)\n\\(\\approxeq 10hours\\) (to the nearest hour)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(622, '2017', '\n\nQuestion\n\nThe table shows the scores of 2000 candidates in an entrance examination into a private secondary school.\n\n\n\n% Mark\n11-20\n21-30\n31-40\n41-50\n51-60\n61-70\n71-80\n81-90\n\n\n\nNo of\npupils\n\n68\n184\n294\n402\n480\n310\n164\n98\n\n\n\n(a) Prepare a cumulative frequency table and draw the cumulative frequency curve for the distribution.\n(b) Use your curve to estimate the : (i) cut off mark, if 300 candidates are to be offered admission ; (ii) probability that a candidate picked at random, scored at least 45%.\n\n\n\nThe table shows the scores of 2000 candidates in an entrance examination into a private secondary school.\n\n\n\n% Mark\n11-20\n21-30\n31-40\n41-50\n51-60\n61-70\n71-80\n81-90\n\n\n\nNo of\npupils\n\n68\n184\n294\n402\n480\n310\n164\n98\n\n\n', '', 'The table shows the scores of 2000 candidates in an entrance examination into a private secondary school.\n\n\n\n% Mark\n11-20\n21-30\n31-40\n41-50\n51-60\n61-70\n71-80\n81-90\n\n\n\nNo of\npupils\n\n68\n184\n294\n402\n480\n310\n164\n98\n\n\n\n(a) Prepare a cumulative frequency table and draw the cumulative frequency curve for the distribution.\n(b) Use your curve to estimate the : (i) cut off mark, if 300 candidates are to be offered admission ; (ii) probability that a candidate picked at random, scored at least 45%.\n\n\n\nThe table shows the scores of 2000 candidates in an entrance examination into a private secondary school.\n\n\n\n% Mark\n11-20\n21-30\n31-40\n41-50\n51-60\n61-70\n71-80\n81-90\n\n\n\nNo of\npupils\n\n68\n184\n294\n402\n480\n310\n164\n98\n', 'The table shows the scores of 2000 candidates in an entrance examination into a private secondary school.\n\n\n\n% Mark\n11-20\n21-30\n31-40\n41-50\n51-60\n61-70\n71-80\n81-90\n\n\n\nNo of\npupils\n\n68\n184\n294\n402\n480\n310\n164\n98\n\n\n\n(a) Prepare a cumulative frequency table and draw the cumulative frequency curve for the distribution.~(b) Use your curve to estimate the : (i) cut off mark, if 300 candidates are to be offered admission ; (ii) probability that a candidate picked at random, scored at least 45%.\n\n\n\nThe table shows the scores of 2000 candidates in an entrance examination into a private secondary school.\n\n\n\n% Mark\n11-20\n21-30\n31-40\n41-50\n51-60\n61-70\n71-80\n81-90\n\n\n\nNo of\npupils\n\n68\n184\n294\n402\n480\n310\n164\n98\n', NULL, '\n\nExplanation\n(a)(i)\n\n\n\nMarks (x)\nNo of pupils (f)\nCum Freq\n\n\n11 - 20\n68\n68\n\n\n21 - 30\n184\n252\n\n\n31 - 40\n294\n546\n\n\n41 - 50\n402\n948\n\n\n51 - 60\n480\n1428\n\n\n61 - 70\n310\n1738\n\n\n71 - 80\n164\n1902\n\n\n81 - 90\n98\n2000\n\n\n\n(ii)\n(b)(i) Cut off mark = 68 marks for reading at 1700.\n(ii) Candidates scoring more than 45% = 2000 - 720 = 1280\n\\(\\therefore\\) P(scoring at least 45%) = \\(\\frac{1280}{2000} = 0.64\\)\n\n', '', 'Explanation\n(a)(i)\n\n\n\nMarks (x)\nNo of pupils (f)\nCum Freq\n\n\n11 - 20\n68\n68\n\n\n21 - 30\n184\n252\n\n\n31 - 40\n294\n546\n\n\n41 - 50\n402\n948\n\n\n51 - 60\n480\n1428\n\n\n61 - 70\n310\n1738\n\n\n71 - 80\n164\n1902\n\n\n81 - 90\n98\n2000\n\n\n\n(ii)\n(b)(i) Cut off mark = 68 marks for reading at 1700.\n(ii) Candidates scoring more than 45% = 2000 - 720 = 1280\n\\(\\therefore\\) P(scoring at least 45%) = \\(\\frac{1280}{2000} = 0.64\\)\n\n', '(a)(i)\n\n\n\nMarks (x)\nNo of pupils (f)\nCum Freq\n\n\n11 - 20\n68\n68\n\n\n21 - 30\n184\n252\n\n\n31 - 40\n294\n546\n\n\n41 - 50\n402\n948\n\n\n51 - 60\n480\n1428\n\n\n61 - 70\n310\n1738\n\n\n71 - 80\n164\n1902\n\n\n81 - 90\n98\n2000\n\n\n\n(ii)~(b)(i) Cut off mark = 68 marks for reading at 1700.\n(ii) Candidates scoring more than 45% = 2000 - 720 = 1280\n\\(\\therefore\\) P(scoring at least 45%) = \\(\\frac{1280}{2000} = 0.64\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(623, '2017', '\n\nQuestion\n\nA box contains 5 blue balls, 3 black balls and 2 red balls of the same size. A ball is selected at random from the box and then replaced. A second ball is then selected. Find the probability of obtaining\n(a) two red balls ;\n(b) two blue balls or two black balls ;\n(c) one black and one red ball in any order.\n\n', '', 'A box contains 5 blue balls, 3 black balls and 2 red balls of the same size. A ball is selected at random from the box and then replaced. A second ball is then selected. Find the probability of obtaining\n(a) two red balls ;\n(b) two blue balls or two black balls ;\n(c) one black and one red ball in any order.', 'A box contains 5 blue balls, 3 black balls and 2 red balls of the same size. A ball is selected at random from the box and then replaced. A second ball is then selected. Find the probability of obtaining\n(a) two red balls ;~(b) two blue balls or two black balls ;\n(c) one black and one red ball in any order.~(c) one black and one red ball in any order.', NULL, '\n\nExplanation\nNumber of blue balls = 5\nNumber of black balls = 3\nNumber of red balls = 2\n= 10 balls.\n(a) P(two red balls) = \\(\\frac{2}{10} \\times \\frac{2}{10} = \\frac{1}{25}\\)\n(b) P(2 blue or 2 black balls) = P(2 blue) or P(2 black)\n= \\((\\frac{5}{10} \\times \\frac{5}{10}) + (\\frac{3}{10} \\times \\frac{3}{10})\\)\n= \\(\\frac{25}{100} + \\frac{9}{100}\\)\n= \\(\\frac{34}{100} = \\frac{17}{50}\\)\n(c) P( one red and one black in any order) = P(one red, one black) or P(one black, one red)\n= \\((\\frac{2}{10} \\times \\frac{3}{10}) + (\\frac{3}{10} \\times \\frac{2}{10})\\)\n= \\(\\frac{12}{100} = \\frac{3}{25}\\)\n\n', '', 'Explanation\nNumber of blue balls = 5\nNumber of black balls = 3\nNumber of red balls = 2\n= 10 balls.\n(a) P(two red balls) = \\(\\frac{2}{10} \\times \\frac{2}{10} = \\frac{1}{25}\\)\n(b) P(2 blue or 2 black balls) = P(2 blue) or P(2 black)\n= \\((\\frac{5}{10} \\times \\frac{5}{10}) + (\\frac{3}{10} \\times \\frac{3}{10})\\)\n= \\(\\frac{25}{100} + \\frac{9}{100}\\)\n= \\(\\frac{34}{100} = \\frac{17}{50}\\)\n(c) P( one red and one black in any order) = P(one red, one black) or P(one black, one red)\n= \\((\\frac{2}{10} \\times \\frac{3}{10}) + (\\frac{3}{10} \\times \\frac{2}{10})\\)\n= \\(\\frac{12}{100} = \\frac{3}{25}\\)\n\n', '(a) P(two red balls) = \\(\\frac{2}{10} \\times \\frac{2}{10} = \\frac{1}{25}\\)~(b) P(2 blue or 2 black balls) = P(2 blue) or P(2 black)\n= \\((\\frac{5}{10} \\times \\frac{5}{10}) + (\\frac{3}{10} \\times \\frac{3}{10})\\)\n= \\(\\frac{25}{100} + \\frac{9}{100}\\)\n= \\(\\frac{34}{100} = \\frac{17}{50}\\)\n(c) P( one red and one black in any order) = P(one red, one black) or P(one black, one red)\n= \\((\\frac{2}{10} \\times \\frac{3}{10}) + (\\frac{3}{10} \\times \\frac{2}{10})\\)\n= \\(\\frac{12}{100} = \\frac{3}{25}\\)\n~(c) P( one red and one black in any order) = P(one red, one black) or P(one black, one red)\n= \\((\\frac{2}{10} \\times \\frac{3}{10}) + (\\frac{3}{10} \\times \\frac{2}{10})\\)\n= \\(\\frac{12}{100} = \\frac{3}{25}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(624, '2017', '\n\nQuestion\n\n(a) Given that \\(3 \\times 9^{1 + x} = 27^{-x}\\), find x.\n(b) Evaluate \\(\\log_{10} \\sqrt{35} + \\log_{10} \\sqrt{2} - \\log_{10} \\sqrt{7}\\)\n\n', '', '(a) Given that \\(3 \\times 9^{1 + x} = 27^{-x}\\), find x.\n(b) Evaluate \\(\\log_{10} \\sqrt{35} + \\log_{10} \\sqrt{2} - \\log_{10} \\sqrt{7}\\)', '(a) Given that \\(3 \\times 9^{1 + x} = 27^{-x}\\), find x.~(b) Evaluate \\(\\log_{10} \\sqrt{35} + \\log_{10} \\sqrt{2} - \\log_{10} \\sqrt{7}\\)', NULL, '\n\nExplanation\n(a) \\(3 \\times 9^{1 + x} = 27^{-x}\\)\n\\(3 \\times (3^{2})^{1 + x} = (3^{3})^{-x}\\)\n\\(3 \\times 3^{2 + 2x} = 3^{-3x} \\)\n\\(3^{2x + 2 + 1} = 3^{-3x} \\implies 3^{2x + 3} = 3^{-3x}\\)\n\\(\\implies 2x + 3 = - 3x\\)\n\\(-5x = 3 \\implies x = \\frac{-3}{5}\\)\n(b) \\(\\log_{10} \\sqrt{35} + \\log_{10} \\sqrt{2} - \\log_{10} \\sqrt{7}\\)\n= \\(\\log_{10} (\\frac{\\sqrt{35} \\times \\sqrt{2}}{\\sqrt{7}}\\)\n= \\(\\log_{10} (\\frac{\\sqrt{35 \\times 2}{\\sqrt{7}}\\)\n= \\(\\log_{10} \\sqrt{10}\\)\n= \\(\\log_{10} 10^{\\frac{1}{2}}\\)\n= \\(\\frac{1}{2} \\log_{10} 10 = \\frac{1}{2}\\)\n\n', '', 'Explanation\n(a) \\(3 \\times 9^{1 + x} = 27^{-x}\\)\n\\(3 \\times (3^{2})^{1 + x} = (3^{3})^{-x}\\)\n\\(3 \\times 3^{2 + 2x} = 3^{-3x} \\)\n\\(3^{2x + 2 + 1} = 3^{-3x} \\implies 3^{2x + 3} = 3^{-3x}\\)\n\\(\\implies 2x + 3 = - 3x\\)\n\\(-5x = 3 \\implies x = \\frac{-3}{5}\\)\n(b) \\(\\log_{10} \\sqrt{35} + \\log_{10} \\sqrt{2} - \\log_{10} \\sqrt{7}\\)\n= \\(\\log_{10} (\\frac{\\sqrt{35} \\times \\sqrt{2}}{\\sqrt{7}}\\)\n= \\(\\log_{10} (\\frac{\\sqrt{35 \\times 2}{\\sqrt{7}}\\)\n= \\(\\log_{10} \\sqrt{10}\\)\n= \\(\\log_{10} 10^{\\frac{1}{2}}\\)\n= \\(\\frac{1}{2} \\log_{10} 10 = \\frac{1}{2}\\)\n\n', '(a) \\(3 \\times 9^{1 + x} = 27^{-x}\\)\n\\(3 \\times (3^{2})^{1 + x} = (3^{3})^{-x}\\)\n\\(3 \\times 3^{2 + 2x} = 3^{-3x} \\)\n\\(3^{2x + 2 + 1} = 3^{-3x} \\implies 3^{2x + 3} = 3^{-3x}\\)\n\\(\\implies 2x + 3 = - 3x\\)\n\\(-5x = 3 \\implies x = \\frac{-3}{5}\\)~(b) \\(\\log_{10} \\sqrt{35} + \\log_{10} \\sqrt{2} - \\log_{10} \\sqrt{7}\\)\n= \\(\\log_{10} (\\frac{\\sqrt{35} \\times \\sqrt{2}}{\\sqrt{7}}\\)\n= \\(\\log_{10} (\\frac{\\sqrt{35 \\times 2}{\\sqrt{7}}\\)\n= \\(\\log_{10} \\sqrt{10}\\)\n= \\(\\log_{10} 10^{\\frac{1}{2}}\\)\n= \\(\\frac{1}{2} \\log_{10} 10 = \\frac{1}{2}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(625, '2017', '\n\nQuestion\n\n(a) Evaluate, without using mathematical tables, \\(17.57^{2} - 12.43^{2}\\).\n(b) Prove that angles in the same segment of a circle are equal.\n\n', '', '(a) Evaluate, without using mathematical tables, \\(17.57^{2} - 12.43^{2}\\).\n(b) Prove that angles in the same segment of a circle are equal.', '(a) Evaluate, without using mathematical tables, \\(17.57^{2} - 12.43^{2}\\).~(b) Prove that angles in the same segment of a circle are equal.', NULL, '\n\nExplanation\n(a) \\(17.57^{2} - 12.43^{2}\\)\nUsingthe difference of two squares method,\n= \\((17.57 + 12.43)(17.57 - 12.43)\\)\n= \\((30.00)(5.14)\\)\n= \\(154.2\\)\n(b)\nGiven: D and C are points on the major arc of circle ADCB. To prove that < ADB = < ACB.\nConstruction : Join A and B to O, the centre of the circle .\nProof: With the lettering \\(< AOB = 2x_{1}\\) (angle at the centre is twice that subtended at the circumference)\nBut \\(< AOB = 2x_{2}\\) (the same theorem applies here)\n\\(\\therefore 2x_{1} = 2x_{2} \\implies x_{1} = x_{2}\\)\n\\(\\therefore < ADB = < ACB\\)\n\n', '', 'Explanation\n(a) \\(17.57^{2} - 12.43^{2}\\)\nUsingthe difference of two squares method,\n= \\((17.57 + 12.43)(17.57 - 12.43)\\)\n= \\((30.00)(5.14)\\)\n= \\(154.2\\)\n(b)\nGiven: D and C are points on the major arc of circle ADCB. To prove that < ADB = < ACB.\nConstruction : Join A and B to O, the centre of the circle .\nProof: With the lettering \\(< AOB = 2x_{1}\\) (angle at the centre is twice that subtended at the circumference)\nBut \\(< AOB = 2x_{2}\\) (the same theorem applies here)\n\\(\\therefore 2x_{1} = 2x_{2} \\implies x_{1} = x_{2}\\)\n\\(\\therefore < ADB = < ACB\\)\n\n', '(a) \\(17.57^{2} - 12.43^{2}\\)\nUsingthe difference of two squares method,\n= \\((17.57 + 12.43)(17.57 - 12.43)\\)\n= \\((30.00)(5.14)\\)\n= \\(154.2\\)~(b)\nGiven: D and C are points on the major arc of circle ADCB. To prove that < ADB = < ACB.\nConstruction : Join A and B to O, the centre of the circle .\nProof: With the lettering \\(< AOB = 2x_{1}\\) (angle at the centre is twice that subtended at the circumference)\nBut \\(< AOB = 2x_{2}\\) (the same theorem applies here)\n\\(\\therefore 2x_{1} = 2x_{2} \\implies x_{1} = x_{2}\\)\n\\(\\therefore < ADB = < ACB\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(626, '2017', '\n\nQuestion\n\n(a) A tower and a building stand on the same horizontal level. From the point P at the bottom of the building, the angle of elevation of the top, T of the tower is 65°. From the top Q of the building, the angle of elevation of the point T is 25°. If the building is 20m high, calculate the distance PT.\n(b) Hence or otherwise, calculate the height of the tower. [Give your answers correct to 3 significant figures].\n\n', '', '(a) A tower and a building stand on the same horizontal level. From the point P at the bottom of the building, the angle of elevation of the top, T of the tower is 65°. From the top Q of the building, the angle of elevation of the point T is 25°. If the building is 20m high, calculate the distance PT.\n(b) Hence or otherwise, calculate the height of the tower. [Give your answers correct to 3 significant figures].', '(a) A tower and a building stand on the same horizontal level. From the point P at the bottom of the building, the angle of elevation of the top, T of the tower is 65°. From the top Q of the building, the angle of elevation of the point T is 25°. If the building is 20m high, calculate the distance PT.~(b) Hence or otherwise, calculate the height of the tower. [Give your answers correct to 3 significant figures].', NULL, '\n\nExplanation\n(a)\n\\(< QPB = 90° - 65° = 25°\\)\nAlso, \\(< TQP = 90° + 25° = 115°\\)\n\\(\\therefore < QTP = 180° - (25° + 115°) = 40°\\)\nUsing sine rule, \\(\\frac{PQ}{\\sin 40°} = \\frac{PT}{\\sin 115°}\\)\n\\(PT = \\frac{20 \\sin 115°}{\\sin 40°} = \\frac{20 \\sin 65°}{\\sin 40°}\\)\n= \\(28.199m \\approxeq 28.2m\\) (to 3 sig. fig)\n(b) Height of the tower = ST\n\\(\\sin 65 = \\frac{ST}{PT} = \\frac{ST}{28.199}\\)\n\\(ST = 28.199 \\times \\sin 65° = 25.557m\\)\n\\(\\approxeq 25.6m\\) (to 3 sig. figs)\n\n', '', 'Explanation\n(a)\n\\(< QPB = 90° - 65° = 25°\\)\nAlso, \\(< TQP = 90° + 25° = 115°\\)\n\\(\\therefore < QTP = 180° - (25° + 115°) = 40°\\)\nUsing sine rule, \\(\\frac{PQ}{\\sin 40°} = \\frac{PT}{\\sin 115°}\\)\n\\(PT = \\frac{20 \\sin 115°}{\\sin 40°} = \\frac{20 \\sin 65°}{\\sin 40°}\\)\n= \\(28.199m \\approxeq 28.2m\\) (to 3 sig. fig)\n(b) Height of the tower = ST\n\\(\\sin 65 = \\frac{ST}{PT} = \\frac{ST}{28.199}\\)\n\\(ST = 28.199 \\times \\sin 65° = 25.557m\\)\n\\(\\approxeq 25.6m\\) (to 3 sig. figs)\n\n', '(a)\n\\(< QPB = 90° - 65° = 25°\\)\nAlso, \\(< TQP = 90° + 25° = 115°\\)\n\\(\\therefore < QTP = 180° - (25° + 115°) = 40°\\)\nUsing sine rule, \\(\\frac{PQ}{\\sin 40°} = \\frac{PT}{\\sin 115°}\\)\n\\(PT = \\frac{20 \\sin 115°}{\\sin 40°} = \\frac{20 \\sin 65°}{\\sin 40°}\\)\n= \\(28.199m \\approxeq 28.2m\\) (to 3 sig. fig)~(b) Height of the tower = ST\n\\(\\sin 65 = \\frac{ST}{PT} = \\frac{ST}{28.199}\\)\n\\(ST = 28.199 \\times \\sin 65° = 25.557m\\)\n\\(\\approxeq 25.6m\\) (to 3 sig. figs)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(627, '2017', '\n\nQuestion\n\n(a) Divide \\(11111111_{two}\\) by \\(101_{two}\\)\n(b) A sector of radius 6 cm has an angle of 105° at the centre. Calculate its:\n(i) perimeter ; (ii) area . [Take \\(\\pi = \\frac{22}{7}\\)]\n\n', '', '(a) Divide \\(11111111_{two}\\) by \\(101_{two}\\)\n(b) A sector of radius 6 cm has an angle of 105° at the centre. Calculate its:\n(i) perimeter ; (ii) area . [Take \\(\\pi = \\frac{22}{7}\\)]', '(a) Divide \\(11111111_{two}\\) by \\(101_{two}\\)~(b) A sector of radius 6 cm has an angle of 105° at the centre. Calculate its:\n(i) perimeter ; (ii) area . [Take \\(\\pi = \\frac{22}{7}\\)]', NULL, '\n\nExplanation\n(a) This can either be done using the long division method (easier) or convert everything to base ten and divide, then re-convert to base two.\nI\'ll be using the latter method.\n\\(11111111_{two} = 1 \\times 2^{7} + 1 \\times 2^{6} + 1 \\times 2^{5} + 1 \\times 2^{4} + 1 \\times 2^{3} + 1 \\times 2^{2} + 1 \\times 2^{1} + 1 \\times 2^{0}\\)\n= \\(128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255_{ten}\\)\n\\(101_{two} = 1 \\times 2^{2} + 0 \\times 2^{1} + 1 \\times 2^{0} \\)\n= \\(4 + 0 + 1 = 5_{ten}\\)\n\\(\\frac{255}{5} = 51_{ten}\\)\n\n\n\n2\n51\n\n\n2\n25 r 1\n\n\n2\n12 r 1\n\n\n2\n6 r 0\n\n\n2\n3 r 0\n\n\n2\n1 r 1\n\n\n\n0 r 1\n\n\n\n= \\(110011_{two}\\)\n(b)\n(i) Length of arc AXB = \\(\\frac{105}{360} \\times 2 \\times \\frac{22}{7} \\times 6cm\\)\n= \\(11cm\\)\n\\(\\therefore Perimeter = 6cm + 11cm + 6cm = 23cm\\)\n(ii) Area of sector = \\(\\frac{105}{360} \\times \\frac{22}{7} \\times 6cm \\times 6cm\\)\n= \\(33 cm^{2}\\)\n\n', '', 'Explanation\n(a) This can either be done using the long division method (easier) or convert everything to base ten and divide, then re-convert to base two.\nI\'ll be using the latter method.\n\\(11111111_{two} = 1 \\times 2^{7} + 1 \\times 2^{6} + 1 \\times 2^{5} + 1 \\times 2^{4} + 1 \\times 2^{3} + 1 \\times 2^{2} + 1 \\times 2^{1} + 1 \\times 2^{0}\\)\n= \\(128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255_{ten}\\)\n\\(101_{two} = 1 \\times 2^{2} + 0 \\times 2^{1} + 1 \\times 2^{0} \\)\n= \\(4 + 0 + 1 = 5_{ten}\\)\n\\(\\frac{255}{5} = 51_{ten}\\)\n\n\n\n2\n51\n\n\n2\n25 r 1\n\n\n2\n12 r 1\n\n\n2\n6 r 0\n\n\n2\n3 r 0\n\n\n2\n1 r 1\n\n\n\n0 r 1\n\n\n\n= \\(110011_{two}\\)\n(b)\n(i) Length of arc AXB = \\(\\frac{105}{360} \\times 2 \\times \\frac{22}{7} \\times 6cm\\)\n= \\(11cm\\)\n\\(\\therefore Perimeter = 6cm + 11cm + 6cm = 23cm\\)\n(ii) Area of sector = \\(\\frac{105}{360} \\times \\frac{22}{7} \\times 6cm \\times 6cm\\)\n= \\(33 cm^{2}\\)\n\n', '(a) This can either be done using the long division method (easier) or convert everything to base ten and divide, then re-convert to base two.\nI\'ll be using the latter method.\n\\(11111111_{two} = 1 \\times 2^{7} + 1 \\times 2^{6} + 1 \\times 2^{5} + 1 \\times 2^{4} + 1 \\times 2^{3} + 1 \\times 2^{2} + 1 \\times 2^{1} + 1 \\times 2^{0}\\)\n= \\(128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255_{ten}\\)\n\\(101_{two} = 1 \\times 2^{2} + 0 \\times 2^{1} + 1 \\times 2^{0} \\)\n= \\(4 + 0 + 1 = 5_{ten}\\)\n\\(\\frac{255}{5} = 51_{ten}\\)\n\n\n\n2\n51\n\n\n2\n25 r 1\n\n\n2\n12 r 1\n\n\n2\n6 r 0\n\n\n2\n3 r 0\n\n\n2\n1 r 1\n\n\n\n0 r 1\n\n\n\n= \\(110011_{two}\\)~(b)\n(i) Length of arc AXB = \\(\\frac{105}{360} \\times 2 \\times \\frac{22}{7} \\times 6cm\\)\n= \\(11cm\\)\n\\(\\therefore Perimeter = 6cm + 11cm + 6cm = 23cm\\)\n(ii) Area of sector = \\(\\frac{105}{360} \\times \\frac{22}{7} \\times 6cm \\times 6cm\\)\n= \\(33 cm^{2}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(628, '2017', '\n\nQuestion\n\nThe table below gives the frequency distribution of the marks obtained by some students in a scholarship examination.\n\n\n\nScores (x)\n15\n25\n35\n45\n55\n65\n75\n\n\nFreq (f)\n1\n4\n12\n24\n18\n8\n3\n\n\n\n(a) Calculate, correct to 3 significant figures, the mean mark.\n(b) Find the : (i) mode ; (ii) range of the distribution.\n\n\n\n\nThe table below gives the frequency distribution of the marks obtained by some students in a scholarship examination.\n\n\n\nScores (x)\n15\n25\n35\n45\n55\n65\n75\n\n\nFreq (f)\n1\n4\n12\n24\n18\n8\n3\n\n\n\n(a) Calculate, correct to 3 significant figures, the mean mark.\n(b) Find the : (i) mode ; (ii) range of the distribution.\n\n', '', 'The table below gives the frequency distribution of the marks obtained by some students in a scholarship examination.\n\n\n\nScores (x)\n15\n25\n35\n45\n55\n65\n75\n\n\nFreq (f)\n1\n4\n12\n24\n18\n8\n3\n\n\n\n(a) Calculate, correct to 3 significant figures, the mean mark.\n(b) Find the : (i) mode ; (ii) range of the distribution.\n\n\n\n\nThe table below gives the frequency distribution of the marks obtained by some students in a scholarship examination.\n\n\n\nScores (x)\n15\n25\n35\n45\n55\n65\n75\n\n\nFreq (f)\n1\n4\n12\n24\n18\n8\n3\n\n\n\n(a) Calculate, correct to 3 significant figures, the mean mark.\n(b) Find the : (i) mode ; (ii) range of the distribution.', 'The table below gives the frequency distribution of the marks obtained by some students in a scholarship examination.\n\n\n\nScores (x)\n15\n25\n35\n45\n55\n65\n75\n\n\nFreq (f)\n1\n4\n12\n24\n18\n8\n3\n\n\n\n(a) Calculate, correct to 3 significant figures, the mean mark.\n(b) Find the : (i) mode ; (ii) range of the distribution.\n\n\n\n\nThe table below gives the frequency distribution of the marks obtained by some students in a scholarship examination.\n\n\n\nScores (x)\n15\n25\n35\n45\n55\n65\n75\n\n\nFreq (f)\n1\n4\n12\n24\n18\n8\n3\n\n\n\n(a) Calculate, correct to 3 significant figures, the mean mark.~(b) Find the : (i) mode ; (ii) range of the distribution.\n\n\n\n\nThe table below gives the frequency distribution of the marks obtained by some students in a scholarship examination.\n\n\n\nScores (x)\n15\n25\n35\n45\n55\n65\n75\n\n\nFreq (f)\n1\n4\n12\n24\n18\n8\n3\n\n\n\n(a) Calculate, correct to 3 significant figures, the mean mark.\n(b) Find the : (i) mode ; (ii) range of the distribution.', NULL, '\n\nExplanation\n(a)\n\n\n\nx\nf\nfx\n\n\n15\n1\n15\n\n\n25\n4\n100\n\n\n35\n12\n420\n\n\n45\n24\n1080\n\n\n55\n18\n990\n\n\n65\n8\n520\n\n\n75\n3\n225\n\n\n\n70\n3350\n\n\n\n\\(Mean = \\frac{\\sum fx}{\\sum f} = \\frac{3350}{70} = 47.8571\\)\n\\(\\approxeq 47.9\\) (to 3 sig. figs)\n(b)(i) Mode = 45 (score with the highest frequency)\n(ii) Range = Highest mark - Lowest mark\n= 75 - 15 = 60.\n\n\n', '', 'Explanation\n(a)\n\n\n\nx\nf\nfx\n\n\n15\n1\n15\n\n\n25\n4\n100\n\n\n35\n12\n420\n\n\n45\n24\n1080\n\n\n55\n18\n990\n\n\n65\n8\n520\n\n\n75\n3\n225\n\n\n\n70\n3350\n\n\n\n\\(Mean = \\frac{\\sum fx}{\\sum f} = \\frac{3350}{70} = 47.8571\\)\n\\(\\approxeq 47.9\\) (to 3 sig. figs)\n(b)(i) Mode = 45 (score with the highest frequency)\n(ii) Range = Highest mark - Lowest mark\n= 75 - 15 = 60.\n\n\n', '(a)\n\n\n\nx\nf\nfx\n\n\n15\n1\n15\n\n\n25\n4\n100\n\n\n35\n12\n420\n\n\n45\n24\n1080\n\n\n55\n18\n990\n\n\n65\n8\n520\n\n\n75\n3\n225\n\n\n\n70\n3350\n\n\n\n\\(Mean = \\frac{\\sum fx}{\\sum f} = \\frac{3350}{70} = 47.8571\\)\n\\(\\approxeq 47.9\\) (to 3 sig. figs)~(b)(i) Mode = 45 (score with the highest frequency)\n(ii) Range = Highest mark - Lowest mark\n= 75 - 15 = 60.\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(629, '2017', '\n\nQuestion\n\n(a) If \\(\\log_{10} (3x - 1) - \\log_{10} 2 = 3\\), find the value of x.\n(b) Use logarithm tables to evaluate \\(\\sqrt{\\frac{0.897 \\times 3.536}{0.00249}}\\), correct to 3 significant figures.\n\n', '', '(a) If \\(\\log_{10} (3x - 1) - \\log_{10} 2 = 3\\), find the value of x.\n(b) Use logarithm tables to evaluate \\(\\sqrt{\\frac{0.897 \\times 3.536}{0.00249}}\\), correct to 3 significant figures.', '(a) If \\(\\log_{10} (3x - 1) - \\log_{10} 2 = 3\\), find the value of x.~(b) Use logarithm tables to evaluate \\(\\sqrt{\\frac{0.897 \\times 3.536}{0.00249}}\\), correct to 3 significant figures.', NULL, '\n\nExplanation\n(a) \\(\\log_{10} (3x - 1) - \\log_{10} 2 = 3\\)\n\\(\\log_{10} (\\frac{3x - 1}{2}) = 3\\)\n\\(\\implies \\frac{3x - 1}{2} = 10^{3} = 1000\\)\n\\(\\implies 3x - 1 = 2000\\)\n\\(3x = 2000 + 1 = 2001\\)\n\\(x = \\frac{2001}{3} = 667\\)\n(b)\n\n\n\nNo\nLog\n\n\n0.8972\n\\(\\bar{1}.9528\\) +\n\n\n3.536\n\\(0.5485\\)\n\n\n\n= 0.5013 -\n\n\n0.00249\n\\(\\bar{3}.3962\\)\n\n\nAntilog - 35.70\n\\(3.1051 \\div 2 = 1.5526\\)\n\n\n\n\\(\\therefore \\sqrt{\\frac{0.8972 \\times 3.536}{0.00249}} = 35.70\\)\n\n', '', 'Explanation\n(a) \\(\\log_{10} (3x - 1) - \\log_{10} 2 = 3\\)\n\\(\\log_{10} (\\frac{3x - 1}{2}) = 3\\)\n\\(\\implies \\frac{3x - 1}{2} = 10^{3} = 1000\\)\n\\(\\implies 3x - 1 = 2000\\)\n\\(3x = 2000 + 1 = 2001\\)\n\\(x = \\frac{2001}{3} = 667\\)\n(b)\n\n\n\nNo\nLog\n\n\n0.8972\n\\(\\bar{1}.9528\\) +\n\n\n3.536\n\\(0.5485\\)\n\n\n\n= 0.5013 -\n\n\n0.00249\n\\(\\bar{3}.3962\\)\n\n\nAntilog - 35.70\n\\(3.1051 \\div 2 = 1.5526\\)\n\n\n\n\\(\\therefore \\sqrt{\\frac{0.8972 \\times 3.536}{0.00249}} = 35.70\\)\n\n', '(a) \\(\\log_{10} (3x - 1) - \\log_{10} 2 = 3\\)\n\\(\\log_{10} (\\frac{3x - 1}{2}) = 3\\)\n\\(\\implies \\frac{3x - 1}{2} = 10^{3} = 1000\\)\n\\(\\implies 3x - 1 = 2000\\)\n\\(3x = 2000 + 1 = 2001\\)\n\\(x = \\frac{2001}{3} = 667\\)~(b)\n\n\n\nNo\nLog\n\n\n0.8972\n\\(\\bar{1}.9528\\) +\n\n\n3.536\n\\(0.5485\\)\n\n\n\n= 0.5013 -\n\n\n0.00249\n\\(\\bar{3}.3962\\)\n\n\nAntilog - 35.70\n\\(3.1051 \\div 2 = 1.5526\\)\n\n\n\n\\(\\therefore \\sqrt{\\frac{0.8972 \\times 3.536}{0.00249}} = 35.70\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(630, '2017', '\n\nQuestion\n\nA bag contains 12 white balls and 8 black balls, another contains 10 white balls and 15 black balls. If two balls are drawn, without replacement from each bag, find the probability that :\n(a) all four balls are black ;\n(b) exactly one of thefour balls is white.\n\n', '', 'A bag contains 12 white balls and 8 black balls, another contains 10 white balls and 15 black balls. If two balls are drawn, without replacement from each bag, find the probability that :\n(a) all four balls are black ;\n(b) exactly one of thefour balls is white.', 'A bag contains 12 white balls and 8 black balls, another contains 10 white balls and 15 black balls. If two balls are drawn, without replacement from each bag, find the probability that :\n(a) all four balls are black ;~(b) exactly one of thefour balls is white.', NULL, '\n\nExplanation\n(a) 1st bag contains 12 white (W)balls and 8 black (B) balls; 2nd bag contains 10 white (W) balls and 15 black (B) balls.\nP(two black balls from the 1st bag and two black balls from 2nd bag)\n= \\((\\frac{8}{20} \\times \\frac{7}{19}) \\times (\\frac{15}{25} \\times \\frac{14}{24})\\)\n= \\(\\frac{14}{95} \\times \\frac{7}{20}\\)\n= \\(\\frac{49}{950}\\)\n(b) P(picking exactly one white ball) = P(WB from 1st bag and BB from 2nd bag) + P(BW from 1st bag and BB from 2nd bag) + P(BB from 1st bag and WB from 2nd bag) + P(BB from 1st bag and BW from 2nd bag)\n= \\(\\frac{12}{20} \\times \\frac{8}{19} \\times \\frac{15}{25} \\times \\frac{14}{24} + \\frac{8}{20} \\times \\frac{12}{19} \\times \\frac{15}{25} \\times \\frac{14}{24} + \\frac{8}{20} \\times \\frac{7}{19} \\times \\frac{10}{25} \\times \\frac{15}{24} + \\frac{8}{20} \\times \\frac{7}{19} \\times \\frac{15}{25} \\times \\frac{10}{24}\\)\n= \\(\\frac{42}{475} + \\frac{42}{475} + \\frac{7}{190} + \\frac{7}{190}\\)\n= \\(\\frac{238}{950}\\)\n\n', '', 'Explanation\n(a) 1st bag contains 12 white (W)balls and 8 black (B) balls; 2nd bag contains 10 white (W) balls and 15 black (B) balls.\nP(two black balls from the 1st bag and two black balls from 2nd bag)\n= \\((\\frac{8}{20} \\times \\frac{7}{19}) \\times (\\frac{15}{25} \\times \\frac{14}{24})\\)\n= \\(\\frac{14}{95} \\times \\frac{7}{20}\\)\n= \\(\\frac{49}{950}\\)\n(b) P(picking exactly one white ball) = P(WB from 1st bag and BB from 2nd bag) + P(BW from 1st bag and BB from 2nd bag) + P(BB from 1st bag and WB from 2nd bag) + P(BB from 1st bag and BW from 2nd bag)\n= \\(\\frac{12}{20} \\times \\frac{8}{19} \\times \\frac{15}{25} \\times \\frac{14}{24} + \\frac{8}{20} \\times \\frac{12}{19} \\times \\frac{15}{25} \\times \\frac{14}{24} + \\frac{8}{20} \\times \\frac{7}{19} \\times \\frac{10}{25} \\times \\frac{15}{24} + \\frac{8}{20} \\times \\frac{7}{19} \\times \\frac{15}{25} \\times \\frac{10}{24}\\)\n= \\(\\frac{42}{475} + \\frac{42}{475} + \\frac{7}{190} + \\frac{7}{190}\\)\n= \\(\\frac{238}{950}\\)\n\n', '(a) 1st bag contains 12 white (W)balls and 8 black (B) balls; 2nd bag contains 10 white (W) balls and 15 black (B) balls.\nP(two black balls from the 1st bag and two black balls from 2nd bag)\n= \\((\\frac{8}{20} \\times \\frac{7}{19}) \\times (\\frac{15}{25} \\times \\frac{14}{24})\\)\n= \\(\\frac{14}{95} \\times \\frac{7}{20}\\)\n= \\(\\frac{49}{950}\\)~(b) P(picking exactly one white ball) = P(WB from 1st bag and BB from 2nd bag) + P(BW from 1st bag and BB from 2nd bag) + P(BB from 1st bag and WB from 2nd bag) + P(BB from 1st bag and BW from 2nd bag)\n= \\(\\frac{12}{20} \\times \\frac{8}{19} \\times \\frac{15}{25} \\times \\frac{14}{24} + \\frac{8}{20} \\times \\frac{12}{19} \\times \\frac{15}{25} \\times \\frac{14}{24} + \\frac{8}{20} \\times \\frac{7}{19} \\times \\frac{10}{25} \\times \\frac{15}{24} + \\frac{8}{20} \\times \\frac{7}{19} \\times \\frac{15}{25} \\times \\frac{10}{24}\\)\n= \\(\\frac{42}{475} + \\frac{42}{475} + \\frac{7}{190} + \\frac{7}{190}\\)\n= \\(\\frac{238}{950}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(631, '2017', '\n\nQuestion\n\n(a) Using a ruler and a pair of compasses only, construct (i) a triangle XYZ in which /YZ/ = 8cm, < XYZ = 60° and < XZY = 75°. Measure /XY/; (ii) the locus \\(l_{1}\\) of points equidistant from Y and Z ; (iii) the locus \\(l_{2}\\) of points equidistant from XY and YZ.\n(b) Measure QY where Q is the point of intersection of \\(l_{1}\\) and \\(l_{2}\\).\n\n', '', '(a) Using a ruler and a pair of compasses only, construct (i) a triangle XYZ in which /YZ/ = 8cm, < XYZ = 60° and < XZY = 75°. Measure /XY/; (ii) the locus \\(l_{1}\\) of points equidistant from Y and Z ; (iii) the locus \\(l_{2}\\) of points equidistant from XY and YZ.\n(b) Measure QY where Q is the point of intersection of \\(l_{1}\\) and \\(l_{2}\\).', '(a) Using a ruler and a pair of compasses only, construct (i) a triangle XYZ in which /YZ/ = 8cm, < XYZ = 60° and < XZY = 75°. Measure /XY/; (ii) the locus \\(l_{1}\\) of points equidistant from Y and Z ; (iii) the locus \\(l_{2}\\) of points equidistant from XY and YZ.~(b) Measure QY where Q is the point of intersection of \\(l_{1}\\) and \\(l_{2}\\).', NULL, '\n\nExplanation\n(a)\n(b) /QY/ = 4.8 cm\n\n', '', 'Explanation\n(a)\n(b) /QY/ = 4.8 cm\n\n', '~(b) /QY/ = 4.8 cm\n', 'WAEC-PK-1', 'WAEC-MATH', 1);
INSERT INTO `exam_simulation2` (`id`, `examYear`, `questions`, `parent_que`, `child_que`, `gen_child_que_desc`, `imageDir`, `explanation`, `parent_que_expl`, `child_que_exp`, `gen_child_que_expl`, `package_id`, `subject_id`, `status`) VALUES
(632, '2017', '\n\nQuestion\n\nThe table below gives the ages, to the nearest 5 years of 50 people.\n\n\n\nAge in years\n10\n15\n20\n25\n30\n\n\nNo of people\n8\n19\n10\n7\n6\n\n\n\n(a) Construct a cumulative frequency table for the distribution.\n(b) Draw a cumulative frequency curve (Ogive)\n(c) From your Ogive, find the : (i) median age ; (ii) number of people who are at most 15 years of age ; (iii) number of people who are between 20 and 25 years of age.\n\n\n\nThe table below gives the ages, to the nearest 5 years of 50 people.\n\n\n\nAge in years\n10\n15\n20\n25\n30\n\n\nNo of people\n8\n19\n10\n7\n6\n\n\n', '', 'The table below gives the ages, to the nearest 5 years of 50 people.\n\n\n\nAge in years\n10\n15\n20\n25\n30\n\n\nNo of people\n8\n19\n10\n7\n6\n\n\n\n(a) Construct a cumulative frequency table for the distribution.\n(b) Draw a cumulative frequency curve (Ogive)\n(c) From your Ogive, find the : (i) median age ; (ii) number of people who are at most 15 years of age ; (iii) number of people who are between 20 and 25 years of age.\n\n\n\nThe table below gives the ages, to the nearest 5 years of 50 people.\n\n\n\nAge in years\n10\n15\n20\n25\n30\n\n\nNo of people\n8\n19\n10\n7\n6\n', 'The table below gives the ages, to the nearest 5 years of 50 people.\n\n\n\nAge in years\n10\n15\n20\n25\n30\n\n\nNo of people\n8\n19\n10\n7\n6\n\n\n\n(a) Construct a cumulative frequency table for the distribution.~(b) Draw a cumulative frequency curve (Ogive)\n(c) From your Ogive, find the : (i) median age ; (ii) number of people who are at most 15 years of age ; (iii) number of people who are between 20 and 25 years of age.\n\n\n\nThe table below gives the ages, to the nearest 5 years of 50 people.\n\n\n\nAge in years\n10\n15\n20\n25\n30\n\n\nNo of people\n8\n19\n10\n7\n6\n~(c) From your Ogive, find the : (i) median age ; (ii) number of people who are at most 15 years of age ; (iii) number of people who are between 20 and 25 years of age.\n\n\n\nThe table below gives the ages, to the nearest 5 years of 50 people.\n\n\n\nAge in years\n10\n15\n20\n25\n30\n\n\nNo of people\n8\n19\n10\n7\n6\n', NULL, '\n\nExplanation\n(a)\n\n\n\nAge in years\nNo of people\nCum Freq\n\n\n10\n8\n8\n\n\n15\n19\n27\n\n\n20\n10\n37\n\n\n25\n7\n44\n\n\n30\n6\n50\n\n\n\n(b)\n(c)(i) Median age = 14.5 years\n(ii) Number of people who are at most 15 years of age = 27 people\n(iii) Number of people who are between 20 and 25 years = people who are 25 years - people who are 20 years\n= 44 - 37 = 7 people\n\n', '', 'Explanation\n(a)\n\n\n\nAge in years\nNo of people\nCum Freq\n\n\n10\n8\n8\n\n\n15\n19\n27\n\n\n20\n10\n37\n\n\n25\n7\n44\n\n\n30\n6\n50\n\n\n\n(b)\n(c)(i) Median age = 14.5 years\n(ii) Number of people who are at most 15 years of age = 27 people\n(iii) Number of people who are between 20 and 25 years = people who are 25 years - people who are 20 years\n= 44 - 37 = 7 people\n\n', '(a)\n\n\n\nAge in years\nNo of people\nCum Freq\n\n\n10\n8\n8\n\n\n15\n19\n27\n\n\n20\n10\n37\n\n\n25\n7\n44\n\n\n30\n6\n50\n\n\n~(b)\n(c)(i) Median age = 14.5 years\n(ii) Number of people who are at most 15 years of age = 27 people\n(iii) Number of people who are between 20 and 25 years = people who are 25 years - people who are 20 years\n= 44 - 37 = 7 people\n~(c)(i) Median age = 14.5 years\n(ii) Number of people who are at most 15 years of age = 27 people\n(iii) Number of people who are between 20 and 25 years = people who are 25 years - people who are 20 years\n= 44 - 37 = 7 people\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(633, '2017', '\n\nQuestion\n\n(a) Copy and complete the following table of values for \\(y = 2x^{2} - 9x - 1\\).\n\n\n\nx\n-1\n0\n1\n2\n3\n4\n5\n6\n\n\ny\n\n-1\n-8\n-11\n\n\n\n17\n\n\n\n(b) Using a scale of 2cm to represent 1 unit on the x- axis and 2cm to represent 5 units on the y- axis, draw the graph of \\(y = 2x^{2} - 9x - 1\\).\n(c) Use your graph to find the : (i) roots of the equation \\(2x^{2} - 9x = 4\\), correct to one decimal place ; (ii) gradient of the curve \\(y = 2x^{2} - 9x - 1\\) at x = 3.\n\n\n\n(a) Copy and complete the following table of values for \\(y = 2x^{2} - 9x - 1\\).\n\n\n\nx\n-1\n0\n1\n2\n3\n4\n5\n6\n\n\ny\n\n-1\n-8\n-11\n\n\n\n17\n\n\n', '', '(a) Copy and complete the following table of values for \\(y = 2x^{2} - 9x - 1\\).\n\n\n\nx\n-1\n0\n1\n2\n3\n4\n5\n6\n\n\ny\n\n-1\n-8\n-11\n\n\n\n17\n\n\n\n(b) Using a scale of 2cm to represent 1 unit on the x- axis and 2cm to represent 5 units on the y- axis, draw the graph of \\(y = 2x^{2} - 9x - 1\\).\n(c) Use your graph to find the : (i) roots of the equation \\(2x^{2} - 9x = 4\\), correct to one decimal place ; (ii) gradient of the curve \\(y = 2x^{2} - 9x - 1\\) at x = 3.\n\n\n\n(a) Copy and complete the following table of values for \\(y = 2x^{2} - 9x - 1\\).\n\n\n\nx\n-1\n0\n1\n2\n3\n4\n5\n6\n\n\ny\n\n-1\n-8\n-11\n\n\n\n17\n', '(a) Copy and complete the following table of values for \\(y = 2x^{2} - 9x - 1\\).\n\n\n\nx\n-1\n0\n1\n2\n3\n4\n5\n6\n\n\ny\n\n-1\n-8\n-11\n\n\n\n17\n\n\n~(b) Using a scale of 2cm to represent 1 unit on the x- axis and 2cm to represent 5 units on the y- axis, draw the graph of \\(y = 2x^{2} - 9x - 1\\).\n(c) Use your graph to find the : (i) roots of the equation \\(2x^{2} - 9x = 4\\), correct to one decimal place ; (ii) gradient of the curve \\(y = 2x^{2} - 9x - 1\\) at x = 3.\n\n\n\n(a) Copy and complete the following table of values for \\(y = 2x^{2} - 9x - 1\\).\n\n\n\nx\n-1\n0\n1\n2\n3\n4\n5\n6\n\n\ny\n\n-1\n-8\n-11\n\n\n\n17\n~(c) Use your graph to find the : (i) roots of the equation \\(2x^{2} - 9x = 4\\), correct to one decimal place ; (ii) gradient of the curve \\(y = 2x^{2} - 9x - 1\\) at x = 3.\n\n\n\n(a) Copy and complete the following table of values for \\(y = 2x^{2} - 9x - 1\\).\n\n\n\nx\n-1\n0\n1\n2\n3\n4\n5\n6\n\n\ny\n\n-1\n-8\n-11\n\n\n\n17\n', NULL, '\n\nExplanation\n(a)\n\n\n\nx\n-1\n0\n1\n2\n3\n4\n5\n6\n\n\ny\n10\n-1\n-8\n-11\n-10\n-5\n4\n17\n\n\n\n(b)\n(c)(i)\\(2x^{2} - 9x = 4\\)\n\\(2x^{2} - 9x - 1 = 4 - 1 = 3\\)\nAt y = 3, x = 0.2 and 4.3.\n(ii) Gradient of \\(y = 2x^{2} - 9x - 1\\) at x = 3\n= \\(\\frac{BC}{AB} = \\frac{9.5}{4.3}\\)\n= \\(2.2\\)\n\n', '', 'Explanation\n(a)\n\n\n\nx\n-1\n0\n1\n2\n3\n4\n5\n6\n\n\ny\n10\n-1\n-8\n-11\n-10\n-5\n4\n17\n\n\n\n(b)\n(c)(i)\\(2x^{2} - 9x = 4\\)\n\\(2x^{2} - 9x - 1 = 4 - 1 = 3\\)\nAt y = 3, x = 0.2 and 4.3.\n(ii) Gradient of \\(y = 2x^{2} - 9x - 1\\) at x = 3\n= \\(\\frac{BC}{AB} = \\frac{9.5}{4.3}\\)\n= \\(2.2\\)\n\n', '(a)\n\n\n\nx\n-1\n0\n1\n2\n3\n4\n5\n6\n\n\ny\n10\n-1\n-8\n-11\n-10\n-5\n4\n17\n\n\n~(b)\n(c)(i)\\(2x^{2} - 9x = 4\\)\n\\(2x^{2} - 9x - 1 = 4 - 1 = 3\\)\nAt y = 3, x = 0.2 and 4.3.\n(ii) Gradient of \\(y = 2x^{2} - 9x - 1\\) at x = 3\n= \\(\\frac{BC}{AB} = \\frac{9.5}{4.3}\\)\n= \\(2.2\\)\n~(c)(i)\\(2x^{2} - 9x = 4\\)\n\\(2x^{2} - 9x - 1 = 4 - 1 = 3\\)\nAt y = 3, x = 0.2 and 4.3.\n(ii) Gradient of \\(y = 2x^{2} - 9x - 1\\) at x = 3\n= \\(\\frac{BC}{AB} = \\frac{9.5}{4.3}\\)\n= \\(2.2\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(634, '2017', '\n\nQuestion\n\n(a) The fourth term of an A.P is 37 and 6th term is 12 more than the fourth term . Find the first and seventh terms.\n(b) If \\(P = {1, 2, 3, 4}\\) and \\(Q = {3, 5, 6}\\), find (i) \\(P \\cap Q\\) ; (ii) \\(P \\cup Q\\) ; (iii) \\((P \\cap Q) \\cup Q\\) ; (iv) \\((P \\cap Q) \\cup P\\).\n\n', '', '(a) The fourth term of an A.P is 37 and 6th term is 12 more than the fourth term . Find the first and seventh terms.\n(b) If \\(P = {1, 2, 3, 4}\\) and \\(Q = {3, 5, 6}\\), find (i) \\(P \\cap Q\\) ; (ii) \\(P \\cup Q\\) ; (iii) \\((P \\cap Q) \\cup Q\\) ; (iv) \\((P \\cap Q) \\cup P\\).', '(a) The fourth term of an A.P is 37 and 6th term is 12 more than the fourth term . Find the first and seventh terms.~(b) If \\(P = {1, 2, 3, 4}\\) and \\(Q = {3, 5, 6}\\), find (i) \\(P \\cap Q\\) ; (ii) \\(P \\cup Q\\) ; (iii) \\((P \\cap Q) \\cup Q\\) ; (iv) \\((P \\cap Q) \\cup P\\).', NULL, '\n\nExplanation\n(a) \\(T_{n} = a + (n - 1)d\\) (nth term of an A.P)\n\\(T_{4} = a + 3d = 37 ... (1)\\)\n\\(T_{6} = a + 5d = 12 + 37 = 49 ... (2)\\)\n(2) - (1) : \\(2d = 12 \\implies d = 6\\)\n\\(a + 3d = 37 \\implies a + 3(6) = 37\\)\n\\(a + 18 = 37 \\implies a = 37 - 18 = 19\\)\n\\(T_{7} = a + 6d = 19 + 6(6) = 19 + 36 = 55\\)\n(b) \\(P = {1, 2, 3, 4} ; Q = {3, 5, 6}\\)\n(i) \\(P \\cap Q = {3}\\)\n(ii) \\(P \\cup Q = {1, 2, 3, 4, 5, 6}\\)\n(iii) \\((P \\cap Q) \\cup Q = {3, 5, 6}\\)\n(iv) \\((P \\cap Q) \\cup P = {1, 2, 3, 4}\\)\n\n', '', 'Explanation\n(a) \\(T_{n} = a + (n - 1)d\\) (nth term of an A.P)\n\\(T_{4} = a + 3d = 37 ... (1)\\)\n\\(T_{6} = a + 5d = 12 + 37 = 49 ... (2)\\)\n(2) - (1) : \\(2d = 12 \\implies d = 6\\)\n\\(a + 3d = 37 \\implies a + 3(6) = 37\\)\n\\(a + 18 = 37 \\implies a = 37 - 18 = 19\\)\n\\(T_{7} = a + 6d = 19 + 6(6) = 19 + 36 = 55\\)\n(b) \\(P = {1, 2, 3, 4} ; Q = {3, 5, 6}\\)\n(i) \\(P \\cap Q = {3}\\)\n(ii) \\(P \\cup Q = {1, 2, 3, 4, 5, 6}\\)\n(iii) \\((P \\cap Q) \\cup Q = {3, 5, 6}\\)\n(iv) \\((P \\cap Q) \\cup P = {1, 2, 3, 4}\\)\n\n', '(a) \\(T_{n} = a + (n - 1)d\\) (nth term of an A.P)\n\\(T_{4} = a + 3d = 37 ... (1)\\)\n\\(T_{6} = a + 5d = 12 + 37 = 49 ... (2)\\)\n(2) - (1) : \\(2d = 12 \\implies d = 6\\)\n\\(a + 3d = 37 \\implies a + 3(6) = 37\\)\n\\(a + 18 = 37 \\implies a = 37 - 18 = 19\\)\n\\(T_{7} = a + 6d = 19 + 6(6) = 19 + 36 = 55\\)~(b) \\(P = {1, 2, 3, 4} ; Q = {3, 5, 6}\\)\n(i) \\(P \\cap Q = {3}\\)\n(ii) \\(P \\cup Q = {1, 2, 3, 4, 5, 6}\\)\n(iii) \\((P \\cap Q) \\cup Q = {3, 5, 6}\\)\n(iv) \\((P \\cap Q) \\cup P = {1, 2, 3, 4}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(635, '2017', '\n\nQuestion\n\n(a) Two points X(32°N, 47°W) and Y(32°N, 25°E) are on the earth\'s surface. If it takes an aeroplane 11 hours to fly from X to Y along the parallel of latitude, calculate its speed, correct to the nearest kilometre per hour. [Radius of the earth = 6400km; \\(\\pi = \\frac{22}{7}\\)]\n(b) Two observers P and Q, 15metres apart observe a kite (K) in the same vertical plane and from the same side of the kite. The angles of elevation of the kite from P and Q are 35° and 45° respectively. Find the height of the kite to the nearest metre.\n\n', '', '(a) Two points X(32°N, 47°W) and Y(32°N, 25°E) are on the earth\'s surface. If it takes an aeroplane 11 hours to fly from X to Y along the parallel of latitude, calculate its speed, correct to the nearest kilometre per hour. [Radius of the earth = 6400km; \\(\\pi = \\frac{22}{7}\\)]\n(b) Two observers P and Q, 15metres apart observe a kite (K) in the same vertical plane and from the same side of the kite. The angles of elevation of the kite from P and Q are 35° and 45° respectively. Find the height of the kite to the nearest metre.', '(a) Two points X(32°N, 47°W) and Y(32°N, 25°E) are on the earth\'s surface. If it takes an aeroplane 11 hours to fly from X to Y along the parallel of latitude, calculate its speed, correct to the nearest kilometre per hour. [Radius of the earth = 6400km; \\(\\pi = \\frac{22}{7}\\)]~(b) Two observers P and Q, 15metres apart observe a kite (K) in the same vertical plane and from the same side of the kite. The angles of elevation of the kite from P and Q are 35° and 45° respectively. Find the height of the kite to the nearest metre.', NULL, '\n\nExplanation\n(a)\nDifference in longitude = 47° + 25° = 72°\nDistance /XY/ = \\(\\frac{\\theta}{360°} \\times 2\\pi r\\)\nWhere \\(r = R \\cos 32°\\)\n\\(\\therefore /XY/ = \\frac{72}{360} \\times 2 \\times \\frac{22}{7} \\times 6400 \\times \\cos 32\\)\n= \\(2 \\times \\frac{22}{7} \\times 1280 \\times 0.8480\\)\n= \\(6822.765km\\)\n\\(Speed = \\frac{Distance}{Time}\\)\n= \\(\\frac{6822.766}{11} = 620.25 km/h\\)\n(b)\\\n\\( <TQP = 180° - 45° = 135°\\) (angles on a straight line)\n\\(< QTP = 180° - (135° + 35°) = 10°\\) (sum of angles in a triangle)\n\\(\\frac{TQ}{\\sin 35°} = \\frac{15}{\\sin 10°}\\)\n\\(\\therefore TQ = \\frac{15 \\sin 35°}{\\sin 10°}\\)\n= \\(49.546m\\)\n\\(\\therefore \\sin 45° = \\frac{TA}{TQ}\\)\n\\(\\sin 45° = \\frac{TA}{49.546}\\)\n\\(TA = 49.546 \\times 0.7071 = 35.03m\\)\n\\(\\therefore\\) The height of the kite = 35m (correct to the nearest whole number)\n\n', '', 'Explanation\n(a)\nDifference in longitude = 47° + 25° = 72°\nDistance /XY/ = \\(\\frac{\\theta}{360°} \\times 2\\pi r\\)\nWhere \\(r = R \\cos 32°\\)\n\\(\\therefore /XY/ = \\frac{72}{360} \\times 2 \\times \\frac{22}{7} \\times 6400 \\times \\cos 32\\)\n= \\(2 \\times \\frac{22}{7} \\times 1280 \\times 0.8480\\)\n= \\(6822.765km\\)\n\\(Speed = \\frac{Distance}{Time}\\)\n= \\(\\frac{6822.766}{11} = 620.25 km/h\\)\n(b)\\\n\\( <TQP = 180° - 45° = 135°\\) (angles on a straight line)\n\\(< QTP = 180° - (135° + 35°) = 10°\\) (sum of angles in a triangle)\n\\(\\frac{TQ}{\\sin 35°} = \\frac{15}{\\sin 10°}\\)\n\\(\\therefore TQ = \\frac{15 \\sin 35°}{\\sin 10°}\\)\n= \\(49.546m\\)\n\\(\\therefore \\sin 45° = \\frac{TA}{TQ}\\)\n\\(\\sin 45° = \\frac{TA}{49.546}\\)\n\\(TA = 49.546 \\times 0.7071 = 35.03m\\)\n\\(\\therefore\\) The height of the kite = 35m (correct to the nearest whole number)\n\n', '(a)\nDifference in longitude = 47° + 25° = 72°\nDistance /XY/ = \\(\\frac{\\theta}{360°} \\times 2\\pi r\\)\nWhere \\(r = R \\cos 32°\\)\n\\(\\therefore /XY/ = \\frac{72}{360} \\times 2 \\times \\frac{22}{7} \\times 6400 \\times \\cos 32\\)\n= \\(2 \\times \\frac{22}{7} \\times 1280 \\times 0.8480\\)\n= \\(6822.765km\\)\n\\(Speed = \\frac{Distance}{Time}\\)\n= \\(\\frac{6822.766}{11} = 620.25 km/h\\)~(b)\\\n\\( <TQP = 180° - 45° = 135°\\) (angles on a straight line)\n\\(< QTP = 180° - (135° + 35°) = 10°\\) (sum of angles in a triangle)\n\\(\\frac{TQ}{\\sin 35°} = \\frac{15}{\\sin 10°}\\)\n\\(\\therefore TQ = \\frac{15 \\sin 35°}{\\sin 10°}\\)\n= \\(49.546m\\)\n\\(\\therefore \\sin 45° = \\frac{TA}{TQ}\\)\n\\(\\sin 45° = \\frac{TA}{49.546}\\)\n\\(TA = 49.546 \\times 0.7071 = 35.03m\\)\n\\(\\therefore\\) The height of the kite = 35m (correct to the nearest whole number)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(636, '2017', '\n\nQuestion\n\n(a) Simplify, without using Mathematical tables: \\(\\log_{10} (\\frac{30}{16}) - 2 \\log_{10} (\\frac{5}{9}) + \\log_{10} (\\frac{400}{243})\\)\n(b) Without using Mathematical tables, calculate \\(\\sqrt{\\frac{P}{Q}}\\) where \\(P = 3.6 \\times 10^{-3}\\) and \\(Q = 2.25 \\times 10^{6}\\), leaving your answer in standard form.\n\n', '', '(a) Simplify, without using Mathematical tables: \\(\\log_{10} (\\frac{30}{16}) - 2 \\log_{10} (\\frac{5}{9}) + \\log_{10} (\\frac{400}{243})\\)\n(b) Without using Mathematical tables, calculate \\(\\sqrt{\\frac{P}{Q}}\\) where \\(P = 3.6 \\times 10^{-3}\\) and \\(Q = 2.25 \\times 10^{6}\\), leaving your answer in standard form.', '(a) Simplify, without using Mathematical tables: \\(\\log_{10} (\\frac{30}{16}) - 2 \\log_{10} (\\frac{5}{9}) + \\log_{10} (\\frac{400}{243})\\)~(b) Without using Mathematical tables, calculate \\(\\sqrt{\\frac{P}{Q}}\\) where \\(P = 3.6 \\times 10^{-3}\\) and \\(Q = 2.25 \\times 10^{6}\\), leaving your answer in standard form.', NULL, '\n\nExplanation\n(a) \\(\\log_{10} (\\frac{30}{16}) - 2 \\log_{10} (\\frac{5}{9}) + \\log_{10} (\\frac{400}{243})\\)\n= \\(\\log_{10} (\\frac{30}{16}) - \\log_{10} (\\frac{5}{9})^{2} + \\log_{10} (\\frac{400}{243})\\)\n= \\(\\log_{10} (\\frac{30}{16} \\times \\frac{400}{243} \\div \\frac{25}{81})\\)\n= \\(\\log_{10} (\\frac{30}{16} \\times \\frac{400}{243} \\times \\frac{81}{25})\\)\n= \\(\\log_{10} (10) = 1\\)\n(b) \\(\\sqrt{\\frac{P}{Q}} = \\sqrt{\\frac{3.6 \\times 10^{-3}}{2.25 \\times 10^{6}}\\)\n= \\(\\sqrt{\\frac{36 \\times 10^{-4}}{2.25 \\times 10^{6}}\\)\n= \\(\\frac{6 \\times 10^{-2}}{1.5 \\times 10^{3}}\\)\n= \\(4 \\times 10^{-2- 3} = 4 \\times 10^{-5}\\)\n\n', '', 'Explanation\n(a) \\(\\log_{10} (\\frac{30}{16}) - 2 \\log_{10} (\\frac{5}{9}) + \\log_{10} (\\frac{400}{243})\\)\n= \\(\\log_{10} (\\frac{30}{16}) - \\log_{10} (\\frac{5}{9})^{2} + \\log_{10} (\\frac{400}{243})\\)\n= \\(\\log_{10} (\\frac{30}{16} \\times \\frac{400}{243} \\div \\frac{25}{81})\\)\n= \\(\\log_{10} (\\frac{30}{16} \\times \\frac{400}{243} \\times \\frac{81}{25})\\)\n= \\(\\log_{10} (10) = 1\\)\n(b) \\(\\sqrt{\\frac{P}{Q}} = \\sqrt{\\frac{3.6 \\times 10^{-3}}{2.25 \\times 10^{6}}\\)\n= \\(\\sqrt{\\frac{36 \\times 10^{-4}}{2.25 \\times 10^{6}}\\)\n= \\(\\frac{6 \\times 10^{-2}}{1.5 \\times 10^{3}}\\)\n= \\(4 \\times 10^{-2- 3} = 4 \\times 10^{-5}\\)\n\n', '(a) \\(\\log_{10} (\\frac{30}{16}) - 2 \\log_{10} (\\frac{5}{9}) + \\log_{10} (\\frac{400}{243})\\)\n= \\(\\log_{10} (\\frac{30}{16}) - \\log_{10} (\\frac{5}{9})^{2} + \\log_{10} (\\frac{400}{243})\\)\n= \\(\\log_{10} (\\frac{30}{16} \\times \\frac{400}{243} \\div \\frac{25}{81})\\)\n= \\(\\log_{10} (\\frac{30}{16} \\times \\frac{400}{243} \\times \\frac{81}{25})\\)\n= \\(\\log_{10} (10) = 1\\)~(b) \\(\\sqrt{\\frac{P}{Q}} = \\sqrt{\\frac{3.6 \\times 10^{-3}}{2.25 \\times 10^{6}}\\)\n= \\(\\sqrt{\\frac{36 \\times 10^{-4}}{2.25 \\times 10^{6}}\\)\n= \\(\\frac{6 \\times 10^{-2}}{1.5 \\times 10^{3}}\\)\n= \\(4 \\times 10^{-2- 3} = 4 \\times 10^{-5}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(637, '2017', '\n\nQuestion\n\nThe universal set \\(\\varepsilon\\) is the set of all integers and the subset P, Q, R of \\(\\varepsilon\\) are given by:\n\\(P = {x : x < 0} ; Q = {... , -5, -3, -1, 1, 3, 5} ; R = {x : -2 \\leq x < 7}\\)\n(a) Find \\(Q \\cap R\\).\n(b) Find \\(R\'\\) where R\' is the complement of R with respect to \\(\\varepsilon\\).\n(c) Find \\(P\' \\cup R\'\\)\n(d) List the members of \\((P \\cap Q)\'\\).\n\n', '', 'The universal set \\(\\varepsilon\\) is the set of all integers and the subset P, Q, R of \\(\\varepsilon\\) are given by:\n\\(P = {x : x < 0} ; Q = {... , -5, -3, -1, 1, 3, 5} ; R = {x : -2 \\leq x < 7}\\)\n(a) Find \\(Q \\cap R\\).\n(b) Find \\(R\'\\) where R\' is the complement of R with respect to \\(\\varepsilon\\).\n(c) Find \\(P\' \\cup R\'\\)\n(d) List the members of \\((P \\cap Q)\'\\).', 'The universal set \\(\\varepsilon\\) is the set of all integers and the subset P, Q, R of \\(\\varepsilon\\) are given by:\n\\(P = {x : x < 0} ; Q = {... , -5, -3, -1, 1, 3, 5} ; R = {x : -2 \\leq x < 7}\\)\n(a) Find \\(Q \\cap R\\).~(b) Find \\(R\'\\) where R\' is the complement of R with respect to \\(\\varepsilon\\).\n(c) Find \\(P\' \\cup R\'\\)\n(d) List the members of \\((P \\cap Q)\'\\).~(c) Find \\(P\' \\cup R\'\\)\n(d) List the members of \\((P \\cap Q)\'\\).~(d) List the members of \\((P \\cap Q)\'\\).', NULL, '\n\nExplanation\n\\(P = {..., -5, -4, -3, -2, -1}\\)\n\\(Q = {..., -5, -3, -1, 1, 3, 5, ...}\\)\n\\(R = {-2, -1, 0, 1, 2, 3, 4, 5, 6}\\)\n(a) \\(Q \\cap R = {-1, 1, 3, 5}\\)\n(b) \\(R\' = {..., -5, -4, -3, 7, 8, ...}\\)\n(c) \\(P\' = {0, 1, 2, 3, ...}\\)\n\\(P\' \\cup R\' = {-5, -4, -3, 0, 1, 2, 3, ...}\\)\n(d) \\(Q = {..., -5, -3, -1, 1, 3, 5,...}\\)\n\\(P \\cap Q = {..., -7, -5, -3, -1}\\)\n\\((P \\cap Q)\' = {..., -8, -6, -4, -2, 0, 1, 2, 3, ...}\\)\n\n', '', 'Explanation\n\\(P = {..., -5, -4, -3, -2, -1}\\)\n\\(Q = {..., -5, -3, -1, 1, 3, 5, ...}\\)\n\\(R = {-2, -1, 0, 1, 2, 3, 4, 5, 6}\\)\n(a) \\(Q \\cap R = {-1, 1, 3, 5}\\)\n(b) \\(R\' = {..., -5, -4, -3, 7, 8, ...}\\)\n(c) \\(P\' = {0, 1, 2, 3, ...}\\)\n\\(P\' \\cup R\' = {-5, -4, -3, 0, 1, 2, 3, ...}\\)\n(d) \\(Q = {..., -5, -3, -1, 1, 3, 5,...}\\)\n\\(P \\cap Q = {..., -7, -5, -3, -1}\\)\n\\((P \\cap Q)\' = {..., -8, -6, -4, -2, 0, 1, 2, 3, ...}\\)\n\n', '(a) \\(Q \\cap R = {-1, 1, 3, 5}\\)~(b) \\(R\' = {..., -5, -4, -3, 7, 8, ...}\\)\n(c) \\(P\' = {0, 1, 2, 3, ...}\\)\n\\(P\' \\cup R\' = {-5, -4, -3, 0, 1, 2, 3, ...}\\)\n(d) \\(Q = {..., -5, -3, -1, 1, 3, 5,...}\\)\n\\(P \\cap Q = {..., -7, -5, -3, -1}\\)\n\\((P \\cap Q)\' = {..., -8, -6, -4, -2, 0, 1, 2, 3, ...}\\)\n~(c) \\(P\' = {0, 1, 2, 3, ...}\\)\n\\(P\' \\cup R\' = {-5, -4, -3, 0, 1, 2, 3, ...}\\)\n(d) \\(Q = {..., -5, -3, -1, 1, 3, 5,...}\\)\n\\(P \\cap Q = {..., -7, -5, -3, -1}\\)\n\\((P \\cap Q)\' = {..., -8, -6, -4, -2, 0, 1, 2, 3, ...}\\)\n~(d) \\(Q = {..., -5, -3, -1, 1, 3, 5,...}\\)\n\\(P \\cap Q = {..., -7, -5, -3, -1}\\)\n\\((P \\cap Q)\' = {..., -8, -6, -4, -2, 0, 1, 2, 3, ...}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(638, '2017', '\n\nQuestion\n\nA simple measuring device is used at points X and Y on the same horizontal level to measure the angles of elevation of the peak P of a certain mountain. If X is known to 5,200m above sea level, /XY/ = 4,000m and the measurements of the angles of elevation of P at X and Y are 15° and 35° respectively, find the height of the mountain. (Take \\(\\tan 15 = 0.3\\) and \\(\\tan 35 = 0.7\\))\n\n', 'A simple measuring device is used at points X and Y on the same horizontal level to measure the angles of elevation of the peak P of a certain mountain. If X is known to 5,200m above sea level, /XY/ = 4,000m and the measurements of the angles of elevation of P at X and Y are 15° and 35° respectively, find the height of the mountain. (Take \\(\\tan 15 = 0.3\\) and \\(\\tan 35 = 0.7\\))', '', '', NULL, '\n\nExplanation\nIn \\(\\Delta\\) BPY, \\(\\tan 35° = \\frac{h}{x}\\)\n\\(h = x \\tan 35 = 0.7 x ... (1)\\)\nIn \\(\\Delta\\) BPX, \\(\\frac{h}{x + 4000} = \\tan 15\\)\n\\(\\frac{h}{x + 4000} = 0.3\\)\n\\(h = 0.3 (x + 4000) = 0.3x + 1200 ... (2)\\)\nEquating (1) and (2) as the values of h, we have\n\\(0.7x = 0.3x + 1200 \\implies 0.7x - 0.3x = 1200\\)\n\\(0.4x = 1200 \\times x = \\frac{1200}{0.4} = 3000m\\)\n\\(\\therefore h = 0.7x = 0.7 (3000) = 2100m\\)\n\\(\\text{The total height = } 5200m + 2100m = 7300m\\)\n\n', 'In \\(\\Delta\\) BPY, \\(\\tan 35° = \\frac{h}{x}\\)\n\\(h = x \\tan 35 = 0.7 x ... (1)\\)\nIn \\(\\Delta\\) BPX, \\(\\frac{h}{x + 4000} = \\tan 15\\)\n\\(\\frac{h}{x + 4000} = 0.3\\)\n\\(h = 0.3 (x + 4000) = 0.3x + 1200 ... (2)\\)\nEquating (1) and (2) as the values of h, we have\n\\(0.7x = 0.3x + 1200 \\implies 0.7x - 0.3x = 1200\\)\n\\(0.4x = 1200 \\times x = \\frac{1200}{0.4} = 3000m\\)\n\\(\\therefore h = 0.7x = 0.7 (3000) = 2100m\\)\n\\(\\text{The total height = } 5200m + 2100m = 7300m\\)\n\n', '', '', 'WAEC-PK-1', 'WAEC-MATH', 1),
(639, '2017', '\n\nQuestion\n\n(a) Simplify \\(\\frac{3}{m + 2n} - \\frac{2}{m - 3n}\\)\n(b) A number is made up of two digits. The sum of the digits is 11. If the digits are interchanged, the original number is increased by 9. Find the number.\n\n', '', '(a) Simplify \\(\\frac{3}{m + 2n} - \\frac{2}{m - 3n}\\)\n(b) A number is made up of two digits. The sum of the digits is 11. If the digits are interchanged, the original number is increased by 9. Find the number.', '(a) Simplify \\(\\frac{3}{m + 2n} - \\frac{2}{m - 3n}\\)~(b) A number is made up of two digits. The sum of the digits is 11. If the digits are interchanged, the original number is increased by 9. Find the number.', NULL, '\n\nExplanation\n(a) \\(\\frac{3}{m + 2n} - \\frac{2}{m - 3n}\\)\n= \\(\\frac{3(m - 3n) - 2(m + 2n)}{(m + 2n)(m - 3n)}\\)\n= \\(\\frac{3m - 9n - 2m - 4n}{(m + 2n)(m - 3n)}\\)\n= \\(\\frac{m - 13n}{(m + 2n)(m - 3n)}\\)\n(b) Let the numbers in the digit be x and z.\nHence the number is \\(10x + z\\).\nWhen it is interchanged, we have \\(10z + x\\).\n\\(10z + x = 10x + z + 9\\)\n\\(10z - z + x - 10x = 9\\)\n\\(9z - 9x = 9 \\implies z - x = 1 ... (1)\\)\n\\(x + z = 11 ... (2)\\)\nFrom (1), \\(z = 1 + x\\)\n\\(\\therefore x + 1 + x = 11\\)\n\\(2x + 1 = 11 \\implies 2x = 11 - 1 = 10\\)\n\\(\\therefore x = 5\\)\n\\(z = 1 + x = 1 + 5 = 6\\)\n\\(\\therefore\\) The original number = 56.\n\n', '', 'Explanation\n(a) \\(\\frac{3}{m + 2n} - \\frac{2}{m - 3n}\\)\n= \\(\\frac{3(m - 3n) - 2(m + 2n)}{(m + 2n)(m - 3n)}\\)\n= \\(\\frac{3m - 9n - 2m - 4n}{(m + 2n)(m - 3n)}\\)\n= \\(\\frac{m - 13n}{(m + 2n)(m - 3n)}\\)\n(b) Let the numbers in the digit be x and z.\nHence the number is \\(10x + z\\).\nWhen it is interchanged, we have \\(10z + x\\).\n\\(10z + x = 10x + z + 9\\)\n\\(10z - z + x - 10x = 9\\)\n\\(9z - 9x = 9 \\implies z - x = 1 ... (1)\\)\n\\(x + z = 11 ... (2)\\)\nFrom (1), \\(z = 1 + x\\)\n\\(\\therefore x + 1 + x = 11\\)\n\\(2x + 1 = 11 \\implies 2x = 11 - 1 = 10\\)\n\\(\\therefore x = 5\\)\n\\(z = 1 + x = 1 + 5 = 6\\)\n\\(\\therefore\\) The original number = 56.\n\n', '(a) \\(\\frac{3}{m + 2n} - \\frac{2}{m - 3n}\\)\n= \\(\\frac{3(m - 3n) - 2(m + 2n)}{(m + 2n)(m - 3n)}\\)\n= \\(\\frac{3m - 9n - 2m - 4n}{(m + 2n)(m - 3n)}\\)\n= \\(\\frac{m - 13n}{(m + 2n)(m - 3n)}\\)~(b) Let the numbers in the digit be x and z.\nHence the number is \\(10x + z\\).\nWhen it is interchanged, we have \\(10z + x\\).\n\\(10z + x = 10x + z + 9\\)\n\\(10z - z + x - 10x = 9\\)\n\\(9z - 9x = 9 \\implies z - x = 1 ... (1)\\)\n\\(x + z = 11 ... (2)\\)\nFrom (1), \\(z = 1 + x\\)\n\\(\\therefore x + 1 + x = 11\\)\n\\(2x + 1 = 11 \\implies 2x = 11 - 1 = 10\\)\n\\(\\therefore x = 5\\)\n\\(z = 1 + x = 1 + 5 = 6\\)\n\\(\\therefore\\) The original number = 56.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(640, '2017', '\n\nQuestion\n\nA box contains identical balls of which 12 are red, 16 white and 8 blue. Three balls are drawn from the box one after the other without replacement. Find the probability that :\n(a) three are red;\n(b) the first is blue and the other two are red;\n(c) two are white and one is blue.\n\n', '', 'A box contains identical balls of which 12 are red, 16 white and 8 blue. Three balls are drawn from the box one after the other without replacement. Find the probability that :\n(a) three are red;\n(b) the first is blue and the other two are red;\n(c) two are white and one is blue.', 'A box contains identical balls of which 12 are red, 16 white and 8 blue. Three balls are drawn from the box one after the other without replacement. Find the probability that :\n(a) three are red;~(b) the first is blue and the other two are red;\n(c) two are white and one is blue.~(c) two are white and one is blue.', NULL, '\n\nExplanation\nn(R) = number of red balls = 12;\nn(W) = number of white balls = 16;\nn(B) = number of blue balls = 8\nTotal number of balls = 36\n(a) P(three are red) = \\(\\frac{12}{36} \\times \\frac{11}{35} \\times \\frac{10}{34}\\)\n= \\(\\frac{11}{357}\\)\n(b) P(first is blue and other two red) = \\(\\frac{8}{36} \\times \\frac{12}{35} \\times \\frac{11}{34}\\)\n= \\(\\frac{44}{1785}\\)\n(c) P(two are white and one is blue) = \\(\\frac{16}{36} \\times \\frac{15}{35} \\times \\frac{8}{34}\\)\n= \\(\\frac{16}{357}\\)\n\n', '', 'Explanation\nn(R) = number of red balls = 12;\nn(W) = number of white balls = 16;\nn(B) = number of blue balls = 8\nTotal number of balls = 36\n(a) P(three are red) = \\(\\frac{12}{36} \\times \\frac{11}{35} \\times \\frac{10}{34}\\)\n= \\(\\frac{11}{357}\\)\n(b) P(first is blue and other two red) = \\(\\frac{8}{36} \\times \\frac{12}{35} \\times \\frac{11}{34}\\)\n= \\(\\frac{44}{1785}\\)\n(c) P(two are white and one is blue) = \\(\\frac{16}{36} \\times \\frac{15}{35} \\times \\frac{8}{34}\\)\n= \\(\\frac{16}{357}\\)\n\n', '(a) P(three are red) = \\(\\frac{12}{36} \\times \\frac{11}{35} \\times \\frac{10}{34}\\)\n= \\(\\frac{11}{357}\\)~(b) P(first is blue and other two red) = \\(\\frac{8}{36} \\times \\frac{12}{35} \\times \\frac{11}{34}\\)\n= \\(\\frac{44}{1785}\\)\n(c) P(two are white and one is blue) = \\(\\frac{16}{36} \\times \\frac{15}{35} \\times \\frac{8}{34}\\)\n= \\(\\frac{16}{357}\\)\n~(c) P(two are white and one is blue) = \\(\\frac{16}{36} \\times \\frac{15}{35} \\times \\frac{8}{34}\\)\n= \\(\\frac{16}{357}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(641, '2017', '\n\nQuestion\n\n(a)(i) Given that \\(\\log_{10} 5 = 0.699\\) and \\(\\log_{10} 3 = 0.477\\), find \\(\\log_{10} 45\\), without using Mathematical tables.\n(ii) Hence, solve \\(x^{0.8265} = 45\\).\n(b) Use Mathematical tables to evaluate \\(\\sqrt{\\frac{2.067}{0.0348 \\times 0.538}}\\)\n\n', '', '(a)(i) Given that \\(\\log_{10} 5 = 0.699\\) and \\(\\log_{10} 3 = 0.477\\), find \\(\\log_{10} 45\\), without using Mathematical tables.\n(ii) Hence, solve \\(x^{0.8265} = 45\\).\n(b) Use Mathematical tables to evaluate \\(\\sqrt{\\frac{2.067}{0.0348 \\times 0.538}}\\)', '(a)(i) Given that \\(\\log_{10} 5 = 0.699\\) and \\(\\log_{10} 3 = 0.477\\), find \\(\\log_{10} 45\\), without using Mathematical tables.\n(ii) Hence, solve \\(x^{0.8265} = 45\\).~(b) Use Mathematical tables to evaluate \\(\\sqrt{\\frac{2.067}{0.0348 \\times 0.538}}\\)', NULL, '\n\nExplanation\n(a)(i) \\(\\log_{10} 45 = \\log_{10} (3 \\times 3 \\times 5)\\)\n= \\(\\log_{10} (3^{2} \\times 5)\\)\n= \\(\\log_{10} 3^{2} + \\log_{10} 5\\)\n= \\(2 \\log_{10} 3 + \\log_{10} 5\\)\n= \\(2(0.477) + 0.699\\)\n= \\(0.954 + 0.699 = 1.653\\)\n(ii) \\(x^{0.8265} = 45\\)\nTaking the log of both sides,\n\\(\\log_{10} x^{0.8265} = \\log_{10} 45\\)\n\\(0.8265 \\log_{10} x = \\log_{10} 45\\)\n\\(\\log_{10} x = \\frac{1.653}{0.8265}\\)\n\\(\\log_{10} x = 2\\)\n\\(x = 10^{2} = 100\\)\n(b) \\(\\sqrt{\\frac{2.067}{0.0348 \\times 0.538}}\\)\n\n\n\nNo\nLog\n\n\n2.067\n\\(0.0348\\)   = 0.3513 -\n\n\n0.0348\n\n\\(\\bar{2}.5416 +\\)\n\n\n\n0.538\n\\(\\bar{1}.7308\\)\n\n\n\n\\(\\bar{2}.2724\\) = \\(\\bar{2}.2724\\)\n\n\n\n            = \\(2.0789 \\div 2 = 1.0395\\)\n\n\nAntilog - 10.95\n\n\n\n\n\\(\\therefore \\sqrt{\\frac{2.067}{0.0348 \\times 0.538}} = 10.95\\)\n\n\n', '', 'Explanation\n(a)(i) \\(\\log_{10} 45 = \\log_{10} (3 \\times 3 \\times 5)\\)\n= \\(\\log_{10} (3^{2} \\times 5)\\)\n= \\(\\log_{10} 3^{2} + \\log_{10} 5\\)\n= \\(2 \\log_{10} 3 + \\log_{10} 5\\)\n= \\(2(0.477) + 0.699\\)\n= \\(0.954 + 0.699 = 1.653\\)\n(ii) \\(x^{0.8265} = 45\\)\nTaking the log of both sides,\n\\(\\log_{10} x^{0.8265} = \\log_{10} 45\\)\n\\(0.8265 \\log_{10} x = \\log_{10} 45\\)\n\\(\\log_{10} x = \\frac{1.653}{0.8265}\\)\n\\(\\log_{10} x = 2\\)\n\\(x = 10^{2} = 100\\)\n(b) \\(\\sqrt{\\frac{2.067}{0.0348 \\times 0.538}}\\)\n\n\n\nNo\nLog\n\n\n2.067\n\\(0.0348\\)   = 0.3513 -\n\n\n0.0348\n\n\\(\\bar{2}.5416 +\\)\n\n\n\n0.538\n\\(\\bar{1}.7308\\)\n\n\n\n\\(\\bar{2}.2724\\) = \\(\\bar{2}.2724\\)\n\n\n\n            = \\(2.0789 \\div 2 = 1.0395\\)\n\n\nAntilog - 10.95\n\n\n\n\n\\(\\therefore \\sqrt{\\frac{2.067}{0.0348 \\times 0.538}} = 10.95\\)\n\n\n', '(a)(i) \\(\\log_{10} 45 = \\log_{10} (3 \\times 3 \\times 5)\\)\n= \\(\\log_{10} (3^{2} \\times 5)\\)\n= \\(\\log_{10} 3^{2} + \\log_{10} 5\\)\n= \\(2 \\log_{10} 3 + \\log_{10} 5\\)\n= \\(2(0.477) + 0.699\\)\n= \\(0.954 + 0.699 = 1.653\\)\n(ii) \\(x^{0.8265} = 45\\)\nTaking the log of both sides,\n\\(\\log_{10} x^{0.8265} = \\log_{10} 45\\)\n\\(0.8265 \\log_{10} x = \\log_{10} 45\\)\n\\(\\log_{10} x = \\frac{1.653}{0.8265}\\)\n\\(\\log_{10} x = 2\\)\n\\(x = 10^{2} = 100\\)~(b) \\(\\sqrt{\\frac{2.067}{0.0348 \\times 0.538}}\\)\n\n\n\nNo\nLog\n\n\n2.067\n\\(0.0348\\)   = 0.3513 -\n\n\n0.0348\n\n\\(\\bar{2}.5416 +\\)\n\n\n\n0.538\n\\(\\bar{1}.7308\\)\n\n\n\n\\(\\bar{2}.2724\\) = \\(\\bar{2}.2724\\)\n\n\n\n            = \\(2.0789 \\div 2 = 1.0395\\)\n\n\nAntilog - 10.95\n\n\n\n\n\\(\\therefore \\sqrt{\\frac{2.067}{0.0348 \\times 0.538}} = 10.95\\)\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(642, '2017', '\n\nQuestion\n\n(a)\nIn the diagram, BA is parallel to DE. Find the value of x.\n(b) Illustrate graphically and shade the region in which inequalities \\(y - 2x < 5 ; 2y + x \\geq 4 ; y + 2x \\leq 10\\) are satisfied.\n\n', '', '(a)\nIn the diagram, BA is parallel to DE. Find the value of x.\n(b) Illustrate graphically and shade the region in which inequalities \\(y - 2x < 5 ; 2y + x \\geq 4 ; y + 2x \\leq 10\\) are satisfied.', '(a)\nIn the diagram, BA is parallel to DE. Find the value of x.~(b) Illustrate graphically and shade the region in which inequalities \\(y - 2x < 5 ; 2y + x \\geq 4 ; y + 2x \\leq 10\\) are satisfied.', NULL, '\n\nExplanation\n(a)\nDraw RC produced to E parallel to DE.\n\\(\\therefore < BCR = 52°\\) (alternate angles BA//RC)\nalso < RCD = 48° (alternate angles RC//DE)\n\\(x° = 360° - (52° + 48°) = 360° - 100° = 260°\\) (angles at a point = 360°)\n(b) \\(y - 2x < 5 ... (i)\\)\n\\(2y + 2x \\geq 4 \\implies y + x \\geq 2 ... (ii)\\)\n\\(y + 2x \\leq 10 ... (iii)\\)\n\nFrom (i), if \\(y - 2x = 5\\), when x = 0, y = 5, when x = 2, y = 9.\nFrom (ii), if \\(y + x = 2\\). when x = 0, y = 2, when x = 3, y = -1\nFrom (iii), if \\(y + 2x = 10\\), when x = 0, y = 10, when x = 2, y = 6.\n\n', '', 'Explanation\n(a)\nDraw RC produced to E parallel to DE.\n\\(\\therefore < BCR = 52°\\) (alternate angles BA//RC)\nalso < RCD = 48° (alternate angles RC//DE)\n\\(x° = 360° - (52° + 48°) = 360° - 100° = 260°\\) (angles at a point = 360°)\n(b) \\(y - 2x < 5 ... (i)\\)\n\\(2y + 2x \\geq 4 \\implies y + x \\geq 2 ... (ii)\\)\n\\(y + 2x \\leq 10 ... (iii)\\)\n\nFrom (i), if \\(y - 2x = 5\\), when x = 0, y = 5, when x = 2, y = 9.\nFrom (ii), if \\(y + x = 2\\). when x = 0, y = 2, when x = 3, y = -1\nFrom (iii), if \\(y + 2x = 10\\), when x = 0, y = 10, when x = 2, y = 6.\n\n', '(a)\nDraw RC produced to E parallel to DE.\n\\(\\therefore < BCR = 52°\\) (alternate angles BA//RC)\nalso < RCD = 48° (alternate angles RC//DE)\n\\(x° = 360° - (52° + 48°) = 360° - 100° = 260°\\) (angles at a point = 360°)~(b) \\(y - 2x < 5 ... (i)\\)\n\\(2y + 2x \\geq 4 \\implies y + x \\geq 2 ... (ii)\\)\n\\(y + 2x \\leq 10 ... (iii)\\)\n\nFrom (i), if \\(y - 2x = 5\\), when x = 0, y = 5, when x = 2, y = 9.\nFrom (ii), if \\(y + x = 2\\). when x = 0, y = 2, when x = 3, y = -1\nFrom (iii), if \\(y + 2x = 10\\), when x = 0, y = 10, when x = 2, y = 6.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(643, '2017', '\n\nQuestion\n\n(a) Prove that the angle which an arc of a circle subtends at the centre is twice that which it subtends at any point on the remaining part of the circumference.\n(b)\nIn the diagram, O is the centre of the circle ACDB. If < CAO = 26° and < AOB = 130°. Calculate : (i) < OBC ; (ii) < COB.\n\n', '', '(a) Prove that the angle which an arc of a circle subtends at the centre is twice that which it subtends at any point on the remaining part of the circumference.\n(b)\nIn the diagram, O is the centre of the circle ACDB. If < CAO = 26° and < AOB = 130°. Calculate : (i) < OBC ; (ii) < COB.', '(a) Prove that the angle which an arc of a circle subtends at the centre is twice that which it subtends at any point on the remaining part of the circumference.~(b)\nIn the diagram, O is the centre of the circle ACDB. If < CAO = 26° and < AOB = 130°. Calculate : (i) < OBC ; (ii) < COB.', NULL, '\n\nExplanation\n(a)\nGiven : Circle ABC, centre O.\nTo prove: < AOB = < ACB\nConstruction : Join CO produced to P.\nProof: With the lettering in the figure, OA = OB (radii)\n\\(x_{1} = x_{2}\\) (base angles of isosceles triangle)\n\\(\\therefore < AOP = x_{1} +x_{2}\\) (exterior angle of triangle AOC)\n\\(\\therefore < AOP = 2x_{2} (x_{1} = x_{2})\\) (base angles of an isosceles triangle)\nSimilarly, \\(< BOP = 2y_{2}\\)\n\\(\\therefore < AOB = 2x_{2} + 2y_{2} \\)\n= \\(2(x_{2} + y_{2})\\)\n\\(\\implies < AOB = 2 \\times < ACB\\) (proved)\n(b) From the figure, \\(< ACB = \\frac{130°}{2} = 65°\\) (angle subtended at the centre of the circle)\n\\(< ACO = 26°\\) (Base angles of isosceles triangle ACO)\n\\(\\therefore < BCO = 65° - 26° = 39°\\)\n\\(< OBC = 39°\\) (base angles ofisosceles triangle OBC)\n\\(\\therefore < COB = 180° - (39° + 39°) = 102°\\)\n\n', '', 'Explanation\n(a)\nGiven : Circle ABC, centre O.\nTo prove: < AOB = < ACB\nConstruction : Join CO produced to P.\nProof: With the lettering in the figure, OA = OB (radii)\n\\(x_{1} = x_{2}\\) (base angles of isosceles triangle)\n\\(\\therefore < AOP = x_{1} +x_{2}\\) (exterior angle of triangle AOC)\n\\(\\therefore < AOP = 2x_{2} (x_{1} = x_{2})\\) (base angles of an isosceles triangle)\nSimilarly, \\(< BOP = 2y_{2}\\)\n\\(\\therefore < AOB = 2x_{2} + 2y_{2} \\)\n= \\(2(x_{2} + y_{2})\\)\n\\(\\implies < AOB = 2 \\times < ACB\\) (proved)\n(b) From the figure, \\(< ACB = \\frac{130°}{2} = 65°\\) (angle subtended at the centre of the circle)\n\\(< ACO = 26°\\) (Base angles of isosceles triangle ACO)\n\\(\\therefore < BCO = 65° - 26° = 39°\\)\n\\(< OBC = 39°\\) (base angles ofisosceles triangle OBC)\n\\(\\therefore < COB = 180° - (39° + 39°) = 102°\\)\n\n', '(a)\nGiven : Circle ABC, centre O.\nTo prove: < AOB = < ACB\nConstruction : Join CO produced to P.\nProof: With the lettering in the figure, OA = OB (radii)\n\\(x_{1} = x_{2}\\) (base angles of isosceles triangle)\n\\(\\therefore < AOP = x_{1} +x_{2}\\) (exterior angle of triangle AOC)\n\\(\\therefore < AOP = 2x_{2} (x_{1} = x_{2})\\) (base angles of an isosceles triangle)\nSimilarly, \\(< BOP = 2y_{2}\\)\n\\(\\therefore < AOB = 2x_{2} + 2y_{2} \\)\n= \\(2(x_{2} + y_{2})\\)\n\\(\\implies < AOB = 2 \\times < ACB\\) (proved)~(b) From the figure, \\(< ACB = \\frac{130°}{2} = 65°\\) (angle subtended at the centre of the circle)\n\\(< ACO = 26°\\) (Base angles of isosceles triangle ACO)\n\\(\\therefore < BCO = 65° - 26° = 39°\\)\n\\(< OBC = 39°\\) (base angles ofisosceles triangle OBC)\n\\(\\therefore < COB = 180° - (39° + 39°) = 102°\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(644, '2017', '\n\nQuestion\n\n(a) What is the 25th term of 5, 9, 13,... ?\n(b) Find the 5th term of \\(\\frac{8}{9}, \\frac{-4}{3}, 2, ...\\).\n(c) The 3rd and 6th terms of a G.P are \\(48\\) and \\(14\\frac{2}{9}\\) respectively. Write down the first four terms of the G.P.\n\n', '', '(a) What is the 25th term of 5, 9, 13,... ?\n(b) Find the 5th term of \\(\\frac{8}{9}, \\frac{-4}{3}, 2, ...\\).\n(c) The 3rd and 6th terms of a G.P are \\(48\\) and \\(14\\frac{2}{9}\\) respectively. Write down the first four terms of the G.P.', '(a) What is the 25th term of 5, 9, 13,... ?~(b) Find the 5th term of \\(\\frac{8}{9}, \\frac{-4}{3}, 2, ...\\).\n(c) The 3rd and 6th terms of a G.P are \\(48\\) and \\(14\\frac{2}{9}\\) respectively. Write down the first four terms of the G.P.~(c) The 3rd and 6th terms of a G.P are \\(48\\) and \\(14\\frac{2}{9}\\) respectively. Write down the first four terms of the G.P.', NULL, '\n\nExplanation\n(a) 5, 9, 13, ... is an A.P with the first term = 5 and common difference = 9 - 5 = 13 - 9 = 4.\n\\(T_{n} = a + (n - 1)d\\) (terms of an A.P)\n\\(T_{25} = 5 + (25 - 1) \\times 4\\)\n= \\(5 + 24 \\times 4\\)\n= \\( 5 + 96 = 101\\)\n(b) \\(\\frac{8}{9}, \\frac{-4}{3}, 2, ...\\) is a G.P with first term = \\(\\frac{8}{9}\\)\n\\(r = \\frac{T_{2}}{T_{1}} = -\\frac{4}{3} \\div \\frac{8}{9} = \\frac{-4}{3} \\times \\frac{9}{8} = -\\frac{3}{2}\\)\n\\(T_{n} = ar^{n - 1}\\) (terms of a G.P)\n\\(T_{5} = (\\frac{8}{9})(-\\frac{3}{2})^{5 - 1}\\)\n= \\(\\frac{8}{9} \\times \\frac{81}{16} = \\frac{9}{2}\\)\n(c) \\(T_{n} = ar^{n - 1}\\) (terms of a G.P)\n\\(T_{3} = ar^{2} = 48 ... (1)\\)\n\\(T_{6} = ar^{5} = 14\\frac{2}{9} ... (2)\\)\n\\((2) \\div (1) : r^{3} = \\frac{128}{9} \\div 48 = \\frac{128}{9} \\times \\frac{1}{48}\\)\n\\(r^{3} = \\frac{8}{27}\\)\n\\(r = \\sqrt[3]{\\frac{8}{27}} = \\frac{2}{3}\\)\nFrom (1), \\(ar^{2} = 48 \\implies a \\times (\\frac{2}{3})^{2} = 48\\)\n\\(4a = 48 \\times 9 \\implies a = \\frac{48 \\times 9}{4} = 108\\)\n\\(T_{2} = 108 \\times \\frac{2}{3} = 72\\)\n\\(T_{4} = 48 \\times \\frac{2}{3} = 32\\)\n\\(\\therefore\\) The first four terms of the sequence are \\(108, 72, 48, 32\\).\n\n', '', 'Explanation\n(a) 5, 9, 13, ... is an A.P with the first term = 5 and common difference = 9 - 5 = 13 - 9 = 4.\n\\(T_{n} = a + (n - 1)d\\) (terms of an A.P)\n\\(T_{25} = 5 + (25 - 1) \\times 4\\)\n= \\(5 + 24 \\times 4\\)\n= \\( 5 + 96 = 101\\)\n(b) \\(\\frac{8}{9}, \\frac{-4}{3}, 2, ...\\) is a G.P with first term = \\(\\frac{8}{9}\\)\n\\(r = \\frac{T_{2}}{T_{1}} = -\\frac{4}{3} \\div \\frac{8}{9} = \\frac{-4}{3} \\times \\frac{9}{8} = -\\frac{3}{2}\\)\n\\(T_{n} = ar^{n - 1}\\) (terms of a G.P)\n\\(T_{5} = (\\frac{8}{9})(-\\frac{3}{2})^{5 - 1}\\)\n= \\(\\frac{8}{9} \\times \\frac{81}{16} = \\frac{9}{2}\\)\n(c) \\(T_{n} = ar^{n - 1}\\) (terms of a G.P)\n\\(T_{3} = ar^{2} = 48 ... (1)\\)\n\\(T_{6} = ar^{5} = 14\\frac{2}{9} ... (2)\\)\n\\((2) \\div (1) : r^{3} = \\frac{128}{9} \\div 48 = \\frac{128}{9} \\times \\frac{1}{48}\\)\n\\(r^{3} = \\frac{8}{27}\\)\n\\(r = \\sqrt[3]{\\frac{8}{27}} = \\frac{2}{3}\\)\nFrom (1), \\(ar^{2} = 48 \\implies a \\times (\\frac{2}{3})^{2} = 48\\)\n\\(4a = 48 \\times 9 \\implies a = \\frac{48 \\times 9}{4} = 108\\)\n\\(T_{2} = 108 \\times \\frac{2}{3} = 72\\)\n\\(T_{4} = 48 \\times \\frac{2}{3} = 32\\)\n\\(\\therefore\\) The first four terms of the sequence are \\(108, 72, 48, 32\\).\n\n', '(a) 5, 9, 13, ... is an A.P with the first term = 5 and common difference = 9 - 5 = 13 - 9 = 4.\n\\(T_{n} = a + (n - 1)d\\) (terms of an A.P)\n\\(T_{25} = 5 + (25 - 1) \\times 4\\)\n= \\(5 + 24 \\times 4\\)\n= \\( 5 + 96 = 101\\)~(b) \\(\\frac{8}{9}, \\frac{-4}{3}, 2, ...\\) is a G.P with first term = \\(\\frac{8}{9}\\)\n\\(r = \\frac{T_{2}}{T_{1}} = -\\frac{4}{3} \\div \\frac{8}{9} = \\frac{-4}{3} \\times \\frac{9}{8} = -\\frac{3}{2}\\)\n\\(T_{n} = ar^{n - 1}\\) (terms of a G.P)\n\\(T_{5} = (\\frac{8}{9})(-\\frac{3}{2})^{5 - 1}\\)\n= \\(\\frac{8}{9} \\times \\frac{81}{16} = \\frac{9}{2}\\)\n(c) \\(T_{n} = ar^{n - 1}\\) (terms of a G.P)\n\\(T_{3} = ar^{2} = 48 ... (1)\\)\n\\(T_{6} = ar^{5} = 14\\frac{2}{9} ... (2)\\)\n\\((2) \\div (1) : r^{3} = \\frac{128}{9} \\div 48 = \\frac{128}{9} \\times \\frac{1}{48}\\)\n\\(r^{3} = \\frac{8}{27}\\)\n\\(r = \\sqrt[3]{\\frac{8}{27}} = \\frac{2}{3}\\)\nFrom (1), \\(ar^{2} = 48 \\implies a \\times (\\frac{2}{3})^{2} = 48\\)\n\\(4a = 48 \\times 9 \\implies a = \\frac{48 \\times 9}{4} = 108\\)\n\\(T_{2} = 108 \\times \\frac{2}{3} = 72\\)\n\\(T_{4} = 48 \\times \\frac{2}{3} = 32\\)\n\\(\\therefore\\) The first four terms of the sequence are \\(108, 72, 48, 32\\).\n~(c) \\(T_{n} = ar^{n - 1}\\) (terms of a G.P)\n\\(T_{3} = ar^{2} = 48 ... (1)\\)\n\\(T_{6} = ar^{5} = 14\\frac{2}{9} ... (2)\\)\n\\((2) \\div (1) : r^{3} = \\frac{128}{9} \\div 48 = \\frac{128}{9} \\times \\frac{1}{48}\\)\n\\(r^{3} = \\frac{8}{27}\\)\n\\(r = \\sqrt[3]{\\frac{8}{27}} = \\frac{2}{3}\\)\nFrom (1), \\(ar^{2} = 48 \\implies a \\times (\\frac{2}{3})^{2} = 48\\)\n\\(4a = 48 \\times 9 \\implies a = \\frac{48 \\times 9}{4} = 108\\)\n\\(T_{2} = 108 \\times \\frac{2}{3} = 72\\)\n\\(T_{4} = 48 \\times \\frac{2}{3} = 32\\)\n\\(\\therefore\\) The first four terms of the sequence are \\(108, 72, 48, 32\\).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(645, '2017', '\n\nQuestion\n\n(a) Copy and complete the following table of values for \\(y = 3\\sin 2\\theta - \\cos \\theta\\).\n\n\n\n\\(\\theta\\)\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n\n\ny\n-1.0\n\n\n0\n\n\n1.0\n\n\n\n(b) Using a scale of 2cm to 30° on the \\(\\theta\\) axis and 2cm to 1 unit on the y- axis, draw the graph of \\(y = 3 \\sin 2\\theta - \\cos \\theta\\) for \\(0° \\leq \\theta \\leq 180°\\).\n(c) Use your graph to find the : (i) solution of the equation \\(3 \\sin 2\\theta - \\cos \\theta = 0\\), correct to the nearest degree; (ii) maximum value of y, correct to one decimal place.\n\n\n\n(a) Copy and complete the following table of values for \\(y = 3\\sin 2\\theta - \\cos \\theta\\).\n\n\n\n\\(\\theta\\)\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n\n\ny\n-1.0\n\n\n0\n\n\n1.0\n\n\n', '', '(a) Copy and complete the following table of values for \\(y = 3\\sin 2\\theta - \\cos \\theta\\).\n\n\n\n\\(\\theta\\)\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n\n\ny\n-1.0\n\n\n0\n\n\n1.0\n\n\n\n(b) Using a scale of 2cm to 30° on the \\(\\theta\\) axis and 2cm to 1 unit on the y- axis, draw the graph of \\(y = 3 \\sin 2\\theta - \\cos \\theta\\) for \\(0° \\leq \\theta \\leq 180°\\).\n(c) Use your graph to find the : (i) solution of the equation \\(3 \\sin 2\\theta - \\cos \\theta = 0\\), correct to the nearest degree; (ii) maximum value of y, correct to one decimal place.\n\n\n\n(a) Copy and complete the following table of values for \\(y = 3\\sin 2\\theta - \\cos \\theta\\).\n\n\n\n\\(\\theta\\)\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n\n\ny\n-1.0\n\n\n0\n\n\n1.0\n', '(a) Copy and complete the following table of values for \\(y = 3\\sin 2\\theta - \\cos \\theta\\).\n\n\n\n\\(\\theta\\)\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n\n\ny\n-1.0\n\n\n0\n\n\n1.0\n\n\n~(b) Using a scale of 2cm to 30° on the \\(\\theta\\) axis and 2cm to 1 unit on the y- axis, draw the graph of \\(y = 3 \\sin 2\\theta - \\cos \\theta\\) for \\(0° \\leq \\theta \\leq 180°\\).\n(c) Use your graph to find the : (i) solution of the equation \\(3 \\sin 2\\theta - \\cos \\theta = 0\\), correct to the nearest degree; (ii) maximum value of y, correct to one decimal place.\n\n\n\n(a) Copy and complete the following table of values for \\(y = 3\\sin 2\\theta - \\cos \\theta\\).\n\n\n\n\\(\\theta\\)\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n\n\ny\n-1.0\n\n\n0\n\n\n1.0\n~(c) Use your graph to find the : (i) solution of the equation \\(3 \\sin 2\\theta - \\cos \\theta = 0\\), correct to the nearest degree; (ii) maximum value of y, correct to one decimal place.\n\n\n\n(a) Copy and complete the following table of values for \\(y = 3\\sin 2\\theta - \\cos \\theta\\).\n\n\n\n\\(\\theta\\)\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n\n\ny\n-1.0\n\n\n0\n\n\n1.0\n', NULL, '\n\nExplanation\n(a)\n\n\n\n\\(\\theta\\)\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n\n\ny\n-1.0\n1.732\n2.098\n0\n-2.098\n-1.732\n1.0\n\n\n\n(b)\n(c)(i) Solution of \\(3 \\sin 2 \\theta - \\cos \\theta = 0\\) is at y = 0\n= \\(9.30°, 80°, 174°\\)\n(ii) Maximum value of y = 2.098 \\(\\approxeq\\) 2.1 (to 1 decimal place)\n\n', '', 'Explanation\n(a)\n\n\n\n\\(\\theta\\)\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n\n\ny\n-1.0\n1.732\n2.098\n0\n-2.098\n-1.732\n1.0\n\n\n\n(b)\n(c)(i) Solution of \\(3 \\sin 2 \\theta - \\cos \\theta = 0\\) is at y = 0\n= \\(9.30°, 80°, 174°\\)\n(ii) Maximum value of y = 2.098 \\(\\approxeq\\) 2.1 (to 1 decimal place)\n\n', '(a)\n\n\n\n\\(\\theta\\)\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n\n\ny\n-1.0\n1.732\n2.098\n0\n-2.098\n-1.732\n1.0\n\n\n~(b)\n(c)(i) Solution of \\(3 \\sin 2 \\theta - \\cos \\theta = 0\\) is at y = 0\n= \\(9.30°, 80°, 174°\\)\n(ii) Maximum value of y = 2.098 \\(\\approxeq\\) 2.1 (to 1 decimal place)\n~(c)(i) Solution of \\(3 \\sin 2 \\theta - \\cos \\theta = 0\\) is at y = 0\n= \\(9.30°, 80°, 174°\\)\n(ii) Maximum value of y = 2.098 \\(\\approxeq\\) 2.1 (to 1 decimal place)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(646, '2017', '\n\nQuestion\n\nThe table below shows the frequency distribution of the marks scored by fifty students in an examination.\n\n\n\nMarks (%)\n0-9\n10-19\n20-29\n30-39\n40-49\n50-59\n60-69\n70-79\n80-89\n90-99\n\n\nFreq\n2\n3\n4\n6\n13\n10\n5\n3\n2\n2\n\n\n\n(a) Draw the cumulative frequency curve for the distribution.\n(b) Use your curve to estimate the : (i) upper quartile; (ii) pass mark if 60% of the students passed.\n\n\n\nThe table below shows the frequency distribution of the marks scored by fifty students in an examination.\n\n\n\nMarks (%)\n0-9\n10-19\n20-29\n30-39\n40-49\n50-59\n60-69\n70-79\n80-89\n90-99\n\n\nFreq\n2\n3\n4\n6\n13\n10\n5\n3\n2\n2\n\n\n', '', 'The table below shows the frequency distribution of the marks scored by fifty students in an examination.\n\n\n\nMarks (%)\n0-9\n10-19\n20-29\n30-39\n40-49\n50-59\n60-69\n70-79\n80-89\n90-99\n\n\nFreq\n2\n3\n4\n6\n13\n10\n5\n3\n2\n2\n\n\n\n(a) Draw the cumulative frequency curve for the distribution.\n(b) Use your curve to estimate the : (i) upper quartile; (ii) pass mark if 60% of the students passed.\n\n\n\nThe table below shows the frequency distribution of the marks scored by fifty students in an examination.\n\n\n\nMarks (%)\n0-9\n10-19\n20-29\n30-39\n40-49\n50-59\n60-69\n70-79\n80-89\n90-99\n\n\nFreq\n2\n3\n4\n6\n13\n10\n5\n3\n2\n2\n', 'The table below shows the frequency distribution of the marks scored by fifty students in an examination.\n\n\n\nMarks (%)\n0-9\n10-19\n20-29\n30-39\n40-49\n50-59\n60-69\n70-79\n80-89\n90-99\n\n\nFreq\n2\n3\n4\n6\n13\n10\n5\n3\n2\n2\n\n\n\n(a) Draw the cumulative frequency curve for the distribution.~(b) Use your curve to estimate the : (i) upper quartile; (ii) pass mark if 60% of the students passed.\n\n\n\nThe table below shows the frequency distribution of the marks scored by fifty students in an examination.\n\n\n\nMarks (%)\n0-9\n10-19\n20-29\n30-39\n40-49\n50-59\n60-69\n70-79\n80-89\n90-99\n\n\nFreq\n2\n3\n4\n6\n13\n10\n5\n3\n2\n2\n', NULL, '\n\nExplanation\nMarks\n\nUpper class\nboundary\n\nFreq\nCum Freq\n\n\n0-9\n9.5\n2\n2\n\n\n10-19\n19.5\n3\n5\n\n\n20-29\n29.5\n4\n9\n\n\n30-39\n39.5\n6\n15\n\n\n40-49\n49.5\n13\n28\n\n\n50-59\n59.5\n10\n38\n\n\n60-69\n69.5\n5\n43\n\n\n70-79\n79.5\n3\n46\n\n\n80-89\n89.5\n2\n48\n\n\n90-99\n99.5\n2\n50\n\n\n\n(a)\n(b)(i) Upper quartile \\(Q_{3}\\) = 57 marks\n(ii) Pass mark if 60% of the class passed = 42 marks\n\n', '', 'Explanation\nMarks\n\nUpper class\nboundary\n\nFreq\nCum Freq\n\n\n0-9\n9.5\n2\n2\n\n\n10-19\n19.5\n3\n5\n\n\n20-29\n29.5\n4\n9\n\n\n30-39\n39.5\n6\n15\n\n\n40-49\n49.5\n13\n28\n\n\n50-59\n59.5\n10\n38\n\n\n60-69\n69.5\n5\n43\n\n\n70-79\n79.5\n3\n46\n\n\n80-89\n89.5\n2\n48\n\n\n90-99\n99.5\n2\n50\n\n\n\n(a)\n(b)(i) Upper quartile \\(Q_{3}\\) = 57 marks\n(ii) Pass mark if 60% of the class passed = 42 marks\n\n', '~(b)(i) Upper quartile \\(Q_{3}\\) = 57 marks\n(ii) Pass mark if 60% of the class passed = 42 marks\n', 'WAEC-PK-1', 'WAEC-MATH', 1);
INSERT INTO `exam_simulation2` (`id`, `examYear`, `questions`, `parent_que`, `child_que`, `gen_child_que_desc`, `imageDir`, `explanation`, `parent_que_expl`, `child_que_exp`, `gen_child_que_expl`, `package_id`, `subject_id`, `status`) VALUES
(647, '2017', '\n\nQuestion\n\nP and Q are two points on latitude 55°N and their longitudes are 33°W and 20°E respectively. Calculate the distance between P and Q measured along\n(a) the parallel of latitude ;\n(b) a great circle.\n[Take \\(\\pi = \\frac{22}{7}\\) and radius of the earth = 6400km].\n\n', '', 'P and Q are two points on latitude 55°N and their longitudes are 33°W and 20°E respectively. Calculate the distance between P and Q measured along\n(a) the parallel of latitude ;\n(b) a great circle.\n[Take \\(\\pi = \\frac{22}{7}\\) and radius of the earth = 6400km].', 'P and Q are two points on latitude 55°N and their longitudes are 33°W and 20°E respectively. Calculate the distance between P and Q measured along\n(a) the parallel of latitude ;~(b) a great circle.\n[Take \\(\\pi = \\frac{22}{7}\\) and radius of the earth = 6400km].', NULL, '\n\nExplanation\n(a) Angular difference = 33° + 23° = 56°\nDistance PQ along parallel of latitude = \\(\\frac{56}{360} \\times 2 \\pi r\\)\nwhere r = R cos 55°\n\\(\\therefore |PQ| = \\frac{56}{360} \\times 2 \\times \\frac{22}{7} \\times 6400 \\cos 55\\)\n= \\(3589.46 km \\approxeq 3589 km\\) (to the nearest degree)\n(b) \\(|PO|^{2} = r^{2} + r^{2} - 2r^{2} \\cos 56° = R^{2} + R^{2} - 2R^{2} \\cos \\theta \\)\n= \\(2r^{2} - 2r^{2} \\cos 56° = 2R^{2} - 2R^{2} \\cos \\theta\\)\nBut \\(r = R cos 55° = 6400 \\times 0.5736 = 3671\\)\n\\(\\therefore 2r^{2} (1 - \\cos 56°) = 2R^{2} (1 - \\cos \\theta)\\)\n\\(2 \\times 3671^{2}( 1 - 0.5592) = 2 \\times 6400^{2} (1 - \\cos \\theta)\\)\n\\(1 - \\cos \\theta = \\frac{2 \\times 3671^{2} \\times 0.4408}{2 \\times 6400^{2}}\\)\n\\(1 - \\cos \\theta = 0.1259\\)\n\\(\\cos \\theta = 1 - 0.1259 = 0.8741\\)\n\\(\\theta = \\cos^{-1} (0.8741) = 29.06°\\)\n\\(\\therefore\\) Distance PQ along great circle = \\(\\frac{29.06}{360} \\times 2 \\times \\frac{22}{7} \\times 6400\\)\n= \\(3247.34 km \\approxeq 3247km\\)\n\n', '', 'Explanation\n(a) Angular difference = 33° + 23° = 56°\nDistance PQ along parallel of latitude = \\(\\frac{56}{360} \\times 2 \\pi r\\)\nwhere r = R cos 55°\n\\(\\therefore |PQ| = \\frac{56}{360} \\times 2 \\times \\frac{22}{7} \\times 6400 \\cos 55\\)\n= \\(3589.46 km \\approxeq 3589 km\\) (to the nearest degree)\n(b) \\(|PO|^{2} = r^{2} + r^{2} - 2r^{2} \\cos 56° = R^{2} + R^{2} - 2R^{2} \\cos \\theta \\)\n= \\(2r^{2} - 2r^{2} \\cos 56° = 2R^{2} - 2R^{2} \\cos \\theta\\)\nBut \\(r = R cos 55° = 6400 \\times 0.5736 = 3671\\)\n\\(\\therefore 2r^{2} (1 - \\cos 56°) = 2R^{2} (1 - \\cos \\theta)\\)\n\\(2 \\times 3671^{2}( 1 - 0.5592) = 2 \\times 6400^{2} (1 - \\cos \\theta)\\)\n\\(1 - \\cos \\theta = \\frac{2 \\times 3671^{2} \\times 0.4408}{2 \\times 6400^{2}}\\)\n\\(1 - \\cos \\theta = 0.1259\\)\n\\(\\cos \\theta = 1 - 0.1259 = 0.8741\\)\n\\(\\theta = \\cos^{-1} (0.8741) = 29.06°\\)\n\\(\\therefore\\) Distance PQ along great circle = \\(\\frac{29.06}{360} \\times 2 \\times \\frac{22}{7} \\times 6400\\)\n= \\(3247.34 km \\approxeq 3247km\\)\n\n', '(a) Angular difference = 33° + 23° = 56°\nDistance PQ along parallel of latitude = \\(\\frac{56}{360} \\times 2 \\pi r\\)\nwhere r = R cos 55°\n\\(\\therefore |PQ| = \\frac{56}{360} \\times 2 \\times \\frac{22}{7} \\times 6400 \\cos 55\\)\n= \\(3589.46 km \\approxeq 3589 km\\) (to the nearest degree)~(b) \\(|PO|^{2} = r^{2} + r^{2} - 2r^{2} \\cos 56° = R^{2} + R^{2} - 2R^{2} \\cos \\theta \\)\n= \\(2r^{2} - 2r^{2} \\cos 56° = 2R^{2} - 2R^{2} \\cos \\theta\\)\nBut \\(r = R cos 55° = 6400 \\times 0.5736 = 3671\\)\n\\(\\therefore 2r^{2} (1 - \\cos 56°) = 2R^{2} (1 - \\cos \\theta)\\)\n\\(2 \\times 3671^{2}( 1 - 0.5592) = 2 \\times 6400^{2} (1 - \\cos \\theta)\\)\n\\(1 - \\cos \\theta = \\frac{2 \\times 3671^{2} \\times 0.4408}{2 \\times 6400^{2}}\\)\n\\(1 - \\cos \\theta = 0.1259\\)\n\\(\\cos \\theta = 1 - 0.1259 = 0.8741\\)\n\\(\\theta = \\cos^{-1} (0.8741) = 29.06°\\)\n\\(\\therefore\\) Distance PQ along great circle = \\(\\frac{29.06}{360} \\times 2 \\times \\frac{22}{7} \\times 6400\\)\n= \\(3247.34 km \\approxeq 3247km\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(648, '2017', '\n\nQuestion\n\n(a) If \\(9^{2x - 1} = \\frac{81^{x - 2}}{3^{x}}\\), find x.\n(b) Without using Mathematical Tables, evaluate: \\(\\sqrt{\\frac{0.81 \\times 10^{-5}}{2.25 \\times 10^{7}}}\\)\n\n', '', '(a) If \\(9^{2x - 1} = \\frac{81^{x - 2}}{3^{x}}\\), find x.\n(b) Without using Mathematical Tables, evaluate: \\(\\sqrt{\\frac{0.81 \\times 10^{-5}}{2.25 \\times 10^{7}}}\\)', '(a) If \\(9^{2x - 1} = \\frac{81^{x - 2}}{3^{x}}\\), find x.~(b) Without using Mathematical Tables, evaluate: \\(\\sqrt{\\frac{0.81 \\times 10^{-5}}{2.25 \\times 10^{7}}}\\)', NULL, '\n\nExplanation\n(a) \\(9^{2x - 1} = \\frac{81^{x - 2}}{3^{x}}\\)\n\\((3^{2})^{2x - 1} = \\frac{(3^{4})^{x - 2}}{3^{x}}\\)\n\\(3^{4x - 2} = \\frac{3^{4x - 8}}{3^{x}}\\)\n\\(3^{4x - 2} = 3^{4x - 8 - x}\\)\n\\(3^{4x - 2} = 3^{3x - 8}\\)\n\\(\\therefore 4x - 2 = 3x - 8 \\implies 4x - 3x = -8 + 2\\)\n\\(x = -6\\)\n(b) \\(\\sqrt{\\frac{0.81 \\times 10^{-5}}{2.25 \\times 10^{7}}}\\)\n= \\(\\sqrt{\\frac{0.81}{2.25}} \\times \\sqrt{10^{-5 - 7}}\\)\n= \\(\\frac{0.9}{1.5} \\times \\sqrt{10^{-12}}\\)\n= \\(0.6 \\times 10^{-6}\\)\n= \\(6 \\times 10^{-7}\\)\n\n', '', 'Explanation\n(a) \\(9^{2x - 1} = \\frac{81^{x - 2}}{3^{x}}\\)\n\\((3^{2})^{2x - 1} = \\frac{(3^{4})^{x - 2}}{3^{x}}\\)\n\\(3^{4x - 2} = \\frac{3^{4x - 8}}{3^{x}}\\)\n\\(3^{4x - 2} = 3^{4x - 8 - x}\\)\n\\(3^{4x - 2} = 3^{3x - 8}\\)\n\\(\\therefore 4x - 2 = 3x - 8 \\implies 4x - 3x = -8 + 2\\)\n\\(x = -6\\)\n(b) \\(\\sqrt{\\frac{0.81 \\times 10^{-5}}{2.25 \\times 10^{7}}}\\)\n= \\(\\sqrt{\\frac{0.81}{2.25}} \\times \\sqrt{10^{-5 - 7}}\\)\n= \\(\\frac{0.9}{1.5} \\times \\sqrt{10^{-12}}\\)\n= \\(0.6 \\times 10^{-6}\\)\n= \\(6 \\times 10^{-7}\\)\n\n', '(a) \\(9^{2x - 1} = \\frac{81^{x - 2}}{3^{x}}\\)\n\\((3^{2})^{2x - 1} = \\frac{(3^{4})^{x - 2}}{3^{x}}\\)\n\\(3^{4x - 2} = \\frac{3^{4x - 8}}{3^{x}}\\)\n\\(3^{4x - 2} = 3^{4x - 8 - x}\\)\n\\(3^{4x - 2} = 3^{3x - 8}\\)\n\\(\\therefore 4x - 2 = 3x - 8 \\implies 4x - 3x = -8 + 2\\)\n\\(x = -6\\)~(b) \\(\\sqrt{\\frac{0.81 \\times 10^{-5}}{2.25 \\times 10^{7}}}\\)\n= \\(\\sqrt{\\frac{0.81}{2.25}} \\times \\sqrt{10^{-5 - 7}}\\)\n= \\(\\frac{0.9}{1.5} \\times \\sqrt{10^{-12}}\\)\n= \\(0.6 \\times 10^{-6}\\)\n= \\(6 \\times 10^{-7}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(649, '2017', '\n\nQuestion\n\n(a) Solve the following pair of simultaneous equations: \\(2x + 5y = 6\\frac{1}{2} ; 5x - 2y = 9\\)\n(b) If \\(\\log_{10} (2x + 1) - \\log_{10} (3x - 2) = 1\\), find x.\n\n', '', '(a) Solve the following pair of simultaneous equations: \\(2x + 5y = 6\\frac{1}{2} ; 5x - 2y = 9\\)\n(b) If \\(\\log_{10} (2x + 1) - \\log_{10} (3x - 2) = 1\\), find x.', '(a) Solve the following pair of simultaneous equations: \\(2x + 5y = 6\\frac{1}{2} ; 5x - 2y = 9\\)~(b) If \\(\\log_{10} (2x + 1) - \\log_{10} (3x - 2) = 1\\), find x.', NULL, '\n\nExplanation\n(a) \\(2x + 5y = \\frac{13}{2} \\implies 4x + 10y = 13 ... (1)\\)\n\\(5x - 2y = 9 ... (2)\\)\n\\((2) \\times 5 : 25x - 10y = 45 ... (2a)\\)\n\\((1) +(2a) : 29x = 58 \\implies x = 2\\)\n\\(4(2) + 10y = 13 \\implies 10y = 13 - 8 = 5\\)\n\\(y = \\frac{5}{10} = 0.5\\)\n(b) \\(\\log_{10} (2x + 1) - \\log_{10} (3x - 2) = 1\\)\n\\(\\log_{10} (\\frac{2x + 1}{3x - 2}) = 1\\)\n\\(\\frac{2x + 1}{3x - 2} = 10^{1} = 10\\)\n\\(2x + 1 = 10(3x - 2) \\implies 2x + 1 = 30x - 20\\)\n\\(21 = 28x \\implies x = \\frac{3}{4}\\)\n\n', '', 'Explanation\n(a) \\(2x + 5y = \\frac{13}{2} \\implies 4x + 10y = 13 ... (1)\\)\n\\(5x - 2y = 9 ... (2)\\)\n\\((2) \\times 5 : 25x - 10y = 45 ... (2a)\\)\n\\((1) +(2a) : 29x = 58 \\implies x = 2\\)\n\\(4(2) + 10y = 13 \\implies 10y = 13 - 8 = 5\\)\n\\(y = \\frac{5}{10} = 0.5\\)\n(b) \\(\\log_{10} (2x + 1) - \\log_{10} (3x - 2) = 1\\)\n\\(\\log_{10} (\\frac{2x + 1}{3x - 2}) = 1\\)\n\\(\\frac{2x + 1}{3x - 2} = 10^{1} = 10\\)\n\\(2x + 1 = 10(3x - 2) \\implies 2x + 1 = 30x - 20\\)\n\\(21 = 28x \\implies x = \\frac{3}{4}\\)\n\n', '(a) \\(2x + 5y = \\frac{13}{2} \\implies 4x + 10y = 13 ... (1)\\)\n\\(5x - 2y = 9 ... (2)\\)\n\\((2) \\times 5 : 25x - 10y = 45 ... (2a)\\)\n\\((1) +(2a) : 29x = 58 \\implies x = 2\\)\n\\(4(2) + 10y = 13 \\implies 10y = 13 - 8 = 5\\)\n\\(y = \\frac{5}{10} = 0.5\\)~(b) \\(\\log_{10} (2x + 1) - \\log_{10} (3x - 2) = 1\\)\n\\(\\log_{10} (\\frac{2x + 1}{3x - 2}) = 1\\)\n\\(\\frac{2x + 1}{3x - 2} = 10^{1} = 10\\)\n\\(2x + 1 = 10(3x - 2) \\implies 2x + 1 = 30x - 20\\)\n\\(21 = 28x \\implies x = \\frac{3}{4}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(650, '2017', '\n\nQuestion\n\n(a) The angle of a sector of a circle radius 7cm is 108°. Calculate the perimeter of the sector. [Take \\(\\pi = \\frac{22}{7}\\)]\n(b) A boat is on the same horizontal level as the foot of a cliff, and the angle of depression of the boat from the top of the cliff is 30°. If the boat is 120m away from the foot of the cliff, find the height of the cliff correct to three significant figures.\n\n', '', '(a) The angle of a sector of a circle radius 7cm is 108°. Calculate the perimeter of the sector. [Take \\(\\pi = \\frac{22}{7}\\)]\n(b) A boat is on the same horizontal level as the foot of a cliff, and the angle of depression of the boat from the top of the cliff is 30°. If the boat is 120m away from the foot of the cliff, find the height of the cliff correct to three significant figures.', '(a) The angle of a sector of a circle radius 7cm is 108°. Calculate the perimeter of the sector. [Take \\(\\pi = \\frac{22}{7}\\)]~(b) A boat is on the same horizontal level as the foot of a cliff, and the angle of depression of the boat from the top of the cliff is 30°. If the boat is 120m away from the foot of the cliff, find the height of the cliff correct to three significant figures.', NULL, '\n\nExplanation\n(a)\nLength of arc AB = \\(\\frac{108°}{360°} \\times 2 \\times \\frac{22}{7} \\times 7 = \\frac{132}{10} =13.2cm\\)\nPerimeter of the sector = \\(13.2cm + 7cm + 7cm = 27.2cm\\)\n(b)\nLet T = top of cliff, C = bottom of cliff, B = the boat.\nSince angle of depression = 30°\nthen < CBT = 30° (alternate angles TR//CB)\n\\(\\therefore \\text{Height of the cliff} (CT) = 120 \\times 0.5774m\\)\n= \\(69.288m \\approxeq 69.3m\\)\n\n', '', 'Explanation\n(a)\nLength of arc AB = \\(\\frac{108°}{360°} \\times 2 \\times \\frac{22}{7} \\times 7 = \\frac{132}{10} =13.2cm\\)\nPerimeter of the sector = \\(13.2cm + 7cm + 7cm = 27.2cm\\)\n(b)\nLet T = top of cliff, C = bottom of cliff, B = the boat.\nSince angle of depression = 30°\nthen < CBT = 30° (alternate angles TR//CB)\n\\(\\therefore \\text{Height of the cliff} (CT) = 120 \\times 0.5774m\\)\n= \\(69.288m \\approxeq 69.3m\\)\n\n', '(a)\nLength of arc AB = \\(\\frac{108°}{360°} \\times 2 \\times \\frac{22}{7} \\times 7 = \\frac{132}{10} =13.2cm\\)\nPerimeter of the sector = \\(13.2cm + 7cm + 7cm = 27.2cm\\)~(b)\nLet T = top of cliff, C = bottom of cliff, B = the boat.\nSince angle of depression = 30°\nthen < CBT = 30° (alternate angles TR//CB)\n\\(\\therefore \\text{Height of the cliff} (CT) = 120 \\times 0.5774m\\)\n= \\(69.288m \\approxeq 69.3m\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(651, '2017', '\n\nQuestion\n\n(a) The sides PQ and PR of \\(\\Delta\\) PQR are produced to T and S respectively, such that TQR = 131° and < QRS = 98°. Find < QPR.\n(b) The circumference of a circular track is 400m. Find its radius, correct to the nearest metre. [Take \\(\\pi = \\frac{22}{7}\\)]\n\n', '', '(a) The sides PQ and PR of \\(\\Delta\\) PQR are produced to T and S respectively, such that TQR = 131° and < QRS = 98°. Find < QPR.\n(b) The circumference of a circular track is 400m. Find its radius, correct to the nearest metre. [Take \\(\\pi = \\frac{22}{7}\\)]', '(a) The sides PQ and PR of \\(\\Delta\\) PQR are produced to T and S respectively, such that TQR = 131° and < QRS = 98°. Find < QPR.~(b) The circumference of a circular track is 400m. Find its radius, correct to the nearest metre. [Take \\(\\pi = \\frac{22}{7}\\)]', NULL, '\n\nExplanation\n(a)\n< PQR = 180° - 131° = 49° (angles on a straight line)\n\\(\\therefore < PQR = 49°\\)\nAlso, \\(< PQR = 180° - 98° = 82°\\)\n\\(\\therefore < QPR = 180° - (49° + 82°) = 180° - 131°\\)\n= \\(49°\\)\n(b) Circumference of the circular track = \\(2\\pi r\\)\n\\(\\therefore 2\\pi r = 400 \\)\nwhere \\(\\pi = \\frac{22}{7}\\)\n\\(\\therefore r = \\frac{400 \\times 1 \\times 7}{22 \\times 2} = \\frac{700}{11}\\)\n= \\(63.63m \\approxeq 64m\\)\n\n', '', 'Explanation\n(a)\n< PQR = 180° - 131° = 49° (angles on a straight line)\n\\(\\therefore < PQR = 49°\\)\nAlso, \\(< PQR = 180° - 98° = 82°\\)\n\\(\\therefore < QPR = 180° - (49° + 82°) = 180° - 131°\\)\n= \\(49°\\)\n(b) Circumference of the circular track = \\(2\\pi r\\)\n\\(\\therefore 2\\pi r = 400 \\)\nwhere \\(\\pi = \\frac{22}{7}\\)\n\\(\\therefore r = \\frac{400 \\times 1 \\times 7}{22 \\times 2} = \\frac{700}{11}\\)\n= \\(63.63m \\approxeq 64m\\)\n\n', '(a)\n< PQR = 180° - 131° = 49° (angles on a straight line)\n\\(\\therefore < PQR = 49°\\)\nAlso, \\(< PQR = 180° - 98° = 82°\\)\n\\(\\therefore < QPR = 180° - (49° + 82°) = 180° - 131°\\)\n= \\(49°\\)~(b) Circumference of the circular track = \\(2\\pi r\\)\n\\(\\therefore 2\\pi r = 400 \\)\nwhere \\(\\pi = \\frac{22}{7}\\)\n\\(\\therefore r = \\frac{400 \\times 1 \\times 7}{22 \\times 2} = \\frac{700}{11}\\)\n= \\(63.63m \\approxeq 64m\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(652, '2017', '\n\nQuestion\n\n(a) In a game, a fair die is rolled once and two unbiased coins are tossed at once. What is the probability of obtaining 3 and a tail?\n(b) A box contains 10 marbles, 7 of which are black and 3 are red. Two marbles are drawn one after the other without replacement. Find the probability of getting:\n(i) a red, then a black marble ; (ii) two black marbles.\n\n', '', '(a) In a game, a fair die is rolled once and two unbiased coins are tossed at once. What is the probability of obtaining 3 and a tail?\n(b) A box contains 10 marbles, 7 of which are black and 3 are red. Two marbles are drawn one after the other without replacement. Find the probability of getting:\n(i) a red, then a black marble ; (ii) two black marbles.', '(a) In a game, a fair die is rolled once and two unbiased coins are tossed at once. What is the probability of obtaining 3 and a tail?~(b) A box contains 10 marbles, 7 of which are black and 3 are red. Two marbles are drawn one after the other without replacement. Find the probability of getting:\n(i) a red, then a black marble ; (ii) two black marbles.', NULL, '\n\nExplanation\n(a) P(obtaining a 3 from a die) = \\(\\frac{1}{6}\\)\nP(obtaining a tail in 2 coins tossed at once) = \\(\\frac{1}{2}\\).\n\\(\\therefore\\) P(obtaining 3 and a tail = \\(\\frac{1}{6} \\times \\frac{1}{2} = \\frac{1}{12}\\)\n(b) No of black marbles = n(B) = 7\nNo of red marbles = n(R) = 3\nTotal marble = 10\n(i) P(red then black) = P(1st red, then black) or P(1st black, then red)\n= \\(\\frac{3}{10} \\times \\frac{7}{9} + \\frac{7}{10} \\times \\frac{3}{9}\\)\n= \\(\\frac{7}{15}\\)\n(ii) Prob (two blacks) = \\(\\frac{7}{10} \\times \\frac{6}{9} = \\frac{7}{15}\\)\n\n\n', '', 'Explanation\n(a) P(obtaining a 3 from a die) = \\(\\frac{1}{6}\\)\nP(obtaining a tail in 2 coins tossed at once) = \\(\\frac{1}{2}\\).\n\\(\\therefore\\) P(obtaining 3 and a tail = \\(\\frac{1}{6} \\times \\frac{1}{2} = \\frac{1}{12}\\)\n(b) No of black marbles = n(B) = 7\nNo of red marbles = n(R) = 3\nTotal marble = 10\n(i) P(red then black) = P(1st red, then black) or P(1st black, then red)\n= \\(\\frac{3}{10} \\times \\frac{7}{9} + \\frac{7}{10} \\times \\frac{3}{9}\\)\n= \\(\\frac{7}{15}\\)\n(ii) Prob (two blacks) = \\(\\frac{7}{10} \\times \\frac{6}{9} = \\frac{7}{15}\\)\n\n\n', '(a) P(obtaining a 3 from a die) = \\(\\frac{1}{6}\\)\nP(obtaining a tail in 2 coins tossed at once) = \\(\\frac{1}{2}\\).\n\\(\\therefore\\) P(obtaining 3 and a tail = \\(\\frac{1}{6} \\times \\frac{1}{2} = \\frac{1}{12}\\)~(b) No of black marbles = n(B) = 7\nNo of red marbles = n(R) = 3\nTotal marble = 10\n(i) P(red then black) = P(1st red, then black) or P(1st black, then red)\n= \\(\\frac{3}{10} \\times \\frac{7}{9} + \\frac{7}{10} \\times \\frac{3}{9}\\)\n= \\(\\frac{7}{15}\\)\n(ii) Prob (two blacks) = \\(\\frac{7}{10} \\times \\frac{6}{9} = \\frac{7}{15}\\)\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(653, '2017', '\n\nQuestion\n\n(a) If \\(17x = 375^{2} - 356^{2}\\), find the exact value of x.\n(b) If \\(4^{x} = 2^{\\frac{1}{2}} \\times 8\\), find x.\n(c) The sum of the first 9 terms of an A.P is 72 and the sum of the next 4 terms is 71, find the A.P.\n\n', '', '(a) If \\(17x = 375^{2} - 356^{2}\\), find the exact value of x.\n(b) If \\(4^{x} = 2^{\\frac{1}{2}} \\times 8\\), find x.\n(c) The sum of the first 9 terms of an A.P is 72 and the sum of the next 4 terms is 71, find the A.P.', '(a) If \\(17x = 375^{2} - 356^{2}\\), find the exact value of x.~(b) If \\(4^{x} = 2^{\\frac{1}{2}} \\times 8\\), find x.\n(c) The sum of the first 9 terms of an A.P is 72 and the sum of the next 4 terms is 71, find the A.P.~(c) The sum of the first 9 terms of an A.P is 72 and the sum of the next 4 terms is 71, find the A.P.', NULL, '\n\nExplanation\n(a) \\(17x = 375^{2} - 356^{2}\\)\n\\(17x = (375 + 356)(375 - 356)\\)\n\\(17x = (731)(19)\\)\n\\(x = \\frac{731 \\times 19}{17} = 817\\)\n(b) \\(4^{x} = 2^{\\frac{1}{2}} \\times 8\\)\n\\(2^{2x} = 2^{\\frac{1}{2}} \\times 2^{3}\\)\n\\(2^{2x} = 2^{3\\frac{1}{2}}\\)\n\\(\\implies 2x = 3\\frac{1}{2} \\implies x = \\frac{7}{4}\\)\n(c) \\(S_{n} = \\frac{n}{2} (2a + (n - 1)d)\\) (sum of terms of an A.P)\n\\(S_{9} = \\frac{9}{2} [2a + (9 - 1) d] = \\frac{9}{2} [2a + 8d]\\)\n\\(72 = 9(a + 4d) \\implies 8 = a + 4d ... (1)\\)\n\\(S_{9 + 4} = S_{13} = \\frac{13}{2} [2a + (13 - 1)d] = \\frac{13}{2} [2a + 12d]\\)\n\\(72 + 71 = 143 = 13(a + 6d) \\implies 11 = a + 6d ... (2)\\)\n\\((2) - (1) : 2d = 3 \\implies d = \\frac{3}{2}\\)\n\\(a + 4(1\\frac{1}{2}) = 8 \\implies a + 6 = 8\\)\n\\(\\implies a = 8 - 6 = 2\\)\n\\(\\therefore \\text{The A.P is } 2, 3\\frac{1}{2}, 5, 6\\frac{1}{2}, 8,...\\)\n\n', '', 'Explanation\n(a) \\(17x = 375^{2} - 356^{2}\\)\n\\(17x = (375 + 356)(375 - 356)\\)\n\\(17x = (731)(19)\\)\n\\(x = \\frac{731 \\times 19}{17} = 817\\)\n(b) \\(4^{x} = 2^{\\frac{1}{2}} \\times 8\\)\n\\(2^{2x} = 2^{\\frac{1}{2}} \\times 2^{3}\\)\n\\(2^{2x} = 2^{3\\frac{1}{2}}\\)\n\\(\\implies 2x = 3\\frac{1}{2} \\implies x = \\frac{7}{4}\\)\n(c) \\(S_{n} = \\frac{n}{2} (2a + (n - 1)d)\\) (sum of terms of an A.P)\n\\(S_{9} = \\frac{9}{2} [2a + (9 - 1) d] = \\frac{9}{2} [2a + 8d]\\)\n\\(72 = 9(a + 4d) \\implies 8 = a + 4d ... (1)\\)\n\\(S_{9 + 4} = S_{13} = \\frac{13}{2} [2a + (13 - 1)d] = \\frac{13}{2} [2a + 12d]\\)\n\\(72 + 71 = 143 = 13(a + 6d) \\implies 11 = a + 6d ... (2)\\)\n\\((2) - (1) : 2d = 3 \\implies d = \\frac{3}{2}\\)\n\\(a + 4(1\\frac{1}{2}) = 8 \\implies a + 6 = 8\\)\n\\(\\implies a = 8 - 6 = 2\\)\n\\(\\therefore \\text{The A.P is } 2, 3\\frac{1}{2}, 5, 6\\frac{1}{2}, 8,...\\)\n\n', '(a) \\(17x = 375^{2} - 356^{2}\\)\n\\(17x = (375 + 356)(375 - 356)\\)\n\\(17x = (731)(19)\\)\n\\(x = \\frac{731 \\times 19}{17} = 817\\)~(b) \\(4^{x} = 2^{\\frac{1}{2}} \\times 8\\)\n\\(2^{2x} = 2^{\\frac{1}{2}} \\times 2^{3}\\)\n\\(2^{2x} = 2^{3\\frac{1}{2}}\\)\n\\(\\implies 2x = 3\\frac{1}{2} \\implies x = \\frac{7}{4}\\)\n(c) \\(S_{n} = \\frac{n}{2} (2a + (n - 1)d)\\) (sum of terms of an A.P)\n\\(S_{9} = \\frac{9}{2} [2a + (9 - 1) d] = \\frac{9}{2} [2a + 8d]\\)\n\\(72 = 9(a + 4d) \\implies 8 = a + 4d ... (1)\\)\n\\(S_{9 + 4} = S_{13} = \\frac{13}{2} [2a + (13 - 1)d] = \\frac{13}{2} [2a + 12d]\\)\n\\(72 + 71 = 143 = 13(a + 6d) \\implies 11 = a + 6d ... (2)\\)\n\\((2) - (1) : 2d = 3 \\implies d = \\frac{3}{2}\\)\n\\(a + 4(1\\frac{1}{2}) = 8 \\implies a + 6 = 8\\)\n\\(\\implies a = 8 - 6 = 2\\)\n\\(\\therefore \\text{The A.P is } 2, 3\\frac{1}{2}, 5, 6\\frac{1}{2}, 8,...\\)\n~(c) \\(S_{n} = \\frac{n}{2} (2a + (n - 1)d)\\) (sum of terms of an A.P)\n\\(S_{9} = \\frac{9}{2} [2a + (9 - 1) d] = \\frac{9}{2} [2a + 8d]\\)\n\\(72 = 9(a + 4d) \\implies 8 = a + 4d ... (1)\\)\n\\(S_{9 + 4} = S_{13} = \\frac{13}{2} [2a + (13 - 1)d] = \\frac{13}{2} [2a + 12d]\\)\n\\(72 + 71 = 143 = 13(a + 6d) \\implies 11 = a + 6d ... (2)\\)\n\\((2) - (1) : 2d = 3 \\implies d = \\frac{3}{2}\\)\n\\(a + 4(1\\frac{1}{2}) = 8 \\implies a + 6 = 8\\)\n\\(\\implies a = 8 - 6 = 2\\)\n\\(\\therefore \\text{The A.P is } 2, 3\\frac{1}{2}, 5, 6\\frac{1}{2}, 8,...\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(654, '2017', '\n\nQuestion\n\n(a) Using a ruler and a pair of compasses only, construct: (i) a triangle ABC such that |AB| = 5cm, |AC| = 7.5cm and < CAB = 120°; (ii) the locus \\(l_{1}\\) of points equidistant from A and B; (iii) the locus \\(l_{2}\\) of points equidistant from AB and AC which passes through triangle ABC .\n(b) Label the point P where \\(l_{1}\\) and \\(l_{2}\\) intersect.\n(c) Measure |CP|.\n\n', '', '(a) Using a ruler and a pair of compasses only, construct: (i) a triangle ABC such that |AB| = 5cm, |AC| = 7.5cm and < CAB = 120°; (ii) the locus \\(l_{1}\\) of points equidistant from A and B; (iii) the locus \\(l_{2}\\) of points equidistant from AB and AC which passes through triangle ABC .\n(b) Label the point P where \\(l_{1}\\) and \\(l_{2}\\) intersect.\n(c) Measure |CP|.', '(a) Using a ruler and a pair of compasses only, construct: (i) a triangle ABC such that |AB| = 5cm, |AC| = 7.5cm and < CAB = 120°; (ii) the locus \\(l_{1}\\) of points equidistant from A and B; (iii) the locus \\(l_{2}\\) of points equidistant from AB and AC which passes through triangle ABC .~(b) Label the point P where \\(l_{1}\\) and \\(l_{2}\\) intersect.\n(c) Measure |CP|.~(c) Measure |CP|.', NULL, '\n\nExplanation\n(c) |CP| = 6.7cm\n\n', '(c) |CP| = 6.7cm\n\n', '', '', 'WAEC-PK-1', 'WAEC-MATH', 1),
(655, '2017', '\n\nQuestion\n\n(a)\nCalculate the area of the shaded segment of the circle shown in the diagram [Take \\(\\pi = \\frac{22}{7}\\)]\n(b) A tin has radius 3cm and height 6cm. Find the (i) total surface area of the tin ; (ii) volume, in litres, that will fill the tin to capacity, correct to two decimal places.\n[Take \\(\\pi = \\frac{22}{7}\\)]\n\n', '', '(a)\nCalculate the area of the shaded segment of the circle shown in the diagram [Take \\(\\pi = \\frac{22}{7}\\)]\n(b) A tin has radius 3cm and height 6cm. Find the (i) total surface area of the tin ; (ii) volume, in litres, that will fill the tin to capacity, correct to two decimal places.\n[Take \\(\\pi = \\frac{22}{7}\\)]', '(a)\nCalculate the area of the shaded segment of the circle shown in the diagram [Take \\(\\pi = \\frac{22}{7}\\)]~(b) A tin has radius 3cm and height 6cm. Find the (i) total surface area of the tin ; (ii) volume, in litres, that will fill the tin to capacity, correct to two decimal places.\n[Take \\(\\pi = \\frac{22}{7}\\)]', NULL, '\n\nExplanation\n(a) Area of theshaded segment = Area of sector - Area of triangle\n= \\(\\frac{\\theta}{360°} \\times \\pi r^{2}- \\frac{1}{2} ab \\sin \\theta\\)\nArea of sector = \\(\\frac{63}{360} \\times \\frac{22}{7} \\times 10^{2}\\)\n= \\(55 cm^{2}\\)\nArea of triangle = \\(\\frac{1}{2} \\times 10 \\times 10 \\times \\sin 63°\\)\n= \\(50 \\times 0.891 = 44.55 cm^{2}\\)\n\\(\\therefore \\text{Area of the shaded segment} = 55 - 44.55 = 10.45 cm^{2}\\)\n(b) (i) Total surface area = \\(2\\pi r (r + h)\\)\n= \\(2 \\times \\frac{22}{7} \\times 3 (3 + 6)\\)\n= \\(\\frac{132}{7} \\times 9 = \\frac{1188}{7}\\)\n= \\(169.71 cm^{2}\\)\n(ii) Volume of tin = \\(\\pi r^{2} h\\)\n= \\(\\frac{22}{7} \\times 3 \\times 3 \\times 6 = \\frac{1188}{7}\\)\n\\(\\approxeq 169.71 cm^{3}\\)\n\n', '', 'Explanation\n(a) Area of theshaded segment = Area of sector - Area of triangle\n= \\(\\frac{\\theta}{360°} \\times \\pi r^{2}- \\frac{1}{2} ab \\sin \\theta\\)\nArea of sector = \\(\\frac{63}{360} \\times \\frac{22}{7} \\times 10^{2}\\)\n= \\(55 cm^{2}\\)\nArea of triangle = \\(\\frac{1}{2} \\times 10 \\times 10 \\times \\sin 63°\\)\n= \\(50 \\times 0.891 = 44.55 cm^{2}\\)\n\\(\\therefore \\text{Area of the shaded segment} = 55 - 44.55 = 10.45 cm^{2}\\)\n(b) (i) Total surface area = \\(2\\pi r (r + h)\\)\n= \\(2 \\times \\frac{22}{7} \\times 3 (3 + 6)\\)\n= \\(\\frac{132}{7} \\times 9 = \\frac{1188}{7}\\)\n= \\(169.71 cm^{2}\\)\n(ii) Volume of tin = \\(\\pi r^{2} h\\)\n= \\(\\frac{22}{7} \\times 3 \\times 3 \\times 6 = \\frac{1188}{7}\\)\n\\(\\approxeq 169.71 cm^{3}\\)\n\n', '(a) Area of theshaded segment = Area of sector - Area of triangle\n= \\(\\frac{\\theta}{360°} \\times \\pi r^{2}- \\frac{1}{2} ab \\sin \\theta\\)\nArea of sector = \\(\\frac{63}{360} \\times \\frac{22}{7} \\times 10^{2}\\)\n= \\(55 cm^{2}\\)\nArea of triangle = \\(\\frac{1}{2} \\times 10 \\times 10 \\times \\sin 63°\\)\n= \\(50 \\times 0.891 = 44.55 cm^{2}\\)\n\\(\\therefore \\text{Area of the shaded segment} = 55 - 44.55 = 10.45 cm^{2}\\)~(b) (i) Total surface area = \\(2\\pi r (r + h)\\)\n= \\(2 \\times \\frac{22}{7} \\times 3 (3 + 6)\\)\n= \\(\\frac{132}{7} \\times 9 = \\frac{1188}{7}\\)\n= \\(169.71 cm^{2}\\)\n(ii) Volume of tin = \\(\\pi r^{2} h\\)\n= \\(\\frac{22}{7} \\times 3 \\times 3 \\times 6 = \\frac{1188}{7}\\)\n\\(\\approxeq 169.71 cm^{3}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(656, '2017', '\n\nQuestion\n\n(a) Copy and complete the following table for the relation \\(y = \\frac{5}{2} + x - 4x^{2}\\)\n\n\n\nx\n-2.0\n-1.5\n-1.0\n-0.5\n0\n0.5\n1\n1.5\n2.0\n\n\ny\n-15.5\n\n\n1\n2.5\n\n\n\n\n\n\n\n(b) Using a scale of 2cm to 1 unit on the x- axis and 2cm to 5 units on the y- axis, draw the graph of the relation for \\(-2.0 \\leq x \\leq 2.0\\).\n(c) What is the maximum value of y?\n(d) From your graph, obtain the roots of the equation \\(8x^{2} - 2x - 5 = 0\\)\n\n\n(a) Copy and complete the following table for the relation \\(y = \\frac{5}{2} + x - 4x^{2}\\)\n\n\n\nx\n-2.0\n-1.5\n-1.0\n-0.5\n0\n0.5\n1\n1.5\n2.0\n\n\ny\n-15.5\n\n\n1\n2.5\n\n\n\n\n\n\n', '', '(a) Copy and complete the following table for the relation \\(y = \\frac{5}{2} + x - 4x^{2}\\)\n\n\n\nx\n-2.0\n-1.5\n-1.0\n-0.5\n0\n0.5\n1\n1.5\n2.0\n\n\ny\n-15.5\n\n\n1\n2.5\n\n\n\n\n\n\n\n(b) Using a scale of 2cm to 1 unit on the x- axis and 2cm to 5 units on the y- axis, draw the graph of the relation for \\(-2.0 \\leq x \\leq 2.0\\).\n(c) What is the maximum value of y?\n(d) From your graph, obtain the roots of the equation \\(8x^{2} - 2x - 5 = 0\\)\n\n\n(a) Copy and complete the following table for the relation \\(y = \\frac{5}{2} + x - 4x^{2}\\)\n\n\n\nx\n-2.0\n-1.5\n-1.0\n-0.5\n0\n0.5\n1\n1.5\n2.0\n\n\ny\n-15.5\n\n\n1\n2.5\n\n\n\n\n', '(a) Copy and complete the following table for the relation \\(y = \\frac{5}{2} + x - 4x^{2}\\)\n\n\n\nx\n-2.0\n-1.5\n-1.0\n-0.5\n0\n0.5\n1\n1.5\n2.0\n\n\ny\n-15.5\n\n\n1\n2.5\n\n\n\n\n\n\n~(b) Using a scale of 2cm to 1 unit on the x- axis and 2cm to 5 units on the y- axis, draw the graph of the relation for \\(-2.0 \\leq x \\leq 2.0\\).\n(c) What is the maximum value of y?\n(d) From your graph, obtain the roots of the equation \\(8x^{2} - 2x - 5 = 0\\)\n\n\n(a) Copy and complete the following table for the relation \\(y = \\frac{5}{2} + x - 4x^{2}\\)\n\n\n\nx\n-2.0\n-1.5\n-1.0\n-0.5\n0\n0.5\n1\n1.5\n2.0\n\n\ny\n-15.5\n\n\n1\n2.5\n\n\n\n\n~(c) What is the maximum value of y?\n(d) From your graph, obtain the roots of the equation \\(8x^{2} - 2x - 5 = 0\\)\n\n\n(a) Copy and complete the following table for the relation \\(y = \\frac{5}{2} + x - 4x^{2}\\)\n\n\n\nx\n-2.0\n-1.5\n-1.0\n-0.5\n0\n0.5\n1\n1.5\n2.0\n\n\ny\n-15.5\n\n\n1\n2.5\n\n\n\n\n~(d) From your graph, obtain the roots of the equation \\(8x^{2} - 2x - 5 = 0\\)\n\n\n(a) Copy and complete the following table for the relation \\(y = \\frac{5}{2} + x - 4x^{2}\\)\n\n\n\nx\n-2.0\n-1.5\n-1.0\n-0.5\n0\n0.5\n1\n1.5\n2.0\n\n\ny\n-15.5\n\n\n1\n2.5\n\n\n\n\n', NULL, '\n\nExplanation\n(a)\n\n\n\nx\n-2.0\n-1.5\n-1.0\n-0.5\n0\n0.5\n1\n1.5\n2.0\n\n\ny\n-15.5\n-8.0\n-2.5\n1\n2.5\n2.0\n-0.5\n-5.0\n-11.5\n\n\n\n(b)\n(c) The maximum value of y = 2.5.\n(d) Obtaining the root of \\(8x^{2} - 2x - 5 = 0\\)\nDivide both sides by -2, i.e \\(-4x^{2} + x + \\frac{5}{2} = 0\\)\n\\(\\therefore \\text{Roots are } x = -0.7 ; x = 0.8.\n\n', '', 'Explanation\n(a)\n\n\n\nx\n-2.0\n-1.5\n-1.0\n-0.5\n0\n0.5\n1\n1.5\n2.0\n\n\ny\n-15.5\n-8.0\n-2.5\n1\n2.5\n2.0\n-0.5\n-5.0\n-11.5\n\n\n\n(b)\n(c) The maximum value of y = 2.5.\n(d) Obtaining the root of \\(8x^{2} - 2x - 5 = 0\\)\nDivide both sides by -2, i.e \\(-4x^{2} + x + \\frac{5}{2} = 0\\)\n\\(\\therefore \\text{Roots are } x = -0.7 ; x = 0.8.\n\n', '(a)\n\n\n\nx\n-2.0\n-1.5\n-1.0\n-0.5\n0\n0.5\n1\n1.5\n2.0\n\n\ny\n-15.5\n-8.0\n-2.5\n1\n2.5\n2.0\n-0.5\n-5.0\n-11.5\n\n\n~(b)\n(c) The maximum value of y = 2.5.\n(d) Obtaining the root of \\(8x^{2} - 2x - 5 = 0\\)\nDivide both sides by -2, i.e \\(-4x^{2} + x + \\frac{5}{2} = 0\\)\n\\(\\therefore \\text{Roots are } x = -0.7 ; x = 0.8.\n~(c) The maximum value of y = 2.5.\n(d) Obtaining the root of \\(8x^{2} - 2x - 5 = 0\\)\nDivide both sides by -2, i.e \\(-4x^{2} + x + \\frac{5}{2} = 0\\)\n\\(\\therefore \\text{Roots are } x = -0.7 ; x = 0.8.\n~(d) Obtaining the root of \\(8x^{2} - 2x - 5 = 0\\)\nDivide both sides by -2, i.e \\(-4x^{2} + x + \\frac{5}{2} = 0\\)\n\\(\\therefore \\text{Roots are } x = -0.7 ; x = 0.8.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(657, '2017', '\n\nQuestion\n\n(a) The distribution of junior workers in an institution is as follows: Clerks - 78, Drivers - 36, Typists - 44, Messengers - 52, Others - 30. Represent the above information by a pie chart.\n(b) The table below shows the frequency distribution of marks scored by 30 candidates in an aptitude test.\n\n\n\nMarks\n4\n5\n6\n7\n8\n9\n\n\nNo of candidates\n5\n8\n5\n6\n4\n2\n\n\n\nFind the mean score to the nearest whole number.\n\n\n(a) The distribution of junior workers in an institution is as follows: Clerks - 78, Drivers - 36, Typists - 44, Messengers - 52, Others - 30. Represent the above information by a pie chart.\n(b) The table below shows the frequency distribution of marks scored by 30 candidates in an aptitude test.\n\n\n\nMarks\n4\n5\n6\n7\n8\n9\n\n\nNo of candidates\n5\n8\n5\n6\n4\n2\n\n\n', '', '(a) The distribution of junior workers in an institution is as follows: Clerks - 78, Drivers - 36, Typists - 44, Messengers - 52, Others - 30. Represent the above information by a pie chart.\n(b) The table below shows the frequency distribution of marks scored by 30 candidates in an aptitude test.\n\n\n\nMarks\n4\n5\n6\n7\n8\n9\n\n\nNo of candidates\n5\n8\n5\n6\n4\n2\n\n\n\nFind the mean score to the nearest whole number.\n\n\n(a) The distribution of junior workers in an institution is as follows: Clerks - 78, Drivers - 36, Typists - 44, Messengers - 52, Others - 30. Represent the above information by a pie chart.\n(b) The table below shows the frequency distribution of marks scored by 30 candidates in an aptitude test.\n\n\n\nMarks\n4\n5\n6\n7\n8\n9\n\n\nNo of candidates\n5\n8\n5\n6\n4\n2\n', '(a) The distribution of junior workers in an institution is as follows: Clerks - 78, Drivers - 36, Typists - 44, Messengers - 52, Others - 30. Represent the above information by a pie chart.\n(b) The table below shows the frequency distribution of marks scored by 30 candidates in an aptitude test.\n\n\n\nMarks\n4\n5\n6\n7\n8\n9\n\n\nNo of candidates\n5\n8\n5\n6\n4\n2\n\n\n\nFind the mean score to the nearest whole number.\n\n\n(a) The distribution of junior workers in an institution is as follows: Clerks - 78, Drivers - 36, Typists - 44, Messengers - 52, Others - 30. Represent the above information by a pie chart.~(b) The table below shows the frequency distribution of marks scored by 30 candidates in an aptitude test.\n\n\n\nMarks\n4\n5\n6\n7\n8\n9\n\n\nNo of candidates\n5\n8\n5\n6\n4\n2\n\n\n\nFind the mean score to the nearest whole number.\n\n\n(a) The distribution of junior workers in an institution is as follows: Clerks - 78, Drivers - 36, Typists - 44, Messengers - 52, Others - 30. Represent the above information by a pie chart.\n(b) The table below shows the frequency distribution of marks scored by 30 candidates in an aptitude test.\n\n\n\nMarks\n4\n5\n6\n7\n8\n9\n\n\nNo of candidates\n5\n8\n5\n6\n4\n2\n', NULL, '\n\nExplanation\n(a) Sector angle represented by the officers:\nTotal number of officers = 240\nClerks = \\(\\frac{78}{240} \\times 360° = 117°\\)\nDrivers = \\(\\frac{36}{240} \\times 360° = 54°\\)\nTypists = \\(\\frac{44}{240} \\times 360° = 66°\\)\nMessengers = \\(\\frac{52}{240} \\times 360° = 78°\\)\nOthers = \\(\\frac{30}{240} \\times 360° = 45°\\)\n\n(b)\n\n\n\nMarks (x)\n4\n5\n6\n7\n8\n9\nTotal\n\n\nNo of candidates (f)\n5\n8\n5\n6\n4\n2\n30\n\n\n\\(fx\\)\n20\n40\n30\n42\n32\n18\n182\n\n\n\nMean score = \\(\\frac{\\sum fx}{\\sum f}\\)\n= \\(\\frac{182}{30}\\)\n= \\(6.067\\) marks \\(\\approxeq\\) 6 marks to the nearest whole number.\n\n\n', '', 'Explanation\n(a) Sector angle represented by the officers:\nTotal number of officers = 240\nClerks = \\(\\frac{78}{240} \\times 360° = 117°\\)\nDrivers = \\(\\frac{36}{240} \\times 360° = 54°\\)\nTypists = \\(\\frac{44}{240} \\times 360° = 66°\\)\nMessengers = \\(\\frac{52}{240} \\times 360° = 78°\\)\nOthers = \\(\\frac{30}{240} \\times 360° = 45°\\)\n\n(b)\n\n\n\nMarks (x)\n4\n5\n6\n7\n8\n9\nTotal\n\n\nNo of candidates (f)\n5\n8\n5\n6\n4\n2\n30\n\n\n\\(fx\\)\n20\n40\n30\n42\n32\n18\n182\n\n\n\nMean score = \\(\\frac{\\sum fx}{\\sum f}\\)\n= \\(\\frac{182}{30}\\)\n= \\(6.067\\) marks \\(\\approxeq\\) 6 marks to the nearest whole number.\n\n\n', '(a) Sector angle represented by the officers:\nTotal number of officers = 240\nClerks = \\(\\frac{78}{240} \\times 360° = 117°\\)\nDrivers = \\(\\frac{36}{240} \\times 360° = 54°\\)\nTypists = \\(\\frac{44}{240} \\times 360° = 66°\\)\nMessengers = \\(\\frac{52}{240} \\times 360° = 78°\\)\nOthers = \\(\\frac{30}{240} \\times 360° = 45°\\)\n~(b)\n\n\n\nMarks (x)\n4\n5\n6\n7\n8\n9\nTotal\n\n\nNo of candidates (f)\n5\n8\n5\n6\n4\n2\n30\n\n\n\\(fx\\)\n20\n40\n30\n42\n32\n18\n182\n\n\n\nMean score = \\(\\frac{\\sum fx}{\\sum f}\\)\n= \\(\\frac{182}{30}\\)\n= \\(6.067\\) marks \\(\\approxeq\\) 6 marks to the nearest whole number.\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(658, '2017', '\n\nQuestion\n\nThree towns P, Q and R are such that the distance between P and Q is 50km and the distance between P and R is 90km. If the bearing of Q from P is 075° and the bearing of R from P is 310°, find the :\n(a) distance between Q and R ;\n(b) baering of R from Q.\n\n', '', 'Three towns P, Q and R are such that the distance between P and Q is 50km and the distance between P and R is 90km. If the bearing of Q from P is 075° and the bearing of R from P is 310°, find the :\n(a) distance between Q and R ;\n(b) baering of R from Q.', 'Three towns P, Q and R are such that the distance between P and Q is 50km and the distance between P and R is 90km. If the bearing of Q from P is 075° and the bearing of R from P is 310°, find the :\n(a) distance between Q and R ;~(b) baering of R from Q.', NULL, '\n\nExplanation\n< RPQ = 50° + 75° = 125°\nBy cosine rule,\n\\(|QR|^{2} = |RP|^{2} + |PQ|^{2} - 2(|PR|)(|PQ|) \\cos 125°\\)\n= \\(90^{2} + 50^{2} +2(90)(50) \\cos 55°\\)\n= \\(8100 + 2500 + 9000 \\times 0.5736\\)\n= \\(10600 + 5162.4\\)\n\\(|QR|^{2} = 15762.4\\)\n\\(|QR| = \\sqrt{15762.4} = 125.548km\\)\n(b) By Sine rule,\n\\(\\frac{RQ}{\\sin 125°} = \\frac{PR}{\\sin <PQR}\\)\n\\(\\frac{125.548}{\\sin 125°} = \\frac{90}{\\sin < PQR}\\)\n\\(\\sin < PQR = \\frac{90 \\times 0.8192}{125.548}\\)\n\\(\\sin < PQR = 0.5782 \\implies < PQR = \\sin^{-1} (0.5872)\\)\n= \\(35.96°\\)\nBearing of R from Q = 180° + 75° + 35.96° = 290.96°\\)\n\\(\\approxeq 291°\\)\n\n', '', 'Explanation\n< RPQ = 50° + 75° = 125°\nBy cosine rule,\n\\(|QR|^{2} = |RP|^{2} + |PQ|^{2} - 2(|PR|)(|PQ|) \\cos 125°\\)\n= \\(90^{2} + 50^{2} +2(90)(50) \\cos 55°\\)\n= \\(8100 + 2500 + 9000 \\times 0.5736\\)\n= \\(10600 + 5162.4\\)\n\\(|QR|^{2} = 15762.4\\)\n\\(|QR| = \\sqrt{15762.4} = 125.548km\\)\n(b) By Sine rule,\n\\(\\frac{RQ}{\\sin 125°} = \\frac{PR}{\\sin <PQR}\\)\n\\(\\frac{125.548}{\\sin 125°} = \\frac{90}{\\sin < PQR}\\)\n\\(\\sin < PQR = \\frac{90 \\times 0.8192}{125.548}\\)\n\\(\\sin < PQR = 0.5782 \\implies < PQR = \\sin^{-1} (0.5872)\\)\n= \\(35.96°\\)\nBearing of R from Q = 180° + 75° + 35.96° = 290.96°\\)\n\\(\\approxeq 291°\\)\n\n', '~(b) By Sine rule,\n\\(\\frac{RQ}{\\sin 125°} = \\frac{PR}{\\sin <PQR}\\)\n\\(\\frac{125.548}{\\sin 125°} = \\frac{90}{\\sin < PQR}\\)\n\\(\\sin < PQR = \\frac{90 \\times 0.8192}{125.548}\\)\n\\(\\sin < PQR = 0.5782 \\implies < PQR = \\sin^{-1} (0.5872)\\)\n= \\(35.96°\\)\nBearing of R from Q = 180° + 75° + 35.96° = 290.96°\\)\n\\(\\approxeq 291°\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(659, '2017', '\n\nQuestion\n\nThe table below shows the weekly profit in naira from a mini-market.\n\n\n\nWeekly profit (N)\n1-10\n11-20\n21-30\n31-40\n41-50\n51-60\n\n\nFreq\n6\n6\n12\n11\n10\n5\n\n\n\n(a) Draw the cumulative frequency curve of the data;\n(b) From your graph, estimate the : (i) median ; (ii) 80th percentile\n(c) What is the modal weekly profit?\n\n\n\nThe table below shows the weekly profit in naira from a mini-market.\n\n\n\nWeekly profit (N)\n1-10\n11-20\n21-30\n31-40\n41-50\n51-60\n\n\nFreq\n6\n6\n12\n11\n10\n5\n\n\n', '', 'The table below shows the weekly profit in naira from a mini-market.\n\n\n\nWeekly profit (N)\n1-10\n11-20\n21-30\n31-40\n41-50\n51-60\n\n\nFreq\n6\n6\n12\n11\n10\n5\n\n\n\n(a) Draw the cumulative frequency curve of the data;\n(b) From your graph, estimate the : (i) median ; (ii) 80th percentile\n(c) What is the modal weekly profit?\n\n\n\nThe table below shows the weekly profit in naira from a mini-market.\n\n\n\nWeekly profit (N)\n1-10\n11-20\n21-30\n31-40\n41-50\n51-60\n\n\nFreq\n6\n6\n12\n11\n10\n5\n', 'The table below shows the weekly profit in naira from a mini-market.\n\n\n\nWeekly profit (N)\n1-10\n11-20\n21-30\n31-40\n41-50\n51-60\n\n\nFreq\n6\n6\n12\n11\n10\n5\n\n\n\n(a) Draw the cumulative frequency curve of the data;~(b) From your graph, estimate the : (i) median ; (ii) 80th percentile\n(c) What is the modal weekly profit?\n\n\n\nThe table below shows the weekly profit in naira from a mini-market.\n\n\n\nWeekly profit (N)\n1-10\n11-20\n21-30\n31-40\n41-50\n51-60\n\n\nFreq\n6\n6\n12\n11\n10\n5\n~(c) What is the modal weekly profit?\n\n\n\nThe table below shows the weekly profit in naira from a mini-market.\n\n\n\nWeekly profit (N)\n1-10\n11-20\n21-30\n31-40\n41-50\n51-60\n\n\nFreq\n6\n6\n12\n11\n10\n5\n', NULL, '\n\nExplanation\n(a)\n\n\n\n\nClass Interval\n(x)\n\n\nFrequency\n(f)\n\nCum. Freq\n\n\n1 - 10\n6\n6\n\n\n11 - 20\n6\n12\n\n\n21 - 30\n12\n24\n\n\n31 - 40\n11\n35\n\n\n41 - 50\n10\n45\n\n\n51 - 60\n5\n50\n\n\n\n\n(b)(i) Median = N30.50\n(ii) 80th percentile = N45\n(c) The modal weekly profit = Lower class boundary of modal class + \\(\\frac{(f- f_{1}) \\times c}{(f - f_{1}) + (f - f_{2})}\\)\nwhere \\(f_{1}\\) = frequency above the modal class ; \\(f_{2}\\) = frequency below modal class; c = class width; f = frequency of modal class.\n\\(\\therefore \\text{Modal profit} = 20.5 + \\frac{(12 - 6) \\times 10}{(12 - 6) + (12 - 11)}\\)\n= \\(20.5 + \\frac{60}{7}\\)\n= \\(N29.07\\)\n\n', '', 'Explanation\n(a)\n\n\n\n\nClass Interval\n(x)\n\n\nFrequency\n(f)\n\nCum. Freq\n\n\n1 - 10\n6\n6\n\n\n11 - 20\n6\n12\n\n\n21 - 30\n12\n24\n\n\n31 - 40\n11\n35\n\n\n41 - 50\n10\n45\n\n\n51 - 60\n5\n50\n\n\n\n\n(b)(i) Median = N30.50\n(ii) 80th percentile = N45\n(c) The modal weekly profit = Lower class boundary of modal class + \\(\\frac{(f- f_{1}) \\times c}{(f - f_{1}) + (f - f_{2})}\\)\nwhere \\(f_{1}\\) = frequency above the modal class ; \\(f_{2}\\) = frequency below modal class; c = class width; f = frequency of modal class.\n\\(\\therefore \\text{Modal profit} = 20.5 + \\frac{(12 - 6) \\times 10}{(12 - 6) + (12 - 11)}\\)\n= \\(20.5 + \\frac{60}{7}\\)\n= \\(N29.07\\)\n\n', '(a)\n\n\n\n\nClass Interval\n(x)\n\n\nFrequency\n(f)\n\nCum. Freq\n\n\n1 - 10\n6\n6\n\n\n11 - 20\n6\n12\n\n\n21 - 30\n12\n24\n\n\n31 - 40\n11\n35\n\n\n41 - 50\n10\n45\n\n\n51 - 60\n5\n50\n\n\n\n~(b)(i) Median = N30.50\n(ii) 80th percentile = N45\n(c) The modal weekly profit = Lower class boundary of modal class + \\(\\frac{(f- f_{1}) \\times c}{(f - f_{1}) + (f - f_{2})}\\)\nwhere \\(f_{1}\\) = frequency above the modal class ; \\(f_{2}\\) = frequency below modal class; c = class width; f = frequency of modal class.\n\\(\\therefore \\text{Modal profit} = 20.5 + \\frac{(12 - 6) \\times 10}{(12 - 6) + (12 - 11)}\\)\n= \\(20.5 + \\frac{60}{7}\\)\n= \\(N29.07\\)\n~(c) The modal weekly profit = Lower class boundary of modal class + \\(\\frac{(f- f_{1}) \\times c}{(f - f_{1}) + (f - f_{2})}\\)\nwhere \\(f_{1}\\) = frequency above the modal class ; \\(f_{2}\\) = frequency below modal class; c = class width; f = frequency of modal class.\n\\(\\therefore \\text{Modal profit} = 20.5 + \\frac{(12 - 6) \\times 10}{(12 - 6) + (12 - 11)}\\)\n= \\(20.5 + \\frac{60}{7}\\)\n= \\(N29.07\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(660, '2017', '\n\nQuestion\n\nSimplify :\n(i) \\(2\\frac{2}{3} - (2\\frac{1}{2} - 1\\frac{4}{5})\\)\n(ii) \\(\\frac{3.25 - 1.64}{2.47 - 2.01}\\)\n\n', '', 'Simplify :\n(i) \\(2\\frac{2}{3} - (2\\frac{1}{2} - 1\\frac{4}{5})\\)\n(ii) \\(\\frac{3.25 - 1.64}{2.47 - 2.01}\\)', '', NULL, '\n\nExplanation\n(i) \\(2\\frac{2}{3} - (2\\frac{1}{2} - 1\\frac{4}{5})\\)\n= \\(2\\frac{2}{3} - (\\frac{5}{2} - \\frac{9}{5})\\)\n= \\(\\frac{8}{3} - (\\frac{25 - 18}{10})\\)\n= \\(\\frac{8}{3} - \\frac{7}{10}\\)\n= \\(\\frac{24 - 21}{30}\\)\n= \\(\\frac{3}{30} = \\frac{1}{10}\\)\n(ii) \\(\\frac{3.25 - 1.64}{2.47 - 2.01}\\)\n= \\(\\frac{1.61}{0.46}\\)\n= \\(\\frac{7}{2}\\)\n= \\(3\\frac{1}{2}\\)\n\n', '', 'Explanation\n(i) \\(2\\frac{2}{3} - (2\\frac{1}{2} - 1\\frac{4}{5})\\)\n= \\(2\\frac{2}{3} - (\\frac{5}{2} - \\frac{9}{5})\\)\n= \\(\\frac{8}{3} - (\\frac{25 - 18}{10})\\)\n= \\(\\frac{8}{3} - \\frac{7}{10}\\)\n= \\(\\frac{24 - 21}{30}\\)\n= \\(\\frac{3}{30} = \\frac{1}{10}\\)\n(ii) \\(\\frac{3.25 - 1.64}{2.47 - 2.01}\\)\n= \\(\\frac{1.61}{0.46}\\)\n= \\(\\frac{7}{2}\\)\n= \\(3\\frac{1}{2}\\)\n\n', '', 'WAEC-PK-1', 'WAEC-MATH', 1),
(661, '2017', '\n\nQuestion\n\n(a) Using logarithm table, evaluate \\(\\frac{\\sqrt[3]{1.376}}{\\sqrt[4]{0.007}}\\) correct to three significant figure.\n(b) Without using Mathematical tables, find the value of \\(\\frac{\\log 81}{\\log \\frac{1}{3}}\\).\n\n', '', '(a) Using logarithm table, evaluate \\(\\frac{\\sqrt[3]{1.376}}{\\sqrt[4]{0.007}}\\) correct to three significant figure.\n(b) Without using Mathematical tables, find the value of \\(\\frac{\\log 81}{\\log \\frac{1}{3}}\\).', '(a) Using logarithm table, evaluate \\(\\frac{\\sqrt[3]{1.376}}{\\sqrt[4]{0.007}}\\) correct to three significant figure.~(b) Without using Mathematical tables, find the value of \\(\\frac{\\log 81}{\\log \\frac{1}{3}}\\).', NULL, '\n\nExplanation\n(a)\n\n\n\nNo\nLog\n\n\n\\(\\sqrt[3]{1.376}\\)\n\\(0.1386 \\div 3 \\) = \\(0.0462\\)\n\n\n\\(\\sqrt[4]{0.007}\\)\n\\(\\bar{3}.8451 \\div 4\\) = \\(\\bar{1}.4613\\)\n\n\n\n\n\n\n\\(\\frac{\\sqrt[3]{1.376}}{\\sqrt[4]{0.007}}\\) =\n\\(0.5849\\)\n\n\nAntilog = 3.845\n\n\n\n\n\\(\\therefore \\frac{\\sqrt[3]{1.376}}{\\sqrt[4]{0.007}} \\approxeq3.85\\) (to 3 s.f)\n(b) \\(\\frac{\\log 81}{\\log \\frac{1}{3}}\\)\n= \\(\\frac{\\log 3^{4}}{\\log 3^{-1}}\\)\n= \\(\\frac{4 \\log 3}{-1 \\log 3}\\)\n= \\(-4\\)\n\n', '', 'Explanation\n(a)\n\n\n\nNo\nLog\n\n\n\\(\\sqrt[3]{1.376}\\)\n\\(0.1386 \\div 3 \\) = \\(0.0462\\)\n\n\n\\(\\sqrt[4]{0.007}\\)\n\\(\\bar{3}.8451 \\div 4\\) = \\(\\bar{1}.4613\\)\n\n\n\n\n\n\n\\(\\frac{\\sqrt[3]{1.376}}{\\sqrt[4]{0.007}}\\) =\n\\(0.5849\\)\n\n\nAntilog = 3.845\n\n\n\n\n\\(\\therefore \\frac{\\sqrt[3]{1.376}}{\\sqrt[4]{0.007}} \\approxeq3.85\\) (to 3 s.f)\n(b) \\(\\frac{\\log 81}{\\log \\frac{1}{3}}\\)\n= \\(\\frac{\\log 3^{4}}{\\log 3^{-1}}\\)\n= \\(\\frac{4 \\log 3}{-1 \\log 3}\\)\n= \\(-4\\)\n\n', '(a)\n\n\n\nNo\nLog\n\n\n\\(\\sqrt[3]{1.376}\\)\n\\(0.1386 \\div 3 \\) = \\(0.0462\\)\n\n\n\\(\\sqrt[4]{0.007}\\)\n\\(\\bar{3}.8451 \\div 4\\) = \\(\\bar{1}.4613\\)\n\n\n\n\n\n\n\\(\\frac{\\sqrt[3]{1.376}}{\\sqrt[4]{0.007}}\\) =\n\\(0.5849\\)\n\n\nAntilog = 3.845\n\n\n\n\n\\(\\therefore \\frac{\\sqrt[3]{1.376}}{\\sqrt[4]{0.007}} \\approxeq3.85\\) (to 3 s.f)~(b) \\(\\frac{\\log 81}{\\log \\frac{1}{3}}\\)\n= \\(\\frac{\\log 3^{4}}{\\log 3^{-1}}\\)\n= \\(\\frac{4 \\log 3}{-1 \\log 3}\\)\n= \\(-4\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(662, '2017', '\n\nQuestion\n\nIn a certain class, 22 pupils take one or more of Chemistry, Economics and Government. 12 take Economics (E), 8 take Government (G) and 7 take Chemistry (C). Nobody takes Economics and Chemistry and 4 pupils take Economics and Government.\n(a)(i) Using set notation and the letters indicated above, write down the two statements in the last sentence; (ii) Draw a Venn diagram to illustrate the information.\n(b) How many pupils take (i) both Chemistry and Government ? (ii) Government only?\n\n', '', 'In a certain class, 22 pupils take one or more of Chemistry, Economics and Government. 12 take Economics (E), 8 take Government (G) and 7 take Chemistry (C). Nobody takes Economics and Chemistry and 4 pupils take Economics and Government.\n(a)(i) Using set notation and the letters indicated above, write down the two statements in the last sentence; (ii) Draw a Venn diagram to illustrate the information.\n(b) How many pupils take (i) both Chemistry and Government ? (ii) Government only?', 'In a certain class, 22 pupils take one or more of Chemistry, Economics and Government. 12 take Economics (E), 8 take Government (G) and 7 take Chemistry (C). Nobody takes Economics and Chemistry and 4 pupils take Economics and Government.\n(a)(i) Using set notation and the letters indicated above, write down the two statements in the last sentence; (ii) Draw a Venn diagram to illustrate the information.~(b) How many pupils take (i) both Chemistry and Government ? (ii) Government only?', NULL, '\n\nExplanation\n(a)\n(i) Nobody takes Economics and Chemistry \\(n(E \\cap C) = \\null \\).\n(ii) 4 pupils take Economics and Government; \\(n(E \\cap G) = 4\\)\n(b) (i) Suppose \\(n(C \\cap G) = x\\)\n\\(22 = 8 + 4 + 8 - 4 - x + x + 7 - x\\)\n\\(22 = 23 - x \\implies x = 1\\)\nOnly 1 person does Chemistry and Government.\n(ii) Number taking only Government = \\(8 - (4 + x)\\)\n\\(8 - (4 + 1) = 3\\)\n\n\n', '', 'Explanation\n(a)\n(i) Nobody takes Economics and Chemistry \\(n(E \\cap C) = \\null \\).\n(ii) 4 pupils take Economics and Government; \\(n(E \\cap G) = 4\\)\n(b) (i) Suppose \\(n(C \\cap G) = x\\)\n\\(22 = 8 + 4 + 8 - 4 - x + x + 7 - x\\)\n\\(22 = 23 - x \\implies x = 1\\)\nOnly 1 person does Chemistry and Government.\n(ii) Number taking only Government = \\(8 - (4 + x)\\)\n\\(8 - (4 + 1) = 3\\)\n\n\n', '(a)\n(i) Nobody takes Economics and Chemistry \\(n(E \\cap C) = \\null \\).\n(ii) 4 pupils take Economics and Government; \\(n(E \\cap G) = 4\\)~(b) (i) Suppose \\(n(C \\cap G) = x\\)\n\\(22 = 8 + 4 + 8 - 4 - x + x + 7 - x\\)\n\\(22 = 23 - x \\implies x = 1\\)\nOnly 1 person does Chemistry and Government.\n(ii) Number taking only Government = \\(8 - (4 + x)\\)\n\\(8 - (4 + 1) = 3\\)\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1);
INSERT INTO `exam_simulation2` (`id`, `examYear`, `questions`, `parent_que`, `child_que`, `gen_child_que_desc`, `imageDir`, `explanation`, `parent_que_expl`, `child_que_exp`, `gen_child_que_expl`, `package_id`, `subject_id`, `status`) VALUES
(663, '2017', '\n\nQuestion\n\nA man has 9 identical balls in a bag. Out of these, 3 are black, 2 are blue and the remaining are red.\n(a) If a ball is drawn at random, what is the probability that it is (i) not blue? (ii) not red?\n(b) If 2 balls are drawn at random, one after the other, what is the probability that both of them will be (i) black, if there is no replacement? (ii) blue, if there is a replacement?\n\n', '', 'A man has 9 identical balls in a bag. Out of these, 3 are black, 2 are blue and the remaining are red.\n(a) If a ball is drawn at random, what is the probability that it is (i) not blue? (ii) not red?\n(b) If 2 balls are drawn at random, one after the other, what is the probability that both of them will be (i) black, if there is no replacement? (ii) blue, if there is a replacement?', 'A man has 9 identical balls in a bag. Out of these, 3 are black, 2 are blue and the remaining are red.\n(a) If a ball is drawn at random, what is the probability that it is (i) not blue? (ii) not red?~(b) If 2 balls are drawn at random, one after the other, what is the probability that both of them will be (i) black, if there is no replacement? (ii) blue, if there is a replacement?', NULL, '\n\nExplanation\nTotal number of balls = 9, No of blue = 2, No of black = 3\n\\(\\therefore\\) No of red = 4.\n(a)(i) P(not blue) = \\( 1 - P(blue)\\)\n= \\(1 - \\frac{2}{9} = \\frac{7}{9}\\)\n(ii) P(not red) = \\(1 - P(red)\\)\n= \\(1 - \\frac{4}{9} = \\frac{5}{9}\\)\n(b) Picking two balls;\n(i) P(black) with no replacement = \\(\\frac{3}{9} \\times \\frac{2}{8} = \\frac{1}{12}\\)\n(ii) P(blue) with replacement = \\(\\frac{2}{9} \\times \\frac{2}{9} = \\frac{4}{81}\\)\n\n', '', 'Explanation\nTotal number of balls = 9, No of blue = 2, No of black = 3\n\\(\\therefore\\) No of red = 4.\n(a)(i) P(not blue) = \\( 1 - P(blue)\\)\n= \\(1 - \\frac{2}{9} = \\frac{7}{9}\\)\n(ii) P(not red) = \\(1 - P(red)\\)\n= \\(1 - \\frac{4}{9} = \\frac{5}{9}\\)\n(b) Picking two balls;\n(i) P(black) with no replacement = \\(\\frac{3}{9} \\times \\frac{2}{8} = \\frac{1}{12}\\)\n(ii) P(blue) with replacement = \\(\\frac{2}{9} \\times \\frac{2}{9} = \\frac{4}{81}\\)\n\n', '(a)(i) P(not blue) = \\( 1 - P(blue)\\)\n= \\(1 - \\frac{2}{9} = \\frac{7}{9}\\)\n(ii) P(not red) = \\(1 - P(red)\\)\n= \\(1 - \\frac{4}{9} = \\frac{5}{9}\\)~(b) Picking two balls;\n(i) P(black) with no replacement = \\(\\frac{3}{9} \\times \\frac{2}{8} = \\frac{1}{12}\\)\n(ii) P(blue) with replacement = \\(\\frac{2}{9} \\times \\frac{2}{9} = \\frac{4}{81}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(664, '2017', '\n\nQuestion\n\nABCDE is a regular pentagon and a rectangle AXYE is drawn on the side AE such that the vertices X and Y lie on the sides BC and CD respectively. Calculate the size of\n(i) an interior angle of the pentagon ;\n(ii) < BXA.\n\n', '', 'ABCDE is a regular pentagon and a rectangle AXYE is drawn on the side AE such that the vertices X and Y lie on the sides BC and CD respectively. Calculate the size of\n(i) an interior angle of the pentagon ;\n(ii) < BXA.', '', NULL, '\n\nExplanation\n(i) Sum of interior angles of a regular polygon = \\((2n - 4) \\times 90°\\)\nFor a pentagon, \\((2(5) - 4) \\times 90° = 6 \\times 90° = 540°\\)\nInterior angle = \\(\\frac{540°}{5} = 108°\\)\n(ii) < BXA = 108° - 90° = 18°\n< ABX = 108° (as shown above)\n\\(\\therefore < BXA = 180° - (108° + 18°) = 54°\\)\n\n', '', 'Explanation\n(i) Sum of interior angles of a regular polygon = \\((2n - 4) \\times 90°\\)\nFor a pentagon, \\((2(5) - 4) \\times 90° = 6 \\times 90° = 540°\\)\nInterior angle = \\(\\frac{540°}{5} = 108°\\)\n(ii) < BXA = 108° - 90° = 18°\n< ABX = 108° (as shown above)\n\\(\\therefore < BXA = 180° - (108° + 18°) = 54°\\)\n\n', '', 'WAEC-PK-1', 'WAEC-MATH', 1),
(665, '2017', '\n\nQuestion\n\n(a) Solve the equation, correct to two decimal places \\(2x^{2} + 7x - 11 = 0\\)\n(b) Using the substitution \\(P = \\frac{1}{x}; Q = \\frac{1}{y}\\), solve the simultaneous equations : \\(\\frac{2}{x} + \\frac{1}{y} = 3 ; \\frac{1}{x} - \\frac{5}{y} = 7\\)\n\n', '', '(a) Solve the equation, correct to two decimal places \\(2x^{2} + 7x - 11 = 0\\)\n(b) Using the substitution \\(P = \\frac{1}{x}; Q = \\frac{1}{y}\\), solve the simultaneous equations : \\(\\frac{2}{x} + \\frac{1}{y} = 3 ; \\frac{1}{x} - \\frac{5}{y} = 7\\)', '(a) Solve the equation, correct to two decimal places \\(2x^{2} + 7x - 11 = 0\\)~(b) Using the substitution \\(P = \\frac{1}{x}; Q = \\frac{1}{y}\\), solve the simultaneous equations : \\(\\frac{2}{x} + \\frac{1}{y} = 3 ; \\frac{1}{x} - \\frac{5}{y} = 7\\)', NULL, '\n\nExplanation\n(a) \\(2x^{2} + 7x - 11 = 0\\)\nUsing the quadratic formula,\n\\(a = 2, b = 7, c = -11\\)\n\\(x = \\frac{-b \\pm \\sqrt{b^{2} - 4ac}}{2a}\\)\n\\(x = \\frac{-7 \\pm \\sqrt{7^{2} - 4(2)(-11)}}{2(2)}\\)\n\\(x = \\frac{-7 \\pm \\sqrt{49 + 88}}{4}\\)\n\\(x = \\frac{-7 \\pm 11.705}{4}\\)\n\\(x = \\frac{-7 + 11.705}{4} ; \\frac{-7 - 11.705}{4}\\)\n\\(x = \\frac{4.705}{4} = 1.17625\\) or \\(x = \\frac{-18.705}{4} = -4.67625\\)\n\\(x \\approxeq 1.18\\) or \\(x \\approxeq -4.68\\)\n(b) \\(\\frac{2}{x} + \\frac{1}{y} = 3\\)\n\\(\\frac{1}{x} - \\frac{5}{y} = 7\\)\nSubstituting using \\(P = \\frac{1}{x}; Q = \\frac{1}{y}\\), we have\n\\(2P + Q = 3 ... (i)\\)\n\\(P - 5Q = 7 ... (ii)\\)\nMultiply (ii) by 2, we have \\(2P - 10Q = 14 ... (iii)\\)\n(iii) - (i) : \\(-10Q - Q = 14 - 3 \\implies -11Q = 11\\)\n\\(Q = -1\\)\nSubstitute Q = -1 in (i),\n\\(2P - 1 = 3 \\implies 2P = 3 + 1 = 4\\)\n\\(\\implies P = 2\\)\nP = 2; Q = -1\nBut \\(P = \\frac{1}{x} \\implies 2 = \\frac{1}{x}\\)\n\\(\\therefore x = \\frac{1}{2}\\)\nAlso, \\(Q = \\frac{1}{y} \\implies -1 = \\frac{1}{y}\\)\n\\(y = -1\\)\n\\(\\therefore x = \\frac{1}{2} ; y = -1\\)\n\n', '', 'Explanation\n(a) \\(2x^{2} + 7x - 11 = 0\\)\nUsing the quadratic formula,\n\\(a = 2, b = 7, c = -11\\)\n\\(x = \\frac{-b \\pm \\sqrt{b^{2} - 4ac}}{2a}\\)\n\\(x = \\frac{-7 \\pm \\sqrt{7^{2} - 4(2)(-11)}}{2(2)}\\)\n\\(x = \\frac{-7 \\pm \\sqrt{49 + 88}}{4}\\)\n\\(x = \\frac{-7 \\pm 11.705}{4}\\)\n\\(x = \\frac{-7 + 11.705}{4} ; \\frac{-7 - 11.705}{4}\\)\n\\(x = \\frac{4.705}{4} = 1.17625\\) or \\(x = \\frac{-18.705}{4} = -4.67625\\)\n\\(x \\approxeq 1.18\\) or \\(x \\approxeq -4.68\\)\n(b) \\(\\frac{2}{x} + \\frac{1}{y} = 3\\)\n\\(\\frac{1}{x} - \\frac{5}{y} = 7\\)\nSubstituting using \\(P = \\frac{1}{x}; Q = \\frac{1}{y}\\), we have\n\\(2P + Q = 3 ... (i)\\)\n\\(P - 5Q = 7 ... (ii)\\)\nMultiply (ii) by 2, we have \\(2P - 10Q = 14 ... (iii)\\)\n(iii) - (i) : \\(-10Q - Q = 14 - 3 \\implies -11Q = 11\\)\n\\(Q = -1\\)\nSubstitute Q = -1 in (i),\n\\(2P - 1 = 3 \\implies 2P = 3 + 1 = 4\\)\n\\(\\implies P = 2\\)\nP = 2; Q = -1\nBut \\(P = \\frac{1}{x} \\implies 2 = \\frac{1}{x}\\)\n\\(\\therefore x = \\frac{1}{2}\\)\nAlso, \\(Q = \\frac{1}{y} \\implies -1 = \\frac{1}{y}\\)\n\\(y = -1\\)\n\\(\\therefore x = \\frac{1}{2} ; y = -1\\)\n\n', '(a) \\(2x^{2} + 7x - 11 = 0\\)\nUsing the quadratic formula,\n\\(a = 2, b = 7, c = -11\\)\n\\(x = \\frac{-b \\pm \\sqrt{b^{2} - 4ac}}{2a}\\)\n\\(x = \\frac{-7 \\pm \\sqrt{7^{2} - 4(2)(-11)}}{2(2)}\\)\n\\(x = \\frac{-7 \\pm \\sqrt{49 + 88}}{4}\\)\n\\(x = \\frac{-7 \\pm 11.705}{4}\\)\n\\(x = \\frac{-7 + 11.705}{4} ; \\frac{-7 - 11.705}{4}\\)\n\\(x = \\frac{4.705}{4} = 1.17625\\) or \\(x = \\frac{-18.705}{4} = -4.67625\\)\n\\(x \\approxeq 1.18\\) or \\(x \\approxeq -4.68\\)~(b) \\(\\frac{2}{x} + \\frac{1}{y} = 3\\)\n\\(\\frac{1}{x} - \\frac{5}{y} = 7\\)\nSubstituting using \\(P = \\frac{1}{x}; Q = \\frac{1}{y}\\), we have\n\\(2P + Q = 3 ... (i)\\)\n\\(P - 5Q = 7 ... (ii)\\)\nMultiply (ii) by 2, we have \\(2P - 10Q = 14 ... (iii)\\)\n(iii) - (i) : \\(-10Q - Q = 14 - 3 \\implies -11Q = 11\\)\n\\(Q = -1\\)\nSubstitute Q = -1 in (i),\n\\(2P - 1 = 3 \\implies 2P = 3 + 1 = 4\\)\n\\(\\implies P = 2\\)\nP = 2; Q = -1\nBut \\(P = \\frac{1}{x} \\implies 2 = \\frac{1}{x}\\)\n\\(\\therefore x = \\frac{1}{2}\\)\nAlso, \\(Q = \\frac{1}{y} \\implies -1 = \\frac{1}{y}\\)\n\\(y = -1\\)\n\\(\\therefore x = \\frac{1}{2} ; y = -1\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(666, '2017', '\n\nQuestion\n\nA man bought 5 reams of duplicating paper, each of which are supposed to contain 480 sheets. The actual number of sheets in the packets were : 435, 420, 405, 415 and 440.\n(a) Calculate, correct to the nearest whole number, the percentage error for the packets of paper;\n(b) If the agreed price for a full ream was N35.00, find, correct to the nearest naira, the amount by which the buyer was cheated.\n\n', '', 'A man bought 5 reams of duplicating paper, each of which are supposed to contain 480 sheets. The actual number of sheets in the packets were : 435, 420, 405, 415 and 440.\n(a) Calculate, correct to the nearest whole number, the percentage error for the packets of paper;\n(b) If the agreed price for a full ream was N35.00, find, correct to the nearest naira, the amount by which the buyer was cheated.', 'A man bought 5 reams of duplicating paper, each of which are supposed to contain 480 sheets. The actual number of sheets in the packets were : 435, 420, 405, 415 and 440.\n(a) Calculate, correct to the nearest whole number, the percentage error for the packets of paper;~(b) If the agreed price for a full ream was N35.00, find, correct to the nearest naira, the amount by which the buyer was cheated.', NULL, '\n\nExplanation\n(a) Sheets expected from 5 packets = \\(480 \\times 5 = 2400\\)\nActual sheet contained in the packets = \\(435 + 420 + 405 + 415 + 440 = 2115\\)\nError in supply = 2400 - 2115 = 285 sheets.\n\\(\\therefore\\) Percentage error = \\(\\frac{285}{2400} \\times 100% = 11.875%\\)\n\\(\\approxeq 12%\\)\n(b) If 1 packet = 480 sheets cost N35.00\n\\(\\therefore \\text{2400 sheets cost } \\frac{2400}{480} \\times N35.00 = N175.00\\)\nAmount of buyer was cheated = \\(\\frac{Error \\times N175}{\\text{Expected no of sheets}}\\)\n\\(\\frac{N285 \\times 175}{2400} = N20.78\\)\nThe buyer was cheated by N21 to the nearest naira.\n\n', '', 'Explanation\n(a) Sheets expected from 5 packets = \\(480 \\times 5 = 2400\\)\nActual sheet contained in the packets = \\(435 + 420 + 405 + 415 + 440 = 2115\\)\nError in supply = 2400 - 2115 = 285 sheets.\n\\(\\therefore\\) Percentage error = \\(\\frac{285}{2400} \\times 100% = 11.875%\\)\n\\(\\approxeq 12%\\)\n(b) If 1 packet = 480 sheets cost N35.00\n\\(\\therefore \\text{2400 sheets cost } \\frac{2400}{480} \\times N35.00 = N175.00\\)\nAmount of buyer was cheated = \\(\\frac{Error \\times N175}{\\text{Expected no of sheets}}\\)\n\\(\\frac{N285 \\times 175}{2400} = N20.78\\)\nThe buyer was cheated by N21 to the nearest naira.\n\n', '(a) Sheets expected from 5 packets = \\(480 \\times 5 = 2400\\)\nActual sheet contained in the packets = \\(435 + 420 + 405 + 415 + 440 = 2115\\)\nError in supply = 2400 - 2115 = 285 sheets.\n\\(\\therefore\\) Percentage error = \\(\\frac{285}{2400} \\times 100% = 11.875%\\)\n\\(\\approxeq 12%\\)~(b) If 1 packet = 480 sheets cost N35.00\n\\(\\therefore \\text{2400 sheets cost } \\frac{2400}{480} \\times N35.00 = N175.00\\)\nAmount of buyer was cheated = \\(\\frac{Error \\times N175}{\\text{Expected no of sheets}}\\)\n\\(\\frac{N285 \\times 175}{2400} = N20.78\\)\nThe buyer was cheated by N21 to the nearest naira.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(667, '2017', '\n\nQuestion\n\nUsing a scale of 2cm to 1 unit on the x- axis and 1cm to 1 unit on the y- axis, draw on the same axes the graphs of \\(y = 3 + 2x - x^{2}; y = 2x - 3\\) for \\(-3 \\leq x \\leq 4\\). Using your graph:\n(i) solve the equation \\(6 - x^{2} = 0\\);\n(ii) find the maximum value of \\(3 + 2x - x^{2}\\);\n(iii) find the range of x for which \\(3 + 2x - x^{2} \\leq 1\\), expressing all your answers correct to one decimal place.\n\n', '', 'Using a scale of 2cm to 1 unit on the x- axis and 1cm to 1 unit on the y- axis, draw on the same axes the graphs of \\(y = 3 + 2x - x^{2}; y = 2x - 3\\) for \\(-3 \\leq x \\leq 4\\). Using your graph:\n(i) solve the equation \\(6 - x^{2} = 0\\);\n(ii) find the maximum value of \\(3 + 2x - x^{2}\\);\n(iii) find the range of x for which \\(3 + 2x - x^{2} \\leq 1\\), expressing all your answers correct to one decimal place.', '', NULL, '\n\nExplanation\n\\(y = 3 + 2x - x^{2}\\) and \\(y = 2x - 3\\)\nTable of values for the equation for \\(-3 \\leq x \\leq 4\\)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\n\\(y = 3 + 2x - x^{2}\\)\n-12\n-5\n0\n3\n4\n3\n0\n5\n\n\n\\(y = 2x - 3\\)\n-9\n-7\n-5\n-3\n-1\n1\n3\n5\n\n\n\n\n(i) \\(6 - x^{2} = 0\\)\n\\(6 - 3 + 2x - x^{2} = 2x - 3\\)\n\\(3 + 2x - x^{2} = 2x - 3\\)\n\\(\\therefore y = 2x - 3\\)\nRead the point where the two equations intersect on the graph.\nx = -2.6 and x = 2.5.\n(ii) Maximum value of \\(3 + 2x - x^{2}\\) is at y = 4.\n(iii) Range for which \\(3 + 2x - x^{2} \\leq 1\\) is represented by the shaded portion in the graph.\n\n', '', 'Explanation\n\\(y = 3 + 2x - x^{2}\\) and \\(y = 2x - 3\\)\nTable of values for the equation for \\(-3 \\leq x \\leq 4\\)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\n\\(y = 3 + 2x - x^{2}\\)\n-12\n-5\n0\n3\n4\n3\n0\n5\n\n\n\\(y = 2x - 3\\)\n-9\n-7\n-5\n-3\n-1\n1\n3\n5\n\n\n\n\n(i) \\(6 - x^{2} = 0\\)\n\\(6 - 3 + 2x - x^{2} = 2x - 3\\)\n\\(3 + 2x - x^{2} = 2x - 3\\)\n\\(\\therefore y = 2x - 3\\)\nRead the point where the two equations intersect on the graph.\nx = -2.6 and x = 2.5.\n(ii) Maximum value of \\(3 + 2x - x^{2}\\) is at y = 4.\n(iii) Range for which \\(3 + 2x - x^{2} \\leq 1\\) is represented by the shaded portion in the graph.\n\n', '', 'WAEC-PK-1', 'WAEC-MATH', 1),
(668, '2017', '\n\nQuestion\n\n(a) Prove that the angle which an arc of a circle subtends at the centre is twice that which it subtends at any point on the remaining part of the circumference.\n(b)\nIn the diagram, O is the centre of the circle, < OQR = 32° and < MPQ = 15°. Calculate (i) <QPR ; (ii) < MQO.\n\n', '', '(a) Prove that the angle which an arc of a circle subtends at the centre is twice that which it subtends at any point on the remaining part of the circumference.\n(b)\nIn the diagram, O is the centre of the circle, < OQR = 32° and < MPQ = 15°. Calculate (i) <QPR ; (ii) < MQO.', '(a) Prove that the angle which an arc of a circle subtends at the centre is twice that which it subtends at any point on the remaining part of the circumference.~(b)\nIn the diagram, O is the centre of the circle, < OQR = 32° and < MPQ = 15°. Calculate (i) <QPR ; (ii) < MQO.', NULL, '\n\nExplanation\n(a)\nGiven: Circle ABC with centre O\nTo prove : < AOB = 2 < ACB\nConstruction: Join CO produced to P.\nProof: With lettering as in the figure,\nOA = OB (radii) ; \\(x_{1} = x_{2}\\) (base angles of an isosceles triangle)\n\\(\\therefore < AOP = x_{1} + x_{2}\\) (exterior angle of triangle AOC)\n\\(\\therefore < AOP = 2x_{2} (x_{1} = x_{2})\\)\nAlso, \\(< BOP = 2y_{2}\\) (similar proof as the earlier done ones)\n\\(\\therefore < AOB = 2x_{2} + 2y_{2} = 2(x_{2} + y_{2})\\)\n\\(< AOB = 2 \\times < ABC\\) (proven)\n(b) From the diagram, < OQR = 32° and < MPQ = 15°.\n(i) \\(\\therefore < ORQ = 32°\\) (base angles of an isosceles triangle)\n\\(< OQR = 180° - (32° + 32°) = 180° - 64° = 116°\\)\n\\(\\therefore < QPR = \\frac{116°}{2} = 58°\\) (angle subtended at the centre)\n(ii) Join MO and MQ,\nSince < MPQ = 15°\nthen < MOQ = 2 < MPQ (angle at the centre)\n\\(\\therefore < MOQ = 2 \\times 15° = 30°\\)\n\\(\\therefore < OMQ = < MQO \\) (base angles of an isosceles triangle)\n\\(\\therefore < MQO = \\frac{180° - 30°}{2} = 75°\\)\n\n\n', '', 'Explanation\n(a)\nGiven: Circle ABC with centre O\nTo prove : < AOB = 2 < ACB\nConstruction: Join CO produced to P.\nProof: With lettering as in the figure,\nOA = OB (radii) ; \\(x_{1} = x_{2}\\) (base angles of an isosceles triangle)\n\\(\\therefore < AOP = x_{1} + x_{2}\\) (exterior angle of triangle AOC)\n\\(\\therefore < AOP = 2x_{2} (x_{1} = x_{2})\\)\nAlso, \\(< BOP = 2y_{2}\\) (similar proof as the earlier done ones)\n\\(\\therefore < AOB = 2x_{2} + 2y_{2} = 2(x_{2} + y_{2})\\)\n\\(< AOB = 2 \\times < ABC\\) (proven)\n(b) From the diagram, < OQR = 32° and < MPQ = 15°.\n(i) \\(\\therefore < ORQ = 32°\\) (base angles of an isosceles triangle)\n\\(< OQR = 180° - (32° + 32°) = 180° - 64° = 116°\\)\n\\(\\therefore < QPR = \\frac{116°}{2} = 58°\\) (angle subtended at the centre)\n(ii) Join MO and MQ,\nSince < MPQ = 15°\nthen < MOQ = 2 < MPQ (angle at the centre)\n\\(\\therefore < MOQ = 2 \\times 15° = 30°\\)\n\\(\\therefore < OMQ = < MQO \\) (base angles of an isosceles triangle)\n\\(\\therefore < MQO = \\frac{180° - 30°}{2} = 75°\\)\n\n\n', '(a)\nGiven: Circle ABC with centre O\nTo prove : < AOB = 2 < ACB\nConstruction: Join CO produced to P.\nProof: With lettering as in the figure,\nOA = OB (radii) ; \\(x_{1} = x_{2}\\) (base angles of an isosceles triangle)\n\\(\\therefore < AOP = x_{1} + x_{2}\\) (exterior angle of triangle AOC)\n\\(\\therefore < AOP = 2x_{2} (x_{1} = x_{2})\\)\nAlso, \\(< BOP = 2y_{2}\\) (similar proof as the earlier done ones)\n\\(\\therefore < AOB = 2x_{2} + 2y_{2} = 2(x_{2} + y_{2})\\)\n\\(< AOB = 2 \\times < ABC\\) (proven)~(b) From the diagram, < OQR = 32° and < MPQ = 15°.\n(i) \\(\\therefore < ORQ = 32°\\) (base angles of an isosceles triangle)\n\\(< OQR = 180° - (32° + 32°) = 180° - 64° = 116°\\)\n\\(\\therefore < QPR = \\frac{116°}{2} = 58°\\) (angle subtended at the centre)\n(ii) Join MO and MQ,\nSince < MPQ = 15°\nthen < MOQ = 2 < MPQ (angle at the centre)\n\\(\\therefore < MOQ = 2 \\times 15° = 30°\\)\n\\(\\therefore < OMQ = < MQO \\) (base angles of an isosceles triangle)\n\\(\\therefore < MQO = \\frac{180° - 30°}{2} = 75°\\)\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(669, '2017', '\n\nQuestion\n\nThe table below shows the distribution of the waiting times for some customers in a certain petrol station.\n\n\n\nWaiting time (in mins)\nNo of customers\n\n\n1.5 - 1.9\n3\n\n\n2.0 - 2.4\n10\n\n\n2.5 - 2.9\n18\n\n\n3.0 - 3.4\n10\n\n\n3.5 - 3.9\n7\n\n\n4.0 - 4.4\n2\n\n\n\n(a) Write down the class boundaries of the distribution.\n(b) Construct a cumulative frequency curve for the data;\n(c) Using your graph, estimate: (i) the interquartile range of the distribution ; (ii) the proportion of customers who could have waited for more than 3 minutes.\n\n\n\nThe table below shows the distribution of the waiting times for some customers in a certain petrol station.\n\n\n\nWaiting time (in mins)\nNo of customers\n\n\n1.5 - 1.9\n3\n\n\n2.0 - 2.4\n10\n\n\n2.5 - 2.9\n18\n\n\n3.0 - 3.4\n10\n\n\n3.5 - 3.9\n7\n\n\n4.0 - 4.4\n2\n\n\n', '', 'The table below shows the distribution of the waiting times for some customers in a certain petrol station.\n\n\n\nWaiting time (in mins)\nNo of customers\n\n\n1.5 - 1.9\n3\n\n\n2.0 - 2.4\n10\n\n\n2.5 - 2.9\n18\n\n\n3.0 - 3.4\n10\n\n\n3.5 - 3.9\n7\n\n\n4.0 - 4.4\n2\n\n\n\n(a) Write down the class boundaries of the distribution.\n(b) Construct a cumulative frequency curve for the data;\n(c) Using your graph, estimate: (i) the interquartile range of the distribution ; (ii) the proportion of customers who could have waited for more than 3 minutes.\n\n\n\nThe table below shows the distribution of the waiting times for some customers in a certain petrol station.\n\n\n\nWaiting time (in mins)\nNo of customers\n\n\n1.5 - 1.9\n3\n\n\n2.0 - 2.4\n10\n\n\n2.5 - 2.9\n18\n\n\n3.0 - 3.4\n10\n\n\n3.5 - 3.9\n7\n\n\n4.0 - 4.4\n2\n', 'The table below shows the distribution of the waiting times for some customers in a certain petrol station.\n\n\n\nWaiting time (in mins)\nNo of customers\n\n\n1.5 - 1.9\n3\n\n\n2.0 - 2.4\n10\n\n\n2.5 - 2.9\n18\n\n\n3.0 - 3.4\n10\n\n\n3.5 - 3.9\n7\n\n\n4.0 - 4.4\n2\n\n\n\n(a) Write down the class boundaries of the distribution.~(b) Construct a cumulative frequency curve for the data;\n(c) Using your graph, estimate: (i) the interquartile range of the distribution ; (ii) the proportion of customers who could have waited for more than 3 minutes.\n\n\n\nThe table below shows the distribution of the waiting times for some customers in a certain petrol station.\n\n\n\nWaiting time (in mins)\nNo of customers\n\n\n1.5 - 1.9\n3\n\n\n2.0 - 2.4\n10\n\n\n2.5 - 2.9\n18\n\n\n3.0 - 3.4\n10\n\n\n3.5 - 3.9\n7\n\n\n4.0 - 4.4\n2\n~(c) Using your graph, estimate: (i) the interquartile range of the distribution ; (ii) the proportion of customers who could have waited for more than 3 minutes.\n\n\n\nThe table below shows the distribution of the waiting times for some customers in a certain petrol station.\n\n\n\nWaiting time (in mins)\nNo of customers\n\n\n1.5 - 1.9\n3\n\n\n2.0 - 2.4\n10\n\n\n2.5 - 2.9\n18\n\n\n3.0 - 3.4\n10\n\n\n3.5 - 3.9\n7\n\n\n4.0 - 4.4\n2\n', NULL, '\n\nExplanation\nWaiting time (in mins)\nClass boundaries\nNo of customers\nCum Freq\n\n\n1.5 - 1.9\n1.45 - 1.95\n3\n3\n\n\n2.0 - 2.4\n1.95 - 2.45\n10\n13\n\n\n2.5 - 2.9\n2.45 - 2.95\n18\n31\n\n\n3.0 - 3.4\n2.95 - 3.45\n10\n41\n\n\n3.5 - 3.9\n3.45 - 3.95\n7\n48\n\n\n4.0 - 4.4\n3.95 - 4.45\n2\n50\n\n\n\n(b)\n(c) Interquartile range of the distribution = \\(Q_{3} - Q_{1}\\)\n\\(Q_{3} = \\frac{3 \\times 51}{4} = 38.25\\)\n\\(Q_{1} = \\frac{1 \\times 51}{4} = 12.75\\)\nFrom the graph, \\(Q_{3} = 3.25 ; Q_{1} = 2.45\\)\n\\(\\therefore \\text{The interquartile range of the distribution } = 3.25 - 2.45 = 0.80\\)\n(ii) Proportion of customers who must have waited for more than 3 minutes = \\(\\frac{16}{50} = \\frac{8}{25}\\)\n\n', '', 'Explanation\nWaiting time (in mins)\nClass boundaries\nNo of customers\nCum Freq\n\n\n1.5 - 1.9\n1.45 - 1.95\n3\n3\n\n\n2.0 - 2.4\n1.95 - 2.45\n10\n13\n\n\n2.5 - 2.9\n2.45 - 2.95\n18\n31\n\n\n3.0 - 3.4\n2.95 - 3.45\n10\n41\n\n\n3.5 - 3.9\n3.45 - 3.95\n7\n48\n\n\n4.0 - 4.4\n3.95 - 4.45\n2\n50\n\n\n\n(b)\n(c) Interquartile range of the distribution = \\(Q_{3} - Q_{1}\\)\n\\(Q_{3} = \\frac{3 \\times 51}{4} = 38.25\\)\n\\(Q_{1} = \\frac{1 \\times 51}{4} = 12.75\\)\nFrom the graph, \\(Q_{3} = 3.25 ; Q_{1} = 2.45\\)\n\\(\\therefore \\text{The interquartile range of the distribution } = 3.25 - 2.45 = 0.80\\)\n(ii) Proportion of customers who must have waited for more than 3 minutes = \\(\\frac{16}{50} = \\frac{8}{25}\\)\n\n', '~(b)\n(c) Interquartile range of the distribution = \\(Q_{3} - Q_{1}\\)\n\\(Q_{3} = \\frac{3 \\times 51}{4} = 38.25\\)\n\\(Q_{1} = \\frac{1 \\times 51}{4} = 12.75\\)\nFrom the graph, \\(Q_{3} = 3.25 ; Q_{1} = 2.45\\)\n\\(\\therefore \\text{The interquartile range of the distribution } = 3.25 - 2.45 = 0.80\\)\n(ii) Proportion of customers who must have waited for more than 3 minutes = \\(\\frac{16}{50} = \\frac{8}{25}\\)\n~(c) Interquartile range of the distribution = \\(Q_{3} - Q_{1}\\)\n\\(Q_{3} = \\frac{3 \\times 51}{4} = 38.25\\)\n\\(Q_{1} = \\frac{1 \\times 51}{4} = 12.75\\)\nFrom the graph, \\(Q_{3} = 3.25 ; Q_{1} = 2.45\\)\n\\(\\therefore \\text{The interquartile range of the distribution } = 3.25 - 2.45 = 0.80\\)\n(ii) Proportion of customers who must have waited for more than 3 minutes = \\(\\frac{16}{50} = \\frac{8}{25}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(670, '2017', '\n\nQuestion\n\n(a) Using a ruler and a pair of compasses only, construct a parallelogram PQRS with diagonals |PR| = 9cm and |QS| = 6cm, intersecting at K and < QKR = 60°.\n(b) Construct a rectangle PABS which is equal in area to PQRS in (a) above and on the same side of PS as PQRS. Measure |PA|.\n\n', '', '(a) Using a ruler and a pair of compasses only, construct a parallelogram PQRS with diagonals |PR| = 9cm and |QS| = 6cm, intersecting at K and < QKR = 60°.\n(b) Construct a rectangle PABS which is equal in area to PQRS in (a) above and on the same side of PS as PQRS. Measure |PA|.', '(a) Using a ruler and a pair of compasses only, construct a parallelogram PQRS with diagonals |PR| = 9cm and |QS| = 6cm, intersecting at K and < QKR = 60°.~(b) Construct a rectangle PABS which is equal in area to PQRS in (a) above and on the same side of PS as PQRS. Measure |PA|.', NULL, '\n\nExplanation\n(a) & (b)\n\n\n', '(a) & (b)\n\n\n', '', '', 'WAEC-PK-1', 'WAEC-MATH', 1),
(671, '2017', '\n\nQuestion\n\nFrom a horizontal distance of 8.5 km, a pilot observes that the angles of depression of the top and the base of a control tower are 30°and 33° respectively. Calculate, correct to three significant figures :\n(a) the shortest distance between the pilot and the base of the control tower;\n(b) the height of the control tower.\n\n', '', 'From a horizontal distance of 8.5 km, a pilot observes that the angles of depression of the top and the base of a control tower are 30°and 33° respectively. Calculate, correct to three significant figures :\n(a) the shortest distance between the pilot and the base of the control tower;\n(b) the height of the control tower.', 'From a horizontal distance of 8.5 km, a pilot observes that the angles of depression of the top and the base of a control tower are 30°and 33° respectively. Calculate, correct to three significant figures :\n(a) the shortest distance between the pilot and the base of the control tower;~(b) the height of the control tower.', NULL, '\n\nExplanation\nLet T and B be the top and base of the control tower respectively; P, the point of pilot\'s observation and PS, the line of observation.\n(a) Shortest distance is |PB|.\n\\(\\therefore \\cos 33° = \\frac{8.5}{PB}\\)\n\\(PB = \\frac{8.5}{\\cos 33} = \\frac{8.5}{0.8387}\\)\n\\(|PB| = 10.135km \\approxeq 10.1km\\)\n(b) Height of the control tower = |SB| - |ST|\n\\(\\therefore \\tan 33° = \\frac{|SB|}{8.5 km} \\implies |SB| = 8.5 \\tan 33°\\)\n\\(|SB| = 5.52 km\\)\n\\(\\tan 30° = \\frac{|ST|}{8.5 km} \\implies |ST| = 8.5 \\tan 30°\\)\n\\(|ST| = 4.907 km\\)\n\\(\\therefore\\) The height of the tower = 5.52 km - 4.907 km = 0.613 km.\n\n', '', 'Explanation\nLet T and B be the top and base of the control tower respectively; P, the point of pilot\'s observation and PS, the line of observation.\n(a) Shortest distance is |PB|.\n\\(\\therefore \\cos 33° = \\frac{8.5}{PB}\\)\n\\(PB = \\frac{8.5}{\\cos 33} = \\frac{8.5}{0.8387}\\)\n\\(|PB| = 10.135km \\approxeq 10.1km\\)\n(b) Height of the control tower = |SB| - |ST|\n\\(\\therefore \\tan 33° = \\frac{|SB|}{8.5 km} \\implies |SB| = 8.5 \\tan 33°\\)\n\\(|SB| = 5.52 km\\)\n\\(\\tan 30° = \\frac{|ST|}{8.5 km} \\implies |ST| = 8.5 \\tan 30°\\)\n\\(|ST| = 4.907 km\\)\n\\(\\therefore\\) The height of the tower = 5.52 km - 4.907 km = 0.613 km.\n\n', '(a) Shortest distance is |PB|.\n\\(\\therefore \\cos 33° = \\frac{8.5}{PB}\\)\n\\(PB = \\frac{8.5}{\\cos 33} = \\frac{8.5}{0.8387}\\)\n\\(|PB| = 10.135km \\approxeq 10.1km\\)~(b) Height of the control tower = |SB| - |ST|\n\\(\\therefore \\tan 33° = \\frac{|SB|}{8.5 km} \\implies |SB| = 8.5 \\tan 33°\\)\n\\(|SB| = 5.52 km\\)\n\\(\\tan 30° = \\frac{|ST|}{8.5 km} \\implies |ST| = 8.5 \\tan 30°\\)\n\\(|ST| = 4.907 km\\)\n\\(\\therefore\\) The height of the tower = 5.52 km - 4.907 km = 0.613 km.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(672, '2017', '\n\nQuestion\n\n(a) Simplify \\(\\frac{0.016 \\times 0.084}{0.48}\\) [Leave your answer in standard form].\n(b) Eight wooden poles are to be used for pillars and the lengths of the poles form an Arithmetic Progression (A.P). If the second pole is 2m and the sixth is 5m, give the lengths of the poles, in order.\n\n', '', '(a) Simplify \\(\\frac{0.016 \\times 0.084}{0.48}\\) [Leave your answer in standard form].\n(b) Eight wooden poles are to be used for pillars and the lengths of the poles form an Arithmetic Progression (A.P). If the second pole is 2m and the sixth is 5m, give the lengths of the poles, in order.', '(a) Simplify \\(\\frac{0.016 \\times 0.084}{0.48}\\) [Leave your answer in standard form].~(b) Eight wooden poles are to be used for pillars and the lengths of the poles form an Arithmetic Progression (A.P). If the second pole is 2m and the sixth is 5m, give the lengths of the poles, in order.', NULL, '\n\nExplanation\n(a) \\(\\frac{0.016 \\times 0.084}{0.48}\\)\n= \\(\\frac{16 \\times 10^{-3} \\times 84 \\times 10^{-3}}{48 \\times 10^{-2}}\\)\n= \\(28 \\times 10^{-3 - 3 - (-2)}\\)\n= \\(28 \\times 10^{-4} = 2.8 \\times 10^{-3}\\)\n(b) \\(T_{n} = a + (n - 1)d\\) (Terms of an A.P)\n\\(T_{2} = a + d = 2 ... (1)\\)\n\\(T_{6} = a + 5d = 5 ... (2)\\)\n\\((2) - (1) : 4d = 3 \\implies d = \\frac{3}{4} = 0.75\\)\nPut the value of d in (1), so we have\n\\(a + 0.75 = 2 \\implies a = 1.25\\)\nTherefore, the length of the poles in metres= \\(1.25, 2, 2.75, 3.5, 4.25, 5, 5.75, 6.5\\)\n\n', '', 'Explanation\n(a) \\(\\frac{0.016 \\times 0.084}{0.48}\\)\n= \\(\\frac{16 \\times 10^{-3} \\times 84 \\times 10^{-3}}{48 \\times 10^{-2}}\\)\n= \\(28 \\times 10^{-3 - 3 - (-2)}\\)\n= \\(28 \\times 10^{-4} = 2.8 \\times 10^{-3}\\)\n(b) \\(T_{n} = a + (n - 1)d\\) (Terms of an A.P)\n\\(T_{2} = a + d = 2 ... (1)\\)\n\\(T_{6} = a + 5d = 5 ... (2)\\)\n\\((2) - (1) : 4d = 3 \\implies d = \\frac{3}{4} = 0.75\\)\nPut the value of d in (1), so we have\n\\(a + 0.75 = 2 \\implies a = 1.25\\)\nTherefore, the length of the poles in metres= \\(1.25, 2, 2.75, 3.5, 4.25, 5, 5.75, 6.5\\)\n\n', '(a) \\(\\frac{0.016 \\times 0.084}{0.48}\\)\n= \\(\\frac{16 \\times 10^{-3} \\times 84 \\times 10^{-3}}{48 \\times 10^{-2}}\\)\n= \\(28 \\times 10^{-3 - 3 - (-2)}\\)\n= \\(28 \\times 10^{-4} = 2.8 \\times 10^{-3}\\)~(b) \\(T_{n} = a + (n - 1)d\\) (Terms of an A.P)\n\\(T_{2} = a + d = 2 ... (1)\\)\n\\(T_{6} = a + 5d = 5 ... (2)\\)\n\\((2) - (1) : 4d = 3 \\implies d = \\frac{3}{4} = 0.75\\)\nPut the value of d in (1), so we have\n\\(a + 0.75 = 2 \\implies a = 1.25\\)\nTherefore, the length of the poles in metres= \\(1.25, 2, 2.75, 3.5, 4.25, 5, 5.75, 6.5\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(673, '2017', '\n\nQuestion\n\nShow on a graph, the area which gives the solution set of the inequalities: \\(y - 2x \\leq 4 ; 3y + x \\geq 6 ; y \\geq 7x - 9\\).\n\n', 'Show on a graph, the area which gives the solution set of the inequalities: \\(y - 2x \\leq 4 ; 3y + x \\geq 6 ; y \\geq 7x - 9\\).', '', '', NULL, '\n\nExplanation\n', '', '', '', 'WAEC-PK-1', 'WAEC-MATH', 1),
(674, '2017', '\n\nQuestion\n\n(a) Prove that the sum of the angles in a triangle is 2 right angles.\n(b) The side AB of a triangle ABC is produced to a point D. The bisector of ACB cuts AB at E. Prove that < CAE + < CBD = 2 < CEB.\n\n', '', '(a) Prove that the sum of the angles in a triangle is 2 right angles.\n(b) The side AB of a triangle ABC is produced to a point D. The bisector of ACB cuts AB at E. Prove that < CAE + < CBD = 2 < CEB.', '(a) Prove that the sum of the angles in a triangle is 2 right angles.~(b) The side AB of a triangle ABC is produced to a point D. The bisector of ACB cuts AB at E. Prove that < CAE + < CBD = 2 < CEB.', NULL, '\n\nExplanation\n(a) Given any triangle ABC\n\nTo prove that < A + < B + < C = 180°\nConstruction: Produce BC to X and draw CP // BA.\nProof: with the lettering \\(a_{1} =a_{2}\\) (alternate angles);\n\\(b_{1} = b_{2}\\) (corresponding angles)\n\\(< BCA + < A + < B = 180° ; < ACB + < A + < B = 180°\\)\n\\(\\therefore < A + < B + < C = 180°\\)\n(b)\nFrom the given triangle and with reference to letterings;\n\\(\\hat{a} + \\hat{c} + \\hat{c} = \\hat{b}\\) (exterior opposite angles of a triangle)\n\\(a + c = e\\) (exterior opposite angles of a triangle)\n\\(a + b = a + a + c + c \\implies a + b = 2(a + c)\\)\n= \\(2e \\text{where e = a + c}\\)\ni.e \\(< CAE + < CBD = 2 < CED\\) (Proved)\n\n', '', 'Explanation\n(a) Given any triangle ABC\n\nTo prove that < A + < B + < C = 180°\nConstruction: Produce BC to X and draw CP // BA.\nProof: with the lettering \\(a_{1} =a_{2}\\) (alternate angles);\n\\(b_{1} = b_{2}\\) (corresponding angles)\n\\(< BCA + < A + < B = 180° ; < ACB + < A + < B = 180°\\)\n\\(\\therefore < A + < B + < C = 180°\\)\n(b)\nFrom the given triangle and with reference to letterings;\n\\(\\hat{a} + \\hat{c} + \\hat{c} = \\hat{b}\\) (exterior opposite angles of a triangle)\n\\(a + c = e\\) (exterior opposite angles of a triangle)\n\\(a + b = a + a + c + c \\implies a + b = 2(a + c)\\)\n= \\(2e \\text{where e = a + c}\\)\ni.e \\(< CAE + < CBD = 2 < CED\\) (Proved)\n\n', '(a) Given any triangle ABC\n\nTo prove that < A + < B + < C = 180°\nConstruction: Produce BC to X and draw CP // BA.\nProof: with the lettering \\(a_{1} =a_{2}\\) (alternate angles);\n\\(b_{1} = b_{2}\\) (corresponding angles)\n\\(< BCA + < A + < B = 180° ; < ACB + < A + < B = 180°\\)\n\\(\\therefore < A + < B + < C = 180°\\)~(b)\nFrom the given triangle and with reference to letterings;\n\\(\\hat{a} + \\hat{c} + \\hat{c} = \\hat{b}\\) (exterior opposite angles of a triangle)\n\\(a + c = e\\) (exterior opposite angles of a triangle)\n\\(a + b = a + a + c + c \\implies a + b = 2(a + c)\\)\n= \\(2e \\text{where e = a + c}\\)\ni.e \\(< CAE + < CBD = 2 < CED\\) (Proved)\n~c)\\)\n= \\(2e \\text{where e = a + c}\\)\ni.e \\(< CAE + < CBD = 2 < CED\\) (Proved)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(675, '2017', '\n\nQuestion\n\nA carpenter was told to make a rectangular desk with top of dimension 50cm by 40cm. The carpenter actually made the desk 60cm by 35cm.\n(a) Calculate the percentage error in the (i) length and the breadth ; (ii) area of the table top.\n(b) Find the product of the two errors in a(i).\n\n', '', 'A carpenter was told to make a rectangular desk with top of dimension 50cm by 40cm. The carpenter actually made the desk 60cm by 35cm.\n(a) Calculate the percentage error in the (i) length and the breadth ; (ii) area of the table top.\n(b) Find the product of the two errors in a(i).', 'A carpenter was told to make a rectangular desk with top of dimension 50cm by 40cm. The carpenter actually made the desk 60cm by 35cm.\n(a) Calculate the percentage error in the (i) length and the breadth ; (ii) area of the table top.~(b) Find the product of the two errors in a(i).', NULL, '\n\nExplanation\nError in length = 60cm - 50cm = 10cm\nError in breadth = 40cm - 35cm = 5cm\n(i) % error in length = \\(\\frac{10}{50} \\times 100% = 20%\\)\n% error in breadth = \\(\\frac{5}{40} \\times 100% = 12.5%\\)\n(ii) Area of the actual table = \\(50cm \\times 40cm = 2000cm^{2}\\)\nArea of the table in error = \\(60cm \\times 35cm = 2100cm^{2}\\)\nError in area = \\(2100cm^{2} - 2000cm^{2} = 100cm^{2}\\)\n% error in area = \\(\\frac{100}{2000} \\times 100% = 5%\\)\n(b) Product of the error in length and in breadth = \\(20 \\times 12.5 = 250%\\)\n\n', '', 'Explanation\nError in length = 60cm - 50cm = 10cm\nError in breadth = 40cm - 35cm = 5cm\n(i) % error in length = \\(\\frac{10}{50} \\times 100% = 20%\\)\n% error in breadth = \\(\\frac{5}{40} \\times 100% = 12.5%\\)\n(ii) Area of the actual table = \\(50cm \\times 40cm = 2000cm^{2}\\)\nArea of the table in error = \\(60cm \\times 35cm = 2100cm^{2}\\)\nError in area = \\(2100cm^{2} - 2000cm^{2} = 100cm^{2}\\)\n% error in area = \\(\\frac{100}{2000} \\times 100% = 5%\\)\n(b) Product of the error in length and in breadth = \\(20 \\times 12.5 = 250%\\)\n\n', '(i) % error in length = \\(\\frac{10}{50} \\times 100% = 20%\\)\n% error in breadth = \\(\\frac{5}{40} \\times 100% = 12.5%\\)\n(ii) Area of the actual table = \\(50cm \\times 40cm = 2000cm^{2}\\)\nArea of the table in error = \\(60cm \\times 35cm = 2100cm^{2}\\)\nError in area = \\(2100cm^{2} - 2000cm^{2} = 100cm^{2}\\)\n% error in area = \\(\\frac{100}{2000} \\times 100% = 5%\\)~(b) Product of the error in length and in breadth = \\(20 \\times 12.5 = 250%\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(676, '2017', '\n\nQuestion\n\nThe feet of two vertical poles of height 3m and 7m are in line with a point P on the ground, the smaller pole being between the taller pole and P and at a distance of 20m from P. The angle of elevation of the top (T) of the taller pole from the top (R) of the smaller pole is 30°. Calculate the :\n(i) distance RT ; (ii) distance of the foot of the taller pole from P, correct to three significant figures ; (iii) angle of elevation of T from P, correct to one decimal place.\n\n', '', 'The feet of two vertical poles of height 3m and 7m are in line with a point P on the ground, the smaller pole being between the taller pole and P and at a distance of 20m from P. The angle of elevation of the top (T) of the taller pole from the top (R) of the smaller pole is 30°. Calculate the :\n(i) distance RT ; (ii) distance of the foot of the taller pole from P, correct to three significant figures ; (iii) angle of elevation of T from P, correct to one decimal place.', '', NULL, '\n\nExplanation\n(i) From the figure,\n\\(\\sin 30° = \\frac{4}{RT} \\implies RT = \\frac{4}{\\sin 30°}\\)\n= \\(\\frac{4}{0.5} = 8m\\)\n(ii) |RQ| = |XY|\n\\(\\therefore |XY|^{2} = 8^{2} - 4^{2} = 64 - 16 = 48\\)\n\\(|XY| = \\sqrt{48} = 4\\sqrt{3} m = 6.928m\\)\nBut |PY| = |PX| + |XY|\n= 20 + 6.928 = 26.928m \\(\\approxeq\\) 26.9m.\n(iii) Let the angle of elevation = \\(\\theta\\)\n\\(\\tan \\theta = \\frac{TY}{PY} = \\frac{7}{26.928}\\)\n\\(\\tan \\theta = 0.26 \\implies \\theta = \\tan^{-1} (0.26)\\)\n\\(\\theta = 14.57° \\approxeq 14.6°\\) (to 1 decimal place)\n\n', '', 'Explanation\n(i) From the figure,\n\\(\\sin 30° = \\frac{4}{RT} \\implies RT = \\frac{4}{\\sin 30°}\\)\n= \\(\\frac{4}{0.5} = 8m\\)\n(ii) |RQ| = |XY|\n\\(\\therefore |XY|^{2} = 8^{2} - 4^{2} = 64 - 16 = 48\\)\n\\(|XY| = \\sqrt{48} = 4\\sqrt{3} m = 6.928m\\)\nBut |PY| = |PX| + |XY|\n= 20 + 6.928 = 26.928m \\(\\approxeq\\) 26.9m.\n(iii) Let the angle of elevation = \\(\\theta\\)\n\\(\\tan \\theta = \\frac{TY}{PY} = \\frac{7}{26.928}\\)\n\\(\\tan \\theta = 0.26 \\implies \\theta = \\tan^{-1} (0.26)\\)\n\\(\\theta = 14.57° \\approxeq 14.6°\\) (to 1 decimal place)\n\n', '', 'WAEC-PK-1', 'WAEC-MATH', 1),
(677, '2017', '\n\nQuestion\n\n(a) Find the volume of a right solid cone of base radius 4cm and perpendicular height 6cm. [\\(\\pi = 3.142\\)]\n(b) A hemispherical tank of diameter which is 10m is filled by water issuing from a pipe of radius 20cm at 2m per second. Calculate, correct to three significant figures, the time, in minutes, it takes to fill the tank.\n\n', '', '(a) Find the volume of a right solid cone of base radius 4cm and perpendicular height 6cm. [\\(\\pi = 3.142\\)]\n(b) A hemispherical tank of diameter which is 10m is filled by water issuing from a pipe of radius 20cm at 2m per second. Calculate, correct to three significant figures, the time, in minutes, it takes to fill the tank.', '(a) Find the volume of a right solid cone of base radius 4cm and perpendicular height 6cm. [\\(\\pi = 3.142\\)]~(b) A hemispherical tank of diameter which is 10m is filled by water issuing from a pipe of radius 20cm at 2m per second. Calculate, correct to three significant figures, the time, in minutes, it takes to fill the tank.', NULL, '\n\nExplanation\n(a)\nFormula : Volume = \\(\\frac{1}{3} \\pi r^{2} h\\)\n\\(\\pi = 3.142; r = 4cm; h = 6cm\\)\n\\(Volume = \\frac{1}{3} \\times 3.142 \\times 4^{2} \\times 6 = 100.544 cm^{3}\\)\n(b)\nVolume of hemisphere = \\(\\frac{1}{2} \\times \\frac{4}{3} \\pi r^{3}\\)\n= \\(\\frac{2}{3} \\times \\frac{22}{7} \\times 5^{3}\\)\n= \\(\\frac{5500}{21} cm^{3} = 261.905cm^{3}\\)\nVolume of water discharged per second = \\(\\pi r^{2} h\\)\n= \\(\\frac{22}{7} \\times 0.2m \\times 0.2m \\times 2m = \\frac{1.76}{7}\\)\n= \\(0.2514 m^{3} / sec\\)\n\\(\\therefore \\text{The time to fill the tank = } \\frac{261.905}{0.2514 \\times 60}\\)\n= \\(\\frac{261.905}{15.084} = 17.363 mins\\)\n\\(\\approxeq 17.4 mins\\)\n\n', '', 'Explanation\n(a)\nFormula : Volume = \\(\\frac{1}{3} \\pi r^{2} h\\)\n\\(\\pi = 3.142; r = 4cm; h = 6cm\\)\n\\(Volume = \\frac{1}{3} \\times 3.142 \\times 4^{2} \\times 6 = 100.544 cm^{3}\\)\n(b)\nVolume of hemisphere = \\(\\frac{1}{2} \\times \\frac{4}{3} \\pi r^{3}\\)\n= \\(\\frac{2}{3} \\times \\frac{22}{7} \\times 5^{3}\\)\n= \\(\\frac{5500}{21} cm^{3} = 261.905cm^{3}\\)\nVolume of water discharged per second = \\(\\pi r^{2} h\\)\n= \\(\\frac{22}{7} \\times 0.2m \\times 0.2m \\times 2m = \\frac{1.76}{7}\\)\n= \\(0.2514 m^{3} / sec\\)\n\\(\\therefore \\text{The time to fill the tank = } \\frac{261.905}{0.2514 \\times 60}\\)\n= \\(\\frac{261.905}{15.084} = 17.363 mins\\)\n\\(\\approxeq 17.4 mins\\)\n\n', '(a)\nFormula : Volume = \\(\\frac{1}{3} \\pi r^{2} h\\)\n\\(\\pi = 3.142; r = 4cm; h = 6cm\\)\n\\(Volume = \\frac{1}{3} \\times 3.142 \\times 4^{2} \\times 6 = 100.544 cm^{3}\\)~(b)\nVolume of hemisphere = \\(\\frac{1}{2} \\times \\frac{4}{3} \\pi r^{3}\\)\n= \\(\\frac{2}{3} \\times \\frac{22}{7} \\times 5^{3}\\)\n= \\(\\frac{5500}{21} cm^{3} = 261.905cm^{3}\\)\nVolume of water discharged per second = \\(\\pi r^{2} h\\)\n= \\(\\frac{22}{7} \\times 0.2m \\times 0.2m \\times 2m = \\frac{1.76}{7}\\)\n= \\(0.2514 m^{3} / sec\\)\n\\(\\therefore \\text{The time to fill the tank = } \\frac{261.905}{0.2514 \\times 60}\\)\n= \\(\\frac{261.905}{15.084} = 17.363 mins\\)\n\\(\\approxeq 17.4 mins\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(678, '2018', '\n\nQuestion\n\nThe weights to the nearest kilogram, of a group of 50 students in a College of Technology are given below:\n65, 70, 60, 46, 51, 55, 59, 63, 68, 53, 47, 53, 72, 53, 67, 62, 64, 70, 57, 56, 73, 56, 48, 51, 58, 63, 65, 62, 49, 64, 53, 59, 63, 50, 48, 72, 67, 56, 61, 64, 66, 52, 49, 62, 71, 58, 53, 69, 63, 59.\n(a) Prepare a grouped fraquency table with class intervals 45 - 49, 50 - 54, 55 - 59 etc.\n(b) Using an assumed mean of 62 or otherwise, calculate the mean and standard deviation of the grouped data, correct to one decimal place.\n\n', '', 'The weights to the nearest kilogram, of a group of 50 students in a College of Technology are given below:\n65, 70, 60, 46, 51, 55, 59, 63, 68, 53, 47, 53, 72, 53, 67, 62, 64, 70, 57, 56, 73, 56, 48, 51, 58, 63, 65, 62, 49, 64, 53, 59, 63, 50, 48, 72, 67, 56, 61, 64, 66, 52, 49, 62, 71, 58, 53, 69, 63, 59.\n(a) Prepare a grouped fraquency table with class intervals 45 - 49, 50 - 54, 55 - 59 etc.\n(b) Using an assumed mean of 62 or otherwise, calculate the mean and standard deviation of the grouped data, correct to one decimal place.', 'The weights to the nearest kilogram, of a group of 50 students in a College of Technology are given below:\n65, 70, 60, 46, 51, 55, 59, 63, 68, 53, 47, 53, 72, 53, 67, 62, 64, 70, 57, 56, 73, 56, 48, 51, 58, 63, 65, 62, 49, 64, 53, 59, 63, 50, 48, 72, 67, 56, 61, 64, 66, 52, 49, 62, 71, 58, 53, 69, 63, 59.\n(a) Prepare a grouped fraquency table with class intervals 45 - 49, 50 - 54, 55 - 59 etc.~(b) Using an assumed mean of 62 or otherwise, calculate the mean and standard deviation of the grouped data, correct to one decimal place.', NULL, '\n\nExplanation\n(a)\n\n\n\nClass Interval\nTally\nFreq\n\n\n45 - 49\n|||||\n6\n\n\n50 - 54\n||||||||\n9\n\n\n55 - 59\n||||||||\n10\n\n\n60 - 64\n||||||||||\n12\n\n\n65 - 69\n||||||\n7\n\n\n70 - 74\n|||||\n6\n\n\n\n(b)\n\n\n\n\nClass\nInterval\n\n\nMid-value\n(x)\n\nx - 62\n\\((x - 62)^{2}\\)\n\\(f\\)\n\\(f(x - 62)\\)\n\\(f(x - 62)^{2}\\)\n\n\n45 - 49\n47\n-15\n225\n6\n-90\n1350\n\n\n50 - 54\n52\n-10\n100\n9\n-90\n900\n\n\n55 - 59\n57\n-5\n25\n10\n-50\n250\n\n\n60 - 64\n62\n0\n0\n12\n0\n0\n\n\n65 - 69\n67\n5\n25\n7\n35\n175\n\n\n70 - 74\n72\n10\n100\n6\n60\n600\n\n\n\\(\\sum\\)\n\n\n\n50\n-135\n3275\n\n\n\n\\(Mean (\\bar{x}) = A + \\frac{\\sum f(x - A)}{\\sum f}\\)\n= \\(62 + \\frac{-135}{50}\\)\n= \\(62 - 2.7 = 49.3\\)\nStandard deviation = \\(\\sqrt{\\frac{\\sum f(x - A)}{\\sum f}}\\)\n= \\(\\sqrt{\\frac{3275}{50}}\\)\n= \\(\\sqrt{65.5}\\)\n= \\(8.093\\approxeq 8.1\\) (to 1 decimal place)\n\n\n', '', 'Explanation\n(a)\n\n\n\nClass Interval\nTally\nFreq\n\n\n45 - 49\n|||||\n6\n\n\n50 - 54\n||||||||\n9\n\n\n55 - 59\n||||||||\n10\n\n\n60 - 64\n||||||||||\n12\n\n\n65 - 69\n||||||\n7\n\n\n70 - 74\n|||||\n6\n\n\n\n(b)\n\n\n\n\nClass\nInterval\n\n\nMid-value\n(x)\n\nx - 62\n\\((x - 62)^{2}\\)\n\\(f\\)\n\\(f(x - 62)\\)\n\\(f(x - 62)^{2}\\)\n\n\n45 - 49\n47\n-15\n225\n6\n-90\n1350\n\n\n50 - 54\n52\n-10\n100\n9\n-90\n900\n\n\n55 - 59\n57\n-5\n25\n10\n-50\n250\n\n\n60 - 64\n62\n0\n0\n12\n0\n0\n\n\n65 - 69\n67\n5\n25\n7\n35\n175\n\n\n70 - 74\n72\n10\n100\n6\n60\n600\n\n\n\\(\\sum\\)\n\n\n\n50\n-135\n3275\n\n\n\n\\(Mean (\\bar{x}) = A + \\frac{\\sum f(x - A)}{\\sum f}\\)\n= \\(62 + \\frac{-135}{50}\\)\n= \\(62 - 2.7 = 49.3\\)\nStandard deviation = \\(\\sqrt{\\frac{\\sum f(x - A)}{\\sum f}}\\)\n= \\(\\sqrt{\\frac{3275}{50}}\\)\n= \\(\\sqrt{65.5}\\)\n= \\(8.093\\approxeq 8.1\\) (to 1 decimal place)\n\n\n', '(a)\n\n\n\nClass Interval\nTally\nFreq\n\n\n45 - 49\n|||||\n6\n\n\n50 - 54\n||||||||\n9\n\n\n55 - 59\n||||||||\n10\n\n\n60 - 64\n||||||||||\n12\n\n\n65 - 69\n||||||\n7\n\n\n70 - 74\n|||||\n6\n\n\n~(b)\n\n\n\n\nClass\nInterval\n\n\nMid-value\n(x)\n\nx - 62\n\\((x - 62)^{2}\\)\n\\(f\\)\n\\(f(x - 62)\\)\n\\(f(x - 62)^{2}\\)\n\n\n45 - 49\n47\n-15\n225\n6\n-90\n1350\n\n\n50 - 54\n52\n-10\n100\n9\n-90\n900\n\n\n55 - 59\n57\n-5\n25\n10\n-50\n250\n\n\n60 - 64\n62\n0\n0\n12\n0\n0\n\n\n65 - 69\n67\n5\n25\n7\n35\n175\n\n\n70 - 74\n72\n10\n100\n6\n60\n600\n\n\n\\(\\sum\\)\n\n\n\n50\n-135\n3275\n\n\n\n\\(Mean (\\bar{x}) = A + \\frac{\\sum f(x - A)}{\\sum f}\\)\n= \\(62 + \\frac{-135}{50}\\)\n= \\(62 - 2.7 = 49.3\\)\nStandard deviation = \\(\\sqrt{\\frac{\\sum f(x - A)}{\\sum f}}\\)\n= \\(\\sqrt{\\frac{3275}{50}}\\)\n= \\(\\sqrt{65.5}\\)\n= \\(8.093\\approxeq 8.1\\) (to 1 decimal place)\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(679, '2018', '\n\nQuestion\n\n(a) ABCD is a trapezium in which AB // DC, |AB| = 8cm, < ABC = 60°, |BC| = 5.5cm and |BD| = 8.3cm. Using a ruler and a pair of compasses only, construct:\n(i) the trapezium ABCD ; (ii) a rectangle PQCD, where P, Q are two points AB;\n(b) Measure |AB| and |QB|.\n\n', '', '(a) ABCD is a trapezium in which AB // DC, |AB| = 8cm, < ABC = 60°, |BC| = 5.5cm and |BD| = 8.3cm. Using a ruler and a pair of compasses only, construct:\n(i) the trapezium ABCD ; (ii) a rectangle PQCD, where P, Q are two points AB;\n(b) Measure |AB| and |QB|.', '(a) ABCD is a trapezium in which AB // DC, |AB| = 8cm, < ABC = 60°, |BC| = 5.5cm and |BD| = 8.3cm. Using a ruler and a pair of compasses only, construct:\n(i) the trapezium ABCD ; (ii) a rectangle PQCD, where P, Q are two points AB;~(b) Measure |AB| and |QB|.', NULL, '\n\nExplanation\n(b) |AC| = 7cm ; |QB| = 2.9cm.\n\n', '(b) |AC| = 7cm ; |QB| = 2.9cm.\n\n', '', '', 'WAEC-PK-1', 'WAEC-MATH', 1),
(680, '2018', '\n\nQuestion\n\n(a) If a number is chosen at random from the integers 5 to 25 inclusive, find the probability that the number is a multiple of 5 or 3.\n(b) A bag contains 10 balls that differ only in colour; 4 are blue and 6 are red. Two balls are picked one after the other, with replacement. What is the probability that:\n(i) both are red? (ii) both are the same colour?\n\n', '', '(a) If a number is chosen at random from the integers 5 to 25 inclusive, find the probability that the number is a multiple of 5 or 3.\n(b) A bag contains 10 balls that differ only in colour; 4 are blue and 6 are red. Two balls are picked one after the other, with replacement. What is the probability that:\n(i) both are red? (ii) both are the same colour?', '(a) If a number is chosen at random from the integers 5 to 25 inclusive, find the probability that the number is a multiple of 5 or 3.~(b) A bag contains 10 balls that differ only in colour; 4 are blue and 6 are red. Two balls are picked one after the other, with replacement. What is the probability that:\n(i) both are red? (ii) both are the same colour?', NULL, '\n\nExplanation\n(a) Let S = sample space = integers from 5 to 25 inclusive.\nF = multiple of 5 ; T = multiple of 3.\nS = {5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25}\nF = {5, 10, 15, 20, 25}; T = {6, 9, 12, 15, 18, 21, 24}\nP(F or T) = \\(\\frac{5}{21} + \\frac{7}{21} = \\frac{12}{21} = \\frac{4}{7}\\)\n(b) Let B = blue balls and R = red balls\nn(B) = 4 ; n(R) = 6.\n(i) P(both are red) = \\(\\frac{6}{10} \\times \\frac{6}{10} = \\frac{36}{100}\\)\n= \\(\\frac{9}{25}\\)\n(ii) P(both are same colour) = P(both are red) or P(both are blue)\n= \\(\\frac{9}{25} + (\\frac{2}{5} \\times \\frac{2}{5})\\)\n= \\(\\frac{9}{25} + \\frac{4}{25}\\)\n= \\(\\frac{13}{25}\\)\n\n', '', 'Explanation\n(a) Let S = sample space = integers from 5 to 25 inclusive.\nF = multiple of 5 ; T = multiple of 3.\nS = {5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25}\nF = {5, 10, 15, 20, 25}; T = {6, 9, 12, 15, 18, 21, 24}\nP(F or T) = \\(\\frac{5}{21} + \\frac{7}{21} = \\frac{12}{21} = \\frac{4}{7}\\)\n(b) Let B = blue balls and R = red balls\nn(B) = 4 ; n(R) = 6.\n(i) P(both are red) = \\(\\frac{6}{10} \\times \\frac{6}{10} = \\frac{36}{100}\\)\n= \\(\\frac{9}{25}\\)\n(ii) P(both are same colour) = P(both are red) or P(both are blue)\n= \\(\\frac{9}{25} + (\\frac{2}{5} \\times \\frac{2}{5})\\)\n= \\(\\frac{9}{25} + \\frac{4}{25}\\)\n= \\(\\frac{13}{25}\\)\n\n', '(a) Let S = sample space = integers from 5 to 25 inclusive.\nF = multiple of 5 ; T = multiple of 3.\nS = {5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25}\nF = {5, 10, 15, 20, 25}; T = {6, 9, 12, 15, 18, 21, 24}\nP(F or T) = \\(\\frac{5}{21} + \\frac{7}{21} = \\frac{12}{21} = \\frac{4}{7}\\)~(b) Let B = blue balls and R = red balls\nn(B) = 4 ; n(R) = 6.\n(i) P(both are red) = \\(\\frac{6}{10} \\times \\frac{6}{10} = \\frac{36}{100}\\)\n= \\(\\frac{9}{25}\\)\n(ii) P(both are same colour) = P(both are red) or P(both are blue)\n= \\(\\frac{9}{25} + (\\frac{2}{5} \\times \\frac{2}{5})\\)\n= \\(\\frac{9}{25} + \\frac{4}{25}\\)\n= \\(\\frac{13}{25}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1);
INSERT INTO `exam_simulation2` (`id`, `examYear`, `questions`, `parent_que`, `child_que`, `gen_child_que_desc`, `imageDir`, `explanation`, `parent_que_expl`, `child_que_exp`, `gen_child_que_expl`, `package_id`, `subject_id`, `status`) VALUES
(681, '2018', '\n\nQuestion\n\nThe following is an incomplete table for the relation \\(y = 2x^{2} - 5x + 1\\)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n\n\n8\n1\n\n-1\n\n\n26\n\n\n\n(a) Copy and complete the table.\n(b) Using a scale of 2cm to 1 unit on the x- axis and 2cm to 10 units on the y- axis, draw the graph of the relation \\(y = 2x^{2} - 5x + 1\\) for \\(-3 \\leq x \\leq 5\\).\n(c) Using the same scale and axes, draw the graph of \\(y = x + 6\\).\n(d) Estimate from your graphs, correct to one decimal place : (i) the least value of y and the value of x for which it occurs ; (ii) the solution of the equation \\(2x^{2} - 5x + 1 = x + 6\\).\n\n\nThe following is an incomplete table for the relation \\(y = 2x^{2} - 5x + 1\\)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n\n\n8\n1\n\n-1\n\n\n26\n\n\n', '', 'The following is an incomplete table for the relation \\(y = 2x^{2} - 5x + 1\\)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n\n\n8\n1\n\n-1\n\n\n26\n\n\n\n(a) Copy and complete the table.\n(b) Using a scale of 2cm to 1 unit on the x- axis and 2cm to 10 units on the y- axis, draw the graph of the relation \\(y = 2x^{2} - 5x + 1\\) for \\(-3 \\leq x \\leq 5\\).\n(c) Using the same scale and axes, draw the graph of \\(y = x + 6\\).\n(d) Estimate from your graphs, correct to one decimal place : (i) the least value of y and the value of x for which it occurs ; (ii) the solution of the equation \\(2x^{2} - 5x + 1 = x + 6\\).\n\n\nThe following is an incomplete table for the relation \\(y = 2x^{2} - 5x + 1\\)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n\n\n8\n1\n\n-1\n\n\n26\n', 'The following is an incomplete table for the relation \\(y = 2x^{2} - 5x + 1\\)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n\n\n8\n1\n\n-1\n\n\n26\n\n\n\n(a) Copy and complete the table.~(b) Using a scale of 2cm to 1 unit on the x- axis and 2cm to 10 units on the y- axis, draw the graph of the relation \\(y = 2x^{2} - 5x + 1\\) for \\(-3 \\leq x \\leq 5\\).\n(c) Using the same scale and axes, draw the graph of \\(y = x + 6\\).\n(d) Estimate from your graphs, correct to one decimal place : (i) the least value of y and the value of x for which it occurs ; (ii) the solution of the equation \\(2x^{2} - 5x + 1 = x + 6\\).\n\n\nThe following is an incomplete table for the relation \\(y = 2x^{2} - 5x + 1\\)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n\n\n8\n1\n\n-1\n\n\n26\n~(c) Using the same scale and axes, draw the graph of \\(y = x + 6\\).\n(d) Estimate from your graphs, correct to one decimal place : (i) the least value of y and the value of x for which it occurs ; (ii) the solution of the equation \\(2x^{2} - 5x + 1 = x + 6\\).\n\n\nThe following is an incomplete table for the relation \\(y = 2x^{2} - 5x + 1\\)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n\n\n8\n1\n\n-1\n\n\n26\n~(d) Estimate from your graphs, correct to one decimal place : (i) the least value of y and the value of x for which it occurs ; (ii) the solution of the equation \\(2x^{2} - 5x + 1 = x + 6\\).\n\n\nThe following is an incomplete table for the relation \\(y = 2x^{2} - 5x + 1\\)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n\n\n8\n1\n\n-1\n\n\n26\n', NULL, '\n\nExplanation\n(a) \\(y = 2x^{2} - 5x + 1\\)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n34\n19\n8\n1\n-2\n-1\n4\n13\n26\n\n\n\n\nTable of value for \\(y = x + 6\\)\n\n\n\nx\n-2\n0\n2\n\n\ny\n4\n6\n8\n\n\n\n(b)\n(d) (i) Least value of y = -2 occurs at x = 1\n(ii) Solution of \\(2x^{2} - 5x + 1 = x + 6\\) are x = -0.7 and x = 3.7.\n\n', '', 'Explanation\n(a) \\(y = 2x^{2} - 5x + 1\\)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n34\n19\n8\n1\n-2\n-1\n4\n13\n26\n\n\n\n\nTable of value for \\(y = x + 6\\)\n\n\n\nx\n-2\n0\n2\n\n\ny\n4\n6\n8\n\n\n\n(b)\n(d) (i) Least value of y = -2 occurs at x = 1\n(ii) Solution of \\(2x^{2} - 5x + 1 = x + 6\\) are x = -0.7 and x = 3.7.\n\n', '(a) \\(y = 2x^{2} - 5x + 1\\)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n34\n19\n8\n1\n-2\n-1\n4\n13\n26\n\n\n\n\nTable of value for \\(y = x + 6\\)\n\n\n\nx\n-2\n0\n2\n\n\ny\n4\n6\n8\n\n\n~(b)\n(d) (i) Least value of y = -2 occurs at x = 1\n(ii) Solution of \\(2x^{2} - 5x + 1 = x + 6\\) are x = -0.7 and x = 3.7.\n~(d) (i) Least value of y = -2 occurs at x = 1\n(ii) Solution of \\(2x^{2} - 5x + 1 = x + 6\\) are x = -0.7 and x = 3.7.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(682, '2018', '\n\nQuestion\n\nIn a class of 40 students, 25 speak Hausa, 16 speak Igbo, 21 speak Yoruba and each of the students speak at least one of the these three languages. If 8 speak Hausa and Igbo, 11 speak Hausa and Yoruba and 6 speak Igbo and Yoruba.\n(a) Draw a Venn diagram to illustrate the information, using x to represent the number of students that speak all three languages.\n(b) calculate the value of x.\n\n', '', 'In a class of 40 students, 25 speak Hausa, 16 speak Igbo, 21 speak Yoruba and each of the students speak at least one of the these three languages. If 8 speak Hausa and Igbo, 11 speak Hausa and Yoruba and 6 speak Igbo and Yoruba.\n(a) Draw a Venn diagram to illustrate the information, using x to represent the number of students that speak all three languages.\n(b) calculate the value of x.', 'In a class of 40 students, 25 speak Hausa, 16 speak Igbo, 21 speak Yoruba and each of the students speak at least one of the these three languages. If 8 speak Hausa and Igbo, 11 speak Hausa and Yoruba and 6 speak Igbo and Yoruba.\n(a) Draw a Venn diagram to illustrate the information, using x to represent the number of students that speak all three languages.~(b) calculate the value of x.', NULL, '\n\nExplanation\n(a) The number that speak Hausa = n(H) = 25\nThe number that speak Igbo = n(I) = 16\nThe number that speak Yoruba = n(Y) = 21\nTotal no of students = U = 40\n\\(n(H \\cap I) = 8 ; n(H \\cap Y) = 11 ; n(Y \\cap I) = 6\\)\n\n(b) \\(U = n(H) + n(I) + n(Y) - n(H \\cap Y) - n(H \\cap I) - n(I \\cap Y) + n(H \\cap Y \\cap I)\\)\n\\(40 = 6 + x + 2 + x + 4 + x + 8 - x + 11 - x + 6 - x + x\\)\n\\(40 = 37+ x \\implies x = 40 - 37= 3\\)\n\n', '', 'Explanation\n(a) The number that speak Hausa = n(H) = 25\nThe number that speak Igbo = n(I) = 16\nThe number that speak Yoruba = n(Y) = 21\nTotal no of students = U = 40\n\\(n(H \\cap I) = 8 ; n(H \\cap Y) = 11 ; n(Y \\cap I) = 6\\)\n\n(b) \\(U = n(H) + n(I) + n(Y) - n(H \\cap Y) - n(H \\cap I) - n(I \\cap Y) + n(H \\cap Y \\cap I)\\)\n\\(40 = 6 + x + 2 + x + 4 + x + 8 - x + 11 - x + 6 - x + x\\)\n\\(40 = 37+ x \\implies x = 40 - 37= 3\\)\n\n', '(a) The number that speak Hausa = n(H) = 25\nThe number that speak Igbo = n(I) = 16\nThe number that speak Yoruba = n(Y) = 21\nTotal no of students = U = 40\n\\(n(H \\cap I) = 8 ; n(H \\cap Y) = 11 ; n(Y \\cap I) = 6\\)\n~(b) \\(U = n(H) + n(I) + n(Y) - n(H \\cap Y) - n(H \\cap I) - n(I \\cap Y) + n(H \\cap Y \\cap I)\\)\n\\(40 = 6 + x + 2 + x + 4 + x + 8 - x + 11 - x + 6 - x + x\\)\n\\(40 = 37+ x \\implies x = 40 - 37= 3\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(683, '2018', '\n\nQuestion\n\nAn aeroplane flies from a town P(lat. 40°N, 38°E) to another town Q(lat. 40°N, 22°W). It later flies to a third town T(28°N, 22°W). Calculate the :\n(a) distance between P and Q along their parallel of latitude ;\n(b) distance between Q and T along their line of longitudes;\n(c) average speed at which the aeroplane will fly from P to T via Q, if the journey takes 12 hours, correct to 3 significant figures. [Take the radius of the earth = 6400km ; \\(\\pi = 3.142\\)]\n\n', '', 'An aeroplane flies from a town P(lat. 40°N, 38°E) to another town Q(lat. 40°N, 22°W). It later flies to a third town T(28°N, 22°W). Calculate the :\n(a) distance between P and Q along their parallel of latitude ;\n(b) distance between Q and T along their line of longitudes;\n(c) average speed at which the aeroplane will fly from P to T via Q, if the journey takes 12 hours, correct to 3 significant figures. [Take the radius of the earth = 6400km ; \\(\\pi = 3.142\\)]', 'An aeroplane flies from a town P(lat. 40°N, 38°E) to another town Q(lat. 40°N, 22°W). It later flies to a third town T(28°N, 22°W). Calculate the :\n(a) distance between P and Q along their parallel of latitude ;~(b) distance between Q and T along their line of longitudes;\n(c) average speed at which the aeroplane will fly from P to T via Q, if the journey takes 12 hours, correct to 3 significant figures. [Take the radius of the earth = 6400km ; \\(\\pi = 3.142\\)]~(c) average speed at which the aeroplane will fly from P to T via Q, if the journey takes 12 hours, correct to 3 significant figures. [Take the radius of the earth = 6400km ; \\(\\pi = 3.142\\)]', NULL, '\n\nExplanation\n(a) Distance along arc PQ = \\(\\frac{\\theta}{360°} \\times 2 \\pi r\\)\nwhere \\(r = R \\cos \\theta\\)\n\\(|PQ| = \\frac{60°}{360°} \\times 2 \\times \\frac{22}{7} \\times 6400 \\times \\cos 40°\\)\n\\(|PQ| = \\frac{1}{6} \\times 2 \\times \\frac{22}{7} \\times 6400 \\times 0.7660\\)\n= \\(5134.74 km \\approxeq 5135km\\) (to the nearest whole number)\n(b) Difference in latitude between Q and T = 40° - 28° = 12°\n\\(\\therefore |QT| \\text{along the line of longitude} = \\frac{12}{360} \\times 2 \\times \\frac{22}{7} \\times 6400\\)\n\\(\\frac{1}{30} \\times 2 \\times \\frac{22}{7} \\times 6400 = 1340.95km \\approxeq 1341km\\)\n(c) Average speed = \\(\\frac{\\text{Total distance covered}}{\\text{Total time taken}}\\)\n= \\(\\frac{5134.74 + 1340.95}{12} = 539.64km/hr\\)\n\\(\\approxeq 540km/hr\\) (to 3 significant figures)\n\n', '', 'Explanation\n(a) Distance along arc PQ = \\(\\frac{\\theta}{360°} \\times 2 \\pi r\\)\nwhere \\(r = R \\cos \\theta\\)\n\\(|PQ| = \\frac{60°}{360°} \\times 2 \\times \\frac{22}{7} \\times 6400 \\times \\cos 40°\\)\n\\(|PQ| = \\frac{1}{6} \\times 2 \\times \\frac{22}{7} \\times 6400 \\times 0.7660\\)\n= \\(5134.74 km \\approxeq 5135km\\) (to the nearest whole number)\n(b) Difference in latitude between Q and T = 40° - 28° = 12°\n\\(\\therefore |QT| \\text{along the line of longitude} = \\frac{12}{360} \\times 2 \\times \\frac{22}{7} \\times 6400\\)\n\\(\\frac{1}{30} \\times 2 \\times \\frac{22}{7} \\times 6400 = 1340.95km \\approxeq 1341km\\)\n(c) Average speed = \\(\\frac{\\text{Total distance covered}}{\\text{Total time taken}}\\)\n= \\(\\frac{5134.74 + 1340.95}{12} = 539.64km/hr\\)\n\\(\\approxeq 540km/hr\\) (to 3 significant figures)\n\n', '(a) Distance along arc PQ = \\(\\frac{\\theta}{360°} \\times 2 \\pi r\\)\nwhere \\(r = R \\cos \\theta\\)\n\\(|PQ| = \\frac{60°}{360°} \\times 2 \\times \\frac{22}{7} \\times 6400 \\times \\cos 40°\\)\n\\(|PQ| = \\frac{1}{6} \\times 2 \\times \\frac{22}{7} \\times 6400 \\times 0.7660\\)\n= \\(5134.74 km \\approxeq 5135km\\) (to the nearest whole number)~(b) Difference in latitude between Q and T = 40° - 28° = 12°\n\\(\\therefore |QT| \\text{along the line of longitude} = \\frac{12}{360} \\times 2 \\times \\frac{22}{7} \\times 6400\\)\n\\(\\frac{1}{30} \\times 2 \\times \\frac{22}{7} \\times 6400 = 1340.95km \\approxeq 1341km\\)\n(c) Average speed = \\(\\frac{\\text{Total distance covered}}{\\text{Total time taken}}\\)\n= \\(\\frac{5134.74 + 1340.95}{12} = 539.64km/hr\\)\n\\(\\approxeq 540km/hr\\) (to 3 significant figures)\n~(c) Average speed = \\(\\frac{\\text{Total distance covered}}{\\text{Total time taken}}\\)\n= \\(\\frac{5134.74 + 1340.95}{12} = 539.64km/hr\\)\n\\(\\approxeq 540km/hr\\) (to 3 significant figures)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(684, '2018', '\n\nQuestion\n\n(a) Without using Mathematical tables, find x, given that \\(6 \\log (x + 4) = \\log 64\\)\n(b) If \\(U = {1, 2, 3,4, 5, 6, 7, 8, 9, 10}, X = {1, 2, 4, 6, 7, 8, 9}, Y = {1, 2, 3, 4, 7, 9}\\) and \\(Z = {2, 3, 4, 7, 9}\\). What is \\(X \\cap Y \\cap Z\' \\)?\n\n', '', '(a) Without using Mathematical tables, find x, given that \\(6 \\log (x + 4) = \\log 64\\)\n(b) If \\(U = {1, 2, 3,4, 5, 6, 7, 8, 9, 10}, X = {1, 2, 4, 6, 7, 8, 9}, Y = {1, 2, 3, 4, 7, 9}\\) and \\(Z = {2, 3, 4, 7, 9}\\). What is \\(X \\cap Y \\cap Z\' \\)?', '(a) Without using Mathematical tables, find x, given that \\(6 \\log (x + 4) = \\log 64\\)~(b) If \\(U = {1, 2, 3,4, 5, 6, 7, 8, 9, 10}, X = {1, 2, 4, 6, 7, 8, 9}, Y = {1, 2, 3, 4, 7, 9}\\) and \\(Z = {2, 3, 4, 7, 9}\\). What is \\(X \\cap Y \\cap Z\' \\)?', NULL, '\n\nExplanation\n(a) \\(6 \\log (x + 4) = \\log 64\\)\n\\(\\implies (x + 4)^{6} = 64\\)\n\\((x + 4)^{6} = 2^{6}\\)\n\\(\\implies x + 4 = 2 ; x = 2 - 4 = -2\\)\n\\(\\therefore x = -2\\)\n(b) \\(U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}\\)\n\\(X = {1, 2, 4, 6, 7, 8, 9}; Y = {1, 2, 3, 4, 7, 9} ; Z = {2, 3, 4, 7,9}\\)\n\\(X \\cap Y = {1, 2, 4, 7, 9}\\)\n\\({1, 2, 4, 7, 9} \\cap Z\'= {1, 2, 4, 7, 9} \\cap {1, 5, 6, 8, 10}\\)\n= {1}\n\n', '', 'Explanation\n(a) \\(6 \\log (x + 4) = \\log 64\\)\n\\(\\implies (x + 4)^{6} = 64\\)\n\\((x + 4)^{6} = 2^{6}\\)\n\\(\\implies x + 4 = 2 ; x = 2 - 4 = -2\\)\n\\(\\therefore x = -2\\)\n(b) \\(U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}\\)\n\\(X = {1, 2, 4, 6, 7, 8, 9}; Y = {1, 2, 3, 4, 7, 9} ; Z = {2, 3, 4, 7,9}\\)\n\\(X \\cap Y = {1, 2, 4, 7, 9}\\)\n\\({1, 2, 4, 7, 9} \\cap Z\'= {1, 2, 4, 7, 9} \\cap {1, 5, 6, 8, 10}\\)\n= {1}\n\n', '(a) \\(6 \\log (x + 4) = \\log 64\\)\n\\(\\implies (x + 4)^{6} = 64\\)\n\\((x + 4)^{6} = 2^{6}\\)\n\\(\\implies x + 4 = 2 ; x = 2 - 4 = -2\\)\n\\(\\therefore x = -2\\)~(b) \\(U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}\\)\n\\(X = {1, 2, 4, 6, 7, 8, 9}; Y = {1, 2, 3, 4, 7, 9} ; Z = {2, 3, 4, 7,9}\\)\n\\(X \\cap Y = {1, 2, 4, 7, 9}\\)\n\\({1, 2, 4, 7, 9} \\cap Z\'= {1, 2, 4, 7, 9} \\cap {1, 5, 6, 8, 10}\\)\n= {1}\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(685, '2018', '\n\nQuestion\n\nThe table shows the weights, to the nearest kilogram, of twelve students in a Further Mathematics class\n\n\n\nWeight in kg\n55\n57\n59\n61\n63\n\n\nNo of students\n2\n1\n2\n4\n3\n\n\n\n(a) Draw a bar chart to illustrate the above information;\n(b) What is (i) the mode; (ii) the median of the distribution?\n(c) Calculate the mean weight correct to the nearest kilogram.\n\n\n\nThe table shows the weights, to the nearest kilogram, of twelve students in a Further Mathematics class\n\n\n\nWeight in kg\n55\n57\n59\n61\n63\n\n\nNo of students\n2\n1\n2\n4\n3\n\n\n', '', 'The table shows the weights, to the nearest kilogram, of twelve students in a Further Mathematics class\n\n\n\nWeight in kg\n55\n57\n59\n61\n63\n\n\nNo of students\n2\n1\n2\n4\n3\n\n\n\n(a) Draw a bar chart to illustrate the above information;\n(b) What is (i) the mode; (ii) the median of the distribution?\n(c) Calculate the mean weight correct to the nearest kilogram.\n\n\n\nThe table shows the weights, to the nearest kilogram, of twelve students in a Further Mathematics class\n\n\n\nWeight in kg\n55\n57\n59\n61\n63\n\n\nNo of students\n2\n1\n2\n4\n3\n', 'The table shows the weights, to the nearest kilogram, of twelve students in a Further Mathematics class\n\n\n\nWeight in kg\n55\n57\n59\n61\n63\n\n\nNo of students\n2\n1\n2\n4\n3\n\n\n\n(a) Draw a bar chart to illustrate the above information;~(b) What is (i) the mode; (ii) the median of the distribution?\n(c) Calculate the mean weight correct to the nearest kilogram.\n\n\n\nThe table shows the weights, to the nearest kilogram, of twelve students in a Further Mathematics class\n\n\n\nWeight in kg\n55\n57\n59\n61\n63\n\n\nNo of students\n2\n1\n2\n4\n3\n~(c) Calculate the mean weight correct to the nearest kilogram.\n\n\n\nThe table shows the weights, to the nearest kilogram, of twelve students in a Further Mathematics class\n\n\n\nWeight in kg\n55\n57\n59\n61\n63\n\n\nNo of students\n2\n1\n2\n4\n3\n', NULL, '\n\nExplanation\n(a)\n(b)(i) The mode is 61kg.\n(ii) From the total number of students = 12, then the median is the average of the 6th and 7th weight.\nFrom the table, the median = 61kg.\n(c) Mean weight = \\(\\frac{\\sum fx}{\\sum f} = \\frac{55 \\times 2 + 57 \\times 1 + 59 \\times 2 + 61 \\times 4 + 63 \\times 3}{12}\\)\n= \\(\\frac{110 + 57 + 118 + 244 + 189}{12} = \\frac{718}{12} = 59.833kg\\)\n\\(\\approxeq 60kg\\)\n\n', '', 'Explanation\n(a)\n(b)(i) The mode is 61kg.\n(ii) From the total number of students = 12, then the median is the average of the 6th and 7th weight.\nFrom the table, the median = 61kg.\n(c) Mean weight = \\(\\frac{\\sum fx}{\\sum f} = \\frac{55 \\times 2 + 57 \\times 1 + 59 \\times 2 + 61 \\times 4 + 63 \\times 3}{12}\\)\n= \\(\\frac{110 + 57 + 118 + 244 + 189}{12} = \\frac{718}{12} = 59.833kg\\)\n\\(\\approxeq 60kg\\)\n\n', '~(b)(i) The mode is 61kg.\n(ii) From the total number of students = 12, then the median is the average of the 6th and 7th weight.\nFrom the table, the median = 61kg.\n(c) Mean weight = \\(\\frac{\\sum fx}{\\sum f} = \\frac{55 \\times 2 + 57 \\times 1 + 59 \\times 2 + 61 \\times 4 + 63 \\times 3}{12}\\)\n= \\(\\frac{110 + 57 + 118 + 244 + 189}{12} = \\frac{718}{12} = 59.833kg\\)\n\\(\\approxeq 60kg\\)\n~(c) Mean weight = \\(\\frac{\\sum fx}{\\sum f} = \\frac{55 \\times 2 + 57 \\times 1 + 59 \\times 2 + 61 \\times 4 + 63 \\times 3}{12}\\)\n= \\(\\frac{110 + 57 + 118 + 244 + 189}{12} = \\frac{718}{12} = 59.833kg\\)\n\\(\\approxeq 60kg\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(686, '2018', '\n\nQuestion\n\nThe diagram shows a wooden structure in the form of a cone, mounted on a hemispherical base. The vertical height of the cone is 24cm and the base height 7cm. Calculate, correct to three significant figures, the surface area of the structure. [Take \\(\\pi = \\frac{22}{7}\\)].\n\n\n', 'The diagram shows a wooden structure in the form of a cone, mounted on a hemispherical base. The vertical height of the cone is 24cm and the base height 7cm. Calculate, correct to three significant figures, the surface area of the structure. [Take \\(\\pi = \\frac{22}{7}\\)].\n', '', '', NULL, '\n\nExplanation\nLet the slant height = L ; height = 24cm ; r = 7cm.\n\\(\\therefore L^{2} = 24^{2} + 7^{2} = 576 + 49 = 625\\)\n\\(L = \\sqrt{625} = 25cm\\)\nSurface area of the structure = Area of cone + Area of hemisphere\n= \\(\\pi r L + 2 \\pi r^{2} = \\pi r (L + 2r)\\)\n= \\(\\frac{22}{7} \\times 7 (25 + 2(7))\\)\n= \\(22 \\times 39 = 858cm^{2}\\)\n\n', 'Let the slant height = L ; height = 24cm ; r = 7cm.\n\\(\\therefore L^{2} = 24^{2} + 7^{2} = 576 + 49 = 625\\)\n\\(L = \\sqrt{625} = 25cm\\)\nSurface area of the structure = Area of cone + Area of hemisphere\n= \\(\\pi r L + 2 \\pi r^{2} = \\pi r (L + 2r)\\)\n= \\(\\frac{22}{7} \\times 7 (25 + 2(7))\\)\n= \\(22 \\times 39 = 858cm^{2}\\)\n\n', '', '', 'WAEC-PK-1', 'WAEC-MATH', 1),
(687, '2018', '\n\nQuestion\n\nUsing a ruler and a pair of compasses only, construct a triangle ABC, given that |AB| = 8.4cm, |BC| = 6.5cm and < ABC = 30°. Construct the locus:\n(a) \\(l_{1}\\) of points equidistant from AB and BC, and within the angle ABC;\n(b) \\(l_{2}\\) of points equidistant from B and C. Locate the point of intersection P of \\(l_{1}\\) and \\(l_{2}\\). Measure |AP|.\n\n', '', 'Using a ruler and a pair of compasses only, construct a triangle ABC, given that |AB| = 8.4cm, |BC| = 6.5cm and < ABC = 30°. Construct the locus:\n(a) \\(l_{1}\\) of points equidistant from AB and BC, and within the angle ABC;\n(b) \\(l_{2}\\) of points equidistant from B and C. Locate the point of intersection P of \\(l_{1}\\) and \\(l_{2}\\). Measure |AP|.', 'Using a ruler and a pair of compasses only, construct a triangle ABC, given that |AB| = 8.4cm, |BC| = 6.5cm and < ABC = 30°. Construct the locus:\n(a) \\(l_{1}\\) of points equidistant from AB and BC, and within the angle ABC;~(b) \\(l_{2}\\) of points equidistant from B and C. Locate the point of intersection P of \\(l_{1}\\) and \\(l_{2}\\). Measure |AP|.', NULL, '\n\nExplanation\n', '', '', '', 'WAEC-PK-1', 'WAEC-MATH', 1),
(688, '2018', '\n\nQuestion\n\n(a) If \\(\\cos \\alpha = 0.6717\\), use mathematical tables to find (i) \\(\\alpha\\) ; (ii) \\(\\sin \\alpha\\)\n(b) The angle of depression of a point P on the ground, from the top T of a building is 23.6°. If the distance of P from the foot of the building is 50m, calculate the height of the building, correct to the nearest metre.\n\n', '', '(a) If \\(\\cos \\alpha = 0.6717\\), use mathematical tables to find (i) \\(\\alpha\\) ; (ii) \\(\\sin \\alpha\\)\n(b) The angle of depression of a point P on the ground, from the top T of a building is 23.6°. If the distance of P from the foot of the building is 50m, calculate the height of the building, correct to the nearest metre.', '(a) If \\(\\cos \\alpha = 0.6717\\), use mathematical tables to find (i) \\(\\alpha\\) ; (ii) \\(\\sin \\alpha\\)~(b) The angle of depression of a point P on the ground, from the top T of a building is 23.6°. If the distance of P from the foot of the building is 50m, calculate the height of the building, correct to the nearest metre.', NULL, '\n\nExplanation\n(a) (i) Given \\(\\cos \\alpha = 0.6717\\)\n\\(\\alpha = \\cos^{-1} 0.6717 = 47.8°\\)\n(From cosine of angle table)\n(ii) \\(\\sin 47.8° = 0.7408\\)\n(b)\nFrom the diagram, \\(< BPT = < PTR = 23.6°\\)\n\\(\\therefore |BT| = |BP| \\tan 23.6°\\)\n= \\(50 \\tan 23.6° = 50 \\times 0.4369\\)\n= \\(21.8469 m\\)\n\\(\\therefore\\) The height of the building = 22m (to the nearest metre).\n\n', '', 'Explanation\n(a) (i) Given \\(\\cos \\alpha = 0.6717\\)\n\\(\\alpha = \\cos^{-1} 0.6717 = 47.8°\\)\n(From cosine of angle table)\n(ii) \\(\\sin 47.8° = 0.7408\\)\n(b)\nFrom the diagram, \\(< BPT = < PTR = 23.6°\\)\n\\(\\therefore |BT| = |BP| \\tan 23.6°\\)\n= \\(50 \\tan 23.6° = 50 \\times 0.4369\\)\n= \\(21.8469 m\\)\n\\(\\therefore\\) The height of the building = 22m (to the nearest metre).\n\n', '(a) (i) Given \\(\\cos \\alpha = 0.6717\\)\n\\(\\alpha = \\cos^{-1} 0.6717 = 47.8°\\)\n(From cosine of angle table)\n(ii) \\(\\sin 47.8° = 0.7408\\)~(b)\nFrom the diagram, \\(< BPT = < PTR = 23.6°\\)\n\\(\\therefore |BT| = |BP| \\tan 23.6°\\)\n= \\(50 \\tan 23.6° = 50 \\times 0.4369\\)\n= \\(21.8469 m\\)\n\\(\\therefore\\) The height of the building = 22m (to the nearest metre).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(689, '2018', '\n\nQuestion\n\n(a)\nIn the diagram, O is the centre of the circle radius 3.2cm. If < PRQ = 42°, calculate, correct to two decimal places, the area of the:\n(i) minor sector POQ ; (ii) shaded part.\n(b) If the sector POQ in (a) is used to form the curved surface of a cone with vertex O, calculate the base radius of the cone, correct to one decimal place.\n\n', '', '(a)\nIn the diagram, O is the centre of the circle radius 3.2cm. If < PRQ = 42°, calculate, correct to two decimal places, the area of the:\n(i) minor sector POQ ; (ii) shaded part.\n(b) If the sector POQ in (a) is used to form the curved surface of a cone with vertex O, calculate the base radius of the cone, correct to one decimal place.', '(a)\nIn the diagram, O is the centre of the circle radius 3.2cm. If < PRQ = 42°, calculate, correct to two decimal places, the area of the:\n(i) minor sector POQ ; (ii) shaded part.~(b) If the sector POQ in (a) is used to form the curved surface of a cone with vertex O, calculate the base radius of the cone, correct to one decimal place.', NULL, '\n\nExplanation\n (a) (i) Area of minor segment POQ = \\(\\frac{84}{360} \\times \\frac{22}{7} \\times (3.2)^{2}\\)\n= \\(\\frac{11 \\times 10.24}{15}\\)\n= \\(7.5093 cm^{2} \\approxeq 7.51 cm^{2}\\) (2 decimal places)\n(ii) Area of \\(\\Delta\\) in POQ = \\(\\frac{1}{2} \\times 3.2 \\times 3.2 \\times \\sin 84°\\)\n= \\(\\frac{127.296}{25} = 5.0918 cm^{2}\\)\nArea of segment PQ (shaded part) = Area of sector POQ - Area of triangle\n= \\(7.5093 cm^{2} - 5.0918 cm^{2} = 2.4175 cm^{2}\\)\n\\(\\approxeq 2.42 cm^{2}\\)\n(b) Area of sector POQ = Curved surface area of the cone\n\\(7.5093 cm^{2} = \\pi rl\\)\nwhere r = base radius, L = slant height = 3.2cm\n\\(7.5093 = \\frac{22}{7} \\times r \\times 3.2 = \\frac{70.4r}{7}\\)\n\\(70.4r = 7.5093 \\times 7 \\implies r = \\frac{7.5093 \\times 7}{70.4} = 0.7467 cm\\)\n\\(\\approxeq 0.7cm\\)\n\n', '', 'Explanation\n (a) (i) Area of minor segment POQ = \\(\\frac{84}{360} \\times \\frac{22}{7} \\times (3.2)^{2}\\)\n= \\(\\frac{11 \\times 10.24}{15}\\)\n= \\(7.5093 cm^{2} \\approxeq 7.51 cm^{2}\\) (2 decimal places)\n(ii) Area of \\(\\Delta\\) in POQ = \\(\\frac{1}{2} \\times 3.2 \\times 3.2 \\times \\sin 84°\\)\n= \\(\\frac{127.296}{25} = 5.0918 cm^{2}\\)\nArea of segment PQ (shaded part) = Area of sector POQ - Area of triangle\n= \\(7.5093 cm^{2} - 5.0918 cm^{2} = 2.4175 cm^{2}\\)\n\\(\\approxeq 2.42 cm^{2}\\)\n(b) Area of sector POQ = Curved surface area of the cone\n\\(7.5093 cm^{2} = \\pi rl\\)\nwhere r = base radius, L = slant height = 3.2cm\n\\(7.5093 = \\frac{22}{7} \\times r \\times 3.2 = \\frac{70.4r}{7}\\)\n\\(70.4r = 7.5093 \\times 7 \\implies r = \\frac{7.5093 \\times 7}{70.4} = 0.7467 cm\\)\n\\(\\approxeq 0.7cm\\)\n\n', '(a) (i) Area of minor segment POQ = \\(\\frac{84}{360} \\times \\frac{22}{7} \\times (3.2)^{2}\\)\n= \\(\\frac{11 \\times 10.24}{15}\\)\n= \\(7.5093 cm^{2} \\approxeq 7.51 cm^{2}\\) (2 decimal places)\n(ii) Area of \\(\\Delta\\) in POQ = \\(\\frac{1}{2} \\times 3.2 \\times 3.2 \\times \\sin 84°\\)\n= \\(\\frac{127.296}{25} = 5.0918 cm^{2}\\)\nArea of segment PQ (shaded part) = Area of sector POQ - Area of triangle\n= \\(7.5093 cm^{2} - 5.0918 cm^{2} = 2.4175 cm^{2}\\)\n\\(\\approxeq 2.42 cm^{2}\\)~(b) Area of sector POQ = Curved surface area of the cone\n\\(7.5093 cm^{2} = \\pi rl\\)\nwhere r = base radius, L = slant height = 3.2cm\n\\(7.5093 = \\frac{22}{7} \\times r \\times 3.2 = \\frac{70.4r}{7}\\)\n\\(70.4r = 7.5093 \\times 7 \\implies r = \\frac{7.5093 \\times 7}{70.4} = 0.7467 cm\\)\n\\(\\approxeq 0.7cm\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(690, '2018', '\n\nQuestion\n\nWhen a stone is thrown vertically upwards, its distance d metres after t seconds is given by the formula \\(d = 60t - 10t^{2}\\). Draw the graph of \\(d = 60t - 10t^{2}\\) for values of t from 1 to 5 seconds using 2cm to 1 unit on the t- axis and 2cm to 20 units on the d- axis.\n(a) Using your graph, (i) how long does it take to reach a height of 70 metres? (ii) determine the height of the stone after 5 seconds. (iii) after how many seconds does it reach its maximum height.\n(b) Determine the slope of the graph when t = 4 seconds.\n\n', '', 'When a stone is thrown vertically upwards, its distance d metres after t seconds is given by the formula \\(d = 60t - 10t^{2}\\). Draw the graph of \\(d = 60t - 10t^{2}\\) for values of t from 1 to 5 seconds using 2cm to 1 unit on the t- axis and 2cm to 20 units on the d- axis.\n(a) Using your graph, (i) how long does it take to reach a height of 70 metres? (ii) determine the height of the stone after 5 seconds. (iii) after how many seconds does it reach its maximum height.\n(b) Determine the slope of the graph when t = 4 seconds.', 'When a stone is thrown vertically upwards, its distance d metres after t seconds is given by the formula \\(d = 60t - 10t^{2}\\). Draw the graph of \\(d = 60t - 10t^{2}\\) for values of t from 1 to 5 seconds using 2cm to 1 unit on the t- axis and 2cm to 20 units on the d- axis.\n(a) Using your graph, (i) how long does it take to reach a height of 70 metres? (ii) determine the height of the stone after 5 seconds. (iii) after how many seconds does it reach its maximum height.~(b) Determine the slope of the graph when t = 4 seconds.', NULL, '\n\nExplanation\nTable of values of \\(d = 60t - 10t^{2}\\)\n\n\n\nt\n1\n2\n3\n4\n5\n\n\n\\(60t\\)\n60\n120\n180\n240\n300\n\n\n\\(- 10t^{2}\\)\n-10\n-40\n-90\n-160\n-250\n\n\nd\n50\n80\n90\n80\n50\n\n\n\n\n(a)(i) It takes 1.6 seconds to reach a height of 70 metres.\n(ii) Height of stone at 5 seconds was 50 metres but after 5 seconds, d < 50 metres.\n(iii) It reaches its maximum height after 2.9 seconds.\n(b) From the graph, the gradient of the slope = \\(\\frac{AB}{BC}\\)\n= \\(\\frac{56}{2.8} = 20 m/s\\)\nBut it is a negative slope, hence the slope = -20 m/s.\n\n\n', '', 'Explanation\nTable of values of \\(d = 60t - 10t^{2}\\)\n\n\n\nt\n1\n2\n3\n4\n5\n\n\n\\(60t\\)\n60\n120\n180\n240\n300\n\n\n\\(- 10t^{2}\\)\n-10\n-40\n-90\n-160\n-250\n\n\nd\n50\n80\n90\n80\n50\n\n\n\n\n(a)(i) It takes 1.6 seconds to reach a height of 70 metres.\n(ii) Height of stone at 5 seconds was 50 metres but after 5 seconds, d < 50 metres.\n(iii) It reaches its maximum height after 2.9 seconds.\n(b) From the graph, the gradient of the slope = \\(\\frac{AB}{BC}\\)\n= \\(\\frac{56}{2.8} = 20 m/s\\)\nBut it is a negative slope, hence the slope = -20 m/s.\n\n\n', '(a)(i) It takes 1.6 seconds to reach a height of 70 metres.\n(ii) Height of stone at 5 seconds was 50 metres but after 5 seconds, d < 50 metres.\n(iii) It reaches its maximum height after 2.9 seconds.~(b) From the graph, the gradient of the slope = \\(\\frac{AB}{BC}\\)\n= \\(\\frac{56}{2.8} = 20 m/s\\)\nBut it is a negative slope, hence the slope = -20 m/s.\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(691, '2018', '\n\nQuestion\n\n(a) (i) Prove that the angle which an arc of a circle subtends at the centre is twice that which it subtends at any point on the remaining part of the circumference.\n\n(ii) In the diagram above, O is the centre of the circle and PT is a diameter. If < PTQ = 22° and < TOR = 98°, calculate < QRS.\n(b) ABCD is a cyclic quadrilateral and the diagonals AC and BD intersect at H. If < DAC = 41° and < AHB = 70°, calculate < ABC.\n\n', '', '(a) (i) Prove that the angle which an arc of a circle subtends at the centre is twice that which it subtends at any point on the remaining part of the circumference.\n\n(ii) In the diagram above, O is the centre of the circle and PT is a diameter. If < PTQ = 22° and < TOR = 98°, calculate < QRS.\n(b) ABCD is a cyclic quadrilateral and the diagonals AC and BD intersect at H. If < DAC = 41° and < AHB = 70°, calculate < ABC.', '(a) (i) Prove that the angle which an arc of a circle subtends at the centre is twice that which it subtends at any point on the remaining part of the circumference.\n\n(ii) In the diagram above, O is the centre of the circle and PT is a diameter. If < PTQ = 22° and < TOR = 98°, calculate < QRS.~(b) ABCD is a cyclic quadrilateral and the diagonals AC and BD intersect at H. If < DAC = 41° and < AHB = 70°, calculate < ABC.', NULL, '\n\nExplanation\n(a)(i)\nGiven a circle APB with centre O.\nTo prove that < AOB = 2 < APB.\nConstruction: Draw PO and produce it to Q.\nProof : Referring to the lettering in the figure, |OA| = |OB| (radius)\n\\(\\therefore x_{1} = x_{2}\\) (base angles of isosceles triangle)\n\\(< AOQ = x_{1} + x_{2}\\) (exterior angles of triangle AOP)\n\\(\\therefore < AOQ = 2x_{2} (x_{1} = x_{2})\\)\nSimilarly, \\(< BOQ = 2y_{2}\\)\n\\(< AOB = < AOQ + < BOQ \\)\nHence, \\(< AOB = 2x_{2} + 2y_{2} = 2(x_{2} + y_{2})\\)\n= \\(2 \\times < APB\\)\n(ii)\nFrom the figure, \\(|OQ| = |QT|\\) (radius)\n\\(< OQT = 22° = < QTO\\) (base angles of isosceles triangle)\n\\(\\therefore < POQ = 44° \\) (exterior angle of \\(\\Delta\\) QOT)\n\\(< QOR = 180° - (44° + 98°) = 180° - 142° = 38°\\) (Angles on a straight line)\n\\(< QOR = 2 \\times < QSR = 38°\\)\n\\(\\therefore < QSR = \\frac{38°}{2} = 19°\\)\n(b)\n\\(< BHC = (180° - 70°) = 110°\\) (angle on a straight line)\n\\(< DAC = 41° = < DBC\\) (angles on the same segment)\nBut \\(< DBC = < HBC\\)\n\\(< AHB = < HBC + < HCB \\) (exterior angles of \\(\\Delta\\) BHC)\n\\(70° = 41° + < HCB \\implies < HCB = 70° - 41° = 29°\\)\n\\(< ACB = < HCB = 29°\\)\n\n', '', 'Explanation\n(a)(i)\nGiven a circle APB with centre O.\nTo prove that < AOB = 2 < APB.\nConstruction: Draw PO and produce it to Q.\nProof : Referring to the lettering in the figure, |OA| = |OB| (radius)\n\\(\\therefore x_{1} = x_{2}\\) (base angles of isosceles triangle)\n\\(< AOQ = x_{1} + x_{2}\\) (exterior angles of triangle AOP)\n\\(\\therefore < AOQ = 2x_{2} (x_{1} = x_{2})\\)\nSimilarly, \\(< BOQ = 2y_{2}\\)\n\\(< AOB = < AOQ + < BOQ \\)\nHence, \\(< AOB = 2x_{2} + 2y_{2} = 2(x_{2} + y_{2})\\)\n= \\(2 \\times < APB\\)\n(ii)\nFrom the figure, \\(|OQ| = |QT|\\) (radius)\n\\(< OQT = 22° = < QTO\\) (base angles of isosceles triangle)\n\\(\\therefore < POQ = 44° \\) (exterior angle of \\(\\Delta\\) QOT)\n\\(< QOR = 180° - (44° + 98°) = 180° - 142° = 38°\\) (Angles on a straight line)\n\\(< QOR = 2 \\times < QSR = 38°\\)\n\\(\\therefore < QSR = \\frac{38°}{2} = 19°\\)\n(b)\n\\(< BHC = (180° - 70°) = 110°\\) (angle on a straight line)\n\\(< DAC = 41° = < DBC\\) (angles on the same segment)\nBut \\(< DBC = < HBC\\)\n\\(< AHB = < HBC + < HCB \\) (exterior angles of \\(\\Delta\\) BHC)\n\\(70° = 41° + < HCB \\implies < HCB = 70° - 41° = 29°\\)\n\\(< ACB = < HCB = 29°\\)\n\n', '(a)(i)\nGiven a circle APB with centre O.\nTo prove that < AOB = 2 < APB.\nConstruction: Draw PO and produce it to Q.\nProof : Referring to the lettering in the figure, |OA| = |OB| (radius)\n\\(\\therefore x_{1} = x_{2}\\) (base angles of isosceles triangle)\n\\(< AOQ = x_{1} + x_{2}\\) (exterior angles of triangle AOP)\n\\(\\therefore < AOQ = 2x_{2} (x_{1} = x_{2})\\)\nSimilarly, \\(< BOQ = 2y_{2}\\)\n\\(< AOB = < AOQ + < BOQ \\)\nHence, \\(< AOB = 2x_{2} + 2y_{2} = 2(x_{2} + y_{2})\\)\n= \\(2 \\times < APB\\)\n(ii)\nFrom the figure, \\(|OQ| = |QT|\\) (radius)\n\\(< OQT = 22° = < QTO\\) (base angles of isosceles triangle)\n\\(\\therefore < POQ = 44° \\) (exterior angle of \\(\\Delta\\) QOT)\n\\(< QOR = 180° - (44° + 98°) = 180° - 142° = 38°\\) (Angles on a straight line)\n\\(< QOR = 2 \\times < QSR = 38°\\)\n\\(\\therefore < QSR = \\frac{38°}{2} = 19°\\)~(b)\n\\(< BHC = (180° - 70°) = 110°\\) (angle on a straight line)\n\\(< DAC = 41° = < DBC\\) (angles on the same segment)\nBut \\(< DBC = < HBC\\)\n\\(< AHB = < HBC + < HCB \\) (exterior angles of \\(\\Delta\\) BHC)\n\\(70° = 41° + < HCB \\implies < HCB = 70° - 41° = 29°\\)\n\\(< ACB = < HCB = 29°\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(692, '2018', '\n\nQuestion\n\n(a) In an A.P, the difference between the 8th and 4th terms is 20 and the 8th term is \\(1\\frac{1}{2}\\) times the 4th term. What is the:\n(i) common difference ; (ii) first term of the sequence?\n(b) The value of a machine depreciates each year by 5% of its value at the beginning of that year. If its value when new on 1st January 1980 was N10,250.00, what was its value in January 1989 when it was 9 years old? Give your answer correct to three significant figures.\n\n', '', '(a) In an A.P, the difference between the 8th and 4th terms is 20 and the 8th term is \\(1\\frac{1}{2}\\) times the 4th term. What is the:\n(i) common difference ; (ii) first term of the sequence?\n(b) The value of a machine depreciates each year by 5% of its value at the beginning of that year. If its value when new on 1st January 1980 was N10,250.00, what was its value in January 1989 when it was 9 years old? Give your answer correct to three significant figures.', '(a) In an A.P, the difference between the 8th and 4th terms is 20 and the 8th term is \\(1\\frac{1}{2}\\) times the 4th term. What is the:\n(i) common difference ; (ii) first term of the sequence?~(b) The value of a machine depreciates each year by 5% of its value at the beginning of that year. If its value when new on 1st January 1980 was N10,250.00, what was its value in January 1989 when it was 9 years old? Give your answer correct to three significant figures.', NULL, '\n\nExplanation\n(a) \\(T_{n} = a + (n - 1) d\\) (terms of an A.P)\nGiven: \\(T_{8} - T_{4} = 20\\)\n\\(T_{8} = 1\\frac{1}{2} \\times T_{4}\\)\n\\(\\therefore a + 7d - (a + 3d) = 20\\)\n\\(4d = 20 \\implies d = 5\\)\n(ii) Put d = 5 in the equation\n\\(a + 7d = 1\\frac{1}{2} \\times (a + 3d)\\)\n\\(a + 7d = 1.5a + 4.5d\\)\n\\(7d - 4.5d = 1.5a - a \\implies 2.5d = 0.5a\\)\n\\(a = \\frac{2.5 \\times 5}{0.5} = 25\\)\n(b) \\(A = P(1- \\frac{r}{100})^{n}\\)\nwhere P = N10,250.00\nr = 5%\nn = 9\n\\(\\therefore A = (10,250) (1 - \\frac{5}{100})^{9}\\)\n= \\(10,250 \\times (0.95)^{9} = 10,250 \\times 0.6302\\)\n= \\(N6,460.056 \\approxeq N6,460\\) (to 3 sig. figs)\n\n\n', '', 'Explanation\n(a) \\(T_{n} = a + (n - 1) d\\) (terms of an A.P)\nGiven: \\(T_{8} - T_{4} = 20\\)\n\\(T_{8} = 1\\frac{1}{2} \\times T_{4}\\)\n\\(\\therefore a + 7d - (a + 3d) = 20\\)\n\\(4d = 20 \\implies d = 5\\)\n(ii) Put d = 5 in the equation\n\\(a + 7d = 1\\frac{1}{2} \\times (a + 3d)\\)\n\\(a + 7d = 1.5a + 4.5d\\)\n\\(7d - 4.5d = 1.5a - a \\implies 2.5d = 0.5a\\)\n\\(a = \\frac{2.5 \\times 5}{0.5} = 25\\)\n(b) \\(A = P(1- \\frac{r}{100})^{n}\\)\nwhere P = N10,250.00\nr = 5%\nn = 9\n\\(\\therefore A = (10,250) (1 - \\frac{5}{100})^{9}\\)\n= \\(10,250 \\times (0.95)^{9} = 10,250 \\times 0.6302\\)\n= \\(N6,460.056 \\approxeq N6,460\\) (to 3 sig. figs)\n\n\n', '(a) \\(T_{n} = a + (n - 1) d\\) (terms of an A.P)\nGiven: \\(T_{8} - T_{4} = 20\\)\n\\(T_{8} = 1\\frac{1}{2} \\times T_{4}\\)\n\\(\\therefore a + 7d - (a + 3d) = 20\\)\n\\(4d = 20 \\implies d = 5\\)\n(ii) Put d = 5 in the equation\n\\(a + 7d = 1\\frac{1}{2} \\times (a + 3d)\\)\n\\(a + 7d = 1.5a + 4.5d\\)\n\\(7d - 4.5d = 1.5a - a \\implies 2.5d = 0.5a\\)\n\\(a = \\frac{2.5 \\times 5}{0.5} = 25\\)~(b) \\(A = P(1- \\frac{r}{100})^{n}\\)\nwhere P = N10,250.00\nr = 5%\nn = 9\n\\(\\therefore A = (10,250) (1 - \\frac{5}{100})^{9}\\)\n= \\(10,250 \\times (0.95)^{9} = 10,250 \\times 0.6302\\)\n= \\(N6,460.056 \\approxeq N6,460\\) (to 3 sig. figs)\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(693, '2018', '\n\nQuestion\n\n(a) A pair of fair dice each numbered 1 to 6 is tossed. Find the probability of getting a sum of at least 9.\n(b) If the probability that a civil servant owns a car is \\(\\frac{1}{6}\\), find the probability that:\n(i) two civil servants, A and B, selected at random each owns a car ; (ii) of two civil servants, C and D selected at random, only one owns a car ; (iii) of three civil servants, X, Y and Z, selected at random, only one owns a car.\n\n', '', '(a) A pair of fair dice each numbered 1 to 6 is tossed. Find the probability of getting a sum of at least 9.\n(b) If the probability that a civil servant owns a car is \\(\\frac{1}{6}\\), find the probability that:\n(i) two civil servants, A and B, selected at random each owns a car ; (ii) of two civil servants, C and D selected at random, only one owns a car ; (iii) of three civil servants, X, Y and Z, selected at random, only one owns a car.', '(a) A pair of fair dice each numbered 1 to 6 is tossed. Find the probability of getting a sum of at least 9.~(b) If the probability that a civil servant owns a car is \\(\\frac{1}{6}\\), find the probability that:\n(i) two civil servants, A and B, selected at random each owns a car ; (ii) of two civil servants, C and D selected at random, only one owns a car ; (iii) of three civil servants, X, Y and Z, selected at random, only one owns a car.', NULL, '\n\nExplanation\n(a) Sample space for a pair of fair dice = 36 results\n\n\n\n+\n1\n2\n3\n4\n5\n6\n\n\n1\n2\n3\n4\n5\n6\n7\n\n\n2\n3\n4\n5\n6\n7\n8\n\n\n3\n4\n5\n6\n7\n8\n9\n\n\n4\n5\n6\n7\n8\n9\n10\n\n\n5\n6\n7\n8\n9\n10\n11\n\n\n6\n7\n8\n9\n10\n11\n12\n\n\n\nTotal with sum at least 9 = 10\nP(sum at least 9) = \\(\\frac{10}{36} = \\frac{5}{18}\\)\n(b) P(civil servant owns a car) = \\(\\frac{1}{6}\\)\nP(civil servant owns no car) = \\(\\frac{5}{6}\\)\n(i) P(A and B own a car) = \\(\\frac{1}{6} \\times \\frac{1}{6} = \\frac{1}{36}\\)\n(ii) P(C and D, only one owns a car) = P(C owns, not D) + P(not C but D owns)\n= \\((\\frac{1}{6} \\times \\frac{5}{6}) + (\\frac{5}{6} \\times \\frac{1}{6})\\)\n= \\(\\frac{5}{36} + \\frac{5}{36}\\)\n= \\(\\frac{10}{36} = \\frac{5}{18}\\)\n(iii) P(of X, Y and Z, only one owns a car) = P(X owns a car, not Y, not Z) + P(not X, Y owns a car, not Z) + P(not X, not Y, Z owns a car)\n= \\(3(\\frac{1}{6} \\times \\frac{5}{6} \\times \\frac{5}{6})\\)\n= \\(3 \\times \\frac{25}{216}\\)\n= \\(\\frac{25}{72}\\)\n\n', '', 'Explanation\n(a) Sample space for a pair of fair dice = 36 results\n\n\n\n+\n1\n2\n3\n4\n5\n6\n\n\n1\n2\n3\n4\n5\n6\n7\n\n\n2\n3\n4\n5\n6\n7\n8\n\n\n3\n4\n5\n6\n7\n8\n9\n\n\n4\n5\n6\n7\n8\n9\n10\n\n\n5\n6\n7\n8\n9\n10\n11\n\n\n6\n7\n8\n9\n10\n11\n12\n\n\n\nTotal with sum at least 9 = 10\nP(sum at least 9) = \\(\\frac{10}{36} = \\frac{5}{18}\\)\n(b) P(civil servant owns a car) = \\(\\frac{1}{6}\\)\nP(civil servant owns no car) = \\(\\frac{5}{6}\\)\n(i) P(A and B own a car) = \\(\\frac{1}{6} \\times \\frac{1}{6} = \\frac{1}{36}\\)\n(ii) P(C and D, only one owns a car) = P(C owns, not D) + P(not C but D owns)\n= \\((\\frac{1}{6} \\times \\frac{5}{6}) + (\\frac{5}{6} \\times \\frac{1}{6})\\)\n= \\(\\frac{5}{36} + \\frac{5}{36}\\)\n= \\(\\frac{10}{36} = \\frac{5}{18}\\)\n(iii) P(of X, Y and Z, only one owns a car) = P(X owns a car, not Y, not Z) + P(not X, Y owns a car, not Z) + P(not X, not Y, Z owns a car)\n= \\(3(\\frac{1}{6} \\times \\frac{5}{6} \\times \\frac{5}{6})\\)\n= \\(3 \\times \\frac{25}{216}\\)\n= \\(\\frac{25}{72}\\)\n\n', '(a) Sample space for a pair of fair dice = 36 results\n\n\n\n+\n1\n2\n3\n4\n5\n6\n\n\n1\n2\n3\n4\n5\n6\n7\n\n\n2\n3\n4\n5\n6\n7\n8\n\n\n3\n4\n5\n6\n7\n8\n9\n\n\n4\n5\n6\n7\n8\n9\n10\n\n\n5\n6\n7\n8\n9\n10\n11\n\n\n6\n7\n8\n9\n10\n11\n12\n\n\n\nTotal with sum at least 9 = 10\nP(sum at least 9) = \\(\\frac{10}{36} = \\frac{5}{18}\\)~(b) P(civil servant owns a car) = \\(\\frac{1}{6}\\)\nP(civil servant owns no car) = \\(\\frac{5}{6}\\)\n(i) P(A and B own a car) = \\(\\frac{1}{6} \\times \\frac{1}{6} = \\frac{1}{36}\\)\n(ii) P(C and D, only one owns a car) = P(C owns, not D) + P(not C but D owns)\n= \\((\\frac{1}{6} \\times \\frac{5}{6}) + (\\frac{5}{6} \\times \\frac{1}{6})\\)\n= \\(\\frac{5}{36} + \\frac{5}{36}\\)\n= \\(\\frac{10}{36} = \\frac{5}{18}\\)\n(iii) P(of X, Y and Z, only one owns a car) = P(X owns a car, not Y, not Z) + P(not X, Y owns a car, not Z) + P(not X, not Y, Z owns a car)\n= \\(3(\\frac{1}{6} \\times \\frac{5}{6} \\times \\frac{5}{6})\\)\n= \\(3 \\times \\frac{25}{216}\\)\n= \\(\\frac{25}{72}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(694, '2018', '\n\nQuestion\n\n(a) Triangle PQR is right-angled at Q. PQ = 3a cm and QR = 4a cm. Determine PR in terms of a.\n(b) Ayo travels a distance of 24km from X on a bearing of 060° to Y. He then travels a distance of 18km to a point Z and Z is 30km from X.\n(i) Draw the diagram to show the positions of X, Y and Z ; (ii) What is the bearing of Z from Y ; (iii) Calculate the bearing of X from Z.\n\n', '', '(a) Triangle PQR is right-angled at Q. PQ = 3a cm and QR = 4a cm. Determine PR in terms of a.\n(b) Ayo travels a distance of 24km from X on a bearing of 060° to Y. He then travels a distance of 18km to a point Z and Z is 30km from X.\n(i) Draw the diagram to show the positions of X, Y and Z ; (ii) What is the bearing of Z from Y ; (iii) Calculate the bearing of X from Z.', '(a) Triangle PQR is right-angled at Q. PQ = 3a cm and QR = 4a cm. Determine PR in terms of a.~(b) Ayo travels a distance of 24km from X on a bearing of 060° to Y. He then travels a distance of 18km to a point Z and Z is 30km from X.\n(i) Draw the diagram to show the positions of X, Y and Z ; (ii) What is the bearing of Z from Y ; (iii) Calculate the bearing of X from Z.', NULL, '\n\nExplanation\n(a)\nFrom the diagram and by Pythagoras theorem,\n\\(|PR|^{2} = |PQ|^{2} + |QR|^{2}\\)\n\\(|PR|^{2} = (3a)^{2} + (4a)^{2}\\)\n\\(|PR|^{2} = 9a^{2} + 16a^{2} = 25a^{2}\\)\n\\(|PR| = \\sqrt{25a^{2}} = 5a\\)\n(b)\n(ii) From the figure, Bearing of Y from X = 060°\nUsing cosine rule,\n\\(\\cos XYZ = \\frac{24^{2} + 18^{2} - 30^{2}}{2 \\times 24 \\times 18}\\)\n\\(\\cos XYZ = 0 \\implies < XYZ = 90°\\)\n\\(< RYZ = 90° - 60° = 30°\\)\n\\(< N_{2} YZ = 180° - 30° = 150°\\) (angles on a straight line)\nThe bearing of Z from Y is 150°.\n(iii) \\(\\cos XZY = \\frac{30^{2} + 18^{2} - 24^{2}}{2 \\times 30 \\times 18}\\)\n= \\(\\frac{900 + 324 - 576}{1080}\\)\n\\(\\cos XZY = \\frac{648}{1080} = 0.6\\)\n\\(< XZY = \\cos^{-1} (0.6) = 53.13°\\)\n\\(\\therefore\\) The bearing of X from Y = 360° - (53.13° + 30°) = 276.87°.\n\n', '', 'Explanation\n(a)\nFrom the diagram and by Pythagoras theorem,\n\\(|PR|^{2} = |PQ|^{2} + |QR|^{2}\\)\n\\(|PR|^{2} = (3a)^{2} + (4a)^{2}\\)\n\\(|PR|^{2} = 9a^{2} + 16a^{2} = 25a^{2}\\)\n\\(|PR| = \\sqrt{25a^{2}} = 5a\\)\n(b)\n(ii) From the figure, Bearing of Y from X = 060°\nUsing cosine rule,\n\\(\\cos XYZ = \\frac{24^{2} + 18^{2} - 30^{2}}{2 \\times 24 \\times 18}\\)\n\\(\\cos XYZ = 0 \\implies < XYZ = 90°\\)\n\\(< RYZ = 90° - 60° = 30°\\)\n\\(< N_{2} YZ = 180° - 30° = 150°\\) (angles on a straight line)\nThe bearing of Z from Y is 150°.\n(iii) \\(\\cos XZY = \\frac{30^{2} + 18^{2} - 24^{2}}{2 \\times 30 \\times 18}\\)\n= \\(\\frac{900 + 324 - 576}{1080}\\)\n\\(\\cos XZY = \\frac{648}{1080} = 0.6\\)\n\\(< XZY = \\cos^{-1} (0.6) = 53.13°\\)\n\\(\\therefore\\) The bearing of X from Y = 360° - (53.13° + 30°) = 276.87°.\n\n', '(a)\nFrom the diagram and by Pythagoras theorem,\n\\(|PR|^{2} = |PQ|^{2} + |QR|^{2}\\)\n\\(|PR|^{2} = (3a)^{2} + (4a)^{2}\\)\n\\(|PR|^{2} = 9a^{2} + 16a^{2} = 25a^{2}\\)\n\\(|PR| = \\sqrt{25a^{2}} = 5a\\)~(b)\n(ii) From the figure, Bearing of Y from X = 060°\nUsing cosine rule,\n\\(\\cos XYZ = \\frac{24^{2} + 18^{2} - 30^{2}}{2 \\times 24 \\times 18}\\)\n\\(\\cos XYZ = 0 \\implies < XYZ = 90°\\)\n\\(< RYZ = 90° - 60° = 30°\\)\n\\(< N_{2} YZ = 180° - 30° = 150°\\) (angles on a straight line)\nThe bearing of Z from Y is 150°.\n(iii) \\(\\cos XZY = \\frac{30^{2} + 18^{2} - 24^{2}}{2 \\times 30 \\times 18}\\)\n= \\(\\frac{900 + 324 - 576}{1080}\\)\n\\(\\cos XZY = \\frac{648}{1080} = 0.6\\)\n\\(< XZY = \\cos^{-1} (0.6) = 53.13°\\)\n\\(\\therefore\\) The bearing of X from Y = 360° - (53.13° + 30°) = 276.87°.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(695, '2018', '\n\nQuestion\n\n(a) Derive the smallestequation whose coefficients are integers and which has roots of \\(\\frac{1}{2}\\) and -7.\n(b) Three years ago, a father was four times as old as his daughter is now. The product of their present ages is 430. Calculate the ages of the father and daughter.\n\n', '', '(a) Derive the smallestequation whose coefficients are integers and which has roots of \\(\\frac{1}{2}\\) and -7.\n(b) Three years ago, a father was four times as old as his daughter is now. The product of their present ages is 430. Calculate the ages of the father and daughter.', '(a) Derive the smallestequation whose coefficients are integers and which has roots of \\(\\frac{1}{2}\\) and -7.~(b) Three years ago, a father was four times as old as his daughter is now. The product of their present ages is 430. Calculate the ages of the father and daughter.', NULL, '\n\nExplanation\n(a) Roots of the equation = \\(\\frac{1}{2}\\) and -7\n\\(\\therefore (x - \\frac{1}{2})(x + 7) = 0\\)\n\\(x^{2} - \\frac{1}{2} x + 7x - \\frac{7}{2} = 0\\)\n\\(x^{2} + \\frac{13}{2} x - \\frac{7}{2} = 0\\)\nMultiply through by 2 to get integers as coefficients,\n\\(2x^{2} + 13x - 7 = 0\\)\n(b) Let the father and daughter\'s present ages be r and t respectively.\nThree years ago, father = (r - 3) years\n\\((r - 3) = 4t \\implies r = (4t + 3) ... (1)\\)\n\\(r \\times t = 430 ... (2)\\)\nSubstitute (1) in (2),\n\\((4t + 3) \\times t = 430 \\implies 4t^{2} + 3t = 430\\)\n\\(\\implies 4t^{2} + 3t - 430 = 0\\)\n\\(4t^{2} - 40t + 43t - 430 = 0\\)\n\\(4t(t - 10) + 43(t - 10) = 0\\)\n\\(\\therefore (4t + 43)(t - 10) = 0\\)\n\\(t = \\frac{-43}{4}; t = 10\\)\nSince t is age, it cannot be negative. Hence, t = 10 years.\nRecall : \\(r = 4t + 3\\)\n\\(\\therefore r = 4(10) + 3 = 40 + 3 = 43\\)\nTherefore, the father\'s present age is 43 years while the daughter\'s age is 10 years.\n\n', '', 'Explanation\n(a) Roots of the equation = \\(\\frac{1}{2}\\) and -7\n\\(\\therefore (x - \\frac{1}{2})(x + 7) = 0\\)\n\\(x^{2} - \\frac{1}{2} x + 7x - \\frac{7}{2} = 0\\)\n\\(x^{2} + \\frac{13}{2} x - \\frac{7}{2} = 0\\)\nMultiply through by 2 to get integers as coefficients,\n\\(2x^{2} + 13x - 7 = 0\\)\n(b) Let the father and daughter\'s present ages be r and t respectively.\nThree years ago, father = (r - 3) years\n\\((r - 3) = 4t \\implies r = (4t + 3) ... (1)\\)\n\\(r \\times t = 430 ... (2)\\)\nSubstitute (1) in (2),\n\\((4t + 3) \\times t = 430 \\implies 4t^{2} + 3t = 430\\)\n\\(\\implies 4t^{2} + 3t - 430 = 0\\)\n\\(4t^{2} - 40t + 43t - 430 = 0\\)\n\\(4t(t - 10) + 43(t - 10) = 0\\)\n\\(\\therefore (4t + 43)(t - 10) = 0\\)\n\\(t = \\frac{-43}{4}; t = 10\\)\nSince t is age, it cannot be negative. Hence, t = 10 years.\nRecall : \\(r = 4t + 3\\)\n\\(\\therefore r = 4(10) + 3 = 40 + 3 = 43\\)\nTherefore, the father\'s present age is 43 years while the daughter\'s age is 10 years.\n\n', '(a) Roots of the equation = \\(\\frac{1}{2}\\) and -7\n\\(\\therefore (x - \\frac{1}{2})(x + 7) = 0\\)\n\\(x^{2} - \\frac{1}{2} x + 7x - \\frac{7}{2} = 0\\)\n\\(x^{2} + \\frac{13}{2} x - \\frac{7}{2} = 0\\)\nMultiply through by 2 to get integers as coefficients,\n\\(2x^{2} + 13x - 7 = 0\\)~(b) Let the father and daughter\'s present ages be r and t respectively.\nThree years ago, father = (r - 3) years\n\\((r - 3) = 4t \\implies r = (4t + 3) ... (1)\\)\n\\(r \\times t = 430 ... (2)\\)\nSubstitute (1) in (2),\n\\((4t + 3) \\times t = 430 \\implies 4t^{2} + 3t = 430\\)\n\\(\\implies 4t^{2} + 3t - 430 = 0\\)\n\\(4t^{2} - 40t + 43t - 430 = 0\\)\n\\(4t(t - 10) + 43(t - 10) = 0\\)\n\\(\\therefore (4t + 43)(t - 10) = 0\\)\n\\(t = \\frac{-43}{4}; t = 10\\)\nSince t is age, it cannot be negative. Hence, t = 10 years.\nRecall : \\(r = 4t + 3\\)\n\\(\\therefore r = 4(10) + 3 = 40 + 3 = 43\\)\nTherefore, the father\'s present age is 43 years while the daughter\'s age is 10 years.\n', 'WAEC-PK-1', 'WAEC-MATH', 1);
INSERT INTO `exam_simulation2` (`id`, `examYear`, `questions`, `parent_que`, `child_que`, `gen_child_que_desc`, `imageDir`, `explanation`, `parent_que_expl`, `child_que_exp`, `gen_child_que_expl`, `package_id`, `subject_id`, `status`) VALUES
(696, '2018', '\n\nQuestion\n\nThe number of items produced by a company over a five- year period is given below:\n\n\n\nYear\n1978\n1979\n1980\n1981\n1982\n\n\nNo produced\n4100\n2500\n1500\n1800\n9200\n\n\n\n(i) Plot a bar chart for this information; (ii) What is the average production for the five- year period.\n\n\n\nThe number of items produced by a company over a five- year period is given below:\n\n\n\nYear\n1978\n1979\n1980\n1981\n1982\n\n\nNo produced\n4100\n2500\n1500\n1800\n9200\n\n\n', '', 'The number of items produced by a company over a five- year period is given below:\n\n\n\nYear\n1978\n1979\n1980\n1981\n1982\n\n\nNo produced\n4100\n2500\n1500\n1800\n9200\n\n\n\n(i) Plot a bar chart for this information; (ii) What is the average production for the five- year period.\n\n\n\nThe number of items produced by a company over a five- year period is given below:\n\n\n\nYear\n1978\n1979\n1980\n1981\n1982\n\n\nNo produced\n4100\n2500\n1500\n1800\n9200\n', '', NULL, '\n\nExplanation\n(i)\n(ii) Average production in the period\n= \\(\\frac{4100 + 2500 + 1500 + 1800 + 9200}{5}\\)\n= \\(\\frac{19100}{5}\\)\n= \\(3820\\)\n\n', '', 'Explanation\n(i)\n(ii) Average production in the period\n= \\(\\frac{4100 + 2500 + 1500 + 1800 + 9200}{5}\\)\n= \\(\\frac{19100}{5}\\)\n= \\(3820\\)\n\n', '', 'WAEC-PK-1', 'WAEC-MATH', 1),
(697, '2018', '\n\nQuestion\n\nIn the diagram, < PQR = < PSQ = 90°, |PS| = 9 cm, |SR| = 16 cm and |SQ| = x cm.\n(a) Find the value of x using a trigonometric ratio.\n(b) Calculate : (i) the size of < QRS to the nearest degree; (ii) |PQ|.\n\n\n', '', 'In the diagram, < PQR = < PSQ = 90°, |PS| = 9 cm, |SR| = 16 cm and |SQ| = x cm.\n(a) Find the value of x using a trigonometric ratio.\n(b) Calculate : (i) the size of < QRS to the nearest degree; (ii) |PQ|.\n', 'In the diagram, < PQR = < PSQ = 90°, |PS| = 9 cm, |SR| = 16 cm and |SQ| = x cm.\n(a) Find the value of x using a trigonometric ratio.~(b) Calculate : (i) the size of < QRS to the nearest degree; (ii) |PQ|.\n', NULL, '\n\nExplanation\n(a)\nLet \\(< SQR = \\theta ; < QRS = 90° - \\theta ; < QPS = \\theta\\)\nIn \\(\\Delta\\) PQS, \\(\\tan \\theta = \\frac{x}{9} ... (1)\\)\nIn \\(\\Delta\\) QRS, \\(\\tan \\theta = \\frac{16}{x} ... (2)\\)\nFrom (1) and (2), \\(\\frac{x}{9} = \\frac{16}{x}\\)\n\\(x^{2} = 16 \\times 9 \\implies x = \\sqrt{16 \\times 9} = 12 cm\\)\n(b) (i) From \\(\\Delta\\) QRS,\n\\(\\tan \\theta = \\frac{16}{12} = 1.333\\)\n\\(\\theta = \\tan^{-1} (1.333) = 53.1°\\)\n\\(\\therefore < QRS = 90° - 53.1° = 36.9°\\)\n\\(\\approxeq 37°\\) (to the nearest degree)\n(ii) In \\(\\Delta\\) PQS,\n\\(|PQ|^{2} = 12^{2} + 9^{2}\\)\n= \\(144 + 81 = 225\\)\n\\(|PQ| = \\sqrt{225} = 15 cm\\)\n\n', '', 'Explanation\n(a)\nLet \\(< SQR = \\theta ; < QRS = 90° - \\theta ; < QPS = \\theta\\)\nIn \\(\\Delta\\) PQS, \\(\\tan \\theta = \\frac{x}{9} ... (1)\\)\nIn \\(\\Delta\\) QRS, \\(\\tan \\theta = \\frac{16}{x} ... (2)\\)\nFrom (1) and (2), \\(\\frac{x}{9} = \\frac{16}{x}\\)\n\\(x^{2} = 16 \\times 9 \\implies x = \\sqrt{16 \\times 9} = 12 cm\\)\n(b) (i) From \\(\\Delta\\) QRS,\n\\(\\tan \\theta = \\frac{16}{12} = 1.333\\)\n\\(\\theta = \\tan^{-1} (1.333) = 53.1°\\)\n\\(\\therefore < QRS = 90° - 53.1° = 36.9°\\)\n\\(\\approxeq 37°\\) (to the nearest degree)\n(ii) In \\(\\Delta\\) PQS,\n\\(|PQ|^{2} = 12^{2} + 9^{2}\\)\n= \\(144 + 81 = 225\\)\n\\(|PQ| = \\sqrt{225} = 15 cm\\)\n\n', '(a)\nLet \\(< SQR = \\theta ; < QRS = 90° - \\theta ; < QPS = \\theta\\)\nIn \\(\\Delta\\) PQS, \\(\\tan \\theta = \\frac{x}{9} ... (1)\\)\nIn \\(\\Delta\\) QRS, \\(\\tan \\theta = \\frac{16}{x} ... (2)\\)\nFrom (1) and (2), \\(\\frac{x}{9} = \\frac{16}{x}\\)\n\\(x^{2} = 16 \\times 9 \\implies x = \\sqrt{16 \\times 9} = 12 cm\\)~(b) (i) From \\(\\Delta\\) QRS,\n\\(\\tan \\theta = \\frac{16}{12} = 1.333\\)\n\\(\\theta = \\tan^{-1} (1.333) = 53.1°\\)\n\\(\\therefore < QRS = 90° - 53.1° = 36.9°\\)\n\\(\\approxeq 37°\\) (to the nearest degree)\n(ii) In \\(\\Delta\\) PQS,\n\\(|PQ|^{2} = 12^{2} + 9^{2}\\)\n= \\(144 + 81 = 225\\)\n\\(|PQ| = \\sqrt{225} = 15 cm\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(698, '2018', '\n\nQuestion\n\n(a) Given that \\(\\sin \\alpha = 0.3907\\), use tables to find the value of : (i) \\(\\tan \\alpha\\) ; (ii) \\(\\cos \\alpha\\).\n(b) A ladder of length 4.5m leans against a vertical wall making an angle of 50° with the horizontal ground. If the bottom of a window is 4m above the ground, what is the distance between the top of the ladder and the bottom of the window? (Answer correct to the nearest cm)\n\n', '', '(a) Given that \\(\\sin \\alpha = 0.3907\\), use tables to find the value of : (i) \\(\\tan \\alpha\\) ; (ii) \\(\\cos \\alpha\\).\n(b) A ladder of length 4.5m leans against a vertical wall making an angle of 50° with the horizontal ground. If the bottom of a window is 4m above the ground, what is the distance between the top of the ladder and the bottom of the window? (Answer correct to the nearest cm)', '(a) Given that \\(\\sin \\alpha = 0.3907\\), use tables to find the value of : (i) \\(\\tan \\alpha\\) ; (ii) \\(\\cos \\alpha\\).~(b) A ladder of length 4.5m leans against a vertical wall making an angle of 50° with the horizontal ground. If the bottom of a window is 4m above the ground, what is the distance between the top of the ladder and the bottom of the window? (Answer correct to the nearest cm)', NULL, '\n\nExplanation\n(a) Given \\(\\sin \\alpha = 0.3907\\)\nFrom the sine table, \\(\\alpha = 23°\\)\n(i) \\(\\tan \\alpha = \\tan 23° = 0.4245\\)\n(ii) \\(\\cos \\alpha = \\cos 23° = 0.9205\\)\n(b) Let T be the top of the ladder and B, the bottom of the vertical wall.\n\nFrom the figure, \\(\\sin 50° = \\frac{BT}{4.5m}\\)\n\\(BT = 4.5 \\sin 50° = 4.5 \\times 0.7660\\)\n\\(BT = 3.447m\\)\nDistance between the top of the ladder and bottom of the window\n= \\(4 - 3.447 = 0.553m\\)\n= \\( 0.553m \\times 100\\)\n= \\(55.3 cm\\)\n\n', '', 'Explanation\n(a) Given \\(\\sin \\alpha = 0.3907\\)\nFrom the sine table, \\(\\alpha = 23°\\)\n(i) \\(\\tan \\alpha = \\tan 23° = 0.4245\\)\n(ii) \\(\\cos \\alpha = \\cos 23° = 0.9205\\)\n(b) Let T be the top of the ladder and B, the bottom of the vertical wall.\n\nFrom the figure, \\(\\sin 50° = \\frac{BT}{4.5m}\\)\n\\(BT = 4.5 \\sin 50° = 4.5 \\times 0.7660\\)\n\\(BT = 3.447m\\)\nDistance between the top of the ladder and bottom of the window\n= \\(4 - 3.447 = 0.553m\\)\n= \\( 0.553m \\times 100\\)\n= \\(55.3 cm\\)\n\n', '(a) Given \\(\\sin \\alpha = 0.3907\\)\nFrom the sine table, \\(\\alpha = 23°\\)\n(i) \\(\\tan \\alpha = \\tan 23° = 0.4245\\)\n(ii) \\(\\cos \\alpha = \\cos 23° = 0.9205\\)~(b) Let T be the top of the ladder and B, the bottom of the vertical wall.\n\nFrom the figure, \\(\\sin 50° = \\frac{BT}{4.5m}\\)\n\\(BT = 4.5 \\sin 50° = 4.5 \\times 0.7660\\)\n\\(BT = 3.447m\\)\nDistance between the top of the ladder and bottom of the window\n= \\(4 - 3.447 = 0.553m\\)\n= \\( 0.553m \\times 100\\)\n= \\(55.3 cm\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(699, '2018', '\n\nQuestion\n\nIn a certain school, the lesson periods for each week are as itemised below: English 10, Mathematics 7, Biology 3, Statistics 4, Ibo 3, others 9. Draw a pie chart to illustrate this information.\n\n', 'In a certain school, the lesson periods for each week are as itemised below: English 10, Mathematics 7, Biology 3, Statistics 4, Ibo 3, others 9. Draw a pie chart to illustrate this information.', '', '', NULL, '\n\nExplanation\nSubject\nPeriod\nAngle of sector in pie chart\n\n\nEnglish\n10\n\\(\\frac{10}{36} \\times 360° = 100°\\)\n\n\nMathematics\n7\n\\(\\frac{7}{36} \\times 360° = 70°\\)\n\n\nBiology\n3\n\\(\\frac{3}{36} \\times 360° = 30°\\)\n\n\nStatistics\n4\n\\(\\frac{4}{36} \\times 360° = 40°\\)\n\n\nIgbo\n3\n\\(\\frac{3}{36} \\times 360° = 30°\\)\n\n\nOthers\n9\n\\(\\frac{9}{36} \\times 360° = 90°\\)\n\n\nTotal\n36\n360°\n\n\n\n\n\n\n\n', 'Subject\nPeriod\nAngle of sector in pie chart\n\n\nEnglish\n10\n\\(\\frac{10}{36} \\times 360° = 100°\\)\n\n\nMathematics\n7\n\\(\\frac{7}{36} \\times 360° = 70°\\)\n\n\nBiology\n3\n\\(\\frac{3}{36} \\times 360° = 30°\\)\n\n\nStatistics\n4\n\\(\\frac{4}{36} \\times 360° = 40°\\)\n\n\nIgbo\n3\n\\(\\frac{3}{36} \\times 360° = 30°\\)\n\n\nOthers\n9\n\\(\\frac{9}{36} \\times 360° = 90°\\)\n\n\nTotal\n36\n360°\n\n\n\n\n\n\n\n', '', '', 'WAEC-PK-1', 'WAEC-MATH', 1),
(700, '2018', '\n\nQuestion\n\nA sector of a circle of radius 7cm subtending an angle of 270° at the centre of the circle is used to form a cone.\n(a) Find the base radius of the cone.\n(b) Calculate the area of the base of the cone to the nearest square centimetre.\n[Take \\(\\pi = \\frac{22}{7}\\)]\n\n', '', 'A sector of a circle of radius 7cm subtending an angle of 270° at the centre of the circle is used to form a cone.\n(a) Find the base radius of the cone.\n(b) Calculate the area of the base of the cone to the nearest square centimetre.\n[Take \\(\\pi = \\frac{22}{7}\\)]', 'A sector of a circle of radius 7cm subtending an angle of 270° at the centre of the circle is used to form a cone.\n(a) Find the base radius of the cone.~(b) Calculate the area of the base of the cone to the nearest square centimetre.\n[Take \\(\\pi = \\frac{22}{7}\\)]', NULL, '\n\nExplanation\n(a) Length of the arc XY = Circumference of the base.\n\\(\\frac{\\theta}{360°} \\times 2\\pi r\\)\nwhere \\(\\theta\\) = 270° ; r = 7cm.\nLength of arc XY = \\(\\frac{270°}{360°} \\times 2 \\times \\frac{22}{7} \\times 7 = 33cm\\)\nLet the radius of the base of the cone = x\n\\(2 \\pi r = 33cm\\)\n\\(2 \\times \\frac{22}{7} \\times x = 33\\)\n\\(\\frac{44}{7} \\times x = 33 \\implies x = \\frac{33 \\times 7}{44}\\)\n\\(x = 5.25cm\\)\n(b) Area of the base of the cone = \\(\\pi r^{2}\\)\n= \\(\\frac{22}{7} \\times \\frac{21}{4} \\times \\frac{21}{4}\\)\n= \\(\\frac{693}{8} cm^{2} = 86.625 cm^{2}\\)\n\\(\\approxeq 87 cm^{2}\\)\n\n', '', 'Explanation\n(a) Length of the arc XY = Circumference of the base.\n\\(\\frac{\\theta}{360°} \\times 2\\pi r\\)\nwhere \\(\\theta\\) = 270° ; r = 7cm.\nLength of arc XY = \\(\\frac{270°}{360°} \\times 2 \\times \\frac{22}{7} \\times 7 = 33cm\\)\nLet the radius of the base of the cone = x\n\\(2 \\pi r = 33cm\\)\n\\(2 \\times \\frac{22}{7} \\times x = 33\\)\n\\(\\frac{44}{7} \\times x = 33 \\implies x = \\frac{33 \\times 7}{44}\\)\n\\(x = 5.25cm\\)\n(b) Area of the base of the cone = \\(\\pi r^{2}\\)\n= \\(\\frac{22}{7} \\times \\frac{21}{4} \\times \\frac{21}{4}\\)\n= \\(\\frac{693}{8} cm^{2} = 86.625 cm^{2}\\)\n\\(\\approxeq 87 cm^{2}\\)\n\n', '(a) Length of the arc XY = Circumference of the base.\n\\(\\frac{\\theta}{360°} \\times 2\\pi r\\)\nwhere \\(\\theta\\) = 270° ; r = 7cm.\nLength of arc XY = \\(\\frac{270°}{360°} \\times 2 \\times \\frac{22}{7} \\times 7 = 33cm\\)\nLet the radius of the base of the cone = x\n\\(2 \\pi r = 33cm\\)\n\\(2 \\times \\frac{22}{7} \\times x = 33\\)\n\\(\\frac{44}{7} \\times x = 33 \\implies x = \\frac{33 \\times 7}{44}\\)\n\\(x = 5.25cm\\)~(b) Area of the base of the cone = \\(\\pi r^{2}\\)\n= \\(\\frac{22}{7} \\times \\frac{21}{4} \\times \\frac{21}{4}\\)\n= \\(\\frac{693}{8} cm^{2} = 86.625 cm^{2}\\)\n\\(\\approxeq 87 cm^{2}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(701, '2018', '\n\nQuestion\n\n(a) The subsets A, B and C of a universal set are defined as follows :\nA = {m, a, p, e} ; B = {a, e, i, o, u} ; C = {l, m, n, o, p, q, r, s, t, u}. List the elements of the following sets.\n(i) \\(A \\cup B\\) ; (ii) \\(A \\cup C\\) ; (iii) \\(A \\cup (B \\cap C)\\).\n(b) Out of the 400 students in the final year in a Senior Secondary School, 300 are offering Biology and 190 are offering Chemistry.\n(i) How many students are offering both Biology and Chemistry, if only 70 students are offering neither Biology nor Chemistry? (ii) How many students are offering at least one of Biology or Chemistry?\n\n', '', '(a) The subsets A, B and C of a universal set are defined as follows :\nA = {m, a, p, e} ; B = {a, e, i, o, u} ; C = {l, m, n, o, p, q, r, s, t, u}. List the elements of the following sets.\n(i) \\(A \\cup B\\) ; (ii) \\(A \\cup C\\) ; (iii) \\(A \\cup (B \\cap C)\\).\n(b) Out of the 400 students in the final year in a Senior Secondary School, 300 are offering Biology and 190 are offering Chemistry.\n(i) How many students are offering both Biology and Chemistry, if only 70 students are offering neither Biology nor Chemistry? (ii) How many students are offering at least one of Biology or Chemistry?', '(a) The subsets A, B and C of a universal set are defined as follows :\nA = {m, a, p, e} ; B = {a, e, i, o, u} ; C = {l, m, n, o, p, q, r, s, t, u}. List the elements of the following sets.\n(i) \\(A \\cup B\\) ; (ii) \\(A \\cup C\\) ; (iii) \\(A \\cup (B \\cap C)\\).~(b) Out of the 400 students in the final year in a Senior Secondary School, 300 are offering Biology and 190 are offering Chemistry.\n(i) How many students are offering both Biology and Chemistry, if only 70 students are offering neither Biology nor Chemistry? (ii) How many students are offering at least one of Biology or Chemistry?', NULL, '\n\nExplanation\n(a) A = {m, a, p, e}; B = {a, e, i, o, u} ; C = {l, m, n, o, p, q, r, s, t, u}\n(i) \\(A \\cup B = {a, e, i, m, o, p, u}\\)\n(ii) \\(A \\cup C = {a, e, l, m, n, o, p, q, r, s, t, u}\\)\n(iii) \\(A \\cup (B \\cap C)\\)\n\\(B \\cap C = {o, u}\\)\n\\(A \\cup (B \\cap C) = {m, a, p, e, o, u}\\)\n\nBiology students = n(B) = 190\nChemistry students = n(A) = 300\nLet \\(n(A \\cap B) = a\\)\n(i) From the Venn diagram,\n\\((300 - a) + a + (190 - a) + 70 = 400\\)\n\\(560 - a = 400 \\implies a = 160\\)\n\\(\\therefore\\) 160 students offer both Biology and Chemistry.\n(ii) Students offering at least one of Biology and Chemistry\n= \\(n(A) + n(B) + n(A \\cap B)\\)\n= \\((300 - 160) + (190 - 160) + 160\\)\n= \\(140 + 30 + 160\\)\n= 330 students\n330 students offer at least one of Biology and Chemistry.\n\n\n', '', 'Explanation\n(a) A = {m, a, p, e}; B = {a, e, i, o, u} ; C = {l, m, n, o, p, q, r, s, t, u}\n(i) \\(A \\cup B = {a, e, i, m, o, p, u}\\)\n(ii) \\(A \\cup C = {a, e, l, m, n, o, p, q, r, s, t, u}\\)\n(iii) \\(A \\cup (B \\cap C)\\)\n\\(B \\cap C = {o, u}\\)\n\\(A \\cup (B \\cap C) = {m, a, p, e, o, u}\\)\n\nBiology students = n(B) = 190\nChemistry students = n(A) = 300\nLet \\(n(A \\cap B) = a\\)\n(i) From the Venn diagram,\n\\((300 - a) + a + (190 - a) + 70 = 400\\)\n\\(560 - a = 400 \\implies a = 160\\)\n\\(\\therefore\\) 160 students offer both Biology and Chemistry.\n(ii) Students offering at least one of Biology and Chemistry\n= \\(n(A) + n(B) + n(A \\cap B)\\)\n= \\((300 - 160) + (190 - 160) + 160\\)\n= \\(140 + 30 + 160\\)\n= 330 students\n330 students offer at least one of Biology and Chemistry.\n\n\n', '', 'WAEC-PK-1', 'WAEC-MATH', 1),
(702, '2018', '\n\nQuestion\n\nIllustrate the following on graph paper and shade the region which satisfies all the three inequalities at the same time :\n\\(- x + 5y \\leq 10 ; 3x - 4y \\leq 8\\) and \\(x > -1\\).\n\n', 'Illustrate the following on graph paper and shade the region which satisfies all the three inequalities at the same time :\n\\(- x + 5y \\leq 10 ; 3x - 4y \\leq 8\\) and \\(x > -1\\).', '', '', NULL, '\n\nExplanation\nDraw the table of value such that \\(-x + 5y \\leq 10 \\implies -x + 5y = 10\\)\n\\(\\implies 5y = 10 + x \\)\n\\(y = 2 + \\frac{x}{5}\\)\nand \\(3x - 4y \\leq 8 \\implies 3x - 4y = 8\\)\n\\(y = -2 + \\frac{3x}{4}\\)\nTable of value for \\(-x + 5y = 10\\)\n\n\n\nx\n-4\n0\n4\n\n\ny\n1.2\n2\n2.8\n\n\n\nTable of value for \\(3x - 4y = 8\\)\n\n\n\nx\n-4\n0\n4\n\n\ny\n-5\n-2\n1\n\n\n\n\n\n\n\n', 'Draw the table of value such that \\(-x + 5y \\leq 10 \\implies -x + 5y = 10\\)\n\\(\\implies 5y = 10 + x \\)\n\\(y = 2 + \\frac{x}{5}\\)\nand \\(3x - 4y \\leq 8 \\implies 3x - 4y = 8\\)\n\\(y = -2 + \\frac{3x}{4}\\)\nTable of value for \\(-x + 5y = 10\\)\n\n\n\nx\n-4\n0\n4\n\n\ny\n1.2\n2\n2.8\n\n\n\nTable of value for \\(3x - 4y = 8\\)\n\n\n\nx\n-4\n0\n4\n\n\ny\n-5\n-2\n1\n\n\n\n\n\n\n\n', '', '', 'WAEC-PK-1', 'WAEC-MATH', 1),
(703, '2018', '\n\nQuestion\n\nThe table below is for the relation \\(y = 2 + x - x^{2}\\)\n\n\n\nx\n-2\n-1.5\n-1\n-0.5\n0\n0.5\n1\n1.5\n2\n2.5\n3\n\n\ny\n-4\n-1.75\n0\n1.25\n2\n2.25\n2\n1.25\n0\n-1.75\n-4\n\n\n\n(a) Using a scale of 2cm to 1 unit on each axis, draw the graph of the relation in the interval \\(-2 \\leq x \\leq 3\\).\n(b) From your graph, find the greatest value of y and the value of x for which this occurs.\n(c) Using the same scale and axes, draw the graph of \\(y = 1 - x\\)\n(d) Use your graphs to solve the equation \\(1 + 2x - x^{2} = 0\\)\n\n\n\nThe table below is for the relation \\(y = 2 + x - x^{2}\\)\n\n\n\nx\n-2\n-1.5\n-1\n-0.5\n0\n0.5\n1\n1.5\n2\n2.5\n3\n\n\ny\n-4\n-1.75\n0\n1.25\n2\n2.25\n2\n1.25\n0\n-1.75\n-4\n\n\n', '', 'The table below is for the relation \\(y = 2 + x - x^{2}\\)\n\n\n\nx\n-2\n-1.5\n-1\n-0.5\n0\n0.5\n1\n1.5\n2\n2.5\n3\n\n\ny\n-4\n-1.75\n0\n1.25\n2\n2.25\n2\n1.25\n0\n-1.75\n-4\n\n\n\n(a) Using a scale of 2cm to 1 unit on each axis, draw the graph of the relation in the interval \\(-2 \\leq x \\leq 3\\).\n(b) From your graph, find the greatest value of y and the value of x for which this occurs.\n(c) Using the same scale and axes, draw the graph of \\(y = 1 - x\\)\n(d) Use your graphs to solve the equation \\(1 + 2x - x^{2} = 0\\)\n\n\n\nThe table below is for the relation \\(y = 2 + x - x^{2}\\)\n\n\n\nx\n-2\n-1.5\n-1\n-0.5\n0\n0.5\n1\n1.5\n2\n2.5\n3\n\n\ny\n-4\n-1.75\n0\n1.25\n2\n2.25\n2\n1.25\n0\n-1.75\n-4\n', 'The table below is for the relation \\(y = 2 + x - x^{2}\\)\n\n\n\nx\n-2\n-1.5\n-1\n-0.5\n0\n0.5\n1\n1.5\n2\n2.5\n3\n\n\ny\n-4\n-1.75\n0\n1.25\n2\n2.25\n2\n1.25\n0\n-1.75\n-4\n\n\n\n(a) Using a scale of 2cm to 1 unit on each axis, draw the graph of the relation in the interval \\(-2 \\leq x \\leq 3\\).~(b) From your graph, find the greatest value of y and the value of x for which this occurs.\n(c) Using the same scale and axes, draw the graph of \\(y = 1 - x\\)\n(d) Use your graphs to solve the equation \\(1 + 2x - x^{2} = 0\\)\n\n\n\nThe table below is for the relation \\(y = 2 + x - x^{2}\\)\n\n\n\nx\n-2\n-1.5\n-1\n-0.5\n0\n0.5\n1\n1.5\n2\n2.5\n3\n\n\ny\n-4\n-1.75\n0\n1.25\n2\n2.25\n2\n1.25\n0\n-1.75\n-4\n~(c) Using the same scale and axes, draw the graph of \\(y = 1 - x\\)\n(d) Use your graphs to solve the equation \\(1 + 2x - x^{2} = 0\\)\n\n\n\nThe table below is for the relation \\(y = 2 + x - x^{2}\\)\n\n\n\nx\n-2\n-1.5\n-1\n-0.5\n0\n0.5\n1\n1.5\n2\n2.5\n3\n\n\ny\n-4\n-1.75\n0\n1.25\n2\n2.25\n2\n1.25\n0\n-1.75\n-4\n~(d) Use your graphs to solve the equation \\(1 + 2x - x^{2} = 0\\)\n\n\n\nThe table below is for the relation \\(y = 2 + x - x^{2}\\)\n\n\n\nx\n-2\n-1.5\n-1\n-0.5\n0\n0.5\n1\n1.5\n2\n2.5\n3\n\n\ny\n-4\n-1.75\n0\n1.25\n2\n2.25\n2\n1.25\n0\n-1.75\n-4\n', NULL, '\n\nExplanation\n(a)\n(b) The greatest value of y = 2.25 and this occurs at x = 0.5.\n(c) Table of values of \\(y = 1 - x\\)\n\n\n\nx\n-2\n2\n\n\ny\n3\n-1\n\n\n\n(d) \\(1 + 2x - x^{2} = 0\\)\n\\(1 + 1 + 2x - x - x^{2} = 1 - x\\)\n\\(2 + x - x^{2} = 1 - x\\)\nThe solution of \\(1 + 2x - x^{2} = 0\\) occurs at the point of intersection of \\(y = 2 + x - x^{2}\\) and \\(y = 1 - x\\). This is at points A (x = -0.8) and B (x = 2.8).\n\n', '', 'Explanation\n(a)\n(b) The greatest value of y = 2.25 and this occurs at x = 0.5.\n(c) Table of values of \\(y = 1 - x\\)\n\n\n\nx\n-2\n2\n\n\ny\n3\n-1\n\n\n\n(d) \\(1 + 2x - x^{2} = 0\\)\n\\(1 + 1 + 2x - x - x^{2} = 1 - x\\)\n\\(2 + x - x^{2} = 1 - x\\)\nThe solution of \\(1 + 2x - x^{2} = 0\\) occurs at the point of intersection of \\(y = 2 + x - x^{2}\\) and \\(y = 1 - x\\). This is at points A (x = -0.8) and B (x = 2.8).\n\n', '~(b) The greatest value of y = 2.25 and this occurs at x = 0.5.\n(c) Table of values of \\(y = 1 - x\\)\n\n\n\nx\n-2\n2\n\n\ny\n3\n-1\n\n\n\n(d) \\(1 + 2x - x^{2} = 0\\)\n\\(1 + 1 + 2x - x - x^{2} = 1 - x\\)\n\\(2 + x - x^{2} = 1 - x\\)\nThe solution of \\(1 + 2x - x^{2} = 0\\) occurs at the point of intersection of \\(y = 2 + x - x^{2}\\) and \\(y = 1 - x\\). This is at points A (x = -0.8) and B (x = 2.8).\n~(c) Table of values of \\(y = 1 - x\\)\n\n\n\nx\n-2\n2\n\n\ny\n3\n-1\n\n\n\n(d) \\(1 + 2x - x^{2} = 0\\)\n\\(1 + 1 + 2x - x - x^{2} = 1 - x\\)\n\\(2 + x - x^{2} = 1 - x\\)\nThe solution of \\(1 + 2x - x^{2} = 0\\) occurs at the point of intersection of \\(y = 2 + x - x^{2}\\) and \\(y = 1 - x\\). This is at points A (x = -0.8) and B (x = 2.8).\n~(d) \\(1 + 2x - x^{2} = 0\\)\n\\(1 + 1 + 2x - x - x^{2} = 1 - x\\)\n\\(2 + x - x^{2} = 1 - x\\)\nThe solution of \\(1 + 2x - x^{2} = 0\\) occurs at the point of intersection of \\(y = 2 + x - x^{2}\\) and \\(y = 1 - x\\). This is at points A (x = -0.8) and B (x = 2.8).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(704, '2018', '\n\nQuestion\n\nTwo towns K and Q are on the parallel of latitude 46°N. The longitude of town K is 130°W and that of town Q is 103°W. A third town P also on latitude 46°N is on longitude 23°E, Calculate:\n(i) the length of the parallel of latitude 46°N, to the nearest 100km;\n(ii) the distance between K and Q, correct to the nearest 100km;\n(iii) the distance between Q and P measured along the parallel of latitude, to the nearest 10km.\n[Take \\(\\pi = 3.142\\); Radius of the earth = 6400km]\n\n', '', 'Two towns K and Q are on the parallel of latitude 46°N. The longitude of town K is 130°W and that of town Q is 103°W. A third town P also on latitude 46°N is on longitude 23°E, Calculate:\n(i) the length of the parallel of latitude 46°N, to the nearest 100km;\n(ii) the distance between K and Q, correct to the nearest 100km;\n(iii) the distance between Q and P measured along the parallel of latitude, to the nearest 10km.\n[Take \\(\\pi = 3.142\\); Radius of the earth = 6400km]', '', NULL, '\n\nExplanation\n\n(i)\nThe length of the parallel of latitude 46°N = \\(2 \\pi R \\cos 46°\\)\n= \\(2 \\times 3.142 \\times 6400 \\times 0.6947\\)\n= \\(27,939.167 km \\approxeq 28000 km\\) (to the nearest 100km)\n(ii) \nLongitudinal difference between K and Q = 130° - 103° = 27°\nDistance between K and Q = length of arc KQ\n= \\(\\frac{27}{360} \\times 2 \\times 3.142 \\times 6400 \\times \\cos 46°\\)\n= \\(2,095.31 km \\approxeq 2100km\\) (to the nearest 100km)\n(iii)\nLongitudinal difference between Q and P = 103° + 23° = 126°\nDistance between Q and P = \\(\\frac{126}{360} \\times 2 \\times 3.142 \\times 6400 \\times \\cos 46°\\)\n= \\(9778.12 km \\approxeq 9780 km\\) (to the nearest 10km)\n\n', '', 'Explanation\n\n(i)\nThe length of the parallel of latitude 46°N = \\(2 \\pi R \\cos 46°\\)\n= \\(2 \\times 3.142 \\times 6400 \\times 0.6947\\)\n= \\(27,939.167 km \\approxeq 28000 km\\) (to the nearest 100km)\n(ii) \nLongitudinal difference between K and Q = 130° - 103° = 27°\nDistance between K and Q = length of arc KQ\n= \\(\\frac{27}{360} \\times 2 \\times 3.142 \\times 6400 \\times \\cos 46°\\)\n= \\(2,095.31 km \\approxeq 2100km\\) (to the nearest 100km)\n(iii)\nLongitudinal difference between Q and P = 103° + 23° = 126°\nDistance between Q and P = \\(\\frac{126}{360} \\times 2 \\times 3.142 \\times 6400 \\times \\cos 46°\\)\n= \\(9778.12 km \\approxeq 9780 km\\) (to the nearest 10km)\n\n', '', 'WAEC-PK-1', 'WAEC-MATH', 1),
(705, '2018', '\n\nQuestion\n\n(a) Prove that the sum of the angles in a triangle is two right angles.\n(b) In a triangle LMN, the side NM is produced to P and the bisector of < LNP meets ML produced at Q. If < LMN = 46°, and < MLN = 80°, calculate < LQN, stating clearly your reasins.\n\n', '', '(a) Prove that the sum of the angles in a triangle is two right angles.\n(b) In a triangle LMN, the side NM is produced to P and the bisector of < LNP meets ML produced at Q. If < LMN = 46°, and < MLN = 80°, calculate < LQN, stating clearly your reasins.', '(a) Prove that the sum of the angles in a triangle is two right angles.~(b) In a triangle LMN, the side NM is produced to P and the bisector of < LNP meets ML produced at Q. If < LMN = 46°, and < MLN = 80°, calculate < LQN, stating clearly your reasins.', NULL, '\n\nExplanation\n(a) Given any \\(\\Delta\\) ABC,\nRequired: That < A + < B + < C = 180°.\nConstruction : Draw CP // BA and produce BC to Q.\n\nProof: Using the above figure,\n\\(a_{1} = a_{2}\\) (alternate angles)\n\\(b_{1} = b_{2}\\) (corresponding angles)\n\\(a_{1} + b_{1} + c = 180°\\) (angles on a straight line)\n\\(\\therefore a_{2} + b_{2} + c = 180°\\)\n\\(< A + < B + < C = 180°\\) (proved)\n(b)\n< LNP = 46° + 80° = 126° (exterior angles of triangle LMN)\n\\(\\theta = \\frac{126°}{2} = 63°\\) (bisector divides into 2 equal halves)\n\\(< MLN = 180° - 80° = 100°\\) (angles on a straight line)\n\\(< LQN = 180° - (63° + 100°) = 17°\\)\n\n', '', 'Explanation\n(a) Given any \\(\\Delta\\) ABC,\nRequired: That < A + < B + < C = 180°.\nConstruction : Draw CP // BA and produce BC to Q.\n\nProof: Using the above figure,\n\\(a_{1} = a_{2}\\) (alternate angles)\n\\(b_{1} = b_{2}\\) (corresponding angles)\n\\(a_{1} + b_{1} + c = 180°\\) (angles on a straight line)\n\\(\\therefore a_{2} + b_{2} + c = 180°\\)\n\\(< A + < B + < C = 180°\\) (proved)\n(b)\n< LNP = 46° + 80° = 126° (exterior angles of triangle LMN)\n\\(\\theta = \\frac{126°}{2} = 63°\\) (bisector divides into 2 equal halves)\n\\(< MLN = 180° - 80° = 100°\\) (angles on a straight line)\n\\(< LQN = 180° - (63° + 100°) = 17°\\)\n\n', '(a) Given any \\(\\Delta\\) ABC,\nRequired: That < A + < B + < C = 180°.\nConstruction : Draw CP // BA and produce BC to Q.\n\nProof: Using the above figure,\n\\(a_{1} = a_{2}\\) (alternate angles)\n\\(b_{1} = b_{2}\\) (corresponding angles)\n\\(a_{1} + b_{1} + c = 180°\\) (angles on a straight line)\n\\(\\therefore a_{2} + b_{2} + c = 180°\\)\n\\(< A + < B + < C = 180°\\) (proved)~(b)\n< LNP = 46° + 80° = 126° (exterior angles of triangle LMN)\n\\(\\theta = \\frac{126°}{2} = 63°\\) (bisector divides into 2 equal halves)\n\\(< MLN = 180° - 80° = 100°\\) (angles on a straight line)\n\\(< LQN = 180° - (63° + 100°) = 17°\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(706, '2018', '\n\nQuestion\n\nUsing a ruler and a pair of compasses only,\n(a) construct (i) a triangle ABC such that |AB| = 5cm, |AC| = 7.5cm and < CAB = 120° ; (ii) the locus \\(L_{1}\\) of points equidistant from A and B ; (iii) the locus \\(L_{2}\\) of points equidistant from Ab and AC, which passes through the triangle ABC.\n(b) Label the point P where \\(L_{1}\\) and \\(L_{2}\\) intersect;\n(c) Measure |CP|.\n\n', '', 'Using a ruler and a pair of compasses only,\n(a) construct (i) a triangle ABC such that |AB| = 5cm, |AC| = 7.5cm and < CAB = 120° ; (ii) the locus \\(L_{1}\\) of points equidistant from A and B ; (iii) the locus \\(L_{2}\\) of points equidistant from Ab and AC, which passes through the triangle ABC.\n(b) Label the point P where \\(L_{1}\\) and \\(L_{2}\\) intersect;\n(c) Measure |CP|.', 'Using a ruler and a pair of compasses only,\n(a) construct (i) a triangle ABC such that |AB| = 5cm, |AC| = 7.5cm and < CAB = 120° ; (ii) the locus \\(L_{1}\\) of points equidistant from A and B ; (iii) the locus \\(L_{2}\\) of points equidistant from Ab and AC, which passes through the triangle ABC.~(b) Label the point P where \\(L_{1}\\) and \\(L_{2}\\) intersect;\n(c) Measure |CP|.~(c) Measure |CP|.', NULL, '\n\nExplanation\n', '', '', '', 'WAEC-PK-1', 'WAEC-MATH', 1),
(707, '2018', '\n\nQuestion\n\nThe table below shows the frequency distribution of the marks of 800 candidates in an examination.\n\n\n\nMarks\n0-9\n10-19\n20-29\n30-39\n40-49\n50-59\n60-69\n70-79\n80-89\n90-99\n\n\nFreq\n10\n40\n80\n140\n170\n130\n100\n70\n40\n20\n\n\n\n(a) (i) Construct a cumulative frequency table ; (ii) Draw the Ogive ; (iii) Use your ogive to determine the 50th percentile.\n(b) The candidates that scored less than 25% are to be withdrawn from the institution, while those that scored than 75% are to be awarded scholarship. Estimate the number of students that will be retained, but will not enjoy the award.\n\n\n\nThe table below shows the frequency distribution of the marks of 800 candidates in an examination.\n\n\n\nMarks\n0-9\n10-19\n20-29\n30-39\n40-49\n50-59\n60-69\n70-79\n80-89\n90-99\n\n\nFreq\n10\n40\n80\n140\n170\n130\n100\n70\n40\n20\n\n\n', '', 'The table below shows the frequency distribution of the marks of 800 candidates in an examination.\n\n\n\nMarks\n0-9\n10-19\n20-29\n30-39\n40-49\n50-59\n60-69\n70-79\n80-89\n90-99\n\n\nFreq\n10\n40\n80\n140\n170\n130\n100\n70\n40\n20\n\n\n\n(a) (i) Construct a cumulative frequency table ; (ii) Draw the Ogive ; (iii) Use your ogive to determine the 50th percentile.\n(b) The candidates that scored less than 25% are to be withdrawn from the institution, while those that scored than 75% are to be awarded scholarship. Estimate the number of students that will be retained, but will not enjoy the award.\n\n\n\nThe table below shows the frequency distribution of the marks of 800 candidates in an examination.\n\n\n\nMarks\n0-9\n10-19\n20-29\n30-39\n40-49\n50-59\n60-69\n70-79\n80-89\n90-99\n\n\nFreq\n10\n40\n80\n140\n170\n130\n100\n70\n40\n20\n', 'The table below shows the frequency distribution of the marks of 800 candidates in an examination.\n\n\n\nMarks\n0-9\n10-19\n20-29\n30-39\n40-49\n50-59\n60-69\n70-79\n80-89\n90-99\n\n\nFreq\n10\n40\n80\n140\n170\n130\n100\n70\n40\n20\n\n\n\n(a) (i) Construct a cumulative frequency table ; (ii) Draw the Ogive ; (iii) Use your ogive to determine the 50th percentile.~(b) The candidates that scored less than 25% are to be withdrawn from the institution, while those that scored than 75% are to be awarded scholarship. Estimate the number of students that will be retained, but will not enjoy the award.\n\n\n\nThe table below shows the frequency distribution of the marks of 800 candidates in an examination.\n\n\n\nMarks\n0-9\n10-19\n20-29\n30-39\n40-49\n50-59\n60-69\n70-79\n80-89\n90-99\n\n\nFreq\n10\n40\n80\n140\n170\n130\n100\n70\n40\n20\n', NULL, '\n\nExplanation\n(a)(i)\n\n\n\nMarks (%)\nFreq\nCum Freq\n\n\n0 - 9\n10\n10\n\n\n10 - 19\n40\n50\n\n\n20 - 29\n80\n130\n\n\n30 - 39\n140\n270\n\n\n40 - 49\n170\n440\n\n\n50 - 59\n130\n570\n\n\n60 - 69\n100\n670\n\n\n70 - 79\n70\n740\n\n\n80 - 89\n40\n780\n\n\n90 - 99\n20\n800\n\n\n\n\n\n\n\n\n\n\n\n(ii)\n(iii) From the Ogive, the 50th percentile is 70 marks.\n(b) Those who scored below 25% (24% and below) = 85students\nThose who scored 75% and below = 710\n\\(\\therefore\\) Those who scored above 75% = 800 - 710 = 90 students\nThose to be retained without award = 800 - (85 + 90)\n= 800 - 175 = 625 students.\n\n', '', 'Explanation\n(a)(i)\n\n\n\nMarks (%)\nFreq\nCum Freq\n\n\n0 - 9\n10\n10\n\n\n10 - 19\n40\n50\n\n\n20 - 29\n80\n130\n\n\n30 - 39\n140\n270\n\n\n40 - 49\n170\n440\n\n\n50 - 59\n130\n570\n\n\n60 - 69\n100\n670\n\n\n70 - 79\n70\n740\n\n\n80 - 89\n40\n780\n\n\n90 - 99\n20\n800\n\n\n\n\n\n\n\n\n\n\n\n(ii)\n(iii) From the Ogive, the 50th percentile is 70 marks.\n(b) Those who scored below 25% (24% and below) = 85students\nThose who scored 75% and below = 710\n\\(\\therefore\\) Those who scored above 75% = 800 - 710 = 90 students\nThose to be retained without award = 800 - (85 + 90)\n= 800 - 175 = 625 students.\n\n', '(a)(i)\n\n\n\nMarks (%)\nFreq\nCum Freq\n\n\n0 - 9\n10\n10\n\n\n10 - 19\n40\n50\n\n\n20 - 29\n80\n130\n\n\n30 - 39\n140\n270\n\n\n40 - 49\n170\n440\n\n\n50 - 59\n130\n570\n\n\n60 - 69\n100\n670\n\n\n70 - 79\n70\n740\n\n\n80 - 89\n40\n780\n\n\n90 - 99\n20\n800\n\n\n\n\n\n\n\n\n\n\n\n(ii)\n(iii) From the Ogive, the 50th percentile is 70 marks.~(b) Those who scored below 25% (24% and below) = 85students\nThose who scored 75% and below = 710\n\\(\\therefore\\) Those who scored above 75% = 800 - 710 = 90 students\nThose to be retained without award = 800 - (85 + 90)\n= 800 - 175 = 625 students.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(708, '2018', '\n\nQuestion\n\n(a) Evaluate without using calculator, (\\(\\frac{1}{4} \\times 9\\frac{1}{7} + \\frac{2}{5} (\\frac{2}{2} + \\frac{3}{4})) \\div (\\frac{2}{5} - \\frac{1}{4})\\)\n(b) A hunter walked 250 m from point P to Q on a bearing 042\\(^o\\). Calculate, correct to the nearest meter the vertical distance he has moved.\n\n', '', '(a) Evaluate without using calculator, (\\(\\frac{1}{4} \\times 9\\frac{1}{7} + \\frac{2}{5} (\\frac{2}{2} + \\frac{3}{4})) \\div (\\frac{2}{5} - \\frac{1}{4})\\)\n(b) A hunter walked 250 m from point P to Q on a bearing 042\\(^o\\). Calculate, correct to the nearest meter the vertical distance he has moved.', '(a) Evaluate without using calculator, (\\(\\frac{1}{4} \\times 9\\frac{1}{7} + \\frac{2}{5} (\\frac{2}{2} + \\frac{3}{4})) \\div (\\frac{2}{5} - \\frac{1}{4})\\)~(b) A hunter walked 250 m from point P to Q on a bearing 042\\(^o\\). Calculate, correct to the nearest meter the vertical distance he has moved.', NULL, '\n\nExplanation\nThe Chief Examiner reported that in part (a), some Candidates\' carried out the requirement of BODMAS correctly tile few couldn\'t. However, in part (b), most Candidates\' could not sketch the diagram correctly and hence failed to earn the marks allocated.\nIn part (a), expressing as improper fraction we have (\\(\\frac{1}{4} \\times 9\\frac{1}{7} + \\frac{2}{5} (\\frac{2}{2} + \\frac{3}{4})) \\div (\\frac{2}{5} - \\frac{1}{4})\\) Next, using BODMAS to get (\\(\\frac{16}{7} + \\frac{2}{5} \\times \\frac{17}{12}) \\div \\frac{17}{12}\\). Simplifying and inverting yields \\(\\frac{1198}{420} \\times \\frac{20}{3} = 19\\frac{1}{63}\\)\nIn part (b), they were expected to sketch\nThen, using trigonometric ratio to have sin 48\\(^o\\) = \\(\\frac{x}{250}\\) or cos 42\\(^o\\) = \\(\\frac{x}{250}\\) and simplifying the vertical distance moved to be 186 m.\n\n\n', 'The Chief Examiner reported that in part (a), some Candidates\' carried out the requirement of BODMAS correctly tile few couldn\'t. However, in part (b), most Candidates\' could not sketch the diagram correctly and hence failed to earn the marks allocated.\nIn part (a), expressing as improper fraction we have (\\(\\frac{1}{4} \\times 9\\frac{1}{7} + \\frac{2}{5} (\\frac{2}{2} + \\frac{3}{4})) \\div (\\frac{2}{5} - \\frac{1}{4})\\) Next, using BODMAS to get (\\(\\frac{16}{7} + \\frac{2}{5} \\times \\frac{17}{12}) \\div \\frac{17}{12}\\). Simplifying and inverting yields \\(\\frac{1198}{420} \\times \\frac{20}{3} = 19\\frac{1}{63}\\)\nIn part (b), they were expected to sketch\nThen, using trigonometric ratio to have sin 48\\(^o\\) = \\(\\frac{x}{250}\\) or cos 42\\(^o\\) = \\(\\frac{x}{250}\\) and simplifying the vertical distance moved to be 186 m.\n\n\n', '', '', 'WAEC-PK-1', 'WAEC-MATH', 1),
(709, '2018', '\n\nQuestion\n\nMusa is three years older than Manya. Seven years ago, Musa was twice as old as Manya.\n(1) How old are they now?\n(2) When  will the sum of them be 45?\n\n', 'Musa is three years older than Manya. Seven years ago, Musa was twice as old as Manya.\n(1) How old are they now?\n(2) When  will the sum of them be 45?', '', '', NULL, '\n\nExplanation\n(a) If Musa\'s age = x and Manya\'s age = y, then x = y + 3.... (1) and (x — 7) = 2(y— 7) .......(2)\nSolving both equations will yield Musa\'s age = 13 years and Manya\'s age = 10 years.\n(b) Let n be the year when the sum of their ages will be 45,\nso that (10 + n) + (13 + n) = 45\nand solving will give n = 11 years.\n\n\n', '', 'Explanation\n(a) If Musa\'s age = x and Manya\'s age = y, then x = y + 3.... (1) and (x — 7) = 2(y— 7) .......(2)\nSolving both equations will yield Musa\'s age = 13 years and Manya\'s age = 10 years.\n(b) Let n be the year when the sum of their ages will be 45,\nso that (10 + n) + (13 + n) = 45\nand solving will give n = 11 years.\n\n\n', '(a) If Musa\'s age = x and Manya\'s age = y, then x = y + 3.... (1) and (x — 7) = 2(y— 7) .......(2)\nSolving both equations will yield Musa\'s age = 13 years and Manya\'s age = 10 years.~(b) Let n be the year when the sum of their ages will be 45,\nso that (10 + n) + (13 + n) = 45\nand solving will give n = 11 years.\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(710, '2018', '\n\nQuestion\n\n\n\nThe diagram shows an athletics track with two parallel sides and two semi-circular ends Each of the parallel sides is 60 metres, long and the diameter of each semi-circular end is 120 metres long.\n(a) Calculate the distance covered by an athlete who runs round the tack the two times. [Take \\(\\pi\\) = \\(\\frac{22}{7}\\)]\n\n(b) If the athlete spends 200 seconds for the race, calculate the speed in km/h.\n\n\n\n\n', '', 'The diagram shows an athletics track with two parallel sides and two semi-circular ends Each of the parallel sides is 60 metres, long and the diameter of each semi-circular end is 120 metres long.\n(a) Calculate the distance covered by an athlete who runs round the tack the two times. [Take \\(\\pi\\) = \\(\\frac{22}{7}\\)]\n\n(b) If the athlete spends 200 seconds for the race, calculate the speed in km/h.\n\n\n', 'The diagram shows an athletics track with two parallel sides and two semi-circular ends Each of the parallel sides is 60 metres, long and the diameter of each semi-circular end is 120 metres long.\n(a) Calculate the distance covered by an athlete who runs round the tack the two times. [Take \\(\\pi\\) = \\(\\frac{22}{7}\\)]\n~(b) If the athlete spends 200 seconds for the race, calculate the speed in km/h.\n\n\n', NULL, '\n\nExplanation\nThe Chief Examiner reported that quite a handful of the Candidates\' got part (a) correctly but couldn\'t establish the relationship between km/h and m/s.\n(a) The distance covered = (2 x 60) + (2 x \\(\\frac{22}{7}\\)) = 120 + 377.142 = 497.1429m so that if the athlete runs two times will covered = 2 x 497.1429\nm = 994.2857m (at least one decimal place).\n(b) The speed m/s = \\(\\frac{994,2857}{200}\\) and converting to km/h we will have \\(\\frac{994,2857}{1000} \\times \\frac{200}{3600}\\) = 17.8973 km/h\n\n', '', 'Explanation\nThe Chief Examiner reported that quite a handful of the Candidates\' got part (a) correctly but couldn\'t establish the relationship between km/h and m/s.\n(a) The distance covered = (2 x 60) + (2 x \\(\\frac{22}{7}\\)) = 120 + 377.142 = 497.1429m so that if the athlete runs two times will covered = 2 x 497.1429\nm = 994.2857m (at least one decimal place).\n(b) The speed m/s = \\(\\frac{994,2857}{200}\\) and converting to km/h we will have \\(\\frac{994,2857}{1000} \\times \\frac{200}{3600}\\) = 17.8973 km/h\n\n', '(a) correctly but couldn\'t establish the relationship between km/h and m/s.\n(a) The distance covered = (2 x 60) + (2 x \\(\\frac{22}{7}\\)) = 120 + 377.142 = 497.1429m so that if the athlete runs two times will covered = 2 x 497.1429\nm = 994.2857m (at least one decimal place).~(b) The speed m/s = \\(\\frac{994,2857}{200}\\) and converting to km/h we will have \\(\\frac{994,2857}{1000} \\times \\frac{200}{3600}\\) = 17.8973 km/h\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(711, '2018', '\n\nQuestion\n\n\n\n(a) The diagram, = IWYI = IXZI and < WXY = 80\\(^o\\). What is the size of < XWZ?\n(b) A man was charged 2 kobo per month for every N1.00 he borrowed from a bank. At what rate per annum was the interest charged?\n\n\n\n\n\n', '', '(a) The diagram, = IWYI = IXZI and < WXY = 80\\(^o\\). What is the size of < XWZ?\n(b) A man was charged 2 kobo per month for every N1.00 he borrowed from a bank. At what rate per annum was the interest charged?\n\n\n\n', '(a) The diagram, = IWYI = IXZI and < WXY = 80\\(^o\\). What is the size of < XWZ?~(b) A man was charged 2 kobo per month for every N1.00 he borrowed from a bank. At what rate per annum was the interest charged?\n\n\n\n', NULL, '\n\nExplanation\n(a) First joining XZ and WY, we have < XWY = < XZY = \\(\\frac{1}{2}\\)(180° — 80°) = 50°. This implies that < ZXY = 50°(base angles of an isosceles triangle). So that < XWZ = 180° - 80° = 100°.\n(b) The rate in every N1.00 = \\(\\frac{2}{100}\\) so that the rate of interest charged per annum will be \\(\\frac{2 \\times 12}{100}\\) x 100 which gives 24%.\n\n\n', '', 'Explanation\n(a) First joining XZ and WY, we have < XWY = < XZY = \\(\\frac{1}{2}\\)(180° — 80°) = 50°. This implies that < ZXY = 50°(base angles of an isosceles triangle). So that < XWZ = 180° - 80° = 100°.\n(b) The rate in every N1.00 = \\(\\frac{2}{100}\\) so that the rate of interest charged per annum will be \\(\\frac{2 \\times 12}{100}\\) x 100 which gives 24%.\n\n\n', '(a) First joining XZ and WY, we have < XWY = < XZY = \\(\\frac{1}{2}\\)(180° — 80°) = 50°. This implies that < ZXY = 50°(base angles of an isosceles triangle). So that < XWZ = 180° - 80° = 100°.~(b) The rate in every N1.00 = \\(\\frac{2}{100}\\) so that the rate of interest charged per annum will be \\(\\frac{2 \\times 12}{100}\\) x 100 which gives 24%.\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(712, '2018', '\n\nQuestion\n\nThe table shows the distribution of sources obtained when a fair diwe was rolled 50 times.\n\n\n\nScore\n1\n2\n3\n4\n5\n6\n\n\nFrequency\n2\n5\n13\n11\n9\n10\n\n\n\n1. Draw a bar chart for the distribution\n2. Calculate the mean score of the distribution\n\n\nThe table shows the distribution of sources obtained when a fair diwe was rolled 50 times.\n\n\n\nScore\n1\n2\n3\n4\n5\n6\n\n\nFrequency\n2\n5\n13\n11\n9\n10\n\n\n', 'The table shows the distribution of sources obtained when a fair diwe was rolled 50 times.\n\n\n\nScore\n1\n2\n3\n4\n5\n6\n\n\nFrequency\n2\n5\n13\n11\n9\n10\n\n\n\n1. Draw a bar chart for the distribution\n2. Calculate the mean score of the distribution\n\n\nThe table shows the distribution of sources obtained when a fair diwe was rolled 50 times.\n\n\n\nScore\n1\n2\n3\n4\n5\n6\n\n\nFrequency\n2\n5\n13\n11\n9\n10\n', '', '', NULL, '\n\nExplanation\n(a) The bar chart for the distribution is as shown\n\n(b) They were able to construct the frequency distribution table as\n\n\n\nScore (x)\n1\n2\n3\n4\n5\n6\nTotal\n\n\nFrequency (f)\n2\n5\n13\n11\n9\n10\n50\n\n\nfx\n2\n10\n39\n44\n45\n60\n200\n\n\n\nMean = \\(\\frac{200}{50}\\) = 4\n\n\n', '', 'Explanation\n(a) The bar chart for the distribution is as shown\n\n(b) They were able to construct the frequency distribution table as\n\n\n\nScore (x)\n1\n2\n3\n4\n5\n6\nTotal\n\n\nFrequency (f)\n2\n5\n13\n11\n9\n10\n50\n\n\nfx\n2\n10\n39\n44\n45\n60\n200\n\n\n\nMean = \\(\\frac{200}{50}\\) = 4\n\n\n', '(a) The bar chart for the distribution is as shown\n~(b) They were able to construct the frequency distribution table as\n\n\n\nScore (x)\n1\n2\n3\n4\n5\n6\nTotal\n\n\nFrequency (f)\n2\n5\n13\n11\n9\n10\n50\n\n\nfx\n2\n10\n39\n44\n45\n60\n200\n\n\n\nMean = \\(\\frac{200}{50}\\) = 4\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(713, '2018', '\n\nQuestion\n\n(a) If tan x = \\(\\frac{5}{12}\\), \\(0^o\\). < x < 90°, evaluate, without using Mathematical tables or calculator, \\(\\frac{sin x}{(sin x)^2 + cosx}\\)\n\n(b) The diagram shows a rectangular lawn measuring 14m by 11m. A path of uniform width \\(x\\)m surrounds it. If the total area of the path is 186 m\\(^2\\), how wide is the path?\n\n\n', '', '(a) If tan x = \\(\\frac{5}{12}\\), \\(0^o\\). < x < 90°, evaluate, without using Mathematical tables or calculator, \\(\\frac{sin x}{(sin x)^2 + cosx}\\)\n\n(b) The diagram shows a rectangular lawn measuring 14m by 11m. A path of uniform width \\(x\\)m surrounds it. If the total area of the path is 186 m\\(^2\\), how wide is the path?\n', '(a) If tan x = \\(\\frac{5}{12}\\), \\(0^o\\). < x < 90°, evaluate, without using Mathematical tables or calculator, \\(\\frac{sin x}{(sin x)^2 + cosx}\\)\n~(b) The diagram shows a rectangular lawn measuring 14m by 11m. A path of uniform width \\(x\\)m surrounds it. If the total area of the path is 186 m\\(^2\\), how wide is the path?\n', NULL, '\n\nExplanation\n(a) By the use of Pythagoras theorem, the hypotenuse side is 13 and also sin x = \\(\\frac{5}{13}\\) andcos x = \\(\\frac{12}{13}\\) Substituting and simplifying yields \\(\\frac{65}{161}\\)\n(b)we have (11 + 2x)(14 + 2x) -(11 x 14) =186.\nExpanding and simplifying to get the quadratic equation 2x\\(^2\\) +25x -93 = 0.\nSolving the quadratic equation results in x = 3 m as the width of the path.\n\n', '', 'Explanation\n(a) By the use of Pythagoras theorem, the hypotenuse side is 13 and also sin x = \\(\\frac{5}{13}\\) andcos x = \\(\\frac{12}{13}\\) Substituting and simplifying yields \\(\\frac{65}{161}\\)\n(b)we have (11 + 2x)(14 + 2x) -(11 x 14) =186.\nExpanding and simplifying to get the quadratic equation 2x\\(^2\\) +25x -93 = 0.\nSolving the quadratic equation results in x = 3 m as the width of the path.\n\n', '(a) By the use of Pythagoras theorem, the hypotenuse side is 13 and also sin x = \\(\\frac{5}{13}\\) andcos x = \\(\\frac{12}{13}\\) Substituting and simplifying yields \\(\\frac{65}{161}\\)~(b)we have (11 + 2x)(14 + 2x) -(11 x 14) =186.\nExpanding and simplifying to get the quadratic equation 2x\\(^2\\) +25x -93 = 0.\nSolving the quadratic equation results in x = 3 m as the width of the path.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(714, '2018', '\n\nQuestion\n\nA shop had two reduction sales during which prices of all items were reduced by 40% in the first sales and 30% in the second.\n\nIf a shirt was sold at GH₵35.00 during the second reduction sales, find the price before the first sale.\nIf the price of an article before the first reduction was GH₵180.00, find the total:\nreduction in the price due to the two sales;\npercentage reduction in the price of the article.\n\n\n\n\n\n\nA shop had two reduction sales during which prices of all items were reduced by 40% in the first sales and 30% in the second.\n\nIf a shirt was sold at GH₵35.00 during the second reduction sales, find the price before the first sale.\nIf the price of an article before the first reduction was GH₵180.00, find the total:\nreduction in the price due to the two sales;\npercentage reduction in the price of the article.\n\n\n\n', '', 'A shop had two reduction sales during which prices of all items were reduced by 40% in the first sales and 30% in the second.\n\nIf a shirt was sold at GH₵35.00 during the second reduction sales, find the price before the first sale.\nIf the price of an article before the first reduction was GH₵180.00, find the total:\nreduction in the price due to the two sales;\npercentage reduction in the price of the article.\n\n\n\n\n\n\nA shop had two reduction sales during which prices of all items were reduced by 40% in the first sales and 30% in the second.\n\nIf a shirt was sold at GH₵35.00 during the second reduction sales, find the price before the first sale.\nIf the price of an article before the first reduction was GH₵180.00, find the total:\nreduction in the price due to the two sales;\npercentage reduction in the price of the article.\n\n', '', NULL, '\n\nExplanation\n(a) let the original price of the shirt be x, then we have two percentages reduction leaving 60% and 70%. This implies that \\(\\frac{60}{100} \\times \\frac{70}{100} \\timesx\\) = 35.\nSolving for x yields x = GH₵ 83.33.\n\n(b) (i), the first reduction = (\\(\\frac{40}{100}\\)) x180 = 72.00 and the second reduction = (\\(\\frac{30}{100}\\) ) x108 = 32.40.\nThen total reduction = (72.00 + 32. 40) = GH₵ 104.40.\n(b) (ii), the percentage reduction in the price of the article = (\\(\\frac{104.40}{180} \\times 100\\)) = 58%\n\n', '', 'Explanation\n(a) let the original price of the shirt be x, then we have two percentages reduction leaving 60% and 70%. This implies that \\(\\frac{60}{100} \\times \\frac{70}{100} \\timesx\\) = 35.\nSolving for x yields x = GH₵ 83.33.\n\n(b) (i), the first reduction = (\\(\\frac{40}{100}\\)) x180 = 72.00 and the second reduction = (\\(\\frac{30}{100}\\) ) x108 = 32.40.\nThen total reduction = (72.00 + 32. 40) = GH₵ 104.40.\n(b) (ii), the percentage reduction in the price of the article = (\\(\\frac{104.40}{180} \\times 100\\)) = 58%\n\n', '(a) let the original price of the shirt be x, then we have two percentages reduction leaving 60% and 70%. This implies that \\(\\frac{60}{100} \\times \\frac{70}{100} \\timesx\\) = 35.\nSolving for x yields x = GH₵ 83.33.\n\n(b) (i), the first reduction = (\\(\\frac{40}{100}\\)) x180 = 72.00 and the second reduction = (\\(\\frac{30}{100}\\) ) x108 = 32.40.\nThen total reduction = (72.00 + 32. 40) = GH₵ 104.40.~(b) (i), the first reduction = (\\(\\frac{40}{100}\\)) x180 = 72.00 and the second reduction = (\\(\\frac{30}{100}\\) ) x108 = 32.40.\nThen total reduction = (72.00 + 32. 40) = GH₵ 104.40.\n(b) (ii), the percentage reduction in the price of the article = (\\(\\frac{104.40}{180} \\times 100\\)) = 58%\n', 'WAEC-PK-1', 'WAEC-MATH', 1);
