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
(444, '2014', '\n\nQuestion\n\n(a) A circle is inscribed in a square. If the sum of the perimeter of the square and the circumference of the circle is 100 cm, calculate the radius of the circle. [Take \\(\\pi = \\frac{22}{7}\\)].\n(b) A rope 60cm long is made to form a rectangle. If the length is 4 times its breadth, calculate, correct to one decimal place, the :\n(i) length ; (ii) diagonal of the rectangle.\n\n', '', '(a) A circle is inscribed in a square. If the sum of the perimeter of the square and the circumference of the circle is 100 cm, calculate the radius of the circle. [Take \\(\\pi = \\frac{22}{7}\\)].\n(b) A rope 60cm long is made to form a rectangle. If the length is 4 times its breadth, calculate, correct to one decimal place, the :\n(i) length ; (ii) diagonal of the rectangle.', '(a) A circle is inscribed in a square. If the sum of the perimeter of the square and the circumference of the circle is 100 cm, calculate the radius of the circle. [Take \\(\\pi = \\frac{22}{7}\\)].~(b) A rope 60cm long is made to form a rectangle. If the length is 4 times its breadth, calculate, correct to one decimal place, the :\n(i) length ; (ii) diagonal of the rectangle.', NULL, '\n\nExplanation\n(a) Perimeter of a circle = \\(2\\pi r\\).\nLength of a side of square = 2r in terms of r.\nSum of perimeter of circle and square = \\((2 \\pi r) + (4 \\times 2r)\\)\n\\(2r (\\pi + 4) = 100\\)\n\\(r ( \\frac{22}{7} + 4 ) = 50\\)\n\\(r = 50 \\div \\frac{50}{7}\\)\n\\(r = 7 cm\\)\n(b) (i) L = \\(4 \\times breadth\\)\nB = breadth.\n60 cm = 2(4B + B)\n2(5B) = 60 cm\n\\(B = \\frac{60}{10} = 6 cm\\)\n\\(L = 4 \\times 6 = 24 cm\\).\n(ii)\nLet the diagonal be d.\n\\(d^{2} = 6^{2} + 24^{2}\\)\n\\(d^{2} = 36 + 576\\)\n\\(d^{2} = 612\\)\n\\(d = \\sqrt{612}\\)\n= \\(24.74 cm \\approxeq 24.7 cm\\).\n\n', '', 'Explanation\n(a) Perimeter of a circle = \\(2\\pi r\\).\nLength of a side of square = 2r in terms of r.\nSum of perimeter of circle and square = \\((2 \\pi r) + (4 \\times 2r)\\)\n\\(2r (\\pi + 4) = 100\\)\n\\(r ( \\frac{22}{7} + 4 ) = 50\\)\n\\(r = 50 \\div \\frac{50}{7}\\)\n\\(r = 7 cm\\)\n(b) (i) L = \\(4 \\times breadth\\)\nB = breadth.\n60 cm = 2(4B + B)\n2(5B) = 60 cm\n\\(B = \\frac{60}{10} = 6 cm\\)\n\\(L = 4 \\times 6 = 24 cm\\).\n(ii)\nLet the diagonal be d.\n\\(d^{2} = 6^{2} + 24^{2}\\)\n\\(d^{2} = 36 + 576\\)\n\\(d^{2} = 612\\)\n\\(d = \\sqrt{612}\\)\n= \\(24.74 cm \\approxeq 24.7 cm\\).\n\n', '(a) Perimeter of a circle = \\(2\\pi r\\).\nLength of a side of square = 2r in terms of r.\nSum of perimeter of circle and square = \\((2 \\pi r) + (4 \\times 2r)\\)\n\\(2r (\\pi + 4) = 100\\)\n\\(r ( \\frac{22}{7} + 4 ) = 50\\)\n\\(r = 50 \\div \\frac{50}{7}\\)\n\\(r = 7 cm\\)~(b) (i) L = \\(4 \\times breadth\\)\nB = breadth.\n60 cm = 2(4B + B)\n2(5B) = 60 cm\n\\(B = \\frac{60}{10} = 6 cm\\)\n\\(L = 4 \\times 6 = 24 cm\\).\n(ii)\nLet the diagonal be d.\n\\(d^{2} = 6^{2} + 24^{2}\\)\n\\(d^{2} = 36 + 576\\)\n\\(d^{2} = 612\\)\n\\(d = \\sqrt{612}\\)\n= \\(24.74 cm \\approxeq 24.7 cm\\).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(445, '2014', '\n\nQuestion\n\n(a) Copy and complete the table of values for \\(y = \\sin x + 2 \\cos x\\), correct to one decimal place.\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n\n240°\n\n\n\ny\n\n2.2\n\n\n\n-1.2\n-2.0\n\n-1.9\n\n\n\n(b) Using a scale of 2 cm to 30° on the x- axis and 2 cm to 0.5 units on the y- axis, draw the graph of \\(y = \\sin x + 2\\cos x\\) for \\(0° \\leq x \\leq 240°\\).\n(c) Use your graph to solve the equation : (i) \\(\\sin x + 2 \\cos x = 0\\) ; (ii) \\(\\sin x = 2.1 - 2\\cos x\\).\n(d) From the graph, find y when x = 171°.\n\n\n\n(a) Copy and complete the table of values for \\(y = \\sin x + 2 \\cos x\\), correct to one decimal place.\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n\n240°\n\n\n\ny\n\n2.2\n\n\n\n-1.2\n-2.0\n\n-1.9\n\n\n', '', '(a) Copy and complete the table of values for \\(y = \\sin x + 2 \\cos x\\), correct to one decimal place.\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n\n240°\n\n\n\ny\n\n2.2\n\n\n\n-1.2\n-2.0\n\n-1.9\n\n\n\n(b) Using a scale of 2 cm to 30° on the x- axis and 2 cm to 0.5 units on the y- axis, draw the graph of \\(y = \\sin x + 2\\cos x\\) for \\(0° \\leq x \\leq 240°\\).\n(c) Use your graph to solve the equation : (i) \\(\\sin x + 2 \\cos x = 0\\) ; (ii) \\(\\sin x = 2.1 - 2\\cos x\\).\n(d) From the graph, find y when x = 171°.\n\n\n\n(a) Copy and complete the table of values for \\(y = \\sin x + 2 \\cos x\\), correct to one decimal place.\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n\n240°\n\n\n\ny\n\n2.2\n\n\n\n-1.2\n-2.0\n\n-1.9\n', '(a) Copy and complete the table of values for \\(y = \\sin x + 2 \\cos x\\), correct to one decimal place.\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n\n240°\n\n\n\ny\n\n2.2\n\n\n\n-1.2\n-2.0\n\n-1.9\n\n\n~(b) Using a scale of 2 cm to 30° on the x- axis and 2 cm to 0.5 units on the y- axis, draw the graph of \\(y = \\sin x + 2\\cos x\\) for \\(0° \\leq x \\leq 240°\\).\n(c) Use your graph to solve the equation : (i) \\(\\sin x + 2 \\cos x = 0\\) ; (ii) \\(\\sin x = 2.1 - 2\\cos x\\).\n(d) From the graph, find y when x = 171°.\n\n\n\n(a) Copy and complete the table of values for \\(y = \\sin x + 2 \\cos x\\), correct to one decimal place.\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n\n240°\n\n\n\ny\n\n2.2\n\n\n\n-1.2\n-2.0\n\n-1.9\n~(c) Use your graph to solve the equation : (i) \\(\\sin x + 2 \\cos x = 0\\) ; (ii) \\(\\sin x = 2.1 - 2\\cos x\\).\n(d) From the graph, find y when x = 171°.\n\n\n\n(a) Copy and complete the table of values for \\(y = \\sin x + 2 \\cos x\\), correct to one decimal place.\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n\n240°\n\n\n\ny\n\n2.2\n\n\n\n-1.2\n-2.0\n\n-1.9\n~(d) From the graph, find y when x = 171°.\n\n\n\n(a) Copy and complete the table of values for \\(y = \\sin x + 2 \\cos x\\), correct to one decimal place.\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n\n240°\n\n\n\ny\n\n2.2\n\n\n\n-1.2\n-2.0\n\n-1.9\n', NULL, '\n\nExplanation\n(a)\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n240°\n\n\ny\n2.0\n2.2\n1.9\n1.0\n-0.1\n-1.2\n-2.0\n-2.2\n-1.9\n\n\n\n(b)\n(c) (i) x = 117° ; (ii) x = 9° , 45°.\n(d) When x = 171°, y = -1.75.\n\n\n', '', 'Explanation\n(a)\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n240°\n\n\ny\n2.0\n2.2\n1.9\n1.0\n-0.1\n-1.2\n-2.0\n-2.2\n-1.9\n\n\n\n(b)\n(c) (i) x = 117° ; (ii) x = 9° , 45°.\n(d) When x = 171°, y = -1.75.\n\n\n', '(a)\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n240°\n\n\ny\n2.0\n2.2\n1.9\n1.0\n-0.1\n-1.2\n-2.0\n-2.2\n-1.9\n\n\n~(b)\n(c) (i) x = 117° ; (ii) x = 9° , 45°.\n(d) When x = 171°, y = -1.75.\n\n~(c) (i) x = 117° ; (ii) x = 9° , 45°.\n(d) When x = 171°, y = -1.75.\n\n~(d) When x = 171°, y = -1.75.\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(446, '2014', '\n\nQuestion\n\n(a) How many numbers between 75 and 500 are divisible by 7?\n(b) The 8th term of an Arithmetic Progression (A.P) is 5 times the 3rd term while the 7th term is 9 greater than the 4th term. Write the first 5 terms of the A.P.\n\n', '', '(a) How many numbers between 75 and 500 are divisible by 7?\n(b) The 8th term of an Arithmetic Progression (A.P) is 5 times the 3rd term while the 7th term is 9 greater than the 4th term. Write the first 5 terms of the A.P.', '(a) How many numbers between 75 and 500 are divisible by 7?~(b) The 8th term of an Arithmetic Progression (A.P) is 5 times the 3rd term while the 7th term is 9 greater than the 4th term. Write the first 5 terms of the A.P.', NULL, '\n\nExplanation\n(a)In the set of numbers from 75, 76, 77, ... , 497, 498, 499, 500, you will find that the first term which is the first number divisible by 7 = 77 and the last term = 497.\nGiven the first and last terms of an A.P, you can use the formula\n\\(n = \\frac{l - a}{d} + 1\\) to find the number of terms in the sequence.\nFirst term = a = 77\nLast term = l = 497\nCommon difference = d = 7\n\\(n = \\frac{497 - 77}{7} + 1\\)\n\\(n = \\frac{420}{7} + 1\\)\n= \\(60 + 1\\)\n= 61.\nThere are 61 numbers between 75 and 500 which are divisible by 7.\n(b)\\(T_{n} = a + (n - 1)d\\) (for an A.P)\n\\(T_{8} = a + 7d\\)\n\\(T_{3} = a + 2d\\)\n\\(\\implies a + 7d = 5(a + 2d)\\)\n\\(T_{7} = a + 6d\\)\n\\(T_{4} = a + 3d\\)\n\\(\\implies a + 6d = a + 3d + 9\\)\n\\(a + 7d = 5a + 10d \\implies 5a - a + 10d - 7d = 0\\)\n\\(4a + 3d = 0 ... (1)\\)\n\\(a + 6d = a + 3d + 9 \\implies a - a + 6d - 3d = 9\\)\n\\(3d = 9 \\implies d = 3\\)\nPutting d = 3 in (1),\n\\(4a + 3(3) = 0 \\implies 4a + 9 = 0\\)\n\\(4a = -9 \\implies a = \\frac{-9}{4}\\)\n\\(\\therefore\\) The secondtermof thesequence = \\(\\frac{-9}{4} + 3 = \\frac{3}{4}\\)\nThird term = \\(\\frac{3}{4} + 3 = \\frac{15}{4}\\)\nFourth term = \\(\\frac{15}{4} + 3 = \\frac{27}{4}\\)\nFifth term = \\(\\frac{27}{4} + 3 = \\frac{39}{4}\\)\nThe first 5 terms of the sequence = \\(\\frac{-9}{4}, \\frac{3}{4}, \\frac{15}{4}, \\frac{27}{4}, \\frac{39}{4}\\)\n\n', '', 'Explanation\n(a)In the set of numbers from 75, 76, 77, ... , 497, 498, 499, 500, you will find that the first term which is the first number divisible by 7 = 77 and the last term = 497.\nGiven the first and last terms of an A.P, you can use the formula\n\\(n = \\frac{l - a}{d} + 1\\) to find the number of terms in the sequence.\nFirst term = a = 77\nLast term = l = 497\nCommon difference = d = 7\n\\(n = \\frac{497 - 77}{7} + 1\\)\n\\(n = \\frac{420}{7} + 1\\)\n= \\(60 + 1\\)\n= 61.\nThere are 61 numbers between 75 and 500 which are divisible by 7.\n(b)\\(T_{n} = a + (n - 1)d\\) (for an A.P)\n\\(T_{8} = a + 7d\\)\n\\(T_{3} = a + 2d\\)\n\\(\\implies a + 7d = 5(a + 2d)\\)\n\\(T_{7} = a + 6d\\)\n\\(T_{4} = a + 3d\\)\n\\(\\implies a + 6d = a + 3d + 9\\)\n\\(a + 7d = 5a + 10d \\implies 5a - a + 10d - 7d = 0\\)\n\\(4a + 3d = 0 ... (1)\\)\n\\(a + 6d = a + 3d + 9 \\implies a - a + 6d - 3d = 9\\)\n\\(3d = 9 \\implies d = 3\\)\nPutting d = 3 in (1),\n\\(4a + 3(3) = 0 \\implies 4a + 9 = 0\\)\n\\(4a = -9 \\implies a = \\frac{-9}{4}\\)\n\\(\\therefore\\) The secondtermof thesequence = \\(\\frac{-9}{4} + 3 = \\frac{3}{4}\\)\nThird term = \\(\\frac{3}{4} + 3 = \\frac{15}{4}\\)\nFourth term = \\(\\frac{15}{4} + 3 = \\frac{27}{4}\\)\nFifth term = \\(\\frac{27}{4} + 3 = \\frac{39}{4}\\)\nThe first 5 terms of the sequence = \\(\\frac{-9}{4}, \\frac{3}{4}, \\frac{15}{4}, \\frac{27}{4}, \\frac{39}{4}\\)\n\n', '(a)In the set of numbers from 75, 76, 77, ... , 497, 498, 499, 500, you will find that the first term which is the first number divisible by 7 = 77 and the last term = 497.\nGiven the first and last terms of an A.P, you can use the formula\n\\(n = \\frac{l - a}{d} + 1\\) to find the number of terms in the sequence.\nFirst term = a = 77\nLast term = l = 497\nCommon difference = d = 7\n\\(n = \\frac{497 - 77}{7} + 1\\)\n\\(n = \\frac{420}{7} + 1\\)\n= \\(60 + 1\\)\n= 61.\nThere are 61 numbers between 75 and 500 which are divisible by 7.~(b)\\(T_{n} = a + (n - 1)d\\) (for an A.P)\n\\(T_{8} = a + 7d\\)\n\\(T_{3} = a + 2d\\)\n\\(\\implies a + 7d = 5(a + 2d)\\)\n\\(T_{7} = a + 6d\\)\n\\(T_{4} = a + 3d\\)\n\\(\\implies a + 6d = a + 3d + 9\\)\n\\(a + 7d = 5a + 10d \\implies 5a - a + 10d - 7d = 0\\)\n\\(4a + 3d = 0 ... (1)\\)\n\\(a + 6d = a + 3d + 9 \\implies a - a + 6d - 3d = 9\\)\n\\(3d = 9 \\implies d = 3\\)\nPutting d = 3 in (1),\n\\(4a + 3(3) = 0 \\implies 4a + 9 = 0\\)\n\\(4a = -9 \\implies a = \\frac{-9}{4}\\)\n\\(\\therefore\\) The secondtermof thesequence = \\(\\frac{-9}{4} + 3 = \\frac{3}{4}\\)\nThird term = \\(\\frac{3}{4} + 3 = \\frac{15}{4}\\)\nFourth term = \\(\\frac{15}{4} + 3 = \\frac{27}{4}\\)\nFifth term = \\(\\frac{27}{4} + 3 = \\frac{39}{4}\\)\nThe first 5 terms of the sequence = \\(\\frac{-9}{4}, \\frac{3}{4}, \\frac{15}{4}, \\frac{27}{4}, \\frac{39}{4}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(447, '2014', '\n\nQuestion\n\n(a) Without using calculator or tables, find the value of \\(\\log 3.6\\) given that \\(\\log 2 = 0.3010, \\log 3 = 0.4771\\) and \\(\\log 5 = 0.6990\\).\n(b) If all numbers in the equation \\(\\frac{y}{y + 101} = \\frac{11}{10010}\\) are in base two, solve for y.\n\n', '', '(a) Without using calculator or tables, find the value of \\(\\log 3.6\\) given that \\(\\log 2 = 0.3010, \\log 3 = 0.4771\\) and \\(\\log 5 = 0.6990\\).\n(b) If all numbers in the equation \\(\\frac{y}{y + 101} = \\frac{11}{10010}\\) are in base two, solve for y.', '(a) Without using calculator or tables, find the value of \\(\\log 3.6\\) given that \\(\\log 2 = 0.3010, \\log 3 = 0.4771\\) and \\(\\log 5 = 0.6990\\).~(b) If all numbers in the equation \\(\\frac{y}{y + 101} = \\frac{11}{10010}\\) are in base two, solve for y.', NULL, '\n\nExplanation\n(a) \\(\\log 3.6 = \\log (\\frac{18}{5})\\)\n\\(\\log 18 - \\log 5 = \\log (2 \\times 3^{2}) - \\log 5\\)\n\\(\\log 2 + \\log 3^{2} - \\log 5 = \\log 2 + 2\\log 3 - \\log 5\\)\n= \\(0.3010 + 2(0.4771) - 0.6990\\)\n= \\(0.3010 + 0.9542 - 0.6990\\)\n= \\(0.5562\\).\n(b) \\(\\frac{y}{y + 101} = \\frac{11}{10010}\\) (all in base 2)\nCross multiplying,\n\\(11(y + 101) = 10010y\\)\n\\(11y + 1111 = 10010y\\)\n\\(10010y - 11y = 1111\\)\n\\(1111y = 1111\\)\n\\(y = 1\\).\n\n', '', 'Explanation\n(a) \\(\\log 3.6 = \\log (\\frac{18}{5})\\)\n\\(\\log 18 - \\log 5 = \\log (2 \\times 3^{2}) - \\log 5\\)\n\\(\\log 2 + \\log 3^{2} - \\log 5 = \\log 2 + 2\\log 3 - \\log 5\\)\n= \\(0.3010 + 2(0.4771) - 0.6990\\)\n= \\(0.3010 + 0.9542 - 0.6990\\)\n= \\(0.5562\\).\n(b) \\(\\frac{y}{y + 101} = \\frac{11}{10010}\\) (all in base 2)\nCross multiplying,\n\\(11(y + 101) = 10010y\\)\n\\(11y + 1111 = 10010y\\)\n\\(10010y - 11y = 1111\\)\n\\(1111y = 1111\\)\n\\(y = 1\\).\n\n', '(a) \\(\\log 3.6 = \\log (\\frac{18}{5})\\)\n\\(\\log 18 - \\log 5 = \\log (2 \\times 3^{2}) - \\log 5\\)\n\\(\\log 2 + \\log 3^{2} - \\log 5 = \\log 2 + 2\\log 3 - \\log 5\\)\n= \\(0.3010 + 2(0.4771) - 0.6990\\)\n= \\(0.3010 + 0.9542 - 0.6990\\)\n= \\(0.5562\\).~(b) \\(\\frac{y}{y + 101} = \\frac{11}{10010}\\) (all in base 2)\nCross multiplying,\n\\(11(y + 101) = 10010y\\)\n\\(11y + 1111 = 10010y\\)\n\\(10010y - 11y = 1111\\)\n\\(1111y = 1111\\)\n\\(y = 1\\).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(448, '2014', '\n\nQuestion\n\n(a) Solve the inequality : \\(\\frac{2}{5}(x - 2) - \\frac{1}{6}(x + 5) \\leq 0\\).\n(b) Given that P = \\(\\frac{x^{2} - y^{2}}{x^{2} + xy}\\),\n(i) express P in its simplest form ; (ii) find the value of P if x = -4 and y = -6.\n\n', '', '(a) Solve the inequality : \\(\\frac{2}{5}(x - 2) - \\frac{1}{6}(x + 5) \\leq 0\\).\n(b) Given that P = \\(\\frac{x^{2} - y^{2}}{x^{2} + xy}\\),\n(i) express P in its simplest form ; (ii) find the value of P if x = -4 and y = -6.', '(a) Solve the inequality : \\(\\frac{2}{5}(x - 2) - \\frac{1}{6}(x + 5) \\leq 0\\).~(b) Given that P = \\(\\frac{x^{2} - y^{2}}{x^{2} + xy}\\),\n(i) express P in its simplest form ; (ii) find the value of P if x = -4 and y = -6.', NULL, '\n\nExplanation\n(a) \\(\\frac{2}{5}(x - 2) - \\frac{1}{6}(x + 5) \\leq 0\\)\nMultiplying through by the LCM of 5 and 6 (i.e 30)\n\\(12(x - 2) - 5(x + 5) \\leq 0\\)\n\\(12x - 24 - 5x - 25 \\leq 0\\)\n\\(7x - 49 \\leq 0 \\implies 7x \\leq 49\\)\n\\(x \\leq 7\\).\n(b) (i) \\(\\frac{x^{2} - y^{2}}{x^{2} + xy}\\)\nP = \\(\\frac{(x - y)(x + y)}{x (x + y)}\\)\nP = \\(\\frac{x - y}{x}\\)\n(ii) When x = -4, y = -6\n\\(P = \\frac{-4 - (-6)}{-4}\\)\n\\(P = \\frac{2}{-4}\\)\n\\(P = - \\frac{1}{2}\\)\n\n', '', 'Explanation\n(a) \\(\\frac{2}{5}(x - 2) - \\frac{1}{6}(x + 5) \\leq 0\\)\nMultiplying through by the LCM of 5 and 6 (i.e 30)\n\\(12(x - 2) - 5(x + 5) \\leq 0\\)\n\\(12x - 24 - 5x - 25 \\leq 0\\)\n\\(7x - 49 \\leq 0 \\implies 7x \\leq 49\\)\n\\(x \\leq 7\\).\n(b) (i) \\(\\frac{x^{2} - y^{2}}{x^{2} + xy}\\)\nP = \\(\\frac{(x - y)(x + y)}{x (x + y)}\\)\nP = \\(\\frac{x - y}{x}\\)\n(ii) When x = -4, y = -6\n\\(P = \\frac{-4 - (-6)}{-4}\\)\n\\(P = \\frac{2}{-4}\\)\n\\(P = - \\frac{1}{2}\\)\n\n', '(a) \\(\\frac{2}{5}(x - 2) - \\frac{1}{6}(x + 5) \\leq 0\\)\nMultiplying through by the LCM of 5 and 6 (i.e 30)\n\\(12(x - 2) - 5(x + 5) \\leq 0\\)\n\\(12x - 24 - 5x - 25 \\leq 0\\)\n\\(7x - 49 \\leq 0 \\implies 7x \\leq 49\\)\n\\(x \\leq 7\\).~(b) (i) \\(\\frac{x^{2} - y^{2}}{x^{2} + xy}\\)\nP = \\(\\frac{(x - y)(x + y)}{x (x + y)}\\)\nP = \\(\\frac{x - y}{x}\\)\n(ii) When x = -4, y = -6\n\\(P = \\frac{-4 - (-6)}{-4}\\)\n\\(P = \\frac{2}{-4}\\)\n\\(P = - \\frac{1}{2}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(449, '2014', '\n\nQuestion\n\nThe table below shows how a man spends his income in a month.\n\n\n\nItems\nAmount Spent\n\n\nFood\nN4500\n\n\nHouse Rent\nN3000\n\n\nProvisions\nN2500\n\n\nElectricity\nN2000\n\n\nTransportation\nN5000\n\n\nOthers\nN3000\n\n\n\n(a) Represent the information on a pie chart.\n(b) What percentage of his income is spent on transportation?\n\n\nThe table below shows how a man spends his income in a month.\n\n\n\nItems\nAmount Spent\n\n\nFood\nN4500\n\n\nHouse Rent\nN3000\n\n\nProvisions\nN2500\n\n\nElectricity\nN2000\n\n\nTransportation\nN5000\n\n\nOthers\nN3000\n\n\n', '', 'The table below shows how a man spends his income in a month.\n\n\n\nItems\nAmount Spent\n\n\nFood\nN4500\n\n\nHouse Rent\nN3000\n\n\nProvisions\nN2500\n\n\nElectricity\nN2000\n\n\nTransportation\nN5000\n\n\nOthers\nN3000\n\n\n\n(a) Represent the information on a pie chart.\n(b) What percentage of his income is spent on transportation?\n\n\nThe table below shows how a man spends his income in a month.\n\n\n\nItems\nAmount Spent\n\n\nFood\nN4500\n\n\nHouse Rent\nN3000\n\n\nProvisions\nN2500\n\n\nElectricity\nN2000\n\n\nTransportation\nN5000\n\n\nOthers\nN3000\n', 'The table below shows how a man spends his income in a month.\n\n\n\nItems\nAmount Spent\n\n\nFood\nN4500\n\n\nHouse Rent\nN3000\n\n\nProvisions\nN2500\n\n\nElectricity\nN2000\n\n\nTransportation\nN5000\n\n\nOthers\nN3000\n\n\n\n(a) Represent the information on a pie chart.~(b) What percentage of his income is spent on transportation?\n\n\nThe table below shows how a man spends his income in a month.\n\n\n\nItems\nAmount Spent\n\n\nFood\nN4500\n\n\nHouse Rent\nN3000\n\n\nProvisions\nN2500\n\n\nElectricity\nN2000\n\n\nTransportation\nN5000\n\n\nOthers\nN3000\n', NULL, '\n\nExplanation\nItem\nAmount Spent (N)\n\n\nFood\n4,500\n\n\nHouse Rent\n3,000\n\n\nProvisions\n2,500\n\n\nElectricity\n2,000\n\n\nTransportation\n5,000\n\n\nOthers\n3,000\n\n\nTotal\n20,000\n\n\n\nItems in degrees :\nFood :\\(\\frac{4500}{20000} \\times 360° = 81°\\)\nHouse rent :\\(\\frac{3000}{20000} \\times 360° = 54°\\)\nProvisions : \\(\\frac{2500}{20000} \\times 360° = 45°\\)\nElectricity : \\(\\frac{2000}{20000} \\times 360° = 36°\\)\nTransportation : \\(\\frac{5000}{20000} \\times 360° = 90°\\)\nOthers : \\(\\frac{3000}{20000} \\times 360° = 54°\\)\n(a) Pie chart :\n\n(b) Percentage of his income spent on transportation :\n\\(\\frac{5000}{20000} \\times 100% = 25%\\)\n\n\n', '', 'Explanation\nItem\nAmount Spent (N)\n\n\nFood\n4,500\n\n\nHouse Rent\n3,000\n\n\nProvisions\n2,500\n\n\nElectricity\n2,000\n\n\nTransportation\n5,000\n\n\nOthers\n3,000\n\n\nTotal\n20,000\n\n\n\nItems in degrees :\nFood :\\(\\frac{4500}{20000} \\times 360° = 81°\\)\nHouse rent :\\(\\frac{3000}{20000} \\times 360° = 54°\\)\nProvisions : \\(\\frac{2500}{20000} \\times 360° = 45°\\)\nElectricity : \\(\\frac{2000}{20000} \\times 360° = 36°\\)\nTransportation : \\(\\frac{5000}{20000} \\times 360° = 90°\\)\nOthers : \\(\\frac{3000}{20000} \\times 360° = 54°\\)\n(a) Pie chart :\n\n(b) Percentage of his income spent on transportation :\n\\(\\frac{5000}{20000} \\times 100% = 25%\\)\n\n\n', '(a) Pie chart :\n~(b) Percentage of his income spent on transportation :\n\\(\\frac{5000}{20000} \\times 100% = 25%\\)\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(450, '2014', '\n\nQuestion\n\n(a) A pentagon is such that one of its exterior sides is 60°. Two others are (90 - m)° each while the remaining angles are (30 + 2m)° each. Find the value of m.\n(b)\nIn the diagram, PQR is a straight line, \\(\\overline{QR} = \\sqrt{3} cm\\) and \\(\\overline{SQ} = 2 cm\\). Calculate, correct to one decimal place, < PQS.\n\n', '', '(a) A pentagon is such that one of its exterior sides is 60°. Two others are (90 - m)° each while the remaining angles are (30 + 2m)° each. Find the value of m.\n(b)\nIn the diagram, PQR is a straight line, \\(\\overline{QR} = \\sqrt{3} cm\\) and \\(\\overline{SQ} = 2 cm\\). Calculate, correct to one decimal place, < PQS.', '(a) A pentagon is such that one of its exterior sides is 60°. Two others are (90 - m)° each while the remaining angles are (30 + 2m)° each. Find the value of m.~(b)\nIn the diagram, PQR is a straight line, \\(\\overline{QR} = \\sqrt{3} cm\\) and \\(\\overline{SQ} = 2 cm\\). Calculate, correct to one decimal place, < PQS.', NULL, '\n\nExplanation\nThe sum of exterior angles = 360°\n\\(60° + 2(90 - m)° + 2(30 + 2m)° = 360°\\)\n\\(60° + 180° - 2m° + 60° + 4m° = 360°\\)\n\\(300° + 2m = 360°\\)\n\\(\\implies 2m = 360° - 300° = 60°\\)\n\\(m = 30°\\)\n(b)\nIn \\(\\Delta QRS, \\cos Q = \\frac{\\sqrt{3}}{2}\\)\n\\(\\cos Q = 0.8660 \\implies Q = 30°\\)\n\\(\\therefore < SQR = 30°^\\)\n\\(< PQS = 180° - 30° \\) (angles on a straight line = 180°)\n\\(< PQS = 150°\\)\n\n', '', 'Explanation\nThe sum of exterior angles = 360°\n\\(60° + 2(90 - m)° + 2(30 + 2m)° = 360°\\)\n\\(60° + 180° - 2m° + 60° + 4m° = 360°\\)\n\\(300° + 2m = 360°\\)\n\\(\\implies 2m = 360° - 300° = 60°\\)\n\\(m = 30°\\)\n(b)\nIn \\(\\Delta QRS, \\cos Q = \\frac{\\sqrt{3}}{2}\\)\n\\(\\cos Q = 0.8660 \\implies Q = 30°\\)\n\\(\\therefore < SQR = 30°^\\)\n\\(< PQS = 180° - 30° \\) (angles on a straight line = 180°)\n\\(< PQS = 150°\\)\n\n', '~(b)\nIn \\(\\Delta QRS, \\cos Q = \\frac{\\sqrt{3}}{2}\\)\n\\(\\cos Q = 0.8660 \\implies Q = 30°\\)\n\\(\\therefore < SQR = 30°^\\)\n\\(< PQS = 180° - 30° \\) (angles on a straight line = 180°)\n\\(< PQS = 150°\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(451, '2014', '\n\nQuestion\n\n(a) A rectangular field is l metres long and b metres wide. Its perimeter is 280 metres. If the length is two and a half times the breadth, find the values of l and b.\n(b) The base of a pyramid is a 4.5 metres rectangle. The height of the pyramid is 4 metres. Calculate its volume.\n\n', '', '(a) A rectangular field is l metres long and b metres wide. Its perimeter is 280 metres. If the length is two and a half times the breadth, find the values of l and b.\n(b) The base of a pyramid is a 4.5 metres rectangle. The height of the pyramid is 4 metres. Calculate its volume.', '(a) A rectangular field is l metres long and b metres wide. Its perimeter is 280 metres. If the length is two and a half times the breadth, find the values of l and b.~(b) The base of a pyramid is a 4.5 metres rectangle. The height of the pyramid is 4 metres. Calculate its volume.', NULL, '\n\nExplanation\n(a) Perimeter of a rectangle = 2(l + b)\n\\(2l + 2b = 280\\)\nBut l = 2.5b,\n\\(2(2.5b) + 2b = 280 \\implies 7b = 280\\)\n\\(b = 40 m\\)\n\\(\\therefore l = 2.5(40) \\)\n= \\(100 m\\)\n(b) Volume of pyramid = \\(\\frac{1}{3}l \\times b \\times h\\)\n= \\(\\frac{1}{3}(4.5) \\times 2.5 \\times 4 \\)\n= \\(15 m^{3}\\).\n\n', '', 'Explanation\n(a) Perimeter of a rectangle = 2(l + b)\n\\(2l + 2b = 280\\)\nBut l = 2.5b,\n\\(2(2.5b) + 2b = 280 \\implies 7b = 280\\)\n\\(b = 40 m\\)\n\\(\\therefore l = 2.5(40) \\)\n= \\(100 m\\)\n(b) Volume of pyramid = \\(\\frac{1}{3}l \\times b \\times h\\)\n= \\(\\frac{1}{3}(4.5) \\times 2.5 \\times 4 \\)\n= \\(15 m^{3}\\).\n\n', '(a) Perimeter of a rectangle = 2(l + b)\n\\(2l + 2b = 280\\)\nBut l = 2.5b,\n\\(2(2.5b) + 2b = 280 \\implies 7b = 280\\)\n\\(b = 40 m\\)\n\\(\\therefore l = 2.5(40) \\)\n= \\(100 m\\)~b)\n\\(2l + 2b = 280\\)\nBut l = 2.5b,\n\\(2(2.5b) + 2b = 280 \\implies 7b = 280\\)\n\\(b = 40 m\\)\n\\(\\therefore l = 2.5(40) \\)\n= \\(100 m\\)\n(b) Volume of pyramid = \\(\\frac{1}{3}l \\times b \\times h\\)\n= \\(\\frac{1}{3}(4.5) \\times 2.5 \\times 4 \\)\n= \\(15 m^{3}\\).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(452, '2014', '\n\nQuestion\n\n(a) If \\(2^{x + y} = 16\\) and \\(4^{x - y} = \\frac{1}{32}\\), find the value of x and y.\n(b) P, Q and R are related in such a way that \\(P \\propto \\frac{Q^{2}}{R}\\). When P = 36, Q = 3 and R = 4. Calculate Q when P = 200 and R = 2.\n\n', '', '(a) If \\(2^{x + y} = 16\\) and \\(4^{x - y} = \\frac{1}{32}\\), find the value of x and y.\n(b) P, Q and R are related in such a way that \\(P \\propto \\frac{Q^{2}}{R}\\). When P = 36, Q = 3 and R = 4. Calculate Q when P = 200 and R = 2.', '(a) If \\(2^{x + y} = 16\\) and \\(4^{x - y} = \\frac{1}{32}\\), find the value of x and y.~(b) P, Q and R are related in such a way that \\(P \\propto \\frac{Q^{2}}{R}\\). When P = 36, Q = 3 and R = 4. Calculate Q when P = 200 and R = 2.', NULL, '\n\nExplanation\n(a) \\(2^{x + y} = 16 \\implies 2^{x + y} = 2^{4}\\)\n\\(\\implies x + y = 4 ..... (1)\\)\n\\(4^{x - y} = \\frac{1}{32} \\implies (2^{2})^{(x - y)} = 2^{-5}\\)\n\\(\\implies 2x - 2y = -5 .... (2)\\)\nFrom (1), x = 4 - y.\n\\(\\therefore 2(4 - y) - 2y = -5\\)\n\\(8 - 2y - 2y = -5 \\implies -4y = - 5 - 8 = -13\\)\n\\(y = \\frac{13}{4}\\)\n\\(x = 4 - \\frac{13}{4} \\)\n= \\(\\frac{3}{4}\\)\n(b) \\(P \\propto \\frac{Q^{2}}{R}\\)\n\\(\\implies P = \\frac{kQ^{2}}{R}\\)\n\\(k = \\frac{PR}{Q^{2}}\\)\n\\(k = \\frac{36 \\times 4}{9} = 16\\)\n\\(\\implies P = \\frac{16Q^{2}}{R}\\)\n\\(Q^{2} = \\frac{PR}{k}\\)\n\\(Q^{2} = \\frac{200 \\times 2}{16}\\)\n\\(Q^{2} = 25\\)\n\\(Q = \\pm 5\\).\n\n', '', 'Explanation\n(a) \\(2^{x + y} = 16 \\implies 2^{x + y} = 2^{4}\\)\n\\(\\implies x + y = 4 ..... (1)\\)\n\\(4^{x - y} = \\frac{1}{32} \\implies (2^{2})^{(x - y)} = 2^{-5}\\)\n\\(\\implies 2x - 2y = -5 .... (2)\\)\nFrom (1), x = 4 - y.\n\\(\\therefore 2(4 - y) - 2y = -5\\)\n\\(8 - 2y - 2y = -5 \\implies -4y = - 5 - 8 = -13\\)\n\\(y = \\frac{13}{4}\\)\n\\(x = 4 - \\frac{13}{4} \\)\n= \\(\\frac{3}{4}\\)\n(b) \\(P \\propto \\frac{Q^{2}}{R}\\)\n\\(\\implies P = \\frac{kQ^{2}}{R}\\)\n\\(k = \\frac{PR}{Q^{2}}\\)\n\\(k = \\frac{36 \\times 4}{9} = 16\\)\n\\(\\implies P = \\frac{16Q^{2}}{R}\\)\n\\(Q^{2} = \\frac{PR}{k}\\)\n\\(Q^{2} = \\frac{200 \\times 2}{16}\\)\n\\(Q^{2} = 25\\)\n\\(Q = \\pm 5\\).\n\n', '(a) \\(2^{x + y} = 16 \\implies 2^{x + y} = 2^{4}\\)\n\\(\\implies x + y = 4 ..... (1)\\)\n\\(4^{x - y} = \\frac{1}{32} \\implies (2^{2})^{(x - y)} = 2^{-5}\\)\n\\(\\implies 2x - 2y = -5 .... (2)\\)\nFrom (1), x = 4 - y.\n\\(\\therefore 2(4 - y) - 2y = -5\\)\n\\(8 - 2y - 2y = -5 \\implies -4y = - 5 - 8 = -13\\)\n\\(y = \\frac{13}{4}\\)\n\\(x = 4 - \\frac{13}{4} \\)\n= \\(\\frac{3}{4}\\)~(b) \\(P \\propto \\frac{Q^{2}}{R}\\)\n\\(\\implies P = \\frac{kQ^{2}}{R}\\)\n\\(k = \\frac{PR}{Q^{2}}\\)\n\\(k = \\frac{36 \\times 4}{9} = 16\\)\n\\(\\implies P = \\frac{16Q^{2}}{R}\\)\n\\(Q^{2} = \\frac{PR}{k}\\)\n\\(Q^{2} = \\frac{200 \\times 2}{16}\\)\n\\(Q^{2} = 25\\)\n\\(Q = \\pm 5\\).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(453, '2014', '\n\nQuestion\n\n(a) Solve, correct to two decimal places, the equation \\(4x^{2} = 11x + 21\\).\n(b) A man invests£1500 for two years at compound interest. After one year, his money amounts to£1560. Find the :\n(i) rate of interest ; (ii) interest for the second year.\n(c) A car costs N300,000.00. It depreciates by 25% in the first year and 20% in the second year. Find its value after 2 years.\n\n\n', '', '(a) Solve, correct to two decimal places, the equation \\(4x^{2} = 11x + 21\\).\n(b) A man invests£1500 for two years at compound interest. After one year, his money amounts to£1560. Find the :\n(i) rate of interest ; (ii) interest for the second year.\n(c) A car costs N300,000.00. It depreciates by 25% in the first year and 20% in the second year. Find its value after 2 years.\n', '(a) Solve, correct to two decimal places, the equation \\(4x^{2} = 11x + 21\\).~(b) A man invests£1500 for two years at compound interest. After one year, his money amounts to£1560. Find the :\n(i) rate of interest ; (ii) interest for the second year.\n(c) A car costs N300,000.00. It depreciates by 25% in the first year and 20% in the second year. Find its value after 2 years.\n~(c) A car costs N300,000.00. It depreciates by 25% in the first year and 20% in the second year. Find its value after 2 years.\n', NULL, '\n\nExplanation\n(a) \\(4x^{2} = 11x + 21 \\implies 4x^{2} - 11x - 21 = 0\\)\n\\(a = 4, b = - 11 , c = -21\\)\n\\(x = \\frac{-b \\pm \\sqrt{b^{2} - 4ac}}{2a}\\)\n\\(x = \\frac{-(-11) \\pm \\sqrt{(-11)^{2} - 4(4)(-21)}}{2(4)}\\)\n\\(x = \\frac{11 \\pm \\sqrt{121 + 336}}{8}\\)\n\\(x = \\frac{11 \\pm \\sqrt{457}}{8}\\)\n\\(x = \\frac{11 \\pm 21.378}{8}\\)\n\\(x = \\frac{11 + 21.378}{8}\\) or \\(x = \\frac{11 - 21.378}{8}\\)\n\\(x = \\frac{32.378}{8}\\) or \\(x = \\frac{-10.378}{8}\\)\\\n\\(x = \\text{4.047or -1.297}\\)\n\\(x \\approxeq \\text{4.05or -1.30}\\).(2 decimal place).\n(b) (i) \\(I = \\frac{PRT}{100}\\)\n\\(I =£(1560 - 1500) =£60\\)\n\\(60 = \\frac{1500 \\times R \\times 1}{100}\\)\n\\(60 = 15R \\implies R = 4%\\)\n(ii) Interest for the second year\n= \\(\\frac{1560 \\times 4 \\times 1}{100}\\)\n= \\(£62.40\\)\n(c)\n\n\n\nInitial price of the car\n\\(N300,000.00\\)\n\n\nDepreciation after first year (25%)\n\\(\\frac{25}{100} \\times 300,000 = N75,000\\)\n\n\nValue after 1st year\nN225,000.00\n\n\nDepreciation after 2nd year(20%)\n\n\\(\\frac{20}{100} \\times 225,000 = N45,000\\)\n\n\n\nValue after 2nd year\nN180,000.00\n\n\n\n\n\n\n', '', 'Explanation\n(a) \\(4x^{2} = 11x + 21 \\implies 4x^{2} - 11x - 21 = 0\\)\n\\(a = 4, b = - 11 , c = -21\\)\n\\(x = \\frac{-b \\pm \\sqrt{b^{2} - 4ac}}{2a}\\)\n\\(x = \\frac{-(-11) \\pm \\sqrt{(-11)^{2} - 4(4)(-21)}}{2(4)}\\)\n\\(x = \\frac{11 \\pm \\sqrt{121 + 336}}{8}\\)\n\\(x = \\frac{11 \\pm \\sqrt{457}}{8}\\)\n\\(x = \\frac{11 \\pm 21.378}{8}\\)\n\\(x = \\frac{11 + 21.378}{8}\\) or \\(x = \\frac{11 - 21.378}{8}\\)\n\\(x = \\frac{32.378}{8}\\) or \\(x = \\frac{-10.378}{8}\\)\\\n\\(x = \\text{4.047or -1.297}\\)\n\\(x \\approxeq \\text{4.05or -1.30}\\).(2 decimal place).\n(b) (i) \\(I = \\frac{PRT}{100}\\)\n\\(I =£(1560 - 1500) =£60\\)\n\\(60 = \\frac{1500 \\times R \\times 1}{100}\\)\n\\(60 = 15R \\implies R = 4%\\)\n(ii) Interest for the second year\n= \\(\\frac{1560 \\times 4 \\times 1}{100}\\)\n= \\(£62.40\\)\n(c)\n\n\n\nInitial price of the car\n\\(N300,000.00\\)\n\n\nDepreciation after first year (25%)\n\\(\\frac{25}{100} \\times 300,000 = N75,000\\)\n\n\nValue after 1st year\nN225,000.00\n\n\nDepreciation after 2nd year(20%)\n\n\\(\\frac{20}{100} \\times 225,000 = N45,000\\)\n\n\n\nValue after 2nd year\nN180,000.00\n\n\n\n\n\n\n', '(a) \\(4x^{2} = 11x + 21 \\implies 4x^{2} - 11x - 21 = 0\\)\n\\(a = 4, b = - 11 , c = -21\\)\n\\(x = \\frac{-b \\pm \\sqrt{b^{2} - 4ac}}{2a}\\)\n\\(x = \\frac{-(-11) \\pm \\sqrt{(-11)^{2} - 4(4)(-21)}}{2(4)}\\)\n\\(x = \\frac{11 \\pm \\sqrt{121 + 336}}{8}\\)\n\\(x = \\frac{11 \\pm \\sqrt{457}}{8}\\)\n\\(x = \\frac{11 \\pm 21.378}{8}\\)\n\\(x = \\frac{11 + 21.378}{8}\\) or \\(x = \\frac{11 - 21.378}{8}\\)\n\\(x = \\frac{32.378}{8}\\) or \\(x = \\frac{-10.378}{8}\\)\\\n\\(x = \\text{4.047or -1.297}\\)\n\\(x \\approxeq \\text{4.05or -1.30}\\).(2 decimal place).~(b) (i) \\(I = \\frac{PRT}{100}\\)\n\\(I =£(1560 - 1500) =£60\\)\n\\(60 = \\frac{1500 \\times R \\times 1}{100}\\)\n\\(60 = 15R \\implies R = 4%\\)\n(ii) Interest for the second year\n= \\(\\frac{1560 \\times 4 \\times 1}{100}\\)\n= \\(£62.40\\)\n(c)\n\n\n\nInitial price of the car\n\\(N300,000.00\\)\n\n\nDepreciation after first year (25%)\n\\(\\frac{25}{100} \\times 300,000 = N75,000\\)\n\n\nValue after 1st year\nN225,000.00\n\n\nDepreciation after 2nd year(20%)\n\n\\(\\frac{20}{100} \\times 225,000 = N45,000\\)\n\n\n\nValue after 2nd year\nN180,000.00\n\n\n\n\n\n~(c)\n\n\n\nInitial price of the car\n\\(N300,000.00\\)\n\n\nDepreciation after first year (25%)\n\\(\\frac{25}{100} \\times 300,000 = N75,000\\)\n\n\nValue after 1st year\nN225,000.00\n\n\nDepreciation after 2nd year(20%)\n\n\\(\\frac{20}{100} \\times 225,000 = N45,000\\)\n\n\n\nValue after 2nd year\nN180,000.00\n\n\n\n\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(454, '2014', '\n\nQuestion\n\nThe ages, in years, of 50 teachers in a school are given below :\n21 37 49 27 49 42 26 33 46 40 50 29 23 24 29 31 36 22 27 38 30 26 42 39 34 23 21 32 41 46 46 31 33 29 28 43 47 40 34 44 26 38 34 49 45 27 25 33 39 40\n(a) Form a frequency distribution table of the data using the intervals : 21 - 25, 26 - 30, 31 - 35 etc.\n(b) Draw the histogram of the distribution\n(c) Use your histogram to estimate the mode\n(d) Calculate the mean age.\n\n', '', 'The ages, in years, of 50 teachers in a school are given below :\n21 37 49 27 49 42 26 33 46 40 50 29 23 24 29 31 36 22 27 38 30 26 42 39 34 23 21 32 41 46 46 31 33 29 28 43 47 40 34 44 26 38 34 49 45 27 25 33 39 40\n(a) Form a frequency distribution table of the data using the intervals : 21 - 25, 26 - 30, 31 - 35 etc.\n(b) Draw the histogram of the distribution\n(c) Use your histogram to estimate the mode\n(d) Calculate the mean age.', 'The ages, in years, of 50 teachers in a school are given below :\n21 37 49 27 49 42 26 33 46 40 50 29 23 24 29 31 36 22 27 38 30 26 42 39 34 23 21 32 41 46 46 31 33 29 28 43 47 40 34 44 26 38 34 49 45 27 25 33 39 40\n(a) Form a frequency distribution table of the data using the intervals : 21 - 25, 26 - 30, 31 - 35 etc.~(b) Draw the histogram of the distribution\n(c) Use your histogram to estimate the mode\n(d) Calculate the mean age.~(c) Use your histogram to estimate the mode\n(d) Calculate the mean age.~(d) Calculate the mean age.', NULL, '\n\nExplanation\nClass\nInterval\n \nTally\n\nClassmark\n(x)\n\n\nFreq\n(f)\n\n\\(fx\\)\n\n\n21 - 25\nIIII||\n23\n7\n161\n\n\n26 - 30\n|||||||||\n28\n11\n308\n\n\n31 - 35\n||||||||\n33\n9\n297\n\n\n36 - 40\n||||||||\n38\n9\n342\n\n\n41 - 45\n|||||\n43\n6\n258\n\n\n46 - 50\n|||||||\n48\n8\n384\n\n\n\\(\\sum\\)\n\n\n50\n1750\n\n\n\n(b)\n(c) Mode : \\(L_{1} + (\\frac{f_{0} - f_{1}}{2f_{0} - f_{1} - f_{2}})t\\)\nWhere \\(L_{1}\\) = lower class boundary of modal class = 25.5\n\\(f_{0}\\) = frequency of modal class = 11\n\\(f_{1}\\) = frequency of pre-modal class = 7\n\\(f_{2}\\) = frequency of post modal class = 9\n\\(t\\) = interval mark = 5.\nMode : \\(25.5 + (\\frac{11 - 7}{22 - 7 - 9})\\times 5 = 25.5 + 3.3\\)\n= 28.8.\n(d) Mean : \\(\\frac{\\sum fx}{\\sum f} = \\frac{1750}{50}\\)\n= 35.\n\n', '', 'Explanation\nClass\nInterval\n \nTally\n\nClassmark\n(x)\n\n\nFreq\n(f)\n\n\\(fx\\)\n\n\n21 - 25\nIIII||\n23\n7\n161\n\n\n26 - 30\n|||||||||\n28\n11\n308\n\n\n31 - 35\n||||||||\n33\n9\n297\n\n\n36 - 40\n||||||||\n38\n9\n342\n\n\n41 - 45\n|||||\n43\n6\n258\n\n\n46 - 50\n|||||||\n48\n8\n384\n\n\n\\(\\sum\\)\n\n\n50\n1750\n\n\n\n(b)\n(c) Mode : \\(L_{1} + (\\frac{f_{0} - f_{1}}{2f_{0} - f_{1} - f_{2}})t\\)\nWhere \\(L_{1}\\) = lower class boundary of modal class = 25.5\n\\(f_{0}\\) = frequency of modal class = 11\n\\(f_{1}\\) = frequency of pre-modal class = 7\n\\(f_{2}\\) = frequency of post modal class = 9\n\\(t\\) = interval mark = 5.\nMode : \\(25.5 + (\\frac{11 - 7}{22 - 7 - 9})\\times 5 = 25.5 + 3.3\\)\n= 28.8.\n(d) Mean : \\(\\frac{\\sum fx}{\\sum f} = \\frac{1750}{50}\\)\n= 35.\n\n', '~(b)\n(c) Mode : \\(L_{1} + (\\frac{f_{0} - f_{1}}{2f_{0} - f_{1} - f_{2}})t\\)\nWhere \\(L_{1}\\) = lower class boundary of modal class = 25.5\n\\(f_{0}\\) = frequency of modal class = 11\n\\(f_{1}\\) = frequency of pre-modal class = 7\n\\(f_{2}\\) = frequency of post modal class = 9\n\\(t\\) = interval mark = 5.\nMode : \\(25.5 + (\\frac{11 - 7}{22 - 7 - 9})\\times 5 = 25.5 + 3.3\\)\n= 28.8.\n(d) Mean : \\(\\frac{\\sum fx}{\\sum f} = \\frac{1750}{50}\\)\n= 35.\n~(c) Mode : \\(L_{1} + (\\frac{f_{0} - f_{1}}{2f_{0} - f_{1} - f_{2}})t\\)\nWhere \\(L_{1}\\) = lower class boundary of modal class = 25.5\n\\(f_{0}\\) = frequency of modal class = 11\n\\(f_{1}\\) = frequency of pre-modal class = 7\n\\(f_{2}\\) = frequency of post modal class = 9\n\\(t\\) = interval mark = 5.\nMode : \\(25.5 + (\\frac{11 - 7}{22 - 7 - 9})\\times 5 = 25.5 + 3.3\\)\n= 28.8.\n(d) Mean : \\(\\frac{\\sum fx}{\\sum f} = \\frac{1750}{50}\\)\n= 35.\n~(d) Mean : \\(\\frac{\\sum fx}{\\sum f} = \\frac{1750}{50}\\)\n= 35.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(455, '2014', '\n\nQuestion\n\n(a) The triangle ABC has sides AB = 17m, BC = 12m and AC = 10m. Calculate the :\n(i) largest angle of the triangle ; (ii) area of the triangle.\n(b) From a point T on a horizontal ground, the angle of elevation of the top R of a tower RS, 38m high is 63°. Calculate, correct to the nearest metre, the distance between T and S.\n\n', '', '(a) The triangle ABC has sides AB = 17m, BC = 12m and AC = 10m. Calculate the :\n(i) largest angle of the triangle ; (ii) area of the triangle.\n(b) From a point T on a horizontal ground, the angle of elevation of the top R of a tower RS, 38m high is 63°. Calculate, correct to the nearest metre, the distance between T and S.', '(a) The triangle ABC has sides AB = 17m, BC = 12m and AC = 10m. Calculate the :\n(i) largest angle of the triangle ; (ii) area of the triangle.~(b) From a point T on a horizontal ground, the angle of elevation of the top R of a tower RS, 38m high is 63°. Calculate, correct to the nearest metre, the distance between T and S.', NULL, '\n\nExplanation\n(a) (i)\n\\(\\cos C = \\frac{a^{2} + b^{2} - c^{2}}{2ab}\\)\n\\(\\cos C = \\frac{12^{2} + 10^{2} - 17^{2}}{2(12)(10)}\\)\n\\(\\cos C = \\frac{-45}{240} = -0.1875\\)\n\\(C = \\cos^{-1} (-0.1875)\\)\n= \\(100.81°\\)\n(ii) Area of \\(\\Delta ABC = \\frac{1}{2} ab \\sin C\\)\n= \\(\\frac{1}{2} \\times 12 \\times 10 \\times \\sin 100.81\\)\n= \\(60 \\times 0.9822\\)\n= \\(58.936 m^{2}\\)\n(b)\\(\\tan 63 = \\frac{38}{x}\\)\n\\(x = \\frac{38}{\\tan 63}\\)\n\\(x = 19.362 m\\)\n\\(\\approxeq 19 m\\)\n\n', '', 'Explanation\n(a) (i)\n\\(\\cos C = \\frac{a^{2} + b^{2} - c^{2}}{2ab}\\)\n\\(\\cos C = \\frac{12^{2} + 10^{2} - 17^{2}}{2(12)(10)}\\)\n\\(\\cos C = \\frac{-45}{240} = -0.1875\\)\n\\(C = \\cos^{-1} (-0.1875)\\)\n= \\(100.81°\\)\n(ii) Area of \\(\\Delta ABC = \\frac{1}{2} ab \\sin C\\)\n= \\(\\frac{1}{2} \\times 12 \\times 10 \\times \\sin 100.81\\)\n= \\(60 \\times 0.9822\\)\n= \\(58.936 m^{2}\\)\n(b)\\(\\tan 63 = \\frac{38}{x}\\)\n\\(x = \\frac{38}{\\tan 63}\\)\n\\(x = 19.362 m\\)\n\\(\\approxeq 19 m\\)\n\n', '(a) (i)\n\\(\\cos C = \\frac{a^{2} + b^{2} - c^{2}}{2ab}\\)\n\\(\\cos C = \\frac{12^{2} + 10^{2} - 17^{2}}{2(12)(10)}\\)\n\\(\\cos C = \\frac{-45}{240} = -0.1875\\)\n\\(C = \\cos^{-1} (-0.1875)\\)\n= \\(100.81°\\)\n(ii) Area of \\(\\Delta ABC = \\frac{1}{2} ab \\sin C\\)\n= \\(\\frac{1}{2} \\times 12 \\times 10 \\times \\sin 100.81\\)\n= \\(60 \\times 0.9822\\)\n= \\(58.936 m^{2}\\)~(b)\\(\\tan 63 = \\frac{38}{x}\\)\n\\(x = \\frac{38}{\\tan 63}\\)\n\\(x = 19.362 m\\)\n\\(\\approxeq 19 m\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(456, '2014', '\n\nQuestion\n\n(a) Using ruler and a pair of compasses only, construct :\n(i) a quadrilateral PQRS such that /PQ/ = 7 cm, < QPS = 60°, /PS/ = 6.5 cm, < PQR = 135° and /QS/ = /QR/ ;\n(ii) locus, \\(l_{1}\\) of points equidistant from P and Q ;\n(iii) locus, \\(l_{2}\\) of points equidistant from P and S.\n(b)(i) Label the point T where \\(l_{1}\\) and \\(l_{2}\\) intersect. (ii) With center T and radius /TP/, construct a circle \\(l_{3}\\).\n\n', '', '(a) Using ruler and a pair of compasses only, construct :\n(i) a quadrilateral PQRS such that /PQ/ = 7 cm, < QPS = 60°, /PS/ = 6.5 cm, < PQR = 135° and /QS/ = /QR/ ;\n(ii) locus, \\(l_{1}\\) of points equidistant from P and Q ;\n(iii) locus, \\(l_{2}\\) of points equidistant from P and S.\n(b)(i) Label the point T where \\(l_{1}\\) and \\(l_{2}\\) intersect. (ii) With center T and radius /TP/, construct a circle \\(l_{3}\\).', '(a) Using ruler and a pair of compasses only, construct :\n(i) a quadrilateral PQRS such that /PQ/ = 7 cm, < QPS = 60°, /PS/ = 6.5 cm, < PQR = 135° and /QS/ = /QR/ ;\n(ii) locus, \\(l_{1}\\) of points equidistant from P and Q ;\n(iii) locus, \\(l_{2}\\) of points equidistant from P and S.~(b)(i) Label the point T where \\(l_{1}\\) and \\(l_{2}\\) intersect. (ii) With center T and radius /TP/, construct a circle \\(l_{3}\\).', NULL, '\n\nExplanation\n', '', '', '', 'WAEC-PK-1', 'WAEC-MATH', 1),
(457, '2014', '\n\nQuestion\n\n(a)\nIn the diagram, AB // CD and BC // FE. \\(\\stackrel\\frown{CDE} = 75°\\) and \\(\\stackrel\\frown{DEF} = 26°\\). Find the angles marked x and y.\n(b)\nThe diagram shows a circle ABCD with centre O and radius 7 cm. The reflex angle AOC = 190° and < DAO = 35°. Find :\n(i) < ABC ; (ii) < ADC.\n(c) Using the diagram in (b) above, calculate, correct to 3 significant figures, the length of : (i) arc ABC ; (ii) the chord AD. [Take \\(\\pi = 3.142\\)].\n\n', '', '(a)\nIn the diagram, AB // CD and BC // FE. \\(\\stackrel\\frown{CDE} = 75°\\) and \\(\\stackrel\\frown{DEF} = 26°\\). Find the angles marked x and y.\n(b)\nThe diagram shows a circle ABCD with centre O and radius 7 cm. The reflex angle AOC = 190° and < DAO = 35°. Find :\n(i) < ABC ; (ii) < ADC.\n(c) Using the diagram in (b) above, calculate, correct to 3 significant figures, the length of : (i) arc ABC ; (ii) the chord AD. [Take \\(\\pi = 3.142\\)].', '(a)\nIn the diagram, AB // CD and BC // FE. \\(\\stackrel\\frown{CDE} = 75°\\) and \\(\\stackrel\\frown{DEF} = 26°\\). Find the angles marked x and y.~(b)\nThe diagram shows a circle ABCD with centre O and radius 7 cm. The reflex angle AOC = 190° and < DAO = 35°. Find :\n(i) < ABC ; (ii) < ADC.\n(c) Using the diagram in (b) above, calculate, correct to 3 significant figures, the length of : (i) arc ABC ; (ii) the chord AD. [Take \\(\\pi = 3.142\\)].~(c) Using the diagram in (b) above, calculate, correct to 3 significant figures, the length of : (i) arc ABC ; (ii) the chord AD. [Take \\(\\pi = 3.142\\)].', NULL, '\n\nExplanation\n(a)105° + 26° + x = 180°\n131° + x = 180°\nx = 180° - 131°\n= 49°.\n131° + x + 131° + 360 - y = 360°\n131° + 49° + 131° + 360 - y = 360°\n671° - y = 360°\ny = 671° - 360°\n= 311°.\n(b)(i) \\(\\stackrel\\frown{ABC} = \\frac{1}{2} \\stackrel\\frown{AOC} reflex\\)\n= \\(\\frac{1}{2} \\times 190° = 95°\\)\n(ii) \\(\\stackrel\\frown{ADC} = \\frac{1}{2} \\stackrel\\frown{AOC}\\)\n= \\(\\frac{1}{2} (360° - 190°)\\)\n= \\(\\frac{1}{2} \\times 170°\\)\n= 85°\n(c) (i) \\(\\stackrel\\frown{ABC} = \\frac{360 - 190}{360} \\times 2\\pi r\\)\n= \\(\\frac{170}{360} \\times 2 \\times 3.142 \\times 7\\)\n= \\(20.77 cm\\)\n\\(\\approxeq 20.8 cm\\) (3 significant figures)\n(ii) \\(AD = 2 \\times 7\\cos 35\\)\n= \\(14 \\times 0.8192\\)\n= \\(11.468 cm \\approxeq 11.5 cm\\) (3 significant figures)\n\n', '', 'Explanation\n(a)105° + 26° + x = 180°\n131° + x = 180°\nx = 180° - 131°\n= 49°.\n131° + x + 131° + 360 - y = 360°\n131° + 49° + 131° + 360 - y = 360°\n671° - y = 360°\ny = 671° - 360°\n= 311°.\n(b)(i) \\(\\stackrel\\frown{ABC} = \\frac{1}{2} \\stackrel\\frown{AOC} reflex\\)\n= \\(\\frac{1}{2} \\times 190° = 95°\\)\n(ii) \\(\\stackrel\\frown{ADC} = \\frac{1}{2} \\stackrel\\frown{AOC}\\)\n= \\(\\frac{1}{2} (360° - 190°)\\)\n= \\(\\frac{1}{2} \\times 170°\\)\n= 85°\n(c) (i) \\(\\stackrel\\frown{ABC} = \\frac{360 - 190}{360} \\times 2\\pi r\\)\n= \\(\\frac{170}{360} \\times 2 \\times 3.142 \\times 7\\)\n= \\(20.77 cm\\)\n\\(\\approxeq 20.8 cm\\) (3 significant figures)\n(ii) \\(AD = 2 \\times 7\\cos 35\\)\n= \\(14 \\times 0.8192\\)\n= \\(11.468 cm \\approxeq 11.5 cm\\) (3 significant figures)\n\n', '(a)105° + 26° + x = 180°\n131° + x = 180°\nx = 180° - 131°\n= 49°.\n131° + x + 131° + 360 - y = 360°\n131° + 49° + 131° + 360 - y = 360°\n671° - y = 360°\ny = 671° - 360°\n= 311°.~(b)(i) \\(\\stackrel\\frown{ABC} = \\frac{1}{2} \\stackrel\\frown{AOC} reflex\\)\n= \\(\\frac{1}{2} \\times 190° = 95°\\)\n(ii) \\(\\stackrel\\frown{ADC} = \\frac{1}{2} \\stackrel\\frown{AOC}\\)\n= \\(\\frac{1}{2} (360° - 190°)\\)\n= \\(\\frac{1}{2} \\times 170°\\)\n= 85°\n(c) (i) \\(\\stackrel\\frown{ABC} = \\frac{360 - 190}{360} \\times 2\\pi r\\)\n= \\(\\frac{170}{360} \\times 2 \\times 3.142 \\times 7\\)\n= \\(20.77 cm\\)\n\\(\\approxeq 20.8 cm\\) (3 significant figures)\n(ii) \\(AD = 2 \\times 7\\cos 35\\)\n= \\(14 \\times 0.8192\\)\n= \\(11.468 cm \\approxeq 11.5 cm\\) (3 significant figures)\n~(c) (i) \\(\\stackrel\\frown{ABC} = \\frac{360 - 190}{360} \\times 2\\pi r\\)\n= \\(\\frac{170}{360} \\times 2 \\times 3.142 \\times 7\\)\n= \\(20.77 cm\\)\n\\(\\approxeq 20.8 cm\\) (3 significant figures)\n(ii) \\(AD = 2 \\times 7\\cos 35\\)\n= \\(14 \\times 0.8192\\)\n= \\(11.468 cm \\approxeq 11.5 cm\\) (3 significant figures)\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
(458, '2014', '\n\nQuestion\n\n(a) Copy and complete the table of values for \\(y = 3\\sin x + 2\\cos x\\) for \\(0° \\leq x \\leq 360°\\).\n\n\n\nx\n0°\n60°\n120°\n180°\n240°\n300°\n360°\n\n\ny\n2.00\n\n\n\n\n\n2.00\n\n\n\n(b) Using a scale of 2 cm to 60° on x- axis and 2 cm to 1 unit on the y- axis, draw the graph of \\(y = 3 \\sin x + 2 \\cos x\\) for \\(0° \\leq x \\leq 360°\\).\n(c) Use your graph to solve the equation : \\(3 \\sin x + 2 \\cos x = 1.5\\).\n(d) Find the range of values of x for which \\(3\\sin x + 2\\cos x < -1\\).\n\n\n(a) Copy and complete the table of values for \\(y = 3\\sin x + 2\\cos x\\) for \\(0° \\leq x \\leq 360°\\).\n\n\n\nx\n0°\n60°\n120°\n180°\n240°\n300°\n360°\n\n\ny\n2.00\n\n\n\n\n\n2.00\n\n\n', '', '(a) Copy and complete the table of values for \\(y = 3\\sin x + 2\\cos x\\) for \\(0° \\leq x \\leq 360°\\).\n\n\n\nx\n0°\n60°\n120°\n180°\n240°\n300°\n360°\n\n\ny\n2.00\n\n\n\n\n\n2.00\n\n\n\n(b) Using a scale of 2 cm to 60° on x- axis and 2 cm to 1 unit on the y- axis, draw the graph of \\(y = 3 \\sin x + 2 \\cos x\\) for \\(0° \\leq x \\leq 360°\\).\n(c) Use your graph to solve the equation : \\(3 \\sin x + 2 \\cos x = 1.5\\).\n(d) Find the range of values of x for which \\(3\\sin x + 2\\cos x < -1\\).\n\n\n(a) Copy and complete the table of values for \\(y = 3\\sin x + 2\\cos x\\) for \\(0° \\leq x \\leq 360°\\).\n\n\n\nx\n0°\n60°\n120°\n180°\n240°\n300°\n360°\n\n\ny\n2.00\n\n\n\n\n\n2.00\n', '(a) Copy and complete the table of values for \\(y = 3\\sin x + 2\\cos x\\) for \\(0° \\leq x \\leq 360°\\).\n\n\n\nx\n0°\n60°\n120°\n180°\n240°\n300°\n360°\n\n\ny\n2.00\n\n\n\n\n\n2.00\n\n\n~(b) Using a scale of 2 cm to 60° on x- axis and 2 cm to 1 unit on the y- axis, draw the graph of \\(y = 3 \\sin x + 2 \\cos x\\) for \\(0° \\leq x \\leq 360°\\).\n(c) Use your graph to solve the equation : \\(3 \\sin x + 2 \\cos x = 1.5\\).\n(d) Find the range of values of x for which \\(3\\sin x + 2\\cos x < -1\\).\n\n\n(a) Copy and complete the table of values for \\(y = 3\\sin x + 2\\cos x\\) for \\(0° \\leq x \\leq 360°\\).\n\n\n\nx\n0°\n60°\n120°\n180°\n240°\n300°\n360°\n\n\ny\n2.00\n\n\n\n\n\n2.00\n~(c) Use your graph to solve the equation : \\(3 \\sin x + 2 \\cos x = 1.5\\).\n(d) Find the range of values of x for which \\(3\\sin x + 2\\cos x < -1\\).\n\n\n(a) Copy and complete the table of values for \\(y = 3\\sin x + 2\\cos x\\) for \\(0° \\leq x \\leq 360°\\).\n\n\n\nx\n0°\n60°\n120°\n180°\n240°\n300°\n360°\n\n\ny\n2.00\n\n\n\n\n\n2.00\n~(d) Find the range of values of x for which \\(3\\sin x + 2\\cos x < -1\\).\n\n\n(a) Copy and complete the table of values for \\(y = 3\\sin x + 2\\cos x\\) for \\(0° \\leq x \\leq 360°\\).\n\n\n\nx\n0°\n60°\n120°\n180°\n240°\n300°\n360°\n\n\ny\n2.00\n\n\n\n\n\n2.00\n', NULL, '\n\nExplanation\n(a)\n\n\n\nx\n0°\n60°\n120°\n180°\n240°\n300°\n360°\n\n\n\\(\\sin x\\)\n0\n0.8660\n0.8660\n0\n-0.8660\n-0.8660\n0\n\n\n\\(\\cos x\\)\n1\n0.5\n-0.5\n-1\n-0.5\n-0.5\n\n1\n\n\n\n\\(3\\sin x\\)\n0\n2.598\n2.598\n0\n-2.598\n-2.598\n0\n\n\n\\(2\\cos x\\)\n2\n1\n-1\n-2\n-1\n-1\n2\n\n\ny\n2\n3.598\n1.598\n-2\n-3.598\n-3.598\n2\n\n\ny\n2.00\n3.60\n1.60\n-2.00\n-3.60\n-3.60\n2.00\n\n\n\n(b)\n(c) From graph, \\(x_{1} = 120 + \\frac{1}{5} \\times 15 = 120 + 3 = 123°\\)\n\\(x_{2} = 360° - \\frac{2}{3} \\times 15 = 360 - 10 = 350°\\)\n(d) From graph, \\(180° - 15 < x < 300 + \\frac{3}{5} \\times 15\\)\n\\(165° < x < 300 + 9 = 165° < x < 309°\\)\n\n', '', 'Explanation\n(a)\n\n\n\nx\n0°\n60°\n120°\n180°\n240°\n300°\n360°\n\n\n\\(\\sin x\\)\n0\n0.8660\n0.8660\n0\n-0.8660\n-0.8660\n0\n\n\n\\(\\cos x\\)\n1\n0.5\n-0.5\n-1\n-0.5\n-0.5\n\n1\n\n\n\n\\(3\\sin x\\)\n0\n2.598\n2.598\n0\n-2.598\n-2.598\n0\n\n\n\\(2\\cos x\\)\n2\n1\n-1\n-2\n-1\n-1\n2\n\n\ny\n2\n3.598\n1.598\n-2\n-3.598\n-3.598\n2\n\n\ny\n2.00\n3.60\n1.60\n-2.00\n-3.60\n-3.60\n2.00\n\n\n\n(b)\n(c) From graph, \\(x_{1} = 120 + \\frac{1}{5} \\times 15 = 120 + 3 = 123°\\)\n\\(x_{2} = 360° - \\frac{2}{3} \\times 15 = 360 - 10 = 350°\\)\n(d) From graph, \\(180° - 15 < x < 300 + \\frac{3}{5} \\times 15\\)\n\\(165° < x < 300 + 9 = 165° < x < 309°\\)\n\n', '(a)\n\n\n\nx\n0°\n60°\n120°\n180°\n240°\n300°\n360°\n\n\n\\(\\sin x\\)\n0\n0.8660\n0.8660\n0\n-0.8660\n-0.8660\n0\n\n\n\\(\\cos x\\)\n1\n0.5\n-0.5\n-1\n-0.5\n-0.5\n\n1\n\n\n\n\\(3\\sin x\\)\n0\n2.598\n2.598\n0\n-2.598\n-2.598\n0\n\n\n\\(2\\cos x\\)\n2\n1\n-1\n-2\n-1\n-1\n2\n\n\ny\n2\n3.598\n1.598\n-2\n-3.598\n-3.598\n2\n\n\ny\n2.00\n3.60\n1.60\n-2.00\n-3.60\n-3.60\n2.00\n\n\n~(b)\n(c) From graph, \\(x_{1} = 120 + \\frac{1}{5} \\times 15 = 120 + 3 = 123°\\)\n\\(x_{2} = 360° - \\frac{2}{3} \\times 15 = 360 - 10 = 350°\\)\n(d) From graph, \\(180° - 15 < x < 300 + \\frac{3}{5} \\times 15\\)\n\\(165° < x < 300 + 9 = 165° < x < 309°\\)\n~(c) From graph, \\(x_{1} = 120 + \\frac{1}{5} \\times 15 = 120 + 3 = 123°\\)\n\\(x_{2} = 360° - \\frac{2}{3} \\times 15 = 360 - 10 = 350°\\)\n(d) From graph, \\(180° - 15 < x < 300 + \\frac{3}{5} \\times 15\\)\n\\(165° < x < 300 + 9 = 165° < x < 309°\\)\n~(d) From graph, \\(180° - 15 < x < 300 + \\frac{3}{5} \\times 15\\)\n\\(165° < x < 300 + 9 = 165° < x < 309°\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(459, '2014', '\n\nQuestion\n\n(a) If 3, x, y, 18 are the terms of an Arithmetic Progression (A.P), find the values of x and y.\n(b)(i) The sum of the second and third terms of a grometric progression is six times the fourth term. Find the two possible values of the common ratio.\n(ii) If the second term is 8 and the common ratio is positive, find the first six terms.\n\n', '', '(a) If 3, x, y, 18 are the terms of an Arithmetic Progression (A.P), find the values of x and y.\n(b)(i) The sum of the second and third terms of a grometric progression is six times the fourth term. Find the two possible values of the common ratio.\n(ii) If the second term is 8 and the common ratio is positive, find the first six terms.', '(a) If 3, x, y, 18 are the terms of an Arithmetic Progression (A.P), find the values of x and y.~(b)(i) The sum of the second and third terms of a grometric progression is six times the fourth term. Find the two possible values of the common ratio.\n(ii) If the second term is 8 and the common ratio is positive, find the first six terms.', NULL, '\n\nExplanation\n(a) 3, x, y, 18.\nNote : When given the consecutive terms of an A.P to be \\(a, b, c, d\\), then\n\\(\\frac{a + c}{2} = b ; \\frac{b + d}{2} = c\\).\n\\(\\therefore \\frac{3 + y}{2} = x \\implies 3 + y = 2x .... (1)\\)\nAlso, \\(\\frac{x + 18}{2} = y \\implies x + 18 = 2y ..... (2)\\)\nFrom (2), x = 2y - 18. Putting that in (1), we have\n\\(3 + y = 2(2y - 18) \\implies 3 + y = 4y - 36\\)\n\\(3 + 36 = 4y - y \\implies 3y = 39\\)\n\\(y = 13\\)\n\\(x = 2y - 18\\)\n\\(x = 2(13) - 18 = 26 - 18 = 8\\)\n\\((x, y) = (8, 13)\\).\n(b)(i) G.P\n\\(T_{2} = ar ; T_{3} = ar^{2} ; T_{4} = ar^{3}\\)\n\\(T_{2} + T_{3} = 6T_{4}\\)\n\\(ar + ar^{2} = 6ar^{3}\\)\n\\(6ar^{3} - ar^{2} - ar = 0\\)\n\\(6ar^{3} - 3ar^{2} + 2ar^{2} - ar = 0\\)\n\\(3ar^{2}(2r - 1) + ar(2r - 1) = 0\\)\n\\((3ar^{2} + ar)(2r - 1) = 0\\)\n\\(ar(3r + 1)(2r - 1) = 0\\)\n\\(\\implies 3r + 1 = 0 ; 2r - 1 = 0\\)\n\\(r = \\frac{-1}{3} ; r = \\frac{1}{2}\\)\n(ii) Since r is positive, then \\(r = \\frac{1}{2}\\).\n\\(T_{2} = ar = 8 \\implies \\frac{a}{2} = 8\\)\n\\(a = 16\\)\n\\(\\therefore \\text{The first 6 terms} = 16, 8, 4, 2, 1, \\frac{1}{2}\\)\n\n', '', 'Explanation\n(a) 3, x, y, 18.\nNote : When given the consecutive terms of an A.P to be \\(a, b, c, d\\), then\n\\(\\frac{a + c}{2} = b ; \\frac{b + d}{2} = c\\).\n\\(\\therefore \\frac{3 + y}{2} = x \\implies 3 + y = 2x .... (1)\\)\nAlso, \\(\\frac{x + 18}{2} = y \\implies x + 18 = 2y ..... (2)\\)\nFrom (2), x = 2y - 18. Putting that in (1), we have\n\\(3 + y = 2(2y - 18) \\implies 3 + y = 4y - 36\\)\n\\(3 + 36 = 4y - y \\implies 3y = 39\\)\n\\(y = 13\\)\n\\(x = 2y - 18\\)\n\\(x = 2(13) - 18 = 26 - 18 = 8\\)\n\\((x, y) = (8, 13)\\).\n(b)(i) G.P\n\\(T_{2} = ar ; T_{3} = ar^{2} ; T_{4} = ar^{3}\\)\n\\(T_{2} + T_{3} = 6T_{4}\\)\n\\(ar + ar^{2} = 6ar^{3}\\)\n\\(6ar^{3} - ar^{2} - ar = 0\\)\n\\(6ar^{3} - 3ar^{2} + 2ar^{2} - ar = 0\\)\n\\(3ar^{2}(2r - 1) + ar(2r - 1) = 0\\)\n\\((3ar^{2} + ar)(2r - 1) = 0\\)\n\\(ar(3r + 1)(2r - 1) = 0\\)\n\\(\\implies 3r + 1 = 0 ; 2r - 1 = 0\\)\n\\(r = \\frac{-1}{3} ; r = \\frac{1}{2}\\)\n(ii) Since r is positive, then \\(r = \\frac{1}{2}\\).\n\\(T_{2} = ar = 8 \\implies \\frac{a}{2} = 8\\)\n\\(a = 16\\)\n\\(\\therefore \\text{The first 6 terms} = 16, 8, 4, 2, 1, \\frac{1}{2}\\)\n\n', '(a) 3, x, y, 18.\nNote : When given the consecutive terms of an A.P to be \\(a, b, c, d\\), then\n\\(\\frac{a + c}{2} = b ; \\frac{b + d}{2} = c\\).\n\\(\\therefore \\frac{3 + y}{2} = x \\implies 3 + y = 2x .... (1)\\)\nAlso, \\(\\frac{x + 18}{2} = y \\implies x + 18 = 2y ..... (2)\\)\nFrom (2), x = 2y - 18. Putting that in (1), we have\n\\(3 + y = 2(2y - 18) \\implies 3 + y = 4y - 36\\)\n\\(3 + 36 = 4y - y \\implies 3y = 39\\)\n\\(y = 13\\)\n\\(x = 2y - 18\\)\n\\(x = 2(13) - 18 = 26 - 18 = 8\\)\n\\((x, y) = (8, 13)\\).~(b)(i) G.P\n\\(T_{2} = ar ; T_{3} = ar^{2} ; T_{4} = ar^{3}\\)\n\\(T_{2} + T_{3} = 6T_{4}\\)\n\\(ar + ar^{2} = 6ar^{3}\\)\n\\(6ar^{3} - ar^{2} - ar = 0\\)\n\\(6ar^{3} - 3ar^{2} + 2ar^{2} - ar = 0\\)\n\\(3ar^{2}(2r - 1) + ar(2r - 1) = 0\\)\n\\((3ar^{2} + ar)(2r - 1) = 0\\)\n\\(ar(3r + 1)(2r - 1) = 0\\)\n\\(\\implies 3r + 1 = 0 ; 2r - 1 = 0\\)\n\\(r = \\frac{-1}{3} ; r = \\frac{1}{2}\\)\n(ii) Since r is positive, then \\(r = \\frac{1}{2}\\).\n\\(T_{2} = ar = 8 \\implies \\frac{a}{2} = 8\\)\n\\(a = 16\\)\n\\(\\therefore \\text{The first 6 terms} = 16, 8, 4, 2, 1, \\frac{1}{2}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(460, '2014', '\n\nQuestion\n\n(a) Evaluate, without using mathematical tables or calculator, \\((3.69 \\times 10^{5}) \\div(1.64 \\times 10^{-3})\\), leaving your answer in standard form.\n(b) A man invested N20,000 in bank A and N25,000 in bank B at the beginning of the year. Bank A pays simple interest at a rate of y% per annum and B pays 1.5y% per annum. If his total interest at the end of the year from the two banks was N4,600, find the value of y.\n\n', '', '(a) Evaluate, without using mathematical tables or calculator, \\((3.69 \\times 10^{5}) \\div(1.64 \\times 10^{-3})\\), leaving your answer in standard form.\n(b) A man invested N20,000 in bank A and N25,000 in bank B at the beginning of the year. Bank A pays simple interest at a rate of y% per annum and B pays 1.5y% per annum. If his total interest at the end of the year from the two banks was N4,600, find the value of y.', '(a) Evaluate, without using mathematical tables or calculator, \\((3.69 \\times 10^{5}) \\div(1.64 \\times 10^{-3})\\), leaving your answer in standard form.~(b) A man invested N20,000 in bank A and N25,000 in bank B at the beginning of the year. Bank A pays simple interest at a rate of y% per annum and B pays 1.5y% per annum. If his total interest at the end of the year from the two banks was N4,600, find the value of y.', NULL, '\n\nExplanation\n(a) \\((3.69 \\times 10^{5}) \\div (1.64 \\times 10^{-3})\\)\n= \\(\\frac{3.69}{1.64} \\times 10^{5 - (-3)}\\)\n= \\(2.25 \\times 10^{8}\\)\n(b) Interest from bank A :\\(\\frac{y}{100} \\times N20,000 = N200y\\)\nInterest on bank B :\\(\\frac{1.5y}{100} \\times N25,000 = N375y\\)\n\\(\\therefore N(200y + 375y) = N4,600\\)\n\\(575y = 4,600 \\implies y = 8\\)\n\n', '', 'Explanation\n(a) \\((3.69 \\times 10^{5}) \\div (1.64 \\times 10^{-3})\\)\n= \\(\\frac{3.69}{1.64} \\times 10^{5 - (-3)}\\)\n= \\(2.25 \\times 10^{8}\\)\n(b) Interest from bank A :\\(\\frac{y}{100} \\times N20,000 = N200y\\)\nInterest on bank B :\\(\\frac{1.5y}{100} \\times N25,000 = N375y\\)\n\\(\\therefore N(200y + 375y) = N4,600\\)\n\\(575y = 4,600 \\implies y = 8\\)\n\n', '(a) \\((3.69 \\times 10^{5}) \\div (1.64 \\times 10^{-3})\\)\n= \\(\\frac{3.69}{1.64} \\times 10^{5 - (-3)}\\)\n= \\(2.25 \\times 10^{8}\\)~(b) Interest from bank A :\\(\\frac{y}{100} \\times N20,000 = N200y\\)\nInterest on bank B :\\(\\frac{1.5y}{100} \\times N25,000 = N375y\\)\n\\(\\therefore N(200y + 375y) = N4,600\\)\n\\(575y = 4,600 \\implies y = 8\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(461, '2014', '\n\nQuestion\n\n(a) Simplify : \\(\\frac{x^{2} - 8x + 16}{x^{2} - 7x + 12}\\).\n(b) If \\(\\frac{1}{2}, \\frac{1}{x}, \\frac{1}{3}\\) are successive terms of an arithmetic progression (A.P), show that \\(\\frac{2 - x}{x - 3} = \\frac{2}{3}\\).\n\n', '', '(a) Simplify : \\(\\frac{x^{2} - 8x + 16}{x^{2} - 7x + 12}\\).\n(b) If \\(\\frac{1}{2}, \\frac{1}{x}, \\frac{1}{3}\\) are successive terms of an arithmetic progression (A.P), show that \\(\\frac{2 - x}{x - 3} = \\frac{2}{3}\\).', '(a) Simplify : \\(\\frac{x^{2} - 8x + 16}{x^{2} - 7x + 12}\\).~(b) If \\(\\frac{1}{2}, \\frac{1}{x}, \\frac{1}{3}\\) are successive terms of an arithmetic progression (A.P), show that \\(\\frac{2 - x}{x - 3} = \\frac{2}{3}\\).', NULL, '\n\nExplanation\n(a) \\(\\frac{x^{2} - 8x + 16}{x^{2} - 7x + 12}\\)\n= \\(\\frac{(x - 4)(x - 4)}{(x - 4)(x - 3)}\\)\n= \\(\\frac{(x - 4)}{(x - 3)}\\)\n(b) Arithmetic Progression:\n\\(\\frac{1}{2}, \\frac{1}{x}, \\frac{1}{3}\\)\n\\(\\frac{1}{x} - \\frac{1}{2} = \\frac{1}{3} - \\frac{1}{x}\\)\n\\(\\frac{2 - x}{2x} = \\frac{x - 3}{3x}\\)\n\\(\\frac{2 - x}{x - 3} = \\frac{2x}{3x}\\)\n\\(\\implies \\frac{2 - x}{x - 3} = \\frac{2}{3}\\)\n\n', '', 'Explanation\n(a) \\(\\frac{x^{2} - 8x + 16}{x^{2} - 7x + 12}\\)\n= \\(\\frac{(x - 4)(x - 4)}{(x - 4)(x - 3)}\\)\n= \\(\\frac{(x - 4)}{(x - 3)}\\)\n(b) Arithmetic Progression:\n\\(\\frac{1}{2}, \\frac{1}{x}, \\frac{1}{3}\\)\n\\(\\frac{1}{x} - \\frac{1}{2} = \\frac{1}{3} - \\frac{1}{x}\\)\n\\(\\frac{2 - x}{2x} = \\frac{x - 3}{3x}\\)\n\\(\\frac{2 - x}{x - 3} = \\frac{2x}{3x}\\)\n\\(\\implies \\frac{2 - x}{x - 3} = \\frac{2}{3}\\)\n\n', '(a) \\(\\frac{x^{2} - 8x + 16}{x^{2} - 7x + 12}\\)\n= \\(\\frac{(x - 4)(x - 4)}{(x - 4)(x - 3)}\\)\n= \\(\\frac{(x - 4)}{(x - 3)}\\)~(b) Arithmetic Progression:\n\\(\\frac{1}{2}, \\frac{1}{x}, \\frac{1}{3}\\)\n\\(\\frac{1}{x} - \\frac{1}{2} = \\frac{1}{3} - \\frac{1}{x}\\)\n\\(\\frac{2 - x}{2x} = \\frac{x - 3}{3x}\\)\n\\(\\frac{2 - x}{x - 3} = \\frac{2x}{3x}\\)\n\\(\\implies \\frac{2 - x}{x - 3} = \\frac{2}{3}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(462, '2014', '\n\nQuestion\n\nThe diagram shows the cross- section of a railway tunnel. If |AB| = 100m and the radius of the arc is 56m, calculate, correct to the nearest metre, the perimetre of the cross- section.\n\n\n', 'The diagram shows the cross- section of a railway tunnel. If |AB| = 100m and the radius of the arc is 56m, calculate, correct to the nearest metre, the perimetre of the cross- section.\n', '', '', NULL, '\n\nExplanation\nFrom \\(\\Delta\\) BCD,\n\\(\\sin \\theta = \\frac{50}{56}\\)\n\\(\\theta = \\sin^{-1} (0.8929)\\)\n\\( < BCD = 63.23°\\)\n\\(\\therefore < ACB = 63.23° \\times 2 = 126.46°\\)\n\\(< ACB (reflex) = 360° - 126.46°\\)\n= \\(233.54°\\)\nPerimeter of the arc = \\(\\frac{233.54}{360} \\times 2 \\times \\frac{22}{7} \\times 56\\)\n= \\(228.35 m\\)\nPerimeter of the railway tunnel = Perimeter of the arc + |AB|\n= 228.35 m + 100 m\n= 328.35 m.\n\n', 'From \\(\\Delta\\) BCD,\n\\(\\sin \\theta = \\frac{50}{56}\\)\n\\(\\theta = \\sin^{-1} (0.8929)\\)\n\\( < BCD = 63.23°\\)\n\\(\\therefore < ACB = 63.23° \\times 2 = 126.46°\\)\n\\(< ACB (reflex) = 360° - 126.46°\\)\n= \\(233.54°\\)\nPerimeter of the arc = \\(\\frac{233.54}{360} \\times 2 \\times \\frac{22}{7} \\times 56\\)\n= \\(228.35 m\\)\nPerimeter of the railway tunnel = Perimeter of the arc + |AB|\n= 228.35 m + 100 m\n= 328.35 m.\n\n', '', '', 'WAEC-PK-1', 'WAEC-MATH', 1),
(463, '2014', '\n\nQuestion\n\nY is 60 km away from X on a bearing of 135°. Z is 80 km away from X on a bearing of 225°. Find the :\n(a) distance of Z from Y ;\n(b) bearing of Z from Y.\n\n', '', 'Y is 60 km away from X on a bearing of 135°. Z is 80 km away from X on a bearing of 225°. Find the :\n(a) distance of Z from Y ;\n(b) bearing of Z from Y.', 'Y is 60 km away from X on a bearing of 135°. Z is 80 km away from X on a bearing of 225°. Find the :\n(a) distance of Z from Y ;~(b) bearing of Z from Y.', NULL, '\n\nExplanation\n(a) < ZXY = 225° - 135°\n= 90°.\n\\(\\therefore /ZY/ = 80^{2} + 60^{2}\\)\n= \\(6400 + 3600 = 10000\\)\n\\(/ZY/ = \\sqrt{10000}\\)\n= 100 km.\n(b) From \\(\\Delta\\) XYZ,\n\\(\\tan \\theta = \\frac{opp}{adj}\\)\n\\(\\tan \\theta = \\frac{80}{60} = 1.333\\)\n\\(\\theta = \\tan^{-1} (1.333)\\)\n\\(\\theta = 53.13°\\)\nBearing of Z from Y = \\(360° - (45° + 53.13°)\\)\n= \\(360° - 98.13°\\)\n= \\(261.87° \\approxeq 262°\\) (3 significant figures)\n\n', '', 'Explanation\n(a) < ZXY = 225° - 135°\n= 90°.\n\\(\\therefore /ZY/ = 80^{2} + 60^{2}\\)\n= \\(6400 + 3600 = 10000\\)\n\\(/ZY/ = \\sqrt{10000}\\)\n= 100 km.\n(b) From \\(\\Delta\\) XYZ,\n\\(\\tan \\theta = \\frac{opp}{adj}\\)\n\\(\\tan \\theta = \\frac{80}{60} = 1.333\\)\n\\(\\theta = \\tan^{-1} (1.333)\\)\n\\(\\theta = 53.13°\\)\nBearing of Z from Y = \\(360° - (45° + 53.13°)\\)\n= \\(360° - 98.13°\\)\n= \\(261.87° \\approxeq 262°\\) (3 significant figures)\n\n', '(a) < ZXY = 225° - 135°\n= 90°.\n\\(\\therefore /ZY/ = 80^{2} + 60^{2}\\)\n= \\(6400 + 3600 = 10000\\)\n\\(/ZY/ = \\sqrt{10000}\\)\n= 100 km.~(b) From \\(\\Delta\\) XYZ,\n\\(\\tan \\theta = \\frac{opp}{adj}\\)\n\\(\\tan \\theta = \\frac{80}{60} = 1.333\\)\n\\(\\theta = \\tan^{-1} (1.333)\\)\n\\(\\theta = 53.13°\\)\nBearing of Z from Y = \\(360° - (45° + 53.13°)\\)\n= \\(360° - 98.13°\\)\n= \\(261.87° \\approxeq 262°\\) (3 significant figures)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(464, '2014', '\n\nQuestion\n\nOut of the 24 apples in a box, 6 are bad. If three apples are taken from the box at random, with replacement, find the probability that :\n(a) the first two are good and the third is bad ;\n(b) all three are bad ;\n(c) all the three are good.\n\n', '', 'Out of the 24 apples in a box, 6 are bad. If three apples are taken from the box at random, with replacement, find the probability that :\n(a) the first two are good and the third is bad ;\n(b) all three are bad ;\n(c) all the three are good.', 'Out of the 24 apples in a box, 6 are bad. If three apples are taken from the box at random, with replacement, find the probability that :\n(a) the first two are good and the third is bad ;~(b) all three are bad ;\n(c) all the three are good.~(c) all the three are good.', NULL, '\n\nExplanation\n(a) \\(P(Good) = \\frac{18}{24} = \\frac{3}{4}\\)\n\\(P(Bad) = \\frac{6}{24} = \\frac{1}{4}\\)\nP(2 are good and 1 bad) = \\(\\frac{3}{4} \\times \\frac{3}{4} \\times \\frac{1}{4}\\)\n= \\(\\frac{9}{64}\\)\n(b) P(the 3 are bad) = \\(\\frac{1}{4} \\times \\frac{1}{4} \\times \\frac{1}{4}\\)\n= \\(\\frac{1}{64}\\)\n(c) P(the 3 are good) = \\(\\frac{3}{4} \\times \\frac{3}{4} \\times \\frac{3}{4}\\)\n= \\(\\frac{27}{64}\\)\n\n', '', 'Explanation\n(a) \\(P(Good) = \\frac{18}{24} = \\frac{3}{4}\\)\n\\(P(Bad) = \\frac{6}{24} = \\frac{1}{4}\\)\nP(2 are good and 1 bad) = \\(\\frac{3}{4} \\times \\frac{3}{4} \\times \\frac{1}{4}\\)\n= \\(\\frac{9}{64}\\)\n(b) P(the 3 are bad) = \\(\\frac{1}{4} \\times \\frac{1}{4} \\times \\frac{1}{4}\\)\n= \\(\\frac{1}{64}\\)\n(c) P(the 3 are good) = \\(\\frac{3}{4} \\times \\frac{3}{4} \\times \\frac{3}{4}\\)\n= \\(\\frac{27}{64}\\)\n\n', '(a) \\(P(Good) = \\frac{18}{24} = \\frac{3}{4}\\)\n\\(P(Bad) = \\frac{6}{24} = \\frac{1}{4}\\)\nP(2 are good and 1 bad) = \\(\\frac{3}{4} \\times \\frac{3}{4} \\times \\frac{1}{4}\\)\n= \\(\\frac{9}{64}\\)~(b) P(the 3 are bad) = \\(\\frac{1}{4} \\times \\frac{1}{4} \\times \\frac{1}{4}\\)\n= \\(\\frac{1}{64}\\)\n(c) P(the 3 are good) = \\(\\frac{3}{4} \\times \\frac{3}{4} \\times \\frac{3}{4}\\)\n= \\(\\frac{27}{64}\\)\n~(c) P(the 3 are good) = \\(\\frac{3}{4} \\times \\frac{3}{4} \\times \\frac{3}{4}\\)\n= \\(\\frac{27}{64}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(465, '2014', '\n\nQuestion\n\n(a) Simplify : \\(\\frac{3\\frac{1}{12} + \\frac{7}{8}}{2\\frac{1}{4} - \\frac{1}{6}}\\)\n(b) If \\(p = \\frac{m}{2} - \\frac{n^{2}}{5m}\\) ;\n(i) make n the subject of the relation ; (ii) find, correct to three significant figures, the value of n when p = 14 and m = -8.\n\n', '', '(a) Simplify : \\(\\frac{3\\frac{1}{12} + \\frac{7}{8}}{2\\frac{1}{4} - \\frac{1}{6}}\\)\n(b) If \\(p = \\frac{m}{2} - \\frac{n^{2}}{5m}\\) ;\n(i) make n the subject of the relation ; (ii) find, correct to three significant figures, the value of n when p = 14 and m = -8.', '(a) Simplify : \\(\\frac{3\\frac{1}{12} + \\frac{7}{8}}{2\\frac{1}{4} - \\frac{1}{6}}\\)~(b) If \\(p = \\frac{m}{2} - \\frac{n^{2}}{5m}\\) ;\n(i) make n the subject of the relation ; (ii) find, correct to three significant figures, the value of n when p = 14 and m = -8.', NULL, '\n\nExplanation\n(a) \\(\\frac{3\\frac{1}{12} + \\frac{7}{8}}{2\\frac{1}{4} - \\frac{1}{6}}\\)\n\\(3\\frac{1}{12} + \\frac{7}{8} = \\frac{37}{12} + \\frac{7}{8}\\)\n= \\(\\frac{74 + 21}{24}\\)\n= \\(\\frac{95}{24}\\)\n\\(2\\frac{1}{4} - \\frac{1}{6} = \\frac{9}{4} - \\frac{1}{6}\\)\n= \\(\\frac{27 - 2}{12}\\)\n= \\(\\frac{25}{12}\\)\n\\(\\therefore \\frac{3\\frac{1}{12} + \\frac{7}{8}}{2\\frac{1}{4} - \\frac{1}{6}} = \\frac{95}{24} \\div \\frac{25}{12}\\)\n\\(\\frac{95}{24} \\times \\frac{12}{25} = \\frac{19}{10}\\)\n= \\(1.9\\)\n(b)(i) \\(p = \\frac{m}{2} - \\frac{n^{2}}{5m}\\)\n\\(\\frac{n^{2}}{5m} = \\frac{m}{2} - p\\)\n\\(n^{2} = 5m(\\frac{m}{2} - p)\\)\n\\(n = \\pm \\sqrt{5m(\\frac{m}{2} - p)}\\)\n(ii) When p = 14 and m = -8,\n\\(n = \\sqrt{5(-8)(\\frac{-8}{2} - 14)}\\)\n\\(n = \\sqrt{-40(- 4 - 14)}\\)\n\\(n = \\sqrt{720}\\)\n\\(n = 12\\sqrt{5}\\)\n= \\(\\pm 26.83\\)\n\n', '', 'Explanation\n(a) \\(\\frac{3\\frac{1}{12} + \\frac{7}{8}}{2\\frac{1}{4} - \\frac{1}{6}}\\)\n\\(3\\frac{1}{12} + \\frac{7}{8} = \\frac{37}{12} + \\frac{7}{8}\\)\n= \\(\\frac{74 + 21}{24}\\)\n= \\(\\frac{95}{24}\\)\n\\(2\\frac{1}{4} - \\frac{1}{6} = \\frac{9}{4} - \\frac{1}{6}\\)\n= \\(\\frac{27 - 2}{12}\\)\n= \\(\\frac{25}{12}\\)\n\\(\\therefore \\frac{3\\frac{1}{12} + \\frac{7}{8}}{2\\frac{1}{4} - \\frac{1}{6}} = \\frac{95}{24} \\div \\frac{25}{12}\\)\n\\(\\frac{95}{24} \\times \\frac{12}{25} = \\frac{19}{10}\\)\n= \\(1.9\\)\n(b)(i) \\(p = \\frac{m}{2} - \\frac{n^{2}}{5m}\\)\n\\(\\frac{n^{2}}{5m} = \\frac{m}{2} - p\\)\n\\(n^{2} = 5m(\\frac{m}{2} - p)\\)\n\\(n = \\pm \\sqrt{5m(\\frac{m}{2} - p)}\\)\n(ii) When p = 14 and m = -8,\n\\(n = \\sqrt{5(-8)(\\frac{-8}{2} - 14)}\\)\n\\(n = \\sqrt{-40(- 4 - 14)}\\)\n\\(n = \\sqrt{720}\\)\n\\(n = 12\\sqrt{5}\\)\n= \\(\\pm 26.83\\)\n\n', '(a) \\(\\frac{3\\frac{1}{12} + \\frac{7}{8}}{2\\frac{1}{4} - \\frac{1}{6}}\\)\n\\(3\\frac{1}{12} + \\frac{7}{8} = \\frac{37}{12} + \\frac{7}{8}\\)\n= \\(\\frac{74 + 21}{24}\\)\n= \\(\\frac{95}{24}\\)\n\\(2\\frac{1}{4} - \\frac{1}{6} = \\frac{9}{4} - \\frac{1}{6}\\)\n= \\(\\frac{27 - 2}{12}\\)\n= \\(\\frac{25}{12}\\)\n\\(\\therefore \\frac{3\\frac{1}{12} + \\frac{7}{8}}{2\\frac{1}{4} - \\frac{1}{6}} = \\frac{95}{24} \\div \\frac{25}{12}\\)\n\\(\\frac{95}{24} \\times \\frac{12}{25} = \\frac{19}{10}\\)\n= \\(1.9\\)~(b)(i) \\(p = \\frac{m}{2} - \\frac{n^{2}}{5m}\\)\n\\(\\frac{n^{2}}{5m} = \\frac{m}{2} - p\\)\n\\(n^{2} = 5m(\\frac{m}{2} - p)\\)\n\\(n = \\pm \\sqrt{5m(\\frac{m}{2} - p)}\\)\n(ii) When p = 14 and m = -8,\n\\(n = \\sqrt{5(-8)(\\frac{-8}{2} - 14)}\\)\n\\(n = \\sqrt{-40(- 4 - 14)}\\)\n\\(n = \\sqrt{720}\\)\n\\(n = 12\\sqrt{5}\\)\n= \\(\\pm 26.83\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(466, '2014', '\n\nQuestion\n\n(a) With the aid of four- figure logarithm tables, evaluate \\((0.004592)^{\\frac{1}{3}}\\).\n(b) If \\(\\log_{10} y + 3\\log_{10} x = 2\\), express y in terms of x.\n(c) Solve the equations : \\(3x - 2y = 21\\)\n                    \\(4x + 5y = 5\\).\n\n', '', '(a) With the aid of four- figure logarithm tables, evaluate \\((0.004592)^{\\frac{1}{3}}\\).\n(b) If \\(\\log_{10} y + 3\\log_{10} x = 2\\), express y in terms of x.\n(c) Solve the equations : \\(3x - 2y = 21\\)\n                    \\(4x + 5y = 5\\).', '(a) With the aid of four- figure logarithm tables, evaluate \\((0.004592)^{\\frac{1}{3}}\\).~(b) If \\(\\log_{10} y + 3\\log_{10} x = 2\\), express y in terms of x.\n(c) Solve the equations : \\(3x - 2y = 21\\)\n                    \\(4x + 5y = 5\\).~(c) Solve the equations : \\(3x - 2y = 21\\)\n                    \\(4x + 5y = 5\\).', NULL, '\n\nExplanation\n(a)\n\n\n\nNo\nLog\n\n\n0.004592\n\\(\\bar{3}.6620\\)\n\n\n\\((0.004592)^{\\frac{1}{3}}\\)\n\\(\\frac{\\bar{3}.6620}{3} = \\bar{1}.2207\\)\n\n\n\nAntilog of \\(\\bar{1}.2207 = 0.1663\\)\n(b) \\(\\log_{10} y + 3 \\log_{10} x = 2\\)\n\\(\\log_{10} y + \\log_{10} x^{3} = 2\\)\n\\(\\log_{10} (yx^{3}) = 2\\)\n\\(yx^{3} = 10^{2}\\)\n\\(yx^{3} = 100\\)\n\\(y = \\frac{100}{x^{3}}\\)\n(c) \\(3x - 2y = 21 .... (1)\\)\n\\(4x + 5y = 5 ..... (2)\\)\nMultiply (1) by 4and (2) by 3,\n\\(12x - 8y = 84 .... (1a)\\)\n\\(12x + 15y = 15 ... (2a)\\)\n(1a) - (2a),\n\\(- 8y - 15y = 84 - 15\\)\n\\(- 23y = 69 \\implies y = -3\\)\n\\(3x - 2y = 21 \\implies 3x - 2(-3) = 21\\)\n\\(3x + 6 = 21 \\implies 3x = 15\\)\n\\(x = 5\\)\n\n', '', 'Explanation\n(a)\n\n\n\nNo\nLog\n\n\n0.004592\n\\(\\bar{3}.6620\\)\n\n\n\\((0.004592)^{\\frac{1}{3}}\\)\n\\(\\frac{\\bar{3}.6620}{3} = \\bar{1}.2207\\)\n\n\n\nAntilog of \\(\\bar{1}.2207 = 0.1663\\)\n(b) \\(\\log_{10} y + 3 \\log_{10} x = 2\\)\n\\(\\log_{10} y + \\log_{10} x^{3} = 2\\)\n\\(\\log_{10} (yx^{3}) = 2\\)\n\\(yx^{3} = 10^{2}\\)\n\\(yx^{3} = 100\\)\n\\(y = \\frac{100}{x^{3}}\\)\n(c) \\(3x - 2y = 21 .... (1)\\)\n\\(4x + 5y = 5 ..... (2)\\)\nMultiply (1) by 4and (2) by 3,\n\\(12x - 8y = 84 .... (1a)\\)\n\\(12x + 15y = 15 ... (2a)\\)\n(1a) - (2a),\n\\(- 8y - 15y = 84 - 15\\)\n\\(- 23y = 69 \\implies y = -3\\)\n\\(3x - 2y = 21 \\implies 3x - 2(-3) = 21\\)\n\\(3x + 6 = 21 \\implies 3x = 15\\)\n\\(x = 5\\)\n\n', '(a)\n\n\n\nNo\nLog\n\n\n0.004592\n\\(\\bar{3}.6620\\)\n\n\n\\((0.004592)^{\\frac{1}{3}}\\)\n\\(\\frac{\\bar{3}.6620}{3} = \\bar{1}.2207\\)\n\n\n\nAntilog of \\(\\bar{1}.2207 = 0.1663\\)~(b) \\(\\log_{10} y + 3 \\log_{10} x = 2\\)\n\\(\\log_{10} y + \\log_{10} x^{3} = 2\\)\n\\(\\log_{10} (yx^{3}) = 2\\)\n\\(yx^{3} = 10^{2}\\)\n\\(yx^{3} = 100\\)\n\\(y = \\frac{100}{x^{3}}\\)\n(c) \\(3x - 2y = 21 .... (1)\\)\n\\(4x + 5y = 5 ..... (2)\\)\nMultiply (1) by 4and (2) by 3,\n\\(12x - 8y = 84 .... (1a)\\)\n\\(12x + 15y = 15 ... (2a)\\)\n(1a) - (2a),\n\\(- 8y - 15y = 84 - 15\\)\n\\(- 23y = 69 \\implies y = -3\\)\n\\(3x - 2y = 21 \\implies 3x - 2(-3) = 21\\)\n\\(3x + 6 = 21 \\implies 3x = 15\\)\n\\(x = 5\\)\n~(c) \\(3x - 2y = 21 .... (1)\\)\n\\(4x + 5y = 5 ..... (2)\\)\nMultiply (1) by 4and (2) by 3,\n\\(12x - 8y = 84 .... (1a)\\)\n\\(12x + 15y = 15 ... (2a)\\)\n(1a) - (2a),\n\\(- 8y - 15y = 84 - 15\\)\n\\(- 23y = 69 \\implies y = -3\\)\n\\(3x - 2y = 21 \\implies 3x - 2(-3) = 21\\)\n\\(3x + 6 = 21 \\implies 3x = 15\\)\n\\(x = 5\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(467, '2014', '\n\nQuestion\n\n(a) A cylinder with radius 3.5 cm has its two ends closed, if the total surface area is \\(209 cm^{2}\\), calculate the height of the cylinder. [Take \\(\\pi = \\frac{22}{7}\\)].\n(b)In the diagram, O is the centre of the circle and ABC is a tangent at B. If \\(\\stackrel\\frown{BDF} = 66°\\) and \\(\\stackrel\\frown{DBC} = 57°\\), calculate, (i) \\(\\stackrel\\frown{EBF}\\) and (ii) \\(\\stackrel\\frown{BGF}\\).\n\n', '', '(a) A cylinder with radius 3.5 cm has its two ends closed, if the total surface area is \\(209 cm^{2}\\), calculate the height of the cylinder. [Take \\(\\pi = \\frac{22}{7}\\)].\n(b)In the diagram, O is the centre of the circle and ABC is a tangent at B. If \\(\\stackrel\\frown{BDF} = 66°\\) and \\(\\stackrel\\frown{DBC} = 57°\\), calculate, (i) \\(\\stackrel\\frown{EBF}\\) and (ii) \\(\\stackrel\\frown{BGF}\\).', '(a) A cylinder with radius 3.5 cm has its two ends closed, if the total surface area is \\(209 cm^{2}\\), calculate the height of the cylinder. [Take \\(\\pi = \\frac{22}{7}\\)].~(b)In the diagram, O is the centre of the circle and ABC is a tangent at B. If \\(\\stackrel\\frown{BDF} = 66°\\) and \\(\\stackrel\\frown{DBC} = 57°\\), calculate, (i) \\(\\stackrel\\frown{EBF}\\) and (ii) \\(\\stackrel\\frown{BGF}\\).', NULL, '\n\nExplanation\n(a) TSA of a closed cylinder = \\(2\\pi r^{2} + 2\\pi rh\\)\n= \\(2\\pi r (r + h)\\)\n\\(2 \\times \\frac{22}{7} \\times 3.5 (3.5 + h) = 209\\)\n\\(22(3.5 + h) = 209 \\implies 77 + 22h = 209\\)\n\\(22h = 209 - 77 = 132\\)\n\\(h = \\frac{132}{22} = 6 cm\\)\n(b) \\(< EDB = 90°\\) (angle subtended at the circumference by a semi-circle)\n\\(\\therefore < EDF = 90° - 66° = 24°\\)\n\\(< EDF = < EBF = 24°\\) (angles in the same segment)\n(ii)\n\\(< DEB = < DBC = 57°\\) (Alternate segment theorem)\n\\(\\therefore < BGF = 180 - (57 + 24) = 180 - 81\\)\n\\(< BGF = 99°\\)\n\n', '', 'Explanation\n(a) TSA of a closed cylinder = \\(2\\pi r^{2} + 2\\pi rh\\)\n= \\(2\\pi r (r + h)\\)\n\\(2 \\times \\frac{22}{7} \\times 3.5 (3.5 + h) = 209\\)\n\\(22(3.5 + h) = 209 \\implies 77 + 22h = 209\\)\n\\(22h = 209 - 77 = 132\\)\n\\(h = \\frac{132}{22} = 6 cm\\)\n(b) \\(< EDB = 90°\\) (angle subtended at the circumference by a semi-circle)\n\\(\\therefore < EDF = 90° - 66° = 24°\\)\n\\(< EDF = < EBF = 24°\\) (angles in the same segment)\n(ii)\n\\(< DEB = < DBC = 57°\\) (Alternate segment theorem)\n\\(\\therefore < BGF = 180 - (57 + 24) = 180 - 81\\)\n\\(< BGF = 99°\\)\n\n', '(a) TSA of a closed cylinder = \\(2\\pi r^{2} + 2\\pi rh\\)\n= \\(2\\pi r (r + h)\\)\n\\(2 \\times \\frac{22}{7} \\times 3.5 (3.5 + h) = 209\\)\n\\(22(3.5 + h) = 209 \\implies 77 + 22h = 209\\)\n\\(22h = 209 - 77 = 132\\)\n\\(h = \\frac{132}{22} = 6 cm\\)~(b) \\(< EDB = 90°\\) (angle subtended at the circumference by a semi-circle)\n\\(\\therefore < EDF = 90° - 66° = 24°\\)\n\\(< EDF = < EBF = 24°\\) (angles in the same segment)\n(ii)\n\\(< DEB = < DBC = 57°\\) (Alternate segment theorem)\n\\(\\therefore < BGF = 180 - (57 + 24) = 180 - 81\\)\n\\(< BGF = 99°\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(468, '2014', '\n\nQuestion\n\n(a) Using a ruler and a pair of compasses only, construc :\n(i) a triangle PQR such that /PQ/ = 10 cm, /QR/ = 7 cm and < PQR = 90° ; (ii) the locus \\(l_{1}\\) of points equidistant from Q and R ; (iii) the locus \\(l_{2}\\) of points equidistant from P and Q.\n(b) Locate the point O equidistant from P, Q and R.\n(c) With O as centre, draw the circumcircle of the triangle PQR.\n(d) Measure the radius of the circumcircle.\n\n', '', '(a) Using a ruler and a pair of compasses only, construc :\n(i) a triangle PQR such that /PQ/ = 10 cm, /QR/ = 7 cm and < PQR = 90° ; (ii) the locus \\(l_{1}\\) of points equidistant from Q and R ; (iii) the locus \\(l_{2}\\) of points equidistant from P and Q.\n(b) Locate the point O equidistant from P, Q and R.\n(c) With O as centre, draw the circumcircle of the triangle PQR.\n(d) Measure the radius of the circumcircle.', '(a) Using a ruler and a pair of compasses only, construc :\n(i) a triangle PQR such that /PQ/ = 10 cm, /QR/ = 7 cm and < PQR = 90° ; (ii) the locus \\(l_{1}\\) of points equidistant from Q and R ; (iii) the locus \\(l_{2}\\) of points equidistant from P and Q.~(b) Locate the point O equidistant from P, Q and R.\n(c) With O as centre, draw the circumcircle of the triangle PQR.\n(d) Measure the radius of the circumcircle.~(c) With O as centre, draw the circumcircle of the triangle PQR.\n(d) Measure the radius of the circumcircle.~(d) Measure the radius of the circumcircle.', NULL, '\n\nExplanation\n', '', '', '', 'WAEC-PK-1', 'WAEC-MATH', 1),
(469, '2014', '\n\nQuestion\n\n(a) Simplify : \\(\\frac{x^{2} - y^{2}}{3x + 3y}\\)\n(b)\nIn the diagram, PQRS is a rectangle. /PK/ = 15 cm, /SK/ = /KR/ and <PKS = 30°. Calculate, correct to three significant figures : (i) /PS/ ; (ii) /SK/ and (iii) the area of the shaded portion.\n\n', '', '(a) Simplify : \\(\\frac{x^{2} - y^{2}}{3x + 3y}\\)\n(b)\nIn the diagram, PQRS is a rectangle. /PK/ = 15 cm, /SK/ = /KR/ and <PKS = 30°. Calculate, correct to three significant figures : (i) /PS/ ; (ii) /SK/ and (iii) the area of the shaded portion.', '(a) Simplify : \\(\\frac{x^{2} - y^{2}}{3x + 3y}\\)~(b)\nIn the diagram, PQRS is a rectangle. /PK/ = 15 cm, /SK/ = /KR/ and <PKS = 30°. Calculate, correct to three significant figures : (i) /PS/ ; (ii) /SK/ and (iii) the area of the shaded portion.', NULL, '\n\nExplanation\n(a) \\(\\frac{x^{2} - y^{2}}{3x + 3y}\\)\n\\(\\frac{(x + y)(x - y)}{3(x + y)}\\) (Using difference of two squares)\n= \\(\\frac{x - y}{3}\\)\n(b)(i)\n\\(\\sin 37 = \\frac{/PS/}{15}\\)\n\\(/PS/ = 15 \\times 0.6018\\)\n= \\(9.03 cm\\)\n(ii) \\(\\cos 37 = \\frac{/SK/}{15}\\)\n\\(/SK/ = 15 \\times 0.7986\\)\n= \\(11.98 cm \\)\n\\(\\approxeq 12.0 cm\\)\n(iii) Area of the shaded portion = Area of rectangle PQRS - Area of triangle PSK.\n/SR/ = 2(/SK/) = 2(11.98)\n= 23.96 cm\nArea of rectangle PQRS = \\(23.96 \\times 9.03 \\)\n= \\(216.3588 cm^{2}\\)\nArea of triangle PKS = \\(\\frac{1}{2} \\times 11.98 \\times 9.03\\)\n= \\(54.088 cm^{2}\\)\nArea of shaded portion : \\((216.3588 - 54.088)cm^{2}\\)\n\\(162.2708 cm^{2}\\)\n\\(\\approxeq 162 cm^{2}\\)\n\n', '', 'Explanation\n(a) \\(\\frac{x^{2} - y^{2}}{3x + 3y}\\)\n\\(\\frac{(x + y)(x - y)}{3(x + y)}\\) (Using difference of two squares)\n= \\(\\frac{x - y}{3}\\)\n(b)(i)\n\\(\\sin 37 = \\frac{/PS/}{15}\\)\n\\(/PS/ = 15 \\times 0.6018\\)\n= \\(9.03 cm\\)\n(ii) \\(\\cos 37 = \\frac{/SK/}{15}\\)\n\\(/SK/ = 15 \\times 0.7986\\)\n= \\(11.98 cm \\)\n\\(\\approxeq 12.0 cm\\)\n(iii) Area of the shaded portion = Area of rectangle PQRS - Area of triangle PSK.\n/SR/ = 2(/SK/) = 2(11.98)\n= 23.96 cm\nArea of rectangle PQRS = \\(23.96 \\times 9.03 \\)\n= \\(216.3588 cm^{2}\\)\nArea of triangle PKS = \\(\\frac{1}{2} \\times 11.98 \\times 9.03\\)\n= \\(54.088 cm^{2}\\)\nArea of shaded portion : \\((216.3588 - 54.088)cm^{2}\\)\n\\(162.2708 cm^{2}\\)\n\\(\\approxeq 162 cm^{2}\\)\n\n', '(a) \\(\\frac{x^{2} - y^{2}}{3x + 3y}\\)\n\\(\\frac{(x + y)(x - y)}{3(x + y)}\\) (Using difference of two squares)\n= \\(\\frac{x - y}{3}\\)~(b)(i)\n\\(\\sin 37 = \\frac{/PS/}{15}\\)\n\\(/PS/ = 15 \\times 0.6018\\)\n= \\(9.03 cm\\)\n(ii) \\(\\cos 37 = \\frac{/SK/}{15}\\)\n\\(/SK/ = 15 \\times 0.7986\\)\n= \\(11.98 cm \\)\n\\(\\approxeq 12.0 cm\\)\n(iii) Area of the shaded portion = Area of rectangle PQRS - Area of triangle PSK.\n/SR/ = 2(/SK/) = 2(11.98)\n= 23.96 cm\nArea of rectangle PQRS = \\(23.96 \\times 9.03 \\)\n= \\(216.3588 cm^{2}\\)\nArea of triangle PKS = \\(\\frac{1}{2} \\times 11.98 \\times 9.03\\)\n= \\(54.088 cm^{2}\\)\nArea of shaded portion : \\((216.3588 - 54.088)cm^{2}\\)\n\\(162.2708 cm^{2}\\)\n\\(\\approxeq 162 cm^{2}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(470, '2014', '\n\nQuestion\n\n(a)\nIn the diagram, AOB is a straight line. < AOC = 3(x + y)°, < COB = 45°, < AOD = (5x + y)° and < DOB = y°. Find the values of x and y.\n(b) From two points on opposite sides of a pole 33m high, the angles of elevation of the top of the pole are 53° and 67°. If the two points and the base are on te same horizontal level, calculate, correct to three significant figures, the distance between the two points.\n\n', '', '(a)\nIn the diagram, AOB is a straight line. < AOC = 3(x + y)°, < COB = 45°, < AOD = (5x + y)° and < DOB = y°. Find the values of x and y.\n(b) From two points on opposite sides of a pole 33m high, the angles of elevation of the top of the pole are 53° and 67°. If the two points and the base are on te same horizontal level, calculate, correct to three significant figures, the distance between the two points.', '(a)\nIn the diagram, AOB is a straight line. < AOC = 3(x + y)°, < COB = 45°, < AOD = (5x + y)° and < DOB = y°. Find the values of x and y.~(b) From two points on opposite sides of a pole 33m high, the angles of elevation of the top of the pole are 53° and 67°. If the two points and the base are on te same horizontal level, calculate, correct to three significant figures, the distance between the two points.', NULL, '\n\nExplanation\n(a) \\(3(x + y) + 45 = 180°\\) ( sum of angles on a straight line)\n\\(3x + 3y + 45 = 180 \\implies 3x + 3y = 135\\)\n\\(3x = 135 - 3y \\implies x = 45 - y\\)\nAlso,\n\\(5x + y + y = 180\\)\n\\(5x + 2y = 180..... (*)\\)\nPut x = 45 - y into (*) above,\n\\(5(45 - y) + 2y = 180 \\implies 225 - 5y + 2y = 180\\)\n\\(225 - 3y = 180 \\implies 3y = 45\\)\n\\(y = \\frac{45}{3} = 15°\\)\n\\(x = 45 - y\\)\n\\(x = 45 - 15 = 30°\\)\n(b)\nLet the distance BD = x and CD = y.\nIn \\(\\Delta ABD\\),\n\\(\\tan 53 = \\frac{33}{x}\\)\n\\(x = \\frac{33}{\\tan 53}\\)\n= \\( 24.868 cm\\)\nIn \\(\\Delta ACD\\),\n\\(\\tan 67 = \\frac{33}{y}\\)\n\\(y = \\frac{33}{\\tan 67}\\)\n\\(y = 14.01 cm\\)\n\\(\\therefore BC = x + y \\)\n\\(24.868 + 14.01 = 38.878 cm\\)\n\\(\\approxeq 38.9 cm\\)\n\n', '', 'Explanation\n(a) \\(3(x + y) + 45 = 180°\\) ( sum of angles on a straight line)\n\\(3x + 3y + 45 = 180 \\implies 3x + 3y = 135\\)\n\\(3x = 135 - 3y \\implies x = 45 - y\\)\nAlso,\n\\(5x + y + y = 180\\)\n\\(5x + 2y = 180..... (*)\\)\nPut x = 45 - y into (*) above,\n\\(5(45 - y) + 2y = 180 \\implies 225 - 5y + 2y = 180\\)\n\\(225 - 3y = 180 \\implies 3y = 45\\)\n\\(y = \\frac{45}{3} = 15°\\)\n\\(x = 45 - y\\)\n\\(x = 45 - 15 = 30°\\)\n(b)\nLet the distance BD = x and CD = y.\nIn \\(\\Delta ABD\\),\n\\(\\tan 53 = \\frac{33}{x}\\)\n\\(x = \\frac{33}{\\tan 53}\\)\n= \\( 24.868 cm\\)\nIn \\(\\Delta ACD\\),\n\\(\\tan 67 = \\frac{33}{y}\\)\n\\(y = \\frac{33}{\\tan 67}\\)\n\\(y = 14.01 cm\\)\n\\(\\therefore BC = x + y \\)\n\\(24.868 + 14.01 = 38.878 cm\\)\n\\(\\approxeq 38.9 cm\\)\n\n', '(a) \\(3(x + y) + 45 = 180°\\) ( sum of angles on a straight line)\n\\(3x + 3y + 45 = 180 \\implies 3x + 3y = 135\\)\n\\(3x = 135 - 3y \\implies x = 45 - y\\)\nAlso,\n\\(5x + y + y = 180\\)\n\\(5x + 2y = 180..... (*)\\)\nPut x = 45 - y into (*) above,\n\\(5(45 - y) + 2y = 180 \\implies 225 - 5y + 2y = 180\\)\n\\(225 - 3y = 180 \\implies 3y = 45\\)\n\\(y = \\frac{45}{3} = 15°\\)\n\\(x = 45 - y\\)\n\\(x = 45 - 15 = 30°\\)~(b)\nLet the distance BD = x and CD = y.\nIn \\(\\Delta ABD\\),\n\\(\\tan 53 = \\frac{33}{x}\\)\n\\(x = \\frac{33}{\\tan 53}\\)\n= \\( 24.868 cm\\)\nIn \\(\\Delta ACD\\),\n\\(\\tan 67 = \\frac{33}{y}\\)\n\\(y = \\frac{33}{\\tan 67}\\)\n\\(y = 14.01 cm\\)\n\\(\\therefore BC = x + y \\)\n\\(24.868 + 14.01 = 38.878 cm\\)\n\\(\\approxeq 38.9 cm\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(471, '2014', '\n\nQuestion\n\n(a) The 3rd and 8th terms of an arithmetic progression (A.P) are -9 and 26 respectively. Find the : (i) common difference ; (ii) first term.\n(b)\nIn the diagram \\(\\overline{PQ} || \\overline{YZ}\\), |XP| = 2cm, |PY| = 3 cm, |PQ| = 6 cm and the area of \\(\\Delta\\) XPQ = 24\\(cm^{2}\\).Calculate the area of the trapezium PQZY.\n\n', '', '(a) The 3rd and 8th terms of an arithmetic progression (A.P) are -9 and 26 respectively. Find the : (i) common difference ; (ii) first term.\n(b)\nIn the diagram \\(\\overline{PQ} || \\overline{YZ}\\), |XP| = 2cm, |PY| = 3 cm, |PQ| = 6 cm and the area of \\(\\Delta\\) XPQ = 24\\(cm^{2}\\).Calculate the area of the trapezium PQZY.', '(a) The 3rd and 8th terms of an arithmetic progression (A.P) are -9 and 26 respectively. Find the : (i) common difference ; (ii) first term.~(b)\nIn the diagram \\(\\overline{PQ} || \\overline{YZ}\\), |XP| = 2cm, |PY| = 3 cm, |PQ| = 6 cm and the area of \\(\\Delta\\) XPQ = 24\\(cm^{2}\\).Calculate the area of the trapezium PQZY.', NULL, '\n\nExplanation\n(a) \\(T_{n} = a + (n - 1)d\\) (terms of an A.P)\n3rd term = -9 ; 8th term = 26\n\\(\\implies T_{3} = a + 2d = -9 ... (1)\\)\n\\(\\implies T_{8} = a + 7d = 26 ..... (2)\\)\n(i) Solving (2) - (1), we have\n\\(7d - 2d = 26 - (-9)\\)\n\\(5d = 35 \\implies d = 7\\)\n(ii) Putting d = 7 in (1) above, we have\n\\(a + 2(7) = -9\\)\n\\(a + 14 = -9\\)\n\\(a = - 9 - 14 = -23\\)\n(b) \\(\\frac{XP}{XY} = \\frac{2}{5}\\) (similar triangles)\n\\(\\frac{24}{\\text{Area of \\Delta XYZ} = \\frac{2^{2}}{5^{2}}\\)\n\\(\\text{Area of } \\Delta XYZ = \\frac{24 \\times 25}{4} = 150 cm^{2}\\)\n\\(\\therefore \\text{Area of trapezium PQZY} = 150 - 24 = 126 cm^{2}\\)\n\n\n', '', 'Explanation\n(a) \\(T_{n} = a + (n - 1)d\\) (terms of an A.P)\n3rd term = -9 ; 8th term = 26\n\\(\\implies T_{3} = a + 2d = -9 ... (1)\\)\n\\(\\implies T_{8} = a + 7d = 26 ..... (2)\\)\n(i) Solving (2) - (1), we have\n\\(7d - 2d = 26 - (-9)\\)\n\\(5d = 35 \\implies d = 7\\)\n(ii) Putting d = 7 in (1) above, we have\n\\(a + 2(7) = -9\\)\n\\(a + 14 = -9\\)\n\\(a = - 9 - 14 = -23\\)\n(b) \\(\\frac{XP}{XY} = \\frac{2}{5}\\) (similar triangles)\n\\(\\frac{24}{\\text{Area of \\Delta XYZ} = \\frac{2^{2}}{5^{2}}\\)\n\\(\\text{Area of } \\Delta XYZ = \\frac{24 \\times 25}{4} = 150 cm^{2}\\)\n\\(\\therefore \\text{Area of trapezium PQZY} = 150 - 24 = 126 cm^{2}\\)\n\n\n', '(a) \\(T_{n} = a + (n - 1)d\\) (terms of an A.P)\n3rd term = -9 ; 8th term = 26\n\\(\\implies T_{3} = a + 2d = -9 ... (1)\\)\n\\(\\implies T_{8} = a + 7d = 26 ..... (2)\\)\n(i) Solving (2) - (1), we have\n\\(7d - 2d = 26 - (-9)\\)\n\\(5d = 35 \\implies d = 7\\)\n(ii) Putting d = 7 in (1) above, we have\n\\(a + 2(7) = -9\\)\n\\(a + 14 = -9\\)\n\\(a = - 9 - 14 = -23\\)~(b) \\(\\frac{XP}{XY} = \\frac{2}{5}\\) (similar triangles)\n\\(\\frac{24}{\\text{Area of \\Delta XYZ} = \\frac{2^{2}}{5^{2}}\\)\n\\(\\text{Area of } \\Delta XYZ = \\frac{24 \\times 25}{4} = 150 cm^{2}\\)\n\\(\\therefore \\text{Area of trapezium PQZY} = 150 - 24 = 126 cm^{2}\\)\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(472, '2014', '\n\nQuestion\n\nIn a college, the number of absentees recorded over a period of 30 days was as shown in the frequency distribution table\n\n\n\nNumber of absentees\n0-4\n5-9\n10-14\n15-19\n20-24\n\n\nNumber of Days\n1\n5\n10\n9\n5\n\n\n\nCalculate the : (a) Mean\n(b) Standard deviation , correct to two decimal places.\n\n\n\nIn a college, the number of absentees recorded over a period of 30 days was as shown in the frequency distribution table\n\n\n\nNumber of absentees\n0-4\n5-9\n10-14\n15-19\n20-24\n\n\nNumber of Days\n1\n5\n10\n9\n5\n\n\n', '', 'In a college, the number of absentees recorded over a period of 30 days was as shown in the frequency distribution table\n\n\n\nNumber of absentees\n0-4\n5-9\n10-14\n15-19\n20-24\n\n\nNumber of Days\n1\n5\n10\n9\n5\n\n\n\nCalculate the : (a) Mean\n(b) Standard deviation , correct to two decimal places.\n\n\n\nIn a college, the number of absentees recorded over a period of 30 days was as shown in the frequency distribution table\n\n\n\nNumber of absentees\n0-4\n5-9\n10-14\n15-19\n20-24\n\n\nNumber of Days\n1\n5\n10\n9\n5\n', 'In a college, the number of absentees recorded over a period of 30 days was as shown in the frequency distribution table\n\n\n\nNumber of absentees\n0-4\n5-9\n10-14\n15-19\n20-24\n\n\nNumber of Days\n1\n5\n10\n9\n5\n\n\n\nCalculate the : (a) Mean~(b) Standard deviation , correct to two decimal places.\n\n\n\nIn a college, the number of absentees recorded over a period of 30 days was as shown in the frequency distribution table\n\n\n\nNumber of absentees\n0-4\n5-9\n10-14\n15-19\n20-24\n\n\nNumber of Days\n1\n5\n10\n9\n5\n', NULL, '\n\nExplanation\n\\(x\\)\nmid-value\n\\(f\\)\n\\(fx\\)\n\\((x - \\bar{x})\\)\n\\((x - \\bar{x})^{2}\\)\n\\(f(x - \\bar{x})^{2}\\)\n\n\n0 - 4\n2\n1\n2\n-12\n144\n144\n\n\n5 - 9\n7\n5\n35\n-7\n49\n245\n\n\n10 - 14\n12\n10\n120\n-2\n4\n40\n\n\n15 - 19\n17\n9\n153\n3\n9\n81\n\n\n20 - 24\n22\n5\n110\n8\n64\n320\n\n\n\\(\\sum \\)\n\n30\n420\n\n\n830\n\n\n\n(a) Mean \\(\\bar{x}\\) = \\(\\frac{\\sum fx}{\\sum f}\\)\n= \\(\\frac{420}{30} = 14\\)\n(b) Standard deviation = \\(\\sqrt{\\frac{\\sum f(x - \\bar{x})^{2}}{\\sum f}\\)\n= \\(\\sqrt{\\frac{830}{30}}\\)\n= \\(\\sqrt{27.66}\\)\n= 5.26.\n\n\n', '', 'Explanation\n\\(x\\)\nmid-value\n\\(f\\)\n\\(fx\\)\n\\((x - \\bar{x})\\)\n\\((x - \\bar{x})^{2}\\)\n\\(f(x - \\bar{x})^{2}\\)\n\n\n0 - 4\n2\n1\n2\n-12\n144\n144\n\n\n5 - 9\n7\n5\n35\n-7\n49\n245\n\n\n10 - 14\n12\n10\n120\n-2\n4\n40\n\n\n15 - 19\n17\n9\n153\n3\n9\n81\n\n\n20 - 24\n22\n5\n110\n8\n64\n320\n\n\n\\(\\sum \\)\n\n30\n420\n\n\n830\n\n\n\n(a) Mean \\(\\bar{x}\\) = \\(\\frac{\\sum fx}{\\sum f}\\)\n= \\(\\frac{420}{30} = 14\\)\n(b) Standard deviation = \\(\\sqrt{\\frac{\\sum f(x - \\bar{x})^{2}}{\\sum f}\\)\n= \\(\\sqrt{\\frac{830}{30}}\\)\n= \\(\\sqrt{27.66}\\)\n= 5.26.\n\n\n', '(a) Mean \\(\\bar{x}\\) = \\(\\frac{\\sum fx}{\\sum f}\\)\n= \\(\\frac{420}{30} = 14\\)~(b) Standard deviation = \\(\\sqrt{\\frac{\\sum f(x - \\bar{x})^{2}}{\\sum f}\\)\n= \\(\\sqrt{\\frac{830}{30}}\\)\n= \\(\\sqrt{27.66}\\)\n= 5.26.\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(473, '2014', '\n\nQuestion\n\n(a) Simplify : \\(\\frac{4\\frac{2}{9} - 1\\frac{13}{15}}{2\\frac{1}{5} + \\frac{4}{7} \\times 2\\frac{1}{3}}\\)\n(b) By rationalising the denominator, simplify : \\(\\frac{7\\sqrt{5}}{\\sqrt{7}}\\), leaving your answer in surd form.\n\n', '', '(a) Simplify : \\(\\frac{4\\frac{2}{9} - 1\\frac{13}{15}}{2\\frac{1}{5} + \\frac{4}{7} \\times 2\\frac{1}{3}}\\)\n(b) By rationalising the denominator, simplify : \\(\\frac{7\\sqrt{5}}{\\sqrt{7}}\\), leaving your answer in surd form.', '(a) Simplify : \\(\\frac{4\\frac{2}{9} - 1\\frac{13}{15}}{2\\frac{1}{5} + \\frac{4}{7} \\times 2\\frac{1}{3}}\\)~(b) By rationalising the denominator, simplify : \\(\\frac{7\\sqrt{5}}{\\sqrt{7}}\\), leaving your answer in surd form.', NULL, '\n\nExplanation\n(a) \\(\\frac{4\\frac{2}{9} - 1\\frac{13}{15}}{2\\frac{1}{5} + \\frac{4}{7} \\times 2\\frac{1}{3}}\\)\n\\(4\\frac{2}{9} - 1\\frac{13}{15} = \\frac{38}{9} - \\frac{28}{15}\\)\n= \\(\\frac{190 - 84}{45}\\)\n= \\(\\frac{106}{45}\\)\n\\(2\\frac{1}{5} + \\frac{4}{7} \\times 2\\frac{1}{3} = \\frac{11}{5} + (\\frac{4}{7} \\times \\frac{7}{3})\\)\n= \\(\\frac{11}{5} + \\frac{4}{3}\\)\n= \\(\\frac{33 + 20}{15}\\)\n= \\(\\frac{53}{15}\\)\n\\(\\therefore \\frac{4\\frac{2}{9} - 1\\frac{13}{15}}{2\\frac{1}{5} + \\frac{4}{7} \\times 2\\frac{1}{3}} = \\frac{106}{45} \\div \\frac{53}{15}\\)\n= \\(\\frac{106}{45} \\times \\frac{15}{53}\\)\n= \\(\\frac{2}{3}\\)\n(b) \\(\\frac{7\\sqrt{5}}{\\sqrt{7}}\\)\n= \\(\\frac{7\\sqrt{5}}{\\sqrt{7}} \\times \\frac{\\sqrt{7}}{\\sqrt{7}}\\)\n= \\(\\frac{7\\sqrt{35}}{7}\\)\n= \\(\\sqrt{35}\\)\n\n', '', 'Explanation\n(a) \\(\\frac{4\\frac{2}{9} - 1\\frac{13}{15}}{2\\frac{1}{5} + \\frac{4}{7} \\times 2\\frac{1}{3}}\\)\n\\(4\\frac{2}{9} - 1\\frac{13}{15} = \\frac{38}{9} - \\frac{28}{15}\\)\n= \\(\\frac{190 - 84}{45}\\)\n= \\(\\frac{106}{45}\\)\n\\(2\\frac{1}{5} + \\frac{4}{7} \\times 2\\frac{1}{3} = \\frac{11}{5} + (\\frac{4}{7} \\times \\frac{7}{3})\\)\n= \\(\\frac{11}{5} + \\frac{4}{3}\\)\n= \\(\\frac{33 + 20}{15}\\)\n= \\(\\frac{53}{15}\\)\n\\(\\therefore \\frac{4\\frac{2}{9} - 1\\frac{13}{15}}{2\\frac{1}{5} + \\frac{4}{7} \\times 2\\frac{1}{3}} = \\frac{106}{45} \\div \\frac{53}{15}\\)\n= \\(\\frac{106}{45} \\times \\frac{15}{53}\\)\n= \\(\\frac{2}{3}\\)\n(b) \\(\\frac{7\\sqrt{5}}{\\sqrt{7}}\\)\n= \\(\\frac{7\\sqrt{5}}{\\sqrt{7}} \\times \\frac{\\sqrt{7}}{\\sqrt{7}}\\)\n= \\(\\frac{7\\sqrt{35}}{7}\\)\n= \\(\\sqrt{35}\\)\n\n', '(a) \\(\\frac{4\\frac{2}{9} - 1\\frac{13}{15}}{2\\frac{1}{5} + \\frac{4}{7} \\times 2\\frac{1}{3}}\\)\n\\(4\\frac{2}{9} - 1\\frac{13}{15} = \\frac{38}{9} - \\frac{28}{15}\\)\n= \\(\\frac{190 - 84}{45}\\)\n= \\(\\frac{106}{45}\\)\n\\(2\\frac{1}{5} + \\frac{4}{7} \\times 2\\frac{1}{3} = \\frac{11}{5} + (\\frac{4}{7} \\times \\frac{7}{3})\\)\n= \\(\\frac{11}{5} + \\frac{4}{3}\\)\n= \\(\\frac{33 + 20}{15}\\)\n= \\(\\frac{53}{15}\\)\n\\(\\therefore \\frac{4\\frac{2}{9} - 1\\frac{13}{15}}{2\\frac{1}{5} + \\frac{4}{7} \\times 2\\frac{1}{3}} = \\frac{106}{45} \\div \\frac{53}{15}\\)\n= \\(\\frac{106}{45} \\times \\frac{15}{53}\\)\n= \\(\\frac{2}{3}\\)~(b) \\(\\frac{7\\sqrt{5}}{\\sqrt{7}}\\)\n= \\(\\frac{7\\sqrt{5}}{\\sqrt{7}} \\times \\frac{\\sqrt{7}}{\\sqrt{7}}\\)\n= \\(\\frac{7\\sqrt{35}}{7}\\)\n= \\(\\sqrt{35}\\)\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
(474, '2014', '\n\nQuestion\n\n(a) A dealer sold a car to a man and made a profit of 15%. The man then sold it to a woman for N120,175.00 at a loss of 5%. How much did the dealer buy the car?\n(b) The diameter of the wheel of a car is 36cm. How many revolutions, correct to three significant figures, will it make to cover a distance of 1.05 km? [Take \\(\\pi = \\frac{22}{7}\\)].\n\n', '', '(a) A dealer sold a car to a man and made a profit of 15%. The man then sold it to a woman for N120,175.00 at a loss of 5%. How much did the dealer buy the car?\n(b) The diameter of the wheel of a car is 36cm. How many revolutions, correct to three significant figures, will it make to cover a distance of 1.05 km? [Take \\(\\pi = \\frac{22}{7}\\)].', '(a) A dealer sold a car to a man and made a profit of 15%. The man then sold it to a woman for N120,175.00 at a loss of 5%. How much did the dealer buy the car?~(b) The diameter of the wheel of a car is 36cm. How many revolutions, correct to three significant figures, will it make to cover a distance of 1.05 km? [Take \\(\\pi = \\frac{22}{7}\\)].', NULL, '\n\nExplanation\n(a) Let the cost of the car to the dealer be x.\nSales of car to a man with 15% profit : (115% of x = 1.15 x\\)\nSale of the car to the woman by the man at 5% loss : (95% of 1.15x = N120,175.00)\n\\(0.95 \\times 1.15x = N120,175.00\\)\n\\(1.0925x = N120,175.00\\)\n\\(x = \\frac{120,175.00}{1.0925}\\)\n\\(x = N110,000\\)\nThe original cost of the car to the dealer = N110,000.00.\n(b) Circumference of a circle = \\(2\\pi r\\)\nDistance covered by the wheel = 1.05 km\nRadius of the wheel = 18cm\nCircumference of the wheel = \\(2 \\times \\frac{22}{7} \\times 18\\)\n= \\(113.14 cm\\)\nNote : 1.05 km = 105,000 cm\nNumber of revolution = \\(\\frac{105,000}{113.14}\\)\n= \\(928.05 revs\\)\n\\(\\approxeq 928 revs\\)\n\n\n', '', 'Explanation\n(a) Let the cost of the car to the dealer be x.\nSales of car to a man with 15% profit : (115% of x = 1.15 x\\)\nSale of the car to the woman by the man at 5% loss : (95% of 1.15x = N120,175.00)\n\\(0.95 \\times 1.15x = N120,175.00\\)\n\\(1.0925x = N120,175.00\\)\n\\(x = \\frac{120,175.00}{1.0925}\\)\n\\(x = N110,000\\)\nThe original cost of the car to the dealer = N110,000.00.\n(b) Circumference of a circle = \\(2\\pi r\\)\nDistance covered by the wheel = 1.05 km\nRadius of the wheel = 18cm\nCircumference of the wheel = \\(2 \\times \\frac{22}{7} \\times 18\\)\n= \\(113.14 cm\\)\nNote : 1.05 km = 105,000 cm\nNumber of revolution = \\(\\frac{105,000}{113.14}\\)\n= \\(928.05 revs\\)\n\\(\\approxeq 928 revs\\)\n\n\n', '(a) Let the cost of the car to the dealer be x.\nSales of car to a man with 15% profit : (115% of x = 1.15 x\\)\nSale of the car to the woman by the man at 5% loss : (95% of 1.15x = N120,175.00)\n\\(0.95 \\times 1.15x = N120,175.00\\)\n\\(1.0925x = N120,175.00\\)\n\\(x = \\frac{120,175.00}{1.0925}\\)\n\\(x = N110,000\\)\nThe original cost of the car to the dealer = N110,000.00.~(b) Circumference of a circle = \\(2\\pi r\\)\nDistance covered by the wheel = 1.05 km\nRadius of the wheel = 18cm\nCircumference of the wheel = \\(2 \\times \\frac{22}{7} \\times 18\\)\n= \\(113.14 cm\\)\nNote : 1.05 km = 105,000 cm\nNumber of revolution = \\(\\frac{105,000}{113.14}\\)\n= \\(928.05 revs\\)\n\\(\\approxeq 928 revs\\)\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(475, '2014', '\n\nQuestion\n\n(a) Solve for x and y in the following equations :\n\\(2x - y = \\frac{9}{2}\\)\n\\(x + 4y = 0\\)\n(b)\nIn the diagram, TA is a tangent to the circle at A. If \\(\\stackrel\\frown{BCA} = 40°\\) and \\(\\stackrel\\frown{DAT} = 52°\\), find \\(\\stackrel\\frown{BAD}\\).\n\n', '', '(a) Solve for x and y in the following equations :\n\\(2x - y = \\frac{9}{2}\\)\n\\(x + 4y = 0\\)\n(b)\nIn the diagram, TA is a tangent to the circle at A. If \\(\\stackrel\\frown{BCA} = 40°\\) and \\(\\stackrel\\frown{DAT} = 52°\\), find \\(\\stackrel\\frown{BAD}\\).', '(a) Solve for x and y in the following equations :\n\\(2x - y = \\frac{9}{2}\\)\n\\(x + 4y = 0\\)~(b)\nIn the diagram, TA is a tangent to the circle at A. If \\(\\stackrel\\frown{BCA} = 40°\\) and \\(\\stackrel\\frown{DAT} = 52°\\), find \\(\\stackrel\\frown{BAD}\\).', NULL, '\n\nExplanation\n(a) \\(2x - y = \\frac{9}{2} .... (1)\\)\n\\(x + 4y = 0 ....... (2)\\)\nFrom (2), x = - 4y. Put into (1), we have\n\\(2(- 4y) - y = \\frac{9}{2}\\)\n\\(-8y - y = \\frac{9}{2} \\implies - 9y = \\frac{9}{2}\\)\n\\(y = \\frac{\\frac{9}{2}}{-9} = -\\frac{1}{2}\\)\n\\(x = - 4y = -4(-\\frac{1}{2})\\)\n\\(x = 2\\)\n\\((x, y) = (2, -\\frac{1}{2})\\)\n(b)\nIn \\(\\Delta ACD, \\stackrel\\frown{ACD} = 52° \\) \\(\\stackrel\\frown{ACD} = \\stackrel\\frown{DAT}\\)\n\\(\\therefore \\stackrel\\frown{BCD} = 40° + 52° = 92°\\)\nNote \\(\\stackrel\\frown{BCD}\\) and \\(\\stackrel\\frown{BAD}\\) are supplementary = 180°\n\\(\\therefore \\stackrel\\frown{BAD} = 180° - 92° = 88°\\)\n\n', '', 'Explanation\n(a) \\(2x - y = \\frac{9}{2} .... (1)\\)\n\\(x + 4y = 0 ....... (2)\\)\nFrom (2), x = - 4y. Put into (1), we have\n\\(2(- 4y) - y = \\frac{9}{2}\\)\n\\(-8y - y = \\frac{9}{2} \\implies - 9y = \\frac{9}{2}\\)\n\\(y = \\frac{\\frac{9}{2}}{-9} = -\\frac{1}{2}\\)\n\\(x = - 4y = -4(-\\frac{1}{2})\\)\n\\(x = 2\\)\n\\((x, y) = (2, -\\frac{1}{2})\\)\n(b)\nIn \\(\\Delta ACD, \\stackrel\\frown{ACD} = 52° \\) \\(\\stackrel\\frown{ACD} = \\stackrel\\frown{DAT}\\)\n\\(\\therefore \\stackrel\\frown{BCD} = 40° + 52° = 92°\\)\nNote \\(\\stackrel\\frown{BCD}\\) and \\(\\stackrel\\frown{BAD}\\) are supplementary = 180°\n\\(\\therefore \\stackrel\\frown{BAD} = 180° - 92° = 88°\\)\n\n', '(a) \\(2x - y = \\frac{9}{2} .... (1)\\)\n\\(x + 4y = 0 ....... (2)\\)\nFrom (2), x = - 4y. Put into (1), we have\n\\(2(- 4y) - y = \\frac{9}{2}\\)\n\\(-8y - y = \\frac{9}{2} \\implies - 9y = \\frac{9}{2}\\)\n\\(y = \\frac{\\frac{9}{2}}{-9} = -\\frac{1}{2}\\)\n\\(x = - 4y = -4(-\\frac{1}{2})\\)\n\\(x = 2\\)\n\\((x, y) = (2, -\\frac{1}{2})\\)~(b)\nIn \\(\\Delta ACD, \\stackrel\\frown{ACD} = 52° \\) \\(\\stackrel\\frown{ACD} = \\stackrel\\frown{DAT}\\)\n\\(\\therefore \\stackrel\\frown{BCD} = 40° + 52° = 92°\\)\nNote \\(\\stackrel\\frown{BCD}\\) and \\(\\stackrel\\frown{BAD}\\) are supplementary = 180°\n\\(\\therefore \\stackrel\\frown{BAD} = 180° - 92° = 88°\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(476, '2014', '\n\nQuestion\n\n(a) In a class of 45 students, 32 offered Physics(P), 28 offered Government(G) and 12 did not offer any of the two subjects. (i) Draw the Venn diagram to represent the information ; (ii) How many students offered both subjects? (iii) What is \\(n(P \\cup G)\\)?\n(b) If \\(p = \\frac{2u}{1 - u}\\) and \\(q = \\frac{1 + u}{1 - u}\\) ; express \\(\\frac{p + q}{p - q}\\) in terms of u.\n\n', '', '(a) In a class of 45 students, 32 offered Physics(P), 28 offered Government(G) and 12 did not offer any of the two subjects. (i) Draw the Venn diagram to represent the information ; (ii) How many students offered both subjects? (iii) What is \\(n(P \\cup G)\\)?\n(b) If \\(p = \\frac{2u}{1 - u}\\) and \\(q = \\frac{1 + u}{1 - u}\\) ; express \\(\\frac{p + q}{p - q}\\) in terms of u.', '(a) In a class of 45 students, 32 offered Physics(P), 28 offered Government(G) and 12 did not offer any of the two subjects. (i) Draw the Venn diagram to represent the information ; (ii) How many students offered both subjects? (iii) What is \\(n(P \\cup G)\\)?~(b) If \\(p = \\frac{2u}{1 - u}\\) and \\(q = \\frac{1 + u}{1 - u}\\) ; express \\(\\frac{p + q}{p - q}\\) in terms of u.', NULL, '\n\nExplanation\n(a) (i)\n(ii) \\(32 - x + x + 28 - x + 12 = 45\\)\n\\(72 - x = 45\\)\n\\(x = 72 - 45 = 27\\)\n27 students offered both Physics and Government.\n(iii) \\(n(P \\cup G) = 45 - 12 = 33\\)\n(b) \\(p = \\frac{2u}{1 - u} ; q = \\frac{1 + u}{1 - u}\\)\n\\(\\frac{p + q}{p - q}\\)\n\\(p + q = \\frac{2u}{1 - u} + \\frac{1 + u}{1 - u}\\)\n= \\(\\frac{2u + 1 + u}{1 - u}\\)\n= \\(3u + 1}{1 - u}\\)\n\\(p - q = \\frac{2u - 1 - u}{1 - u}\\)\n= \\(\\frac{u - 1}{1 - u}\\)\n= \\(\\frac{-(1 - u)}{1 - u}\\)\n= \\(-1\\)\n\\(\\frac{p + q}{p - q} = \\frac{\\frac{3u + 1}{1 - u}}{-1}\\)\n= \\(\\frac{-(3u + 1)}{1 - u}\\)\n= \\(\\frac{-(3u + 1)}{-(u - 1)}\\)\n= \\(\\frac{3u + 1}{u - 1}\\)\n\n', '', 'Explanation\n(a) (i)\n(ii) \\(32 - x + x + 28 - x + 12 = 45\\)\n\\(72 - x = 45\\)\n\\(x = 72 - 45 = 27\\)\n27 students offered both Physics and Government.\n(iii) \\(n(P \\cup G) = 45 - 12 = 33\\)\n(b) \\(p = \\frac{2u}{1 - u} ; q = \\frac{1 + u}{1 - u}\\)\n\\(\\frac{p + q}{p - q}\\)\n\\(p + q = \\frac{2u}{1 - u} + \\frac{1 + u}{1 - u}\\)\n= \\(\\frac{2u + 1 + u}{1 - u}\\)\n= \\(3u + 1}{1 - u}\\)\n\\(p - q = \\frac{2u - 1 - u}{1 - u}\\)\n= \\(\\frac{u - 1}{1 - u}\\)\n= \\(\\frac{-(1 - u)}{1 - u}\\)\n= \\(-1\\)\n\\(\\frac{p + q}{p - q} = \\frac{\\frac{3u + 1}{1 - u}}{-1}\\)\n= \\(\\frac{-(3u + 1)}{1 - u}\\)\n= \\(\\frac{-(3u + 1)}{-(u - 1)}\\)\n= \\(\\frac{3u + 1}{u - 1}\\)\n\n', '(a) (i)\n(ii) \\(32 - x + x + 28 - x + 12 = 45\\)\n\\(72 - x = 45\\)\n\\(x = 72 - 45 = 27\\)\n27 students offered both Physics and Government.\n(iii) \\(n(P \\cup G) = 45 - 12 = 33\\)~(b) \\(p = \\frac{2u}{1 - u} ; q = \\frac{1 + u}{1 - u}\\)\n\\(\\frac{p + q}{p - q}\\)\n\\(p + q = \\frac{2u}{1 - u} + \\frac{1 + u}{1 - u}\\)\n= \\(\\frac{2u + 1 + u}{1 - u}\\)\n= \\(3u + 1}{1 - u}\\)\n\\(p - q = \\frac{2u - 1 - u}{1 - u}\\)\n= \\(\\frac{u - 1}{1 - u}\\)\n= \\(\\frac{-(1 - u)}{1 - u}\\)\n= \\(-1\\)\n\\(\\frac{p + q}{p - q} = \\frac{\\frac{3u + 1}{1 - u}}{-1}\\)\n= \\(\\frac{-(3u + 1)}{1 - u}\\)\n= \\(\\frac{-(3u + 1)}{-(u - 1)}\\)\n= \\(\\frac{3u + 1}{u - 1}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(477, '2014', '\n\nQuestion\n\nTwo fair dice are tossed together once.\n(a) Draw a sample space for the possible outcomes ;\n(b) Find the probability of getting a total : (i) of 7 or 8 ; (ii) less than 4.\n\n', '', 'Two fair dice are tossed together once.\n(a) Draw a sample space for the possible outcomes ;\n(b) Find the probability of getting a total : (i) of 7 or 8 ; (ii) less than 4.', 'Two fair dice are tossed together once.\n(a) Draw a sample space for the possible outcomes ;~(b) Find the probability of getting a total : (i) of 7 or 8 ; (ii) less than 4.', NULL, '\n\nExplanation\n(a)\n\n\n\n\n1\n2\n3\n4\n5\n6\n\n\n1\n1, 1\n1, 2\n1, 3\n1, 4\n 1, 5\n1, 6\n\n\n2\n2, 1\n2, 2\n2, 3\n2, 4\n2, 5\n2, 6\n\n\n3\n3, 1\n3, 2\n3, 3\n3, 4\n3, 5\n3, 6\n\n\n4\n4,1\n4, 2\n4, 3\n4, 4\n4, 5\n4, 6\n\n\n5\n5, 1\n5, 2\n5, 3\n5, 4\n5, 5\n5, 6\n\n\n6\n6, 1\n6, 2\n6, 3\n6, 4\n6, 5\n6, 6\n\n\n\n(b) (i) P(getting a total of 7) = \\(\\frac{6}{36}\\)\nP(getting a total of 8) = \\(\\frac{5}{36}\\)\nP(getting a total of 7 or 8) = \\(\\frac{6}{36} + \\frac{5}{36}\\)\n= \\(\\frac{11}{36}\\)\n(ii) P(getting a total less than 4) = \\(\\frac{3}{36}\\)\n= \\(\\frac{1}{12}\\)\n\n\n', '', 'Explanation\n(a)\n\n\n\n\n1\n2\n3\n4\n5\n6\n\n\n1\n1, 1\n1, 2\n1, 3\n1, 4\n 1, 5\n1, 6\n\n\n2\n2, 1\n2, 2\n2, 3\n2, 4\n2, 5\n2, 6\n\n\n3\n3, 1\n3, 2\n3, 3\n3, 4\n3, 5\n3, 6\n\n\n4\n4,1\n4, 2\n4, 3\n4, 4\n4, 5\n4, 6\n\n\n5\n5, 1\n5, 2\n5, 3\n5, 4\n5, 5\n5, 6\n\n\n6\n6, 1\n6, 2\n6, 3\n6, 4\n6, 5\n6, 6\n\n\n\n(b) (i) P(getting a total of 7) = \\(\\frac{6}{36}\\)\nP(getting a total of 8) = \\(\\frac{5}{36}\\)\nP(getting a total of 7 or 8) = \\(\\frac{6}{36} + \\frac{5}{36}\\)\n= \\(\\frac{11}{36}\\)\n(ii) P(getting a total less than 4) = \\(\\frac{3}{36}\\)\n= \\(\\frac{1}{12}\\)\n\n\n', '(a)\n\n\n\n\n1\n2\n3\n4\n5\n6\n\n\n1\n1, 1\n1, 2\n1, 3\n1, 4\n 1, 5\n1, 6\n\n\n2\n2, 1\n2, 2\n2, 3\n2, 4\n2, 5\n2, 6\n\n\n3\n3, 1\n3, 2\n3, 3\n3, 4\n3, 5\n3, 6\n\n\n4\n4,1\n4, 2\n4, 3\n4, 4\n4, 5\n4, 6\n\n\n5\n5, 1\n5, 2\n5, 3\n5, 4\n5, 5\n5, 6\n\n\n6\n6, 1\n6, 2\n6, 3\n6, 4\n6, 5\n6, 6\n\n\n~(b) (i) P(getting a total of 7) = \\(\\frac{6}{36}\\)\nP(getting a total of 8) = \\(\\frac{5}{36}\\)\nP(getting a total of 7 or 8) = \\(\\frac{6}{36} + \\frac{5}{36}\\)\n= \\(\\frac{11}{36}\\)\n(ii) P(getting a total less than 4) = \\(\\frac{3}{36}\\)\n= \\(\\frac{1}{12}\\)\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(478, '2014', '\n\nQuestion\n\n(a) A sector of a circle of radius 8cm subtends an angle of 90° at the centre of the circle. If the sector is folded without overlap to form the curved surface of a cone, find the :\n(i) base radius ; (ii) height ; (iii) volume of the cone. [Take \\(\\pi = \\frac{22}{7}\\)].\n(b) A map is drawn to a scale of 1 : 20,000. Use it to calculate the : (i) distance, in kilometres, represented by 4.5 cm on the map ;\n\n', '', '(a) A sector of a circle of radius 8cm subtends an angle of 90° at the centre of the circle. If the sector is folded without overlap to form the curved surface of a cone, find the :\n(i) base radius ; (ii) height ; (iii) volume of the cone. [Take \\(\\pi = \\frac{22}{7}\\)].\n(b) A map is drawn to a scale of 1 : 20,000. Use it to calculate the : (i) distance, in kilometres, represented by 4.5 cm on the map ;', '(a) A sector of a circle of radius 8cm subtends an angle of 90° at the centre of the circle. If the sector is folded without overlap to form the curved surface of a cone, find the :\n(i) base radius ; (ii) height ; (iii) volume of the cone. [Take \\(\\pi = \\frac{22}{7}\\)].~(b) A map is drawn to a scale of 1 : 20,000. Use it to calculate the : (i) distance, in kilometres, represented by 4.5 cm on the map ;', NULL, '\n\nExplanation\n(a) (i) Length of sector = \\(\\frac{\\theta}{360} \\times 2\\pi R\\)\nLength of the base of the cone formed = \\(2\\pi r\\)\n\\(\\frac{90}{360} \\times 2 \\times \\pi\\times 8 = 2\\pi r\\)\n\\(4\\pi = 2\\pi r\\)\n\\(2r = 4 \\implies r = 2 cm\\)\nThe base radius of the cone = 2 cm.\n(ii) Height of the cone\n\n\\(h^{2} = 8^{2} - 2^{2}\\)\n\\(h^{2} = 64 - 4 = 60\\)\n\\(h = \\sqrt{60}\\)\n= \\(7.746 cm\\)\n(iii) volume of a cone = \\(\\frac{1}{3} \\pi r^{2} l\\)\n= \\(\\frac{1}{3} \\times \\frac{22}{7} \\times 2 \\times 2 \\times 7.746\\)\n= \\(\\frac{88 \\times 7.746}{21}\\)\n= \\(32.46 cm^{3}\\)\n(b)(i) \\(100,000 cm = 1 km\\)\n1 cm represents 20,000 metres\n\\(\\therefore 4.5 cm = (4.5 \\times 20,000) cm\\)\n= 90,000 cm\nConverting to km,\n= \\(\\frac{90,000}{100,000}\\)\n= 0.9km\n(ii) 1 metre represents 20,000 metres\n16km = 16000m\nDistance on the map in metres = \\(\\frac{16000}{20000}\\)\n= 0.8m\n(iii) Area of the forest on the map\n= \\(\\frac{85 \\times (10^{5})^{2} cm^{2}}{(20000)^{2}}\\)\n= \\(\\frac{85 \\times 10^{10}}{4\\times 10^{8}}\\)\n= \\(2125 cm^{2}\\)\n\n', '', 'Explanation\n(a) (i) Length of sector = \\(\\frac{\\theta}{360} \\times 2\\pi R\\)\nLength of the base of the cone formed = \\(2\\pi r\\)\n\\(\\frac{90}{360} \\times 2 \\times \\pi\\times 8 = 2\\pi r\\)\n\\(4\\pi = 2\\pi r\\)\n\\(2r = 4 \\implies r = 2 cm\\)\nThe base radius of the cone = 2 cm.\n(ii) Height of the cone\n\n\\(h^{2} = 8^{2} - 2^{2}\\)\n\\(h^{2} = 64 - 4 = 60\\)\n\\(h = \\sqrt{60}\\)\n= \\(7.746 cm\\)\n(iii) volume of a cone = \\(\\frac{1}{3} \\pi r^{2} l\\)\n= \\(\\frac{1}{3} \\times \\frac{22}{7} \\times 2 \\times 2 \\times 7.746\\)\n= \\(\\frac{88 \\times 7.746}{21}\\)\n= \\(32.46 cm^{3}\\)\n(b)(i) \\(100,000 cm = 1 km\\)\n1 cm represents 20,000 metres\n\\(\\therefore 4.5 cm = (4.5 \\times 20,000) cm\\)\n= 90,000 cm\nConverting to km,\n= \\(\\frac{90,000}{100,000}\\)\n= 0.9km\n(ii) 1 metre represents 20,000 metres\n16km = 16000m\nDistance on the map in metres = \\(\\frac{16000}{20000}\\)\n= 0.8m\n(iii) Area of the forest on the map\n= \\(\\frac{85 \\times (10^{5})^{2} cm^{2}}{(20000)^{2}}\\)\n= \\(\\frac{85 \\times 10^{10}}{4\\times 10^{8}}\\)\n= \\(2125 cm^{2}\\)\n\n', '(a) (i) Length of sector = \\(\\frac{\\theta}{360} \\times 2\\pi R\\)\nLength of the base of the cone formed = \\(2\\pi r\\)\n\\(\\frac{90}{360} \\times 2 \\times \\pi\\times 8 = 2\\pi r\\)\n\\(4\\pi = 2\\pi r\\)\n\\(2r = 4 \\implies r = 2 cm\\)\nThe base radius of the cone = 2 cm.\n(ii) Height of the cone\n\n\\(h^{2} = 8^{2} - 2^{2}\\)\n\\(h^{2} = 64 - 4 = 60\\)\n\\(h = \\sqrt{60}\\)\n= \\(7.746 cm\\)\n(iii) volume of a cone = \\(\\frac{1}{3} \\pi r^{2} l\\)\n= \\(\\frac{1}{3} \\times \\frac{22}{7} \\times 2 \\times 2 \\times 7.746\\)\n= \\(\\frac{88 \\times 7.746}{21}\\)\n= \\(32.46 cm^{3}\\)~(b)(i) \\(100,000 cm = 1 km\\)\n1 cm represents 20,000 metres\n\\(\\therefore 4.5 cm = (4.5 \\times 20,000) cm\\)\n= 90,000 cm\nConverting to km,\n= \\(\\frac{90,000}{100,000}\\)\n= 0.9km\n(ii) 1 metre represents 20,000 metres\n16km = 16000m\nDistance on the map in metres = \\(\\frac{16000}{20000}\\)\n= 0.8m\n(iii) Area of the forest on the map\n= \\(\\frac{85 \\times (10^{5})^{2} cm^{2}}{(20000)^{2}}\\)\n= \\(\\frac{85 \\times 10^{10}}{4\\times 10^{8}}\\)\n= \\(2125 cm^{2}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(479, '2014', '\n\nQuestion\n\n(a)(i) If \\(4x < 2 + 3x\\) and \\(x - 8 < 3x\\), what range of values of x satisfies both inequalities? ; (ii) Represent your result in (i) on the number line.\n(b) A shop is sending out a bill for an amount less than£100. The accountant interchanges the two digits and so overcharges the customer by 45. Given that the sum of the two digits is 9, find how much the bill should be.\n\n', '', '(a)(i) If \\(4x < 2 + 3x\\) and \\(x - 8 < 3x\\), what range of values of x satisfies both inequalities? ; (ii) Represent your result in (i) on the number line.\n(b) A shop is sending out a bill for an amount less than£100. The accountant interchanges the two digits and so overcharges the customer by 45. Given that the sum of the two digits is 9, find how much the bill should be.', '(a)(i) If \\(4x < 2 + 3x\\) and \\(x - 8 < 3x\\), what range of values of x satisfies both inequalities? ; (ii) Represent your result in (i) on the number line.~(b) A shop is sending out a bill for an amount less than£100. The accountant interchanges the two digits and so overcharges the customer by 45. Given that the sum of the two digits is 9, find how much the bill should be.', NULL, '\n\nExplanation\n(a)(i) \\(4x < 2 + 3x ..... (1)\\)\n   \\(4x - 3x < 2\\)\n\\(x < 2\\)\n\\(x - 8 < 3x ..... (2)\\)\n\\(-8 < 2x\\)\n\\(x > -4\\)\n\\(x < 2 ; x > -4\\).\n(ii)\n(b) Let the digits of the number be x and y.\n\\(x + y = 9 .... (1)\\)\n\\((10y + x) - (10x + y) = 45 \\)\n\\(10y - y + x - 10x = 45\\)\n\\(9y - 9x = 45 .... (2)\\)\nFrom (1), y = 9 - x. Putting it into (2), we have\n\\(9(9 - x) - 9x = 45\\)\n\\(81 - 9x - 9x = 45\\)\n\\(36 = 18x \\implies x = 2\\)\n\\(\\therefore y = 9 - x\\)\n= \\(9 - 2 = 7\\)\n\\(\\therefore\\) The bill =£27.\n\n', '', 'Explanation\n(a)(i) \\(4x < 2 + 3x ..... (1)\\)\n   \\(4x - 3x < 2\\)\n\\(x < 2\\)\n\\(x - 8 < 3x ..... (2)\\)\n\\(-8 < 2x\\)\n\\(x > -4\\)\n\\(x < 2 ; x > -4\\).\n(ii)\n(b) Let the digits of the number be x and y.\n\\(x + y = 9 .... (1)\\)\n\\((10y + x) - (10x + y) = 45 \\)\n\\(10y - y + x - 10x = 45\\)\n\\(9y - 9x = 45 .... (2)\\)\nFrom (1), y = 9 - x. Putting it into (2), we have\n\\(9(9 - x) - 9x = 45\\)\n\\(81 - 9x - 9x = 45\\)\n\\(36 = 18x \\implies x = 2\\)\n\\(\\therefore y = 9 - x\\)\n= \\(9 - 2 = 7\\)\n\\(\\therefore\\) The bill =£27.\n\n', '(a)(i) \\(4x < 2 + 3x ..... (1)\\)\n   \\(4x - 3x < 2\\)\n\\(x < 2\\)\n\\(x - 8 < 3x ..... (2)\\)\n\\(-8 < 2x\\)\n\\(x > -4\\)\n\\(x < 2 ; x > -4\\).\n(ii)~(b) Let the digits of the number be x and y.\n\\(x + y = 9 .... (1)\\)\n\\((10y + x) - (10x + y) = 45 \\)\n\\(10y - y + x - 10x = 45\\)\n\\(9y - 9x = 45 .... (2)\\)\nFrom (1), y = 9 - x. Putting it into (2), we have\n\\(9(9 - x) - 9x = 45\\)\n\\(81 - 9x - 9x = 45\\)\n\\(36 = 18x \\implies x = 2\\)\n\\(\\therefore y = 9 - x\\)\n= \\(9 - 2 = 7\\)\n\\(\\therefore\\) The bill =£27.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(480, '2014', '\n\nQuestion\n\nIn the diagram, ABCDEO is two- thirds of a circle centre O. The radius AO is 7cm and /AB/ = /BC/ = /CD/ = /DE/. Calculate, correct to the nearest whole number, the area of the shaded portion. [Take \\(\\pi = \\frac{22}{7}\\)].\n\n\n', 'In the diagram, ABCDEO is two- thirds of a circle centre O. The radius AO is 7cm and /AB/ = /BC/ = /CD/ = /DE/. Calculate, correct to the nearest whole number, the area of the shaded portion. [Take \\(\\pi = \\frac{22}{7}\\)].\n', '', '', NULL, '\n\nExplanation\n\\(\\stackrel\\frown{AOB} = \\stackrel\\frown{BOC} = \\stackrel\\frown{COD} = \\stackrel\\frown{DOE} = 60°\\)\nIn \\(\\Delta AOB\\), Area of sector = \\(\\frac{60}{360} \\times \\frac{22}{7} \\times 7^{2}\\)\n= \\(25.67 cm^{2}\\)\nArea of triangle AOB = \\(\\frac{1}{2} \\times \\7 \\times 7 \\times \\sin 60\\)\n= \\(24.5 \\times 0.866\\)\n= \\(21.22 cm^{2}\\)\nArea of the 4 sectors = \\(4 \\times 25.67 = 102.68 cm^{2}\\)\nArea of the 4 triangles = \\(4 \\times 21.22 = 84.88 cm^{2}\\)\nArea of the shaded portion = \\((102.68 - 84.88)cm^{2}\\)\n= \\(17.80 cm^{2}\\)\n\\(\\approxeq 18 cm^{2}\\)\n\n', '\\(\\stackrel\\frown{AOB} = \\stackrel\\frown{BOC} = \\stackrel\\frown{COD} = \\stackrel\\frown{DOE} = 60°\\)\nIn \\(\\Delta AOB\\), Area of sector = \\(\\frac{60}{360} \\times \\frac{22}{7} \\times 7^{2}\\)\n= \\(25.67 cm^{2}\\)\nArea of triangle AOB = \\(\\frac{1}{2} \\times \\7 \\times 7 \\times \\sin 60\\)\n= \\(24.5 \\times 0.866\\)\n= \\(21.22 cm^{2}\\)\nArea of the 4 sectors = \\(4 \\times 25.67 = 102.68 cm^{2}\\)\nArea of the 4 triangles = \\(4 \\times 21.22 = 84.88 cm^{2}\\)\nArea of the shaded portion = \\((102.68 - 84.88)cm^{2}\\)\n= \\(17.80 cm^{2}\\)\n\\(\\approxeq 18 cm^{2}\\)\n\n', '', '', 'WAEC-PK-1', 'WAEC-MATH', 1),
(481, '2014', '\n\nQuestion\n\n(a) Using a ruler and a pair of compasses only, (i) construct \\(\\Delta\\) XYZ such that |XY| = 8 cm and < YXZ = < ZYX = 45°. (ii) locate a point P inside the triangle equidistant from XY and XZ and also equidistance from YX and YZ. (iii) construct a circle touching the three sides of the triangle (iv) measure the radius of the circle.\n(b) The length of the sides of a hexagon are x - 5, 2x, 2x, 2x + 7, 2x and 2x - 1. If the perimeter is 144 cm, find the value of x.\n\n', '', '(a) Using a ruler and a pair of compasses only, (i) construct \\(\\Delta\\) XYZ such that |XY| = 8 cm and < YXZ = < ZYX = 45°. (ii) locate a point P inside the triangle equidistant from XY and XZ and also equidistance from YX and YZ. (iii) construct a circle touching the three sides of the triangle (iv) measure the radius of the circle.\n(b) The length of the sides of a hexagon are x - 5, 2x, 2x, 2x + 7, 2x and 2x - 1. If the perimeter is 144 cm, find the value of x.', '(a) Using a ruler and a pair of compasses only, (i) construct \\(\\Delta\\) XYZ such that |XY| = 8 cm and < YXZ = < ZYX = 45°. (ii) locate a point P inside the triangle equidistant from XY and XZ and also equidistance from YX and YZ. (iii) construct a circle touching the three sides of the triangle (iv) measure the radius of the circle.~(b) The length of the sides of a hexagon are x - 5, 2x, 2x, 2x + 7, 2x and 2x - 1. If the perimeter is 144 cm, find the value of x.', NULL, '\n\nExplanation\n(a)\n(b) \\(x - 5 + 2x + 2x + 2x + 7 + 2x + 2x - 1 = 144\\)\n\\(11x + 1 = 144\\)\n\\(11x = 143\\)\n\\(x = \\frac{143}{11} = 13cm\\)\n\n', '', 'Explanation\n(a)\n(b) \\(x - 5 + 2x + 2x + 2x + 7 + 2x + 2x - 1 = 144\\)\n\\(11x + 1 = 144\\)\n\\(11x = 143\\)\n\\(x = \\frac{143}{11} = 13cm\\)\n\n', '~(b) \\(x - 5 + 2x + 2x + 2x + 7 + 2x + 2x - 1 = 144\\)\n\\(11x + 1 = 144\\)\n\\(11x = 143\\)\n\\(x = \\frac{143}{11} = 13cm\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(482, '2014', '\n\nQuestion\n\nThe following table shows the distribution of test scores in a class.\n\n\n\nScores\n1\n2\n3\n4\n5\n7\n8\n9\n10\n\n\nNo of pupils\n1\n1\n5\n3\n\\(k^{2} + 1\\)\n6\n2\n3\n4\n\n\n\n(a) If the mean score of the class is 6, find the : (i) value of k (ii) median score.\n(b) Draw a bar chart for the distribution.\n(c) If a pupil is picked at random, what is the probability that he/ she will score less than 6?\n\n\n\nThe following table shows the distribution of test scores in a class.\n\n\n\nScores\n1\n2\n3\n4\n5\n7\n8\n9\n10\n\n\nNo of pupils\n1\n1\n5\n3\n\\(k^{2} + 1\\)\n6\n2\n3\n4\n\n\n', '', 'The following table shows the distribution of test scores in a class.\n\n\n\nScores\n1\n2\n3\n4\n5\n7\n8\n9\n10\n\n\nNo of pupils\n1\n1\n5\n3\n\\(k^{2} + 1\\)\n6\n2\n3\n4\n\n\n\n(a) If the mean score of the class is 6, find the : (i) value of k (ii) median score.\n(b) Draw a bar chart for the distribution.\n(c) If a pupil is picked at random, what is the probability that he/ she will score less than 6?\n\n\n\nThe following table shows the distribution of test scores in a class.\n\n\n\nScores\n1\n2\n3\n4\n5\n7\n8\n9\n10\n\n\nNo of pupils\n1\n1\n5\n3\n\\(k^{2} + 1\\)\n6\n2\n3\n4\n', 'The following table shows the distribution of test scores in a class.\n\n\n\nScores\n1\n2\n3\n4\n5\n7\n8\n9\n10\n\n\nNo of pupils\n1\n1\n5\n3\n\\(k^{2} + 1\\)\n6\n2\n3\n4\n\n\n\n(a) If the mean score of the class is 6, find the : (i) value of k (ii) median score.~(b) Draw a bar chart for the distribution.\n(c) If a pupil is picked at random, what is the probability that he/ she will score less than 6?\n\n\n\nThe following table shows the distribution of test scores in a class.\n\n\n\nScores\n1\n2\n3\n4\n5\n7\n8\n9\n10\n\n\nNo of pupils\n1\n1\n5\n3\n\\(k^{2} + 1\\)\n6\n2\n3\n4\n~(c) If a pupil is picked at random, what is the probability that he/ she will score less than 6?\n\n\n\nThe following table shows the distribution of test scores in a class.\n\n\n\nScores\n1\n2\n3\n4\n5\n7\n8\n9\n10\n\n\nNo of pupils\n1\n1\n5\n3\n\\(k^{2} + 1\\)\n6\n2\n3\n4\n', NULL, '\n\nExplanation\n(a)\n\n\n\nScores (x)\n1\n2\n3\n4\n5\n7\n8\n9\n10\nTotal\n\n\nNo of pupils (f)\n1\n1\n5\n3\n\\(k^{2} + 1\\)\n6\n2\n3\n4\n\\(k^{2} + 26\\)\n\n\nfx\n1\n2\n15\n12\n\\(5k^{2} + 5\\)\n42\n16\n27\n40\n\\(5k^{2} + 160\\)\n\n\n\n\\(\\bar{x} = \\frac{\\sum fx}{\\sum f}\\)\n\\(6 = \\frac{5k^{2} + 160}{k^{2} + 26}\\)\n\\(6(k^{2} + 26) = 5k^{2} + 160\\)\n\\(6k^{2} + 156 = 5k^{2} + 160\\)\n\\(6k^{2} - 5k^{2} = 160 - 156\\)\n\\(k^{2} = 4\\)\n\\( k = \\sqrt{4} = 2\\)\n\\(\\therefore k = 2\\)\n(ii) \\(Median = \\frac{5 + 7}{2}\\)\n= \\(\\frac{12}{2} = 6\\)\n(b)\n(c) Probability of scoring less than 6 = \\(\\frac{15}{30} \\)\n= \\(\\frac{1}{2}\\)\n\n\n', '', 'Explanation\n(a)\n\n\n\nScores (x)\n1\n2\n3\n4\n5\n7\n8\n9\n10\nTotal\n\n\nNo of pupils (f)\n1\n1\n5\n3\n\\(k^{2} + 1\\)\n6\n2\n3\n4\n\\(k^{2} + 26\\)\n\n\nfx\n1\n2\n15\n12\n\\(5k^{2} + 5\\)\n42\n16\n27\n40\n\\(5k^{2} + 160\\)\n\n\n\n\\(\\bar{x} = \\frac{\\sum fx}{\\sum f}\\)\n\\(6 = \\frac{5k^{2} + 160}{k^{2} + 26}\\)\n\\(6(k^{2} + 26) = 5k^{2} + 160\\)\n\\(6k^{2} + 156 = 5k^{2} + 160\\)\n\\(6k^{2} - 5k^{2} = 160 - 156\\)\n\\(k^{2} = 4\\)\n\\( k = \\sqrt{4} = 2\\)\n\\(\\therefore k = 2\\)\n(ii) \\(Median = \\frac{5 + 7}{2}\\)\n= \\(\\frac{12}{2} = 6\\)\n(b)\n(c) Probability of scoring less than 6 = \\(\\frac{15}{30} \\)\n= \\(\\frac{1}{2}\\)\n\n\n', '(a)\n\n\n\nScores (x)\n1\n2\n3\n4\n5\n7\n8\n9\n10\nTotal\n\n\nNo of pupils (f)\n1\n1\n5\n3\n\\(k^{2} + 1\\)\n6\n2\n3\n4\n\\(k^{2} + 26\\)\n\n\nfx\n1\n2\n15\n12\n\\(5k^{2} + 5\\)\n42\n16\n27\n40\n\\(5k^{2} + 160\\)\n\n\n\n\\(\\bar{x} = \\frac{\\sum fx}{\\sum f}\\)\n\\(6 = \\frac{5k^{2} + 160}{k^{2} + 26}\\)\n\\(6(k^{2} + 26) = 5k^{2} + 160\\)\n\\(6k^{2} + 156 = 5k^{2} + 160\\)\n\\(6k^{2} - 5k^{2} = 160 - 156\\)\n\\(k^{2} = 4\\)\n\\( k = \\sqrt{4} = 2\\)\n\\(\\therefore k = 2\\)\n(ii) \\(Median = \\frac{5 + 7}{2}\\)\n= \\(\\frac{12}{2} = 6\\)~(b)\n(c) Probability of scoring less than 6 = \\(\\frac{15}{30} \\)\n= \\(\\frac{1}{2}\\)\n\n~(c) Probability of scoring less than 6 = \\(\\frac{15}{30} \\)\n= \\(\\frac{1}{2}\\)\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(483, '2014', '\n\nQuestion\n\n(a)\nThe diagram shows a pyramid standing on a cuboid. The dimensions of the cuboid are 4m by 3m by 2m and the slant edge of the pyramid is 5m. Calculet the volume of the shape.\n(b) The 2nd, 3rd and 4th terms of an A.P are x - 2, 5 and x + 2 respectively. Calculate the value of x.\n\n', '', '(a)\nThe diagram shows a pyramid standing on a cuboid. The dimensions of the cuboid are 4m by 3m by 2m and the slant edge of the pyramid is 5m. Calculet the volume of the shape.\n(b) The 2nd, 3rd and 4th terms of an A.P are x - 2, 5 and x + 2 respectively. Calculate the value of x.', '(a)\nThe diagram shows a pyramid standing on a cuboid. The dimensions of the cuboid are 4m by 3m by 2m and the slant edge of the pyramid is 5m. Calculet the volume of the shape.~(b) The 2nd, 3rd and 4th terms of an A.P are x - 2, 5 and x + 2 respectively. Calculate the value of x.', NULL, '\n\nExplanation\n(a) From the pyramid, construct \\(\\Delta BFG\\) by drawing a line from B to F.\n\n\\(|BF|^{2} = 4^{2} + 3^{2}\\)\n\\(|BF| = 25 \\implies |BF| = 5m\\)\nIn calculating the height,\n\\(h^{2} = 5^{2} - (\\frac{5}{2})^{2}\\)\n\\(h^{2} = \\frac{75}{4}\\)\n\\(h = \\sqrt{75}{4}\\)\n= \\(4.33 m\\)\nVolume of a pyramid = \\(\\frac{1}{3} \\times 4 \\times 3 \\times 4.33\\)\n= \\(17.32 m^{3}\\)\nVolume of a cuboid = \\(l \\times b \\times h\\)\n= \\(3 \\times 4 \\times 2\\)\n= \\(24 m^{3}\\)\nTotal volume of the shape = \\(17.32 + 24\\)\n= \\(41.32 m^{3}\\)\n(b) \\(5 - (x - 2) = d\\)\n\\((x + 2) - 5 = d\\)\n\\(\\therefore (x + 2) - 5 = 5 - (x - 2)\\)\n\\(x - 3 = 7 - x\\)\n\\(2x = 10\\)\n\\(x = 5\\)\n\n', '', 'Explanation\n(a) From the pyramid, construct \\(\\Delta BFG\\) by drawing a line from B to F.\n\n\\(|BF|^{2} = 4^{2} + 3^{2}\\)\n\\(|BF| = 25 \\implies |BF| = 5m\\)\nIn calculating the height,\n\\(h^{2} = 5^{2} - (\\frac{5}{2})^{2}\\)\n\\(h^{2} = \\frac{75}{4}\\)\n\\(h = \\sqrt{75}{4}\\)\n= \\(4.33 m\\)\nVolume of a pyramid = \\(\\frac{1}{3} \\times 4 \\times 3 \\times 4.33\\)\n= \\(17.32 m^{3}\\)\nVolume of a cuboid = \\(l \\times b \\times h\\)\n= \\(3 \\times 4 \\times 2\\)\n= \\(24 m^{3}\\)\nTotal volume of the shape = \\(17.32 + 24\\)\n= \\(41.32 m^{3}\\)\n(b) \\(5 - (x - 2) = d\\)\n\\((x + 2) - 5 = d\\)\n\\(\\therefore (x + 2) - 5 = 5 - (x - 2)\\)\n\\(x - 3 = 7 - x\\)\n\\(2x = 10\\)\n\\(x = 5\\)\n\n', '(a) From the pyramid, construct \\(\\Delta BFG\\) by drawing a line from B to F.\n\n\\(|BF|^{2} = 4^{2} + 3^{2}\\)\n\\(|BF| = 25 \\implies |BF| = 5m\\)\nIn calculating the height,\n\\(h^{2} = 5^{2} - (\\frac{5}{2})^{2}\\)\n\\(h^{2} = \\frac{75}{4}\\)\n\\(h = \\sqrt{75}{4}\\)\n= \\(4.33 m\\)\nVolume of a pyramid = \\(\\frac{1}{3} \\times 4 \\times 3 \\times 4.33\\)\n= \\(17.32 m^{3}\\)\nVolume of a cuboid = \\(l \\times b \\times h\\)\n= \\(3 \\times 4 \\times 2\\)\n= \\(24 m^{3}\\)\nTotal volume of the shape = \\(17.32 + 24\\)\n= \\(41.32 m^{3}\\)~(b) \\(5 - (x - 2) = d\\)\n\\((x + 2) - 5 = d\\)\n\\(\\therefore (x + 2) - 5 = 5 - (x - 2)\\)\n\\(x - 3 = 7 - x\\)\n\\(2x = 10\\)\n\\(x = 5\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(484, '2014', '\n\nQuestion\n\n(a) Copy and complete the table of the relation \\(y = 2\\sin x - \\cos 2x\\).\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n\n\ny\n\n0.5\n\n\n\n\n-1.0\n\n\n\nUsing a scale of 2 cm to 30° on the x- axis and 2 cm to 0.5 unit on the y- axis, draw the graph of \\(y = 2\\sin x - \\cos 2x\\), for \\(0° \\leq x \\leq 180°\\).\n(b) Using the same axes, draw the graph of \\(y = 1.25\\).\n(c) Use your graphs to find the : (i) values of x for which \\(2\\sin x - \\cos 2x = 0\\) ; (ii) the roots of the equation \\(2\\sin x - \\cos 2x = 1.25\\).\n\n\n\n(a) Copy and complete the table of the relation \\(y = 2\\sin x - \\cos 2x\\).\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n\n\ny\n\n0.5\n\n\n\n\n-1.0\n\n\n', '', '(a) Copy and complete the table of the relation \\(y = 2\\sin x - \\cos 2x\\).\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n\n\ny\n\n0.5\n\n\n\n\n-1.0\n\n\n\nUsing a scale of 2 cm to 30° on the x- axis and 2 cm to 0.5 unit on the y- axis, draw the graph of \\(y = 2\\sin x - \\cos 2x\\), for \\(0° \\leq x \\leq 180°\\).\n(b) Using the same axes, draw the graph of \\(y = 1.25\\).\n(c) Use your graphs to find the : (i) values of x for which \\(2\\sin x - \\cos 2x = 0\\) ; (ii) the roots of the equation \\(2\\sin x - \\cos 2x = 1.25\\).\n\n\n\n(a) Copy and complete the table of the relation \\(y = 2\\sin x - \\cos 2x\\).\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n\n\ny\n\n0.5\n\n\n\n\n-1.0\n', '(a) Copy and complete the table of the relation \\(y = 2\\sin x - \\cos 2x\\).\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n\n\ny\n\n0.5\n\n\n\n\n-1.0\n\n\n\nUsing a scale of 2 cm to 30° on the x- axis and 2 cm to 0.5 unit on the y- axis, draw the graph of \\(y = 2\\sin x - \\cos 2x\\), for \\(0° \\leq x \\leq 180°\\).~(b) Using the same axes, draw the graph of \\(y = 1.25\\).\n(c) Use your graphs to find the : (i) values of x for which \\(2\\sin x - \\cos 2x = 0\\) ; (ii) the roots of the equation \\(2\\sin x - \\cos 2x = 1.25\\).\n\n\n\n(a) Copy and complete the table of the relation \\(y = 2\\sin x - \\cos 2x\\).\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n\n\ny\n\n0.5\n\n\n\n\n-1.0\n~(c) Use your graphs to find the : (i) values of x for which \\(2\\sin x - \\cos 2x = 0\\) ; (ii) the roots of the equation \\(2\\sin x - \\cos 2x = 1.25\\).\n\n\n\n(a) Copy and complete the table of the relation \\(y = 2\\sin x - \\cos 2x\\).\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n\n\ny\n\n0.5\n\n\n\n\n-1.0\n', NULL, '\n\nExplanation\n(a) \\(y = 2\\sin x - \\cos 2x\\)\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n\n\ny\n-1.0\n0.5\n2.2\n3.0\n2.2\n0.5\n-1.0\n\n\n\n(b)\n\n\n', '', 'Explanation\n(a) \\(y = 2\\sin x - \\cos 2x\\)\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n\n\ny\n-1.0\n0.5\n2.2\n3.0\n2.2\n0.5\n-1.0\n\n\n\n(b)\n\n\n', '(a) \\(y = 2\\sin x - \\cos 2x\\)\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n\n\ny\n-1.0\n0.5\n2.2\n3.0\n2.2\n0.5\n-1.0\n\n\n~(b)\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(485, '2014', '\n\nQuestion\n\n(a) Two lines AB and CD intersect at x such that \\(\\stackrel\\frown{CAX}\\) is equal to \\(\\stackrel\\frown{BDX}\\). If |AX| = 6 cm, |XB| = 4 cm and |CX| = 3 cm, find |XD|.\n(b)\nThe diagram shows the positions of three points X, Y and Z on a horizontal plane. The bearing of Y from X is 312° and that of Y from Z is 022°. If |XY| = 32 km and |ZY| = 50 km, calculate, correct to one decimal place : (i) |XZ| ; (ii) the bearing of Zfrom X.\n\n', '', '(a) Two lines AB and CD intersect at x such that \\(\\stackrel\\frown{CAX}\\) is equal to \\(\\stackrel\\frown{BDX}\\). If |AX| = 6 cm, |XB| = 4 cm and |CX| = 3 cm, find |XD|.\n(b)\nThe diagram shows the positions of three points X, Y and Z on a horizontal plane. The bearing of Y from X is 312° and that of Y from Z is 022°. If |XY| = 32 km and |ZY| = 50 km, calculate, correct to one decimal place : (i) |XZ| ; (ii) the bearing of Zfrom X.', '(a) Two lines AB and CD intersect at x such that \\(\\stackrel\\frown{CAX}\\) is equal to \\(\\stackrel\\frown{BDX}\\). If |AX| = 6 cm, |XB| = 4 cm and |CX| = 3 cm, find |XD|.~(b)\nThe diagram shows the positions of three points X, Y and Z on a horizontal plane. The bearing of Y from X is 312° and that of Y from Z is 022°. If |XY| = 32 km and |ZY| = 50 km, calculate, correct to one decimal place : (i) |XZ| ; (ii) the bearing of Zfrom X.', NULL, '\n\nExplanation\n(a)\nNote that triangles AXC and BXD are similar.\n\\(\\therefore \\frac{6}{|XD|} = \\frac{3}{4}\\)\n\\(|XD| = \\frac{6 \\times 4}{3}\\)\n= \\(8 cm\\)\n(b)(i)\nFrom \\(\\Delta ZYX\\),\n\\(\\stackrel\\frown{YZX} = 90° - 22° = 68°\\)\n\\(\\stackrel\\frown{YXZ} = 312° - 270°= 42°\\)\n\\(\\stackrel\\frown{ZYX} = 180° - (68° + 42°)\\)\n= \\(180° - 110°\\)\n= \\(70°\\)\n\\(a^{2} = b^{2} + c^{2} - 2bc \\cos A\\)\n\\(|XZ|^{2} = 50^{2} + 32^{2} - 2(50)(32) \\cos 70\\)\n= \\(2500 +1024 - 3200 \\times 0.342\\)\n= \\(3524 - 1094.46\\)\n\\(|XZ|^{2} = 2429.54\\)\n\\(|XZ| = \\sqrt{2429.54} = 49.29 km\\)\n\\(\\approxeq 49.3 km\\) (to 1 d.p)\n(ii)\n\\(\\frac{a}{\\sin A} = \\frac{b}{\\sin B} = \\frac{c}{\\sin C}\\)\n\\(\\frac{50}{\\sin \\theta} = \\frac{49.3}{\\sin 70}\\)\n\\(\\sin \\theta = \\frac{50 \\times \\sin 70}{49.3}\\)\n\\(\\sin \\theta = 0.9530\\)\n\\(\\theta = \\sin^{-1} (0.9530)\\)\n= \\(72.37°\\)\n\\(\\therefore\\) The bearing of Z from X = \\(312° - 72.37°\\)\n= \\(239.63° \\approxeq 239.6°\\)\n\n', '', 'Explanation\n(a)\nNote that triangles AXC and BXD are similar.\n\\(\\therefore \\frac{6}{|XD|} = \\frac{3}{4}\\)\n\\(|XD| = \\frac{6 \\times 4}{3}\\)\n= \\(8 cm\\)\n(b)(i)\nFrom \\(\\Delta ZYX\\),\n\\(\\stackrel\\frown{YZX} = 90° - 22° = 68°\\)\n\\(\\stackrel\\frown{YXZ} = 312° - 270°= 42°\\)\n\\(\\stackrel\\frown{ZYX} = 180° - (68° + 42°)\\)\n= \\(180° - 110°\\)\n= \\(70°\\)\n\\(a^{2} = b^{2} + c^{2} - 2bc \\cos A\\)\n\\(|XZ|^{2} = 50^{2} + 32^{2} - 2(50)(32) \\cos 70\\)\n= \\(2500 +1024 - 3200 \\times 0.342\\)\n= \\(3524 - 1094.46\\)\n\\(|XZ|^{2} = 2429.54\\)\n\\(|XZ| = \\sqrt{2429.54} = 49.29 km\\)\n\\(\\approxeq 49.3 km\\) (to 1 d.p)\n(ii)\n\\(\\frac{a}{\\sin A} = \\frac{b}{\\sin B} = \\frac{c}{\\sin C}\\)\n\\(\\frac{50}{\\sin \\theta} = \\frac{49.3}{\\sin 70}\\)\n\\(\\sin \\theta = \\frac{50 \\times \\sin 70}{49.3}\\)\n\\(\\sin \\theta = 0.9530\\)\n\\(\\theta = \\sin^{-1} (0.9530)\\)\n= \\(72.37°\\)\n\\(\\therefore\\) The bearing of Z from X = \\(312° - 72.37°\\)\n= \\(239.63° \\approxeq 239.6°\\)\n\n', '(a)\nNote that triangles AXC and BXD are similar.\n\\(\\therefore \\frac{6}{|XD|} = \\frac{3}{4}\\)\n\\(|XD| = \\frac{6 \\times 4}{3}\\)\n= \\(8 cm\\)~(b)(i)\nFrom \\(\\Delta ZYX\\),\n\\(\\stackrel\\frown{YZX} = 90° - 22° = 68°\\)\n\\(\\stackrel\\frown{YXZ} = 312° - 270°= 42°\\)\n\\(\\stackrel\\frown{ZYX} = 180° - (68° + 42°)\\)\n= \\(180° - 110°\\)\n= \\(70°\\)\n\\(a^{2} = b^{2} + c^{2} - 2bc \\cos A\\)\n\\(|XZ|^{2} = 50^{2} + 32^{2} - 2(50)(32) \\cos 70\\)\n= \\(2500 +1024 - 3200 \\times 0.342\\)\n= \\(3524 - 1094.46\\)\n\\(|XZ|^{2} = 2429.54\\)\n\\(|XZ| = \\sqrt{2429.54} = 49.29 km\\)\n\\(\\approxeq 49.3 km\\) (to 1 d.p)\n(ii)\n\\(\\frac{a}{\\sin A} = \\frac{b}{\\sin B} = \\frac{c}{\\sin C}\\)\n\\(\\frac{50}{\\sin \\theta} = \\frac{49.3}{\\sin 70}\\)\n\\(\\sin \\theta = \\frac{50 \\times \\sin 70}{49.3}\\)\n\\(\\sin \\theta = 0.9530\\)\n\\(\\theta = \\sin^{-1} (0.9530)\\)\n= \\(72.37°\\)\n\\(\\therefore\\) The bearing of Z from X = \\(312° - 72.37°\\)\n= \\(239.63° \\approxeq 239.6°\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(486, '2014', '\n\nQuestion\n\n(a) Evaluate without using the mathematical table or calculator, \\(\\log_{10} \\sqrt{30} - \\log_{10} \\sqrt{6} + \\log_{10} \\sqrt{2}\\).\n(b)\n\\(U = {1, 2, 3, ..., 10} ;A = {1, 2, 3, 4, 5} ;B = {2, 3, 5}\\) and \\(C = {6, 8, 10}\\). (i) Given that the Venn diagram represents the sets above, copy and fill in the elements.\n(ii) Find \\(A \\cap C\\) ; (iii) Find \\(A \\cap B\'\\).\n\n', '', '(a) Evaluate without using the mathematical table or calculator, \\(\\log_{10} \\sqrt{30} - \\log_{10} \\sqrt{6} + \\log_{10} \\sqrt{2}\\).\n(b)\n\\(U = {1, 2, 3, ..., 10} ;A = {1, 2, 3, 4, 5} ;B = {2, 3, 5}\\) and \\(C = {6, 8, 10}\\). (i) Given that the Venn diagram represents the sets above, copy and fill in the elements.\n(ii) Find \\(A \\cap C\\) ; (iii) Find \\(A \\cap B\'\\).', '(a) Evaluate without using the mathematical table or calculator, \\(\\log_{10} \\sqrt{30} - \\log_{10} \\sqrt{6} + \\log_{10} \\sqrt{2}\\).~(b)\n\\(U = {1, 2, 3, ..., 10} ;A = {1, 2, 3, 4, 5} ;B = {2, 3, 5}\\) and \\(C = {6, 8, 10}\\). (i) Given that the Venn diagram represents the sets above, copy and fill in the elements.\n(ii) Find \\(A \\cap C\\) ; (iii) Find \\(A \\cap B\'\\).', NULL, '\n\nExplanation\n(a) \\(\\log_{10} \\sqrt{30} - \\log_{10} \\sqrt{6} + \\log_{10} \\sqrt{2}\\)\n= \\(\\log_{10} (\\sqrt{\\frac{30}{6}} \\times \\sqrt{2})\\)\n= \\(\\log_{10} (\\sqrt{5} \\times \\sqrt{2})\\)\n= \\(\\log_{10} \\sqrt{10}\\)\n= \\(\\log_{10} (10^{\\frac{1}{2}})\\)\n= \\(\\frac{1}{2} \\log_{10} 10\\)\n= \\(\\frac{1}{2}\\).\n(b)\n(ii) \\(A \\cap C = { } \\)\n(iii) \\(A \\cap B\' = {1, 4}\\).\n\n', '', 'Explanation\n(a) \\(\\log_{10} \\sqrt{30} - \\log_{10} \\sqrt{6} + \\log_{10} \\sqrt{2}\\)\n= \\(\\log_{10} (\\sqrt{\\frac{30}{6}} \\times \\sqrt{2})\\)\n= \\(\\log_{10} (\\sqrt{5} \\times \\sqrt{2})\\)\n= \\(\\log_{10} \\sqrt{10}\\)\n= \\(\\log_{10} (10^{\\frac{1}{2}})\\)\n= \\(\\frac{1}{2} \\log_{10} 10\\)\n= \\(\\frac{1}{2}\\).\n(b)\n(ii) \\(A \\cap C = { } \\)\n(iii) \\(A \\cap B\' = {1, 4}\\).\n\n', '(a) \\(\\log_{10} \\sqrt{30} - \\log_{10} \\sqrt{6} + \\log_{10} \\sqrt{2}\\)\n= \\(\\log_{10} (\\sqrt{\\frac{30}{6}} \\times \\sqrt{2})\\)\n= \\(\\log_{10} (\\sqrt{5} \\times \\sqrt{2})\\)\n= \\(\\log_{10} \\sqrt{10}\\)\n= \\(\\log_{10} (10^{\\frac{1}{2}})\\)\n= \\(\\frac{1}{2} \\log_{10} 10\\)\n= \\(\\frac{1}{2}\\).~(b)\n(ii) \\(A \\cap C = { } \\)\n(iii) \\(A \\cap B\' = {1, 4}\\).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(487, '2014', '\n\nQuestion\n\n(a) Evaluate : \\(2 \\div (\\frac{64}{125})^{-\\frac{2}{3}}\\)\n(b) The lines \\(y = 3x + 5\\) and \\(y = - 4x - 1\\) intersect at a point k. Find the coordinates of k.\n\n', '', '(a) Evaluate : \\(2 \\div (\\frac{64}{125})^{-\\frac{2}{3}}\\)\n(b) The lines \\(y = 3x + 5\\) and \\(y = - 4x - 1\\) intersect at a point k. Find the coordinates of k.', '(a) Evaluate : \\(2 \\div (\\frac{64}{125})^{-\\frac{2}{3}}\\)~(b) The lines \\(y = 3x + 5\\) and \\(y = - 4x - 1\\) intersect at a point k. Find the coordinates of k.', NULL, '\n\nExplanation\n(a) \\((\\frac{64}{125})^{-\\frac{2}{3}} = (\\frac{125}{64})^{\\frac{2}{3}}\\)\n= \\((\\frac{5}{4})^{3})^{\\frac{2}{3}} \\)\n= \\((\\frac{5}{4})^{2}\\)\n= \\(\\frac{25}{16}\\)\n\\(\\therefore 2 \\div (\\frac{64}{125})^{-\\frac{2}{3}} = 2 \\div \\frac{25}{16}\\)\n= \\(2 \\times \\frac{16}{25}\\)\n= \\(\\frac{32}{25}\\)\n(b) \\(3x + 5 = - 4x - 1\\)\n\\(3x + 4x = - 1 - 5\\)\n\\(7x = - 6\\)\n\\(x = -\\frac{6}{7}\\)\n\\(y = 3x + 5\\) ( You can use any of the given equations to get y)\n\\(y = 3(-\\frac{6}{7}) + 5 = -\\frac{18}{7} + \\frac{35}{7}\\)\n= \\(\\frac{17}{7}\\)\n\\(k = (x, y) = (-\\frac{6}{7}, \\frac{17}{7})\\)\n\n', '', 'Explanation\n(a) \\((\\frac{64}{125})^{-\\frac{2}{3}} = (\\frac{125}{64})^{\\frac{2}{3}}\\)\n= \\((\\frac{5}{4})^{3})^{\\frac{2}{3}} \\)\n= \\((\\frac{5}{4})^{2}\\)\n= \\(\\frac{25}{16}\\)\n\\(\\therefore 2 \\div (\\frac{64}{125})^{-\\frac{2}{3}} = 2 \\div \\frac{25}{16}\\)\n= \\(2 \\times \\frac{16}{25}\\)\n= \\(\\frac{32}{25}\\)\n(b) \\(3x + 5 = - 4x - 1\\)\n\\(3x + 4x = - 1 - 5\\)\n\\(7x = - 6\\)\n\\(x = -\\frac{6}{7}\\)\n\\(y = 3x + 5\\) ( You can use any of the given equations to get y)\n\\(y = 3(-\\frac{6}{7}) + 5 = -\\frac{18}{7} + \\frac{35}{7}\\)\n= \\(\\frac{17}{7}\\)\n\\(k = (x, y) = (-\\frac{6}{7}, \\frac{17}{7})\\)\n\n', '(a) \\((\\frac{64}{125})^{-\\frac{2}{3}} = (\\frac{125}{64})^{\\frac{2}{3}}\\)\n= \\((\\frac{5}{4})^{3})^{\\frac{2}{3}} \\)\n= \\((\\frac{5}{4})^{2}\\)\n= \\(\\frac{25}{16}\\)\n\\(\\therefore 2 \\div (\\frac{64}{125})^{-\\frac{2}{3}} = 2 \\div \\frac{25}{16}\\)\n= \\(2 \\times \\frac{16}{25}\\)\n= \\(\\frac{32}{25}\\)~(b) \\(3x + 5 = - 4x - 1\\)\n\\(3x + 4x = - 1 - 5\\)\n\\(7x = - 6\\)\n\\(x = -\\frac{6}{7}\\)\n\\(y = 3x + 5\\) ( You can use any of the given equations to get y)\n\\(y = 3(-\\frac{6}{7}) + 5 = -\\frac{18}{7} + \\frac{35}{7}\\)\n= \\(\\frac{17}{7}\\)\n\\(k = (x, y) = (-\\frac{6}{7}, \\frac{17}{7})\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(488, '2014', '\n\nQuestion\n\n(a) Simplify : \\(\\frac{5}{8} of 2\\frac{1}{2} - \\frac{3}{4} \\div \\frac{3}{5}\\).\n(b) A cone and a right pyramid have equal heights and volumes. If the area of the base of the pyramid is \\(154 cm^{2}\\), find the base radius of the cone. [Take \\(\\pi = \\frac{22}{7}\\)].\n\n', '', '(a) Simplify : \\(\\frac{5}{8} of 2\\frac{1}{2} - \\frac{3}{4} \\div \\frac{3}{5}\\).\n(b) A cone and a right pyramid have equal heights and volumes. If the area of the base of the pyramid is \\(154 cm^{2}\\), find the base radius of the cone. [Take \\(\\pi = \\frac{22}{7}\\)].', '(a) Simplify : \\(\\frac{5}{8} of 2\\frac{1}{2} - \\frac{3}{4} \\div \\frac{3}{5}\\).~(b) A cone and a right pyramid have equal heights and volumes. If the area of the base of the pyramid is \\(154 cm^{2}\\), find the base radius of the cone. [Take \\(\\pi = \\frac{22}{7}\\)].', NULL, '\n\nExplanation\n(a) \\(\\frac{5}{8} of 2\\frac{1}{2} - \\frac{3}{4} \\div \\frac{3}{5}\\)\n= \\((\\frac{5}{8} \\times \\frac{5}{2}) - (\\frac{3}{4} \\div \\frac{3}{5})\\)\n= \\((\\frac{25}{16}) - (\\frac{5}{4})\\)\n= \\(\\frac{25 - 20}{16}\\)\n= \\(\\frac{5}{16}\\)\n(b)\n\\(V = \\frac{1}{3} \\times A \\times h = \\frac{1}{3} \\pi r^{2} h\\)\n\\(\\therefore \\frac{1}{3} \\times 154 \\times h = \\frac{1}{3} \\times \\pi \\times r^{2} \\times h\\)\nComparing the two equations,\n\\(154 = \\frac{22}{7} \\times r^{2}\\)\n\\(r^{2} = \\frac{154 \\times 7}{22}\\)\n\\(r^{2} = 49\\)\n\\(\\therefore r = 7 cm\\).\n\n', '', 'Explanation\n(a) \\(\\frac{5}{8} of 2\\frac{1}{2} - \\frac{3}{4} \\div \\frac{3}{5}\\)\n= \\((\\frac{5}{8} \\times \\frac{5}{2}) - (\\frac{3}{4} \\div \\frac{3}{5})\\)\n= \\((\\frac{25}{16}) - (\\frac{5}{4})\\)\n= \\(\\frac{25 - 20}{16}\\)\n= \\(\\frac{5}{16}\\)\n(b)\n\\(V = \\frac{1}{3} \\times A \\times h = \\frac{1}{3} \\pi r^{2} h\\)\n\\(\\therefore \\frac{1}{3} \\times 154 \\times h = \\frac{1}{3} \\times \\pi \\times r^{2} \\times h\\)\nComparing the two equations,\n\\(154 = \\frac{22}{7} \\times r^{2}\\)\n\\(r^{2} = \\frac{154 \\times 7}{22}\\)\n\\(r^{2} = 49\\)\n\\(\\therefore r = 7 cm\\).\n\n', '(a) \\(\\frac{5}{8} of 2\\frac{1}{2} - \\frac{3}{4} \\div \\frac{3}{5}\\)\n= \\((\\frac{5}{8} \\times \\frac{5}{2}) - (\\frac{3}{4} \\div \\frac{3}{5})\\)\n= \\((\\frac{25}{16}) - (\\frac{5}{4})\\)\n= \\(\\frac{25 - 20}{16}\\)\n= \\(\\frac{5}{16}\\)~(b)\n\\(V = \\frac{1}{3} \\times A \\times h = \\frac{1}{3} \\pi r^{2} h\\)\n\\(\\therefore \\frac{1}{3} \\times 154 \\times h = \\frac{1}{3} \\times \\pi \\times r^{2} \\times h\\)\nComparing the two equations,\n\\(154 = \\frac{22}{7} \\times r^{2}\\)\n\\(r^{2} = \\frac{154 \\times 7}{22}\\)\n\\(r^{2} = 49\\)\n\\(\\therefore r = 7 cm\\).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(489, '2014', '\n\nQuestion\n\n(a) Two fair die are thrown once. Find the probabitlity of getting : (i) the same digit ; (ii) a total score greater than 5.\n(b) Given that \\(x = \\cos 30°\\) and \\(y = \\sin 30°\\), evaluate without using a mathematical table or calculator : \\(\\frac{x^{2} + y^{2}}{y^{2} - x^{2}}\\).\n\n', '', '(a) Two fair die are thrown once. Find the probabitlity of getting : (i) the same digit ; (ii) a total score greater than 5.\n(b) Given that \\(x = \\cos 30°\\) and \\(y = \\sin 30°\\), evaluate without using a mathematical table or calculator : \\(\\frac{x^{2} + y^{2}}{y^{2} - x^{2}}\\).', '(a) Two fair die are thrown once. Find the probabitlity of getting : (i) the same digit ; (ii) a total score greater than 5.~(b) Given that \\(x = \\cos 30°\\) and \\(y = \\sin 30°\\), evaluate without using a mathematical table or calculator : \\(\\frac{x^{2} + y^{2}}{y^{2} - x^{2}}\\).', NULL, '\n\nExplanation\n(a) Sample - space\n\n\n\n\n1\n2\n3\n4\n5\n6\n\n\n1\n1, 1\n1, 2\n1, 3\n1, 4\n1, 5\n1, 6\n\n\n2\n2, 1\n2, 2\n2, 3\n2, 4\n2, 5\n2, 6\n\n\n3\n3, 1\n3, 2\n3, 3\n3, 4\n3, 5\n3, 6\n\n\n4\n4, 1\n4, 2\n4, 3\n4, 4\n4, 5\n4, 6\n\n\n5\n5, 1\n5, 2\n5, 3\n5, 4\n5, 5\n5, 6\n\n\n6\n6, 1\n6, 2\n6, 3\n6, 4\n6, 5\n6, 6\n\n\n\n(i) Same digit = {(1,1), (2,2),(3,3), (4,4),(5,5), (6,6)}\nP(same digit) = \\(\\frac{6}{36}\\)\n(ii) P(total score greater than 5) = \\(\\frac{26}{36}\\)\n= \\(\\frac{13}{18}\\)\n(b) \\(\\frac{x^{2} + y^{2}}{y^{2} - x^{2}} = \\frac{(\\cos 30)^{2} + (\\sin 30)^{2}}{(\\sin 30)^{2} - (\\cos 30)^{2}}\\)\n= \\(\\frac{(\\frac{\\sqrt{3}}{2})^{2} + (\\frac{1}{2})^{2}}{(\\frac{1}{2})^{2} - (\\frac{\\sqrt{3}}{2})^{2}}\\)\n= \\(\\frac{\\frac{3}{4} + \\frac{1}{4}}{\\frac{1}{4} - \\frac{3}{4}}\\)\n= \\(\\frac{1}{-\\frac{1}{2}}\\)\n= \\(-2\\).\n\n', '', 'Explanation\n(a) Sample - space\n\n\n\n\n1\n2\n3\n4\n5\n6\n\n\n1\n1, 1\n1, 2\n1, 3\n1, 4\n1, 5\n1, 6\n\n\n2\n2, 1\n2, 2\n2, 3\n2, 4\n2, 5\n2, 6\n\n\n3\n3, 1\n3, 2\n3, 3\n3, 4\n3, 5\n3, 6\n\n\n4\n4, 1\n4, 2\n4, 3\n4, 4\n4, 5\n4, 6\n\n\n5\n5, 1\n5, 2\n5, 3\n5, 4\n5, 5\n5, 6\n\n\n6\n6, 1\n6, 2\n6, 3\n6, 4\n6, 5\n6, 6\n\n\n\n(i) Same digit = {(1,1), (2,2),(3,3), (4,4),(5,5), (6,6)}\nP(same digit) = \\(\\frac{6}{36}\\)\n(ii) P(total score greater than 5) = \\(\\frac{26}{36}\\)\n= \\(\\frac{13}{18}\\)\n(b) \\(\\frac{x^{2} + y^{2}}{y^{2} - x^{2}} = \\frac{(\\cos 30)^{2} + (\\sin 30)^{2}}{(\\sin 30)^{2} - (\\cos 30)^{2}}\\)\n= \\(\\frac{(\\frac{\\sqrt{3}}{2})^{2} + (\\frac{1}{2})^{2}}{(\\frac{1}{2})^{2} - (\\frac{\\sqrt{3}}{2})^{2}}\\)\n= \\(\\frac{\\frac{3}{4} + \\frac{1}{4}}{\\frac{1}{4} - \\frac{3}{4}}\\)\n= \\(\\frac{1}{-\\frac{1}{2}}\\)\n= \\(-2\\).\n\n', '(a) Sample - space\n\n\n\n\n1\n2\n3\n4\n5\n6\n\n\n1\n1, 1\n1, 2\n1, 3\n1, 4\n1, 5\n1, 6\n\n\n2\n2, 1\n2, 2\n2, 3\n2, 4\n2, 5\n2, 6\n\n\n3\n3, 1\n3, 2\n3, 3\n3, 4\n3, 5\n3, 6\n\n\n4\n4, 1\n4, 2\n4, 3\n4, 4\n4, 5\n4, 6\n\n\n5\n5, 1\n5, 2\n5, 3\n5, 4\n5, 5\n5, 6\n\n\n6\n6, 1\n6, 2\n6, 3\n6, 4\n6, 5\n6, 6\n\n\n\n(i) Same digit = {(1,1), (2,2),(3,3), (4,4),(5,5), (6,6)}\nP(same digit) = \\(\\frac{6}{36}\\)\n(ii) P(total score greater than 5) = \\(\\frac{26}{36}\\)\n= \\(\\frac{13}{18}\\)~(b) \\(\\frac{x^{2} + y^{2}}{y^{2} - x^{2}} = \\frac{(\\cos 30)^{2} + (\\sin 30)^{2}}{(\\sin 30)^{2} - (\\cos 30)^{2}}\\)\n= \\(\\frac{(\\frac{\\sqrt{3}}{2})^{2} + (\\frac{1}{2})^{2}}{(\\frac{1}{2})^{2} - (\\frac{\\sqrt{3}}{2})^{2}}\\)\n= \\(\\frac{\\frac{3}{4} + \\frac{1}{4}}{\\frac{1}{4} - \\frac{3}{4}}\\)\n= \\(\\frac{1}{-\\frac{1}{2}}\\)\n= \\(-2\\).\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
(490, '2014', '\n\nQuestion\n\n(a)\nIn the diagram, A, B, C and D are points on the circumference of a circle. XY is a tangent at A. Find : (i) < CAX ; (ii) < ABY.\n(b) If (m + 1) and (m - 3) are factors of \\(m^{2} - km + c\\), find the values of k and c.\n\n', '', '(a)\nIn the diagram, A, B, C and D are points on the circumference of a circle. XY is a tangent at A. Find : (i) < CAX ; (ii) < ABY.\n(b) If (m + 1) and (m - 3) are factors of \\(m^{2} - km + c\\), find the values of k and c.', '(a)\nIn the diagram, A, B, C and D are points on the circumference of a circle. XY is a tangent at A. Find : (i) < CAX ; (ii) < ABY.~(b) If (m + 1) and (m - 3) are factors of \\(m^{2} - km + c\\), find the values of k and c.', NULL, '\n\nExplanation\n(a) \\(< ADB = < ACB = 20°\\) (Angle in the same segment).\nIn \\(\\Delta ACY\\),\n\\(\\hat{C} = < ACB\\)\n(i) \\(\\hat{C} + \\hat{A} + \\hat{Y} = 180°\\)\n\\(20° + \\hat{A} + 60° = 180°\\)\n\\(\\hat{A} = 180° - 80° = 100°\\)\n\\(< CAY = \\hat{A} = 100°\\)\n\\(< CAX = 180° - 100° = 80°\\)\n(ii) In \\(\\Delta ACB\\),\n\\(\\hat{B} = \\hat{C} = \\frac{180° - 20°}{2} \\)\n= \\(\\frac{160°}{2}\\)\n= \\(80°\\)\nIn \\(\\Delta ABY\\),\n\\(< ABY = \\hat{B} = 180° - 80°\\)\n= \\(100°\\)\n(b) \\((m +1)(m - 3) \\equiv m^{2} - km + c\\)\n\\(m^{2} - 3m + m - 3 = m^{2} - 2m - 3\\)\n\\(m^{2} - 2m - 3 \\equiv m^{2} - km + c\\)\n\\(\\implies k = 2 ; c = -3\\)\n\n', '', 'Explanation\n(a) \\(< ADB = < ACB = 20°\\) (Angle in the same segment).\nIn \\(\\Delta ACY\\),\n\\(\\hat{C} = < ACB\\)\n(i) \\(\\hat{C} + \\hat{A} + \\hat{Y} = 180°\\)\n\\(20° + \\hat{A} + 60° = 180°\\)\n\\(\\hat{A} = 180° - 80° = 100°\\)\n\\(< CAY = \\hat{A} = 100°\\)\n\\(< CAX = 180° - 100° = 80°\\)\n(ii) In \\(\\Delta ACB\\),\n\\(\\hat{B} = \\hat{C} = \\frac{180° - 20°}{2} \\)\n= \\(\\frac{160°}{2}\\)\n= \\(80°\\)\nIn \\(\\Delta ABY\\),\n\\(< ABY = \\hat{B} = 180° - 80°\\)\n= \\(100°\\)\n(b) \\((m +1)(m - 3) \\equiv m^{2} - km + c\\)\n\\(m^{2} - 3m + m - 3 = m^{2} - 2m - 3\\)\n\\(m^{2} - 2m - 3 \\equiv m^{2} - km + c\\)\n\\(\\implies k = 2 ; c = -3\\)\n\n', '(a) \\(< ADB = < ACB = 20°\\) (Angle in the same segment).\nIn \\(\\Delta ACY\\),\n\\(\\hat{C} = < ACB\\)\n(i) \\(\\hat{C} + \\hat{A} + \\hat{Y} = 180°\\)\n\\(20° + \\hat{A} + 60° = 180°\\)\n\\(\\hat{A} = 180° - 80° = 100°\\)\n\\(< CAY = \\hat{A} = 100°\\)\n\\(< CAX = 180° - 100° = 80°\\)\n(ii) In \\(\\Delta ACB\\),\n\\(\\hat{B} = \\hat{C} = \\frac{180° - 20°}{2} \\)\n= \\(\\frac{160°}{2}\\)\n= \\(80°\\)\nIn \\(\\Delta ABY\\),\n\\(< ABY = \\hat{B} = 180° - 80°\\)\n= \\(100°\\)~(b) \\((m +1)(m - 3) \\equiv m^{2} - km + c\\)\n\\(m^{2} - 3m + m - 3 = m^{2} - 2m - 3\\)\n\\(m^{2} - 2m - 3 \\equiv m^{2} - km + c\\)\n\\(\\implies k = 2 ; c = -3\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(491, '2014', '\n\nQuestion\n\n(a) Without using mathematical table or calculator, evaluate : \\(\\sqrt{\\frac{0.18 \\times 12.5}{0.05 \\times 0.2}}\\).\n(b) Simplify : \\(\\frac{8 - 4\\sqrt{18}}{\\sqrt{50}}\\).\n(c) x, y and z are related such that x varies directly as the cube of y and inversely as the square of z. If x = 108 when y = 3 and z = 4, find z when x = 4000 and y = 10.\n\n', '', '(a) Without using mathematical table or calculator, evaluate : \\(\\sqrt{\\frac{0.18 \\times 12.5}{0.05 \\times 0.2}}\\).\n(b) Simplify : \\(\\frac{8 - 4\\sqrt{18}}{\\sqrt{50}}\\).\n(c) x, y and z are related such that x varies directly as the cube of y and inversely as the square of z. If x = 108 when y = 3 and z = 4, find z when x = 4000 and y = 10.', '(a) Without using mathematical table or calculator, evaluate : \\(\\sqrt{\\frac{0.18 \\times 12.5}{0.05 \\times 0.2}}\\).~(b) Simplify : \\(\\frac{8 - 4\\sqrt{18}}{\\sqrt{50}}\\).\n(c) x, y and z are related such that x varies directly as the cube of y and inversely as the square of z. If x = 108 when y = 3 and z = 4, find z when x = 4000 and y = 10.~(c) x, y and z are related such that x varies directly as the cube of y and inversely as the square of z. If x = 108 when y = 3 and z = 4, find z when x = 4000 and y = 10.', NULL, '\n\nExplanation\n(a) \\(\\sqrt{\\frac{0.18 \\times 12.5}{0.05 \\times 0.2}}\\)\n\\(\\frac{0.18 \\times 12.5}{0.05 \\times 0.2} = \\frac{18 \\times 10^{-2} \\times 125 \\times 10^{-1}}{5 \\times 10^{-2} \\times 2 \\times 10^{-1}}\\)\n= \\(9 \\times 25 \\times 10^{-3 - (-3)}\\)\n= \\(9 \\times 25\\)\n= \\(225\\)\n\\(\\therefore \\sqrt{\\frac{0.18 \\times 12.5}{0.05 \\times 0.2}} = \\sqrt{225}\\)\n= \\(15\\).\n(b) \\(\\frac{8 - 4\\sqrt{18}}{\\sqrt{50}}\\)\n\\(\\sqrt{18} = \\sqrt{9 \\times 2} = 3\\sqrt{2}\\)\n\\(\\sqrt{50} = \\sqrt{25 \\times 2} = 5\\sqrt{2}\\)\n\\(\\frac{8 - 4(3\\sqrt{2})}{5\\sqrt{2}} = \\frac{8 - 12\\sqrt{2}}{5\\sqrt{2}}\\)\nRationalising, we have\n= \\(\\frac{8 - 12\\sqrt{2}}{5\\sqrt{2}} \\times \\frac{\\sqrt{2}}{\\sqrt{2}}\\)\n= \\(\\frac{8\\sqrt{2} - 24}{10}\\)\n= \\(0.8\\sqrt{2} - 2.4\\)\n= \\(-2.4 + 0.8\\sqrt{2}\\)\n(c) \\(x \\propto \\frac{y^{3}}{z^{2}}\\)\n\\(\\implies x = \\frac{ky^{3}}{z^{2}}\\)\n\\(108 = \\frac{k \\times 3^{3}}{4^{2}}\\)\n\\(k = \\frac{108 \\times 16}{27}\\)\n\\(k = 64\\)\n\\(\\therefore x = \\frac{64y^{3}}{z^{2}}\\)\n\\(\\therefore 4000 = \\frac{64 \\times 10^{3}}{z^{2}}\\)\n\\(4 = \\frac{64}{z^{2}}\\)\n\\(z^{2} = \\frac{64}{4} = 16\\)\n\\(z = \\sqrt{16} = \\pm 4\\)\n\n', '', 'Explanation\n(a) \\(\\sqrt{\\frac{0.18 \\times 12.5}{0.05 \\times 0.2}}\\)\n\\(\\frac{0.18 \\times 12.5}{0.05 \\times 0.2} = \\frac{18 \\times 10^{-2} \\times 125 \\times 10^{-1}}{5 \\times 10^{-2} \\times 2 \\times 10^{-1}}\\)\n= \\(9 \\times 25 \\times 10^{-3 - (-3)}\\)\n= \\(9 \\times 25\\)\n= \\(225\\)\n\\(\\therefore \\sqrt{\\frac{0.18 \\times 12.5}{0.05 \\times 0.2}} = \\sqrt{225}\\)\n= \\(15\\).\n(b) \\(\\frac{8 - 4\\sqrt{18}}{\\sqrt{50}}\\)\n\\(\\sqrt{18} = \\sqrt{9 \\times 2} = 3\\sqrt{2}\\)\n\\(\\sqrt{50} = \\sqrt{25 \\times 2} = 5\\sqrt{2}\\)\n\\(\\frac{8 - 4(3\\sqrt{2})}{5\\sqrt{2}} = \\frac{8 - 12\\sqrt{2}}{5\\sqrt{2}}\\)\nRationalising, we have\n= \\(\\frac{8 - 12\\sqrt{2}}{5\\sqrt{2}} \\times \\frac{\\sqrt{2}}{\\sqrt{2}}\\)\n= \\(\\frac{8\\sqrt{2} - 24}{10}\\)\n= \\(0.8\\sqrt{2} - 2.4\\)\n= \\(-2.4 + 0.8\\sqrt{2}\\)\n(c) \\(x \\propto \\frac{y^{3}}{z^{2}}\\)\n\\(\\implies x = \\frac{ky^{3}}{z^{2}}\\)\n\\(108 = \\frac{k \\times 3^{3}}{4^{2}}\\)\n\\(k = \\frac{108 \\times 16}{27}\\)\n\\(k = 64\\)\n\\(\\therefore x = \\frac{64y^{3}}{z^{2}}\\)\n\\(\\therefore 4000 = \\frac{64 \\times 10^{3}}{z^{2}}\\)\n\\(4 = \\frac{64}{z^{2}}\\)\n\\(z^{2} = \\frac{64}{4} = 16\\)\n\\(z = \\sqrt{16} = \\pm 4\\)\n\n', '(a) \\(\\sqrt{\\frac{0.18 \\times 12.5}{0.05 \\times 0.2}}\\)\n\\(\\frac{0.18 \\times 12.5}{0.05 \\times 0.2} = \\frac{18 \\times 10^{-2} \\times 125 \\times 10^{-1}}{5 \\times 10^{-2} \\times 2 \\times 10^{-1}}\\)\n= \\(9 \\times 25 \\times 10^{-3 - (-3)}\\)\n= \\(9 \\times 25\\)\n= \\(225\\)\n\\(\\therefore \\sqrt{\\frac{0.18 \\times 12.5}{0.05 \\times 0.2}} = \\sqrt{225}\\)\n= \\(15\\).~(b) \\(\\frac{8 - 4\\sqrt{18}}{\\sqrt{50}}\\)\n\\(\\sqrt{18} = \\sqrt{9 \\times 2} = 3\\sqrt{2}\\)\n\\(\\sqrt{50} = \\sqrt{25 \\times 2} = 5\\sqrt{2}\\)\n\\(\\frac{8 - 4(3\\sqrt{2})}{5\\sqrt{2}} = \\frac{8 - 12\\sqrt{2}}{5\\sqrt{2}}\\)\nRationalising, we have\n= \\(\\frac{8 - 12\\sqrt{2}}{5\\sqrt{2}} \\times \\frac{\\sqrt{2}}{\\sqrt{2}}\\)\n= \\(\\frac{8\\sqrt{2} - 24}{10}\\)\n= \\(0.8\\sqrt{2} - 2.4\\)\n= \\(-2.4 + 0.8\\sqrt{2}\\)\n(c) \\(x \\propto \\frac{y^{3}}{z^{2}}\\)\n\\(\\implies x = \\frac{ky^{3}}{z^{2}}\\)\n\\(108 = \\frac{k \\times 3^{3}}{4^{2}}\\)\n\\(k = \\frac{108 \\times 16}{27}\\)\n\\(k = 64\\)\n\\(\\therefore x = \\frac{64y^{3}}{z^{2}}\\)\n\\(\\therefore 4000 = \\frac{64 \\times 10^{3}}{z^{2}}\\)\n\\(4 = \\frac{64}{z^{2}}\\)\n\\(z^{2} = \\frac{64}{4} = 16\\)\n\\(z = \\sqrt{16} = \\pm 4\\)\n~(c) \\(x \\propto \\frac{y^{3}}{z^{2}}\\)\n\\(\\implies x = \\frac{ky^{3}}{z^{2}}\\)\n\\(108 = \\frac{k \\times 3^{3}}{4^{2}}\\)\n\\(k = \\frac{108 \\times 16}{27}\\)\n\\(k = 64\\)\n\\(\\therefore x = \\frac{64y^{3}}{z^{2}}\\)\n\\(\\therefore 4000 = \\frac{64 \\times 10^{3}}{z^{2}}\\)\n\\(4 = \\frac{64}{z^{2}}\\)\n\\(z^{2} = \\frac{64}{4} = 16\\)\n\\(z = \\sqrt{16} = \\pm 4\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(492, '2014', '\n\nQuestion\n\n(a) A regular polygon of n sides is such that each interior angle is 120° greater than the exterior angle. Find :\n(i) the value of n ; (ii) the sum of all the interior angles.\n(b) A boy walks 6km from a point P to a point Q on a bearing of 065°. He then walks to a point R, a distance of 13km, on a bearing of 146°.\n(i) Sketch the diagram of his movement. (ii) Calculate, correct to the nearest kilometre, the distance PR.\n\n', '', '(a) A regular polygon of n sides is such that each interior angle is 120° greater than the exterior angle. Find :\n(i) the value of n ; (ii) the sum of all the interior angles.\n(b) A boy walks 6km from a point P to a point Q on a bearing of 065°. He then walks to a point R, a distance of 13km, on a bearing of 146°.\n(i) Sketch the diagram of his movement. (ii) Calculate, correct to the nearest kilometre, the distance PR.', '(a) A regular polygon of n sides is such that each interior angle is 120° greater than the exterior angle. Find :\n(i) the value of n ; (ii) the sum of all the interior angles.~(b) A boy walks 6km from a point P to a point Q on a bearing of 065°. He then walks to a point R, a distance of 13km, on a bearing of 146°.\n(i) Sketch the diagram of his movement. (ii) Calculate, correct to the nearest kilometre, the distance PR.', NULL, '\n\nExplanation\n(a) (i) Let the exterior angle be x\nLet the interior angle be x + 120°.\nExterior angle + Interior angle = 180°\n\\(\\therefore x + x + 120 = 180 \\implies 2x + 120 = 180\\)\n\\(2x = 60 \\implies x = 30°\\)\nNumber of sides = \\(\\frac{360}{\\text{exterior angle}}\\)\n= \\(\\frac{360}{30} = 12 sides\\)\n(ii) Sum of interior angle = \\((2n - 4) \\times 90°\\)\n= \\((2(12) - 4) \\times 90°\\)\n= \\(20 \\times 90\\)\n= \\(1800°\\)\n(b) Note : < PQR = 65° + 34° = 99°\\)\n\n\\(|PR|^{2} = |PQ|^{2} + |QR|^{2} - 2(|QR|)(|PQ|) \\cos 99\\)\n\\(|PR|^{2} = 6^{2} + 13^{2} - 2(6)(13) \\cos 99\\)\n\\(|PR|^{2} = 205 - 156 \\cos 99\\)\n\\(|PR|^{2} = 205 + 24.40 = 229.40\\)\n\\(|PR| = \\sqrt{229.40}\\)\n= \\(15.146 km\\)\n\\(\\approxeq 15 km\\)\n\n', '', 'Explanation\n(a) (i) Let the exterior angle be x\nLet the interior angle be x + 120°.\nExterior angle + Interior angle = 180°\n\\(\\therefore x + x + 120 = 180 \\implies 2x + 120 = 180\\)\n\\(2x = 60 \\implies x = 30°\\)\nNumber of sides = \\(\\frac{360}{\\text{exterior angle}}\\)\n= \\(\\frac{360}{30} = 12 sides\\)\n(ii) Sum of interior angle = \\((2n - 4) \\times 90°\\)\n= \\((2(12) - 4) \\times 90°\\)\n= \\(20 \\times 90\\)\n= \\(1800°\\)\n(b) Note : < PQR = 65° + 34° = 99°\\)\n\n\\(|PR|^{2} = |PQ|^{2} + |QR|^{2} - 2(|QR|)(|PQ|) \\cos 99\\)\n\\(|PR|^{2} = 6^{2} + 13^{2} - 2(6)(13) \\cos 99\\)\n\\(|PR|^{2} = 205 - 156 \\cos 99\\)\n\\(|PR|^{2} = 205 + 24.40 = 229.40\\)\n\\(|PR| = \\sqrt{229.40}\\)\n= \\(15.146 km\\)\n\\(\\approxeq 15 km\\)\n\n', '(a) (i) Let the exterior angle be x\nLet the interior angle be x + 120°.\nExterior angle + Interior angle = 180°\n\\(\\therefore x + x + 120 = 180 \\implies 2x + 120 = 180\\)\n\\(2x = 60 \\implies x = 30°\\)\nNumber of sides = \\(\\frac{360}{\\text{exterior angle}}\\)\n= \\(\\frac{360}{30} = 12 sides\\)\n(ii) Sum of interior angle = \\((2n - 4) \\times 90°\\)\n= \\((2(12) - 4) \\times 90°\\)\n= \\(20 \\times 90\\)\n= \\(1800°\\)~(b) Note : < PQR = 65° + 34° = 99°\\)\n\n\\(|PR|^{2} = |PQ|^{2} + |QR|^{2} - 2(|QR|)(|PQ|) \\cos 99\\)\n\\(|PR|^{2} = 6^{2} + 13^{2} - 2(6)(13) \\cos 99\\)\n\\(|PR|^{2} = 205 - 156 \\cos 99\\)\n\\(|PR|^{2} = 205 + 24.40 = 229.40\\)\n\\(|PR| = \\sqrt{229.40}\\)\n= \\(15.146 km\\)\n\\(\\approxeq 15 km\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(493, '2014', '\n\nQuestion\n\n(a) A plane flies due East from A(lat. 53°N, long. 25°E) to a point B(lat. 53°N, long. 85°E) at an average speed of 400 km/h. The plane then flies South from B to a point C 2000km away. Calculate, correct to the nearest whole number :\n(a) the distance between A and B.\n(b) the time the plane takes to reach point B ;\n(c) the latitude of C.\n[Take radius of the earth = 6400km; \\(\\pi = \\frac{22}{7}\\)].\n\n', '', '(a) A plane flies due East from A(lat. 53°N, long. 25°E) to a point B(lat. 53°N, long. 85°E) at an average speed of 400 km/h. The plane then flies South from B to a point C 2000km away. Calculate, correct to the nearest whole number :\n(a) the distance between A and B.\n(b) the time the plane takes to reach point B ;\n(c) the latitude of C.\n[Take radius of the earth = 6400km; \\(\\pi = \\frac{22}{7}\\)].', '(a) A plane flies due East from A(lat. 53°N, long. 25°E) to a point B(lat. 53°N, long. 85°E) at an average speed of 400 km/h. The plane then flies South from B to a point C 2000km away. Calculate, correct to the nearest whole number :\n(a) the distance between A and B.~(b) the time the plane takes to reach point B ;\n(c) the latitude of C.\n[Take radius of the earth = 6400km; \\(\\pi = \\frac{22}{7}\\)].~(c) the latitude of C.\n[Take radius of the earth = 6400km; \\(\\pi = \\frac{22}{7}\\)].', NULL, '\n\nExplanation\n(a)\nLongitude difference = 85° - 25° = 60°.\nDistance AB along the parallel of latitude = \\(\\frac{\\theta}{360°} \\times 2\\pi R \\cos \\alpha\\)\n\\(AB = \\frac{60}{360} \\times 2 \\times \\frac{22}{7} \\times 6400 \\cos 53\\)\n= \\(\\frac{1}{6} \\times \\frac{44}{7} \\times 3851.62\\)\n= \\(4,035.03 km\\)\n\\(\\approxeq 4035 km\\)\n(b) \\(Speed = \\frac{Distance}{Time}\\)\n\\(\\therefore Time = \\frac{Distance}{Speed}\\)\n= \\(\\frac{4035}{400}\\)\n\\(\\approxeq 10hours\\).\n(c) Distance BC measured along the meridian\n\\(BC = \\frac{\\theta}{360} \\times 2 \\pi R\\)\n\\(2000 = \\frac{\\theta}{360} \\times 2 \\times \\frac{22}{7} \\times 6400\\)\n\\(\\theta = 17.898° \\approxeq 18°\\)\n\\(\\theta\\) = Latitude difference\nLet the latitude of C = x.\n\\(\\theta = 53 - x\\)\n\\(18 = 53 - x\\)\n\\(x = 35°\\)\n\n', '', 'Explanation\n(a)\nLongitude difference = 85° - 25° = 60°.\nDistance AB along the parallel of latitude = \\(\\frac{\\theta}{360°} \\times 2\\pi R \\cos \\alpha\\)\n\\(AB = \\frac{60}{360} \\times 2 \\times \\frac{22}{7} \\times 6400 \\cos 53\\)\n= \\(\\frac{1}{6} \\times \\frac{44}{7} \\times 3851.62\\)\n= \\(4,035.03 km\\)\n\\(\\approxeq 4035 km\\)\n(b) \\(Speed = \\frac{Distance}{Time}\\)\n\\(\\therefore Time = \\frac{Distance}{Speed}\\)\n= \\(\\frac{4035}{400}\\)\n\\(\\approxeq 10hours\\).\n(c) Distance BC measured along the meridian\n\\(BC = \\frac{\\theta}{360} \\times 2 \\pi R\\)\n\\(2000 = \\frac{\\theta}{360} \\times 2 \\times \\frac{22}{7} \\times 6400\\)\n\\(\\theta = 17.898° \\approxeq 18°\\)\n\\(\\theta\\) = Latitude difference\nLet the latitude of C = x.\n\\(\\theta = 53 - x\\)\n\\(18 = 53 - x\\)\n\\(x = 35°\\)\n\n', '(a)\nLongitude difference = 85° - 25° = 60°.\nDistance AB along the parallel of latitude = \\(\\frac{\\theta}{360°} \\times 2\\pi R \\cos \\alpha\\)\n\\(AB = \\frac{60}{360} \\times 2 \\times \\frac{22}{7} \\times 6400 \\cos 53\\)\n= \\(\\frac{1}{6} \\times \\frac{44}{7} \\times 3851.62\\)\n= \\(4,035.03 km\\)\n\\(\\approxeq 4035 km\\)~(b) \\(Speed = \\frac{Distance}{Time}\\)\n\\(\\therefore Time = \\frac{Distance}{Speed}\\)\n= \\(\\frac{4035}{400}\\)\n\\(\\approxeq 10hours\\).\n(c) Distance BC measured along the meridian\n\\(BC = \\frac{\\theta}{360} \\times 2 \\pi R\\)\n\\(2000 = \\frac{\\theta}{360} \\times 2 \\times \\frac{22}{7} \\times 6400\\)\n\\(\\theta = 17.898° \\approxeq 18°\\)\n\\(\\theta\\) = Latitude difference\nLet the latitude of C = x.\n\\(\\theta = 53 - x\\)\n\\(18 = 53 - x\\)\n\\(x = 35°\\)\n~(c) Distance BC measured along the meridian\n\\(BC = \\frac{\\theta}{360} \\times 2 \\pi R\\)\n\\(2000 = \\frac{\\theta}{360} \\times 2 \\times \\frac{22}{7} \\times 6400\\)\n\\(\\theta = 17.898° \\approxeq 18°\\)\n\\(\\theta\\) = Latitude difference\nLet the latitude of C = x.\n\\(\\theta = 53 - x\\)\n\\(18 = 53 - x\\)\n\\(x = 35°\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(494, '2014', '\n\nQuestion\n\n(a) Find the smallest integer that satisfies the inequality \\(x + 8 < 4x - 15\\).\n(b) A sales girl is paid a monthly salary of N2,500 in addition to a commission of 5 kobo in the naira on all sales made by her during the month. If her sales for a month amounts to N200,000.00, calculate her income for that month.\n(c)The diagram shows a window consisting of a rectangular and semi- circular parts. The radius of the semi- circular part is 35 cm and the height of the rectangular part is 50 cm. Find the area of the window. [Take \\(\\pi = \\frac{22}{7}\\)].\n\n', '', '(a) Find the smallest integer that satisfies the inequality \\(x + 8 < 4x - 15\\).\n(b) A sales girl is paid a monthly salary of N2,500 in addition to a commission of 5 kobo in the naira on all sales made by her during the month. If her sales for a month amounts to N200,000.00, calculate her income for that month.\n(c)The diagram shows a window consisting of a rectangular and semi- circular parts. The radius of the semi- circular part is 35 cm and the height of the rectangular part is 50 cm. Find the area of the window. [Take \\(\\pi = \\frac{22}{7}\\)].', '(a) Find the smallest integer that satisfies the inequality \\(x + 8 < 4x - 15\\).~(b) A sales girl is paid a monthly salary of N2,500 in addition to a commission of 5 kobo in the naira on all sales made by her during the month. If her sales for a month amounts to N200,000.00, calculate her income for that month.\n(c)The diagram shows a window consisting of a rectangular and semi- circular parts. The radius of the semi- circular part is 35 cm and the height of the rectangular part is 50 cm. Find the area of the window. [Take \\(\\pi = \\frac{22}{7}\\)].~(c)The diagram shows a window consisting of a rectangular and semi- circular parts. The radius of the semi- circular part is 35 cm and the height of the rectangular part is 50 cm. Find the area of the window. [Take \\(\\pi = \\frac{22}{7}\\)].', NULL, '\n\nExplanation\n(a) \\(x + 8 < 4x - 15 \\implies 4x - 15 > x + 8\\)\n\\(4x - x > 8 + 15 \\implies 3x > 23\\)\n\\(x > \\frac{23}{3}\\)\n\\(x > 7\\frac{2}{3}\\).\n(b) Extra income = 5k of N200,000\n= \\(\\frac{5}{100} \\times N200,000\\)\n= \\(N10,000\\)\nTotal income = Monthly income + commission\n= N(2,500 + 10,000)\n= N12,500.\n(c) Area of semi- circle = \\(\\frac{\\pi r^{2}}{2}\\)\n= \\(\\frac{1}{2} \\times \\frac{22}{7} \\times 35 \\times 35\\)\n= \\(1925 cm^{2}\\)\nLength of rectangular part = 50 cm\nBreadth = Diameter of semi- circle = 2(35 cm) = 70 cm\nArea of rectangular part = \\(length \\times \\breadth\\)\n= \\(50 \\times 70\\)\n= \\(3500 cm^{2}\\)\nTotal area of window = \\((3500 + 1925) cm^{2}\\)\n= \\(5425 cm^{2}\\)\n\n', '', 'Explanation\n(a) \\(x + 8 < 4x - 15 \\implies 4x - 15 > x + 8\\)\n\\(4x - x > 8 + 15 \\implies 3x > 23\\)\n\\(x > \\frac{23}{3}\\)\n\\(x > 7\\frac{2}{3}\\).\n(b) Extra income = 5k of N200,000\n= \\(\\frac{5}{100} \\times N200,000\\)\n= \\(N10,000\\)\nTotal income = Monthly income + commission\n= N(2,500 + 10,000)\n= N12,500.\n(c) Area of semi- circle = \\(\\frac{\\pi r^{2}}{2}\\)\n= \\(\\frac{1}{2} \\times \\frac{22}{7} \\times 35 \\times 35\\)\n= \\(1925 cm^{2}\\)\nLength of rectangular part = 50 cm\nBreadth = Diameter of semi- circle = 2(35 cm) = 70 cm\nArea of rectangular part = \\(length \\times \\breadth\\)\n= \\(50 \\times 70\\)\n= \\(3500 cm^{2}\\)\nTotal area of window = \\((3500 + 1925) cm^{2}\\)\n= \\(5425 cm^{2}\\)\n\n', '(a) \\(x + 8 < 4x - 15 \\implies 4x - 15 > x + 8\\)\n\\(4x - x > 8 + 15 \\implies 3x > 23\\)\n\\(x > \\frac{23}{3}\\)\n\\(x > 7\\frac{2}{3}\\).~(b) Extra income = 5k of N200,000\n= \\(\\frac{5}{100} \\times N200,000\\)\n= \\(N10,000\\)\nTotal income = Monthly income + commission\n= N(2,500 + 10,000)\n= N12,500.\n(c) Area of semi- circle = \\(\\frac{\\pi r^{2}}{2}\\)\n= \\(\\frac{1}{2} \\times \\frac{22}{7} \\times 35 \\times 35\\)\n= \\(1925 cm^{2}\\)\nLength of rectangular part = 50 cm\nBreadth = Diameter of semi- circle = 2(35 cm) = 70 cm\nArea of rectangular part = \\(length \\times \\breadth\\)\n= \\(50 \\times 70\\)\n= \\(3500 cm^{2}\\)\nTotal area of window = \\((3500 + 1925) cm^{2}\\)\n= \\(5425 cm^{2}\\)\n~(c) Area of semi- circle = \\(\\frac{\\pi r^{2}}{2}\\)\n= \\(\\frac{1}{2} \\times \\frac{22}{7} \\times 35 \\times 35\\)\n= \\(1925 cm^{2}\\)\nLength of rectangular part = 50 cm\nBreadth = Diameter of semi- circle = 2(35 cm) = 70 cm\nArea of rectangular part = \\(length \\times \\breadth\\)\n= \\(50 \\times 70\\)\n= \\(3500 cm^{2}\\)\nTotal area of window = \\((3500 + 1925) cm^{2}\\)\n= \\(5425 cm^{2}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(495, '2014', '\n\nQuestion\n\nThe sketch shows a plot of land .\n(a) Using a scale of 1 cm to 10m, draw an accurate diagram of the plot ;\n(b) Construct : (i) The locus \\(l_{1}\\) of points equidistant from AC and BC ; (ii) the locus \\(l_{2}\\) of points 60m from A.\n(c) A tree T inside the plot is on both \\(l_{1}\\) and \\(l_{2}\\). Locate T and find |TC| in metres.\n(d) A flagpole, P is to be placed such that it it is nearer AC than BC and more than 60m from A. Shade the regions where P can be located.\n\n\n', '', 'The sketch shows a plot of land .\n(a) Using a scale of 1 cm to 10m, draw an accurate diagram of the plot ;\n(b) Construct : (i) The locus \\(l_{1}\\) of points equidistant from AC and BC ; (ii) the locus \\(l_{2}\\) of points 60m from A.\n(c) A tree T inside the plot is on both \\(l_{1}\\) and \\(l_{2}\\). Locate T and find |TC| in metres.\n(d) A flagpole, P is to be placed such that it it is nearer AC than BC and more than 60m from A. Shade the regions where P can be located.\n', 'The sketch shows a plot of land .\n(a) Using a scale of 1 cm to 10m, draw an accurate diagram of the plot ;~(b) Construct : (i) The locus \\(l_{1}\\) of points equidistant from AC and BC ; (ii) the locus \\(l_{2}\\) of points 60m from A.\n(c) A tree T inside the plot is on both \\(l_{1}\\) and \\(l_{2}\\). Locate T and find |TC| in metres.\n(d) A flagpole, P is to be placed such that it it is nearer AC than BC and more than 60m from A. Shade the regions where P can be located.\n~(c) A tree T inside the plot is on both \\(l_{1}\\) and \\(l_{2}\\). Locate T and find |TC| in metres.\n(d) A flagpole, P is to be placed such that it it is nearer AC than BC and more than 60m from A. Shade the regions where P can be located.\n~(d) A flagpole, P is to be placed such that it it is nearer AC than BC and more than 60m from A. Shade the regions where P can be located.\n', NULL, '\n\nExplanation\n(a) \\(|AB| = 85m = \\frac{85}{10}\\)\n= \\(8.5cm\\)\n\n\\(|TC| = 6cm\\equiv 60 m\\).\n\n', '(a) \\(|AB| = 85m = \\frac{85}{10}\\)\n= \\(8.5cm\\)\n\n\\(|TC| = 6cm\\equiv 60 m\\).\n\n', '', '', 'WAEC-PK-1', 'WAEC-MATH', 1),
(496, '2014', '\n\nQuestion\n\nThe table shows the age distributions of the members of a club.\n\n\n\nAge (years)\n10-14\n15-19\n20-24\n25-29\n30-34\n35-39\n\n\nFrequency\n7\n18\n25\n17\n9\n4\n\n\n\n(a) Calculate, correct to one decimal place, the mean age.\n(b) (i) Draw a histogram to illustrate the information.\n(ii) Use the histogram to estimate the modal age .\n(c) If a member is selected at random, what is the probability that he/she is in the modal class?\n\n\n\n\nThe table shows the age distributions of the members of a club.\n\n\n\nAge (years)\n10-14\n15-19\n20-24\n25-29\n30-34\n35-39\n\n\nFrequency\n7\n18\n25\n17\n9\n4\n\n\n\n(a) Calculate, correct to one decimal place, the mean age.\n(b) (i) Draw a histogram to illustrate the information.\n(ii) Use the histogram to estimate the modal age .\n(c) If a member is selected at random, what is the probability that he/she is in the modal class?\n\n', '', 'The table shows the age distributions of the members of a club.\n\n\n\nAge (years)\n10-14\n15-19\n20-24\n25-29\n30-34\n35-39\n\n\nFrequency\n7\n18\n25\n17\n9\n4\n\n\n\n(a) Calculate, correct to one decimal place, the mean age.\n(b) (i) Draw a histogram to illustrate the information.\n(ii) Use the histogram to estimate the modal age .\n(c) If a member is selected at random, what is the probability that he/she is in the modal class?\n\n\n\n\nThe table shows the age distributions of the members of a club.\n\n\n\nAge (years)\n10-14\n15-19\n20-24\n25-29\n30-34\n35-39\n\n\nFrequency\n7\n18\n25\n17\n9\n4\n\n\n\n(a) Calculate, correct to one decimal place, the mean age.\n(b) (i) Draw a histogram to illustrate the information.\n(ii) Use the histogram to estimate the modal age .\n(c) If a member is selected at random, what is the probability that he/she is in the modal class?', 'The table shows the age distributions of the members of a club.\n\n\n\nAge (years)\n10-14\n15-19\n20-24\n25-29\n30-34\n35-39\n\n\nFrequency\n7\n18\n25\n17\n9\n4\n\n\n\n(a) Calculate, correct to one decimal place, the mean age.\n(b) (i) Draw a histogram to illustrate the information.\n(ii) Use the histogram to estimate the modal age .\n(c) If a member is selected at random, what is the probability that he/she is in the modal class?\n\n\n\n\nThe table shows the age distributions of the members of a club.\n\n\n\nAge (years)\n10-14\n15-19\n20-24\n25-29\n30-34\n35-39\n\n\nFrequency\n7\n18\n25\n17\n9\n4\n\n\n\n(a) Calculate, correct to one decimal place, the mean age.~(b) (i) Draw a histogram to illustrate the information.\n(ii) Use the histogram to estimate the modal age .\n(c) If a member is selected at random, what is the probability that he/she is in the modal class?\n\n\n\n\nThe table shows the age distributions of the members of a club.\n\n\n\nAge (years)\n10-14\n15-19\n20-24\n25-29\n30-34\n35-39\n\n\nFrequency\n7\n18\n25\n17\n9\n4\n\n\n\n(a) Calculate, correct to one decimal place, the mean age.\n(b) (i) Draw a histogram to illustrate the information.\n(ii) Use the histogram to estimate the modal age .\n(c) If a member is selected at random, what is the probability that he/she is in the modal class?~(c) If a member is selected at random, what is the probability that he/she is in the modal class?\n\n\n\n\nThe table shows the age distributions of the members of a club.\n\n\n\nAge (years)\n10-14\n15-19\n20-24\n25-29\n30-34\n35-39\n\n\nFrequency\n7\n18\n25\n17\n9\n4\n\n\n\n(a) Calculate, correct to one decimal place, the mean age.\n(b) (i) Draw a histogram to illustrate the information.\n(ii) Use the histogram to estimate the modal age .\n(c) If a member is selected at random, what is the probability that he/she is in the modal class?', NULL, '\n\nExplanation\n(a)\n\n\n\nClass Interval\nFreq (f)\nClassmark (x)\n\\(fx\\)\n\n\n10 - 14\n7\n12\n84\n\n\n15 - 19\n18\n17\n306\n\n\n20 - 24\n25\n22\n550\n\n\n25 - 29\n17\n27\n459\n\n\n30 - 34\n9\n32\n288\n\n\n35 - 39\n4\n37\n148\n\n\n\n80\n\n1835\n\n\n\n\\(\\bar{x} = \\frac{\\sum fx}{\\sum f}\\)\n= \\(\\frac{1835}{80}\\)\n\\(\\approxeq 22.9yrs\\)\n(b)(i)\n\n\n\nClass Interval\nFreq (f)\nClass boundaries\n\n\n10 - 14\n7\n9.5 - 14.5\n\n\n15 - 19\n18\n14.5 - 19.5\n\n\n20 - 24\n25\n19.5 - 24.5\n\n\n25 - 29\n17\n24.5 - 29.5\n\n\n30 - 34\n9\n29.5 - 34.5\n\n\n35 - 39\n4\n34.5 - 39.5\n\n\n\n\n(ii) Modal age = \\(\\frac{20 + 24}{2} = 22\\)\n(c) P(picking from the modal class) = \\(\\frac{25}{80}\\)\n= \\(\\frac{5}{16}\\).\n\n', '', 'Explanation\n(a)\n\n\n\nClass Interval\nFreq (f)\nClassmark (x)\n\\(fx\\)\n\n\n10 - 14\n7\n12\n84\n\n\n15 - 19\n18\n17\n306\n\n\n20 - 24\n25\n22\n550\n\n\n25 - 29\n17\n27\n459\n\n\n30 - 34\n9\n32\n288\n\n\n35 - 39\n4\n37\n148\n\n\n\n80\n\n1835\n\n\n\n\\(\\bar{x} = \\frac{\\sum fx}{\\sum f}\\)\n= \\(\\frac{1835}{80}\\)\n\\(\\approxeq 22.9yrs\\)\n(b)(i)\n\n\n\nClass Interval\nFreq (f)\nClass boundaries\n\n\n10 - 14\n7\n9.5 - 14.5\n\n\n15 - 19\n18\n14.5 - 19.5\n\n\n20 - 24\n25\n19.5 - 24.5\n\n\n25 - 29\n17\n24.5 - 29.5\n\n\n30 - 34\n9\n29.5 - 34.5\n\n\n35 - 39\n4\n34.5 - 39.5\n\n\n\n\n(ii) Modal age = \\(\\frac{20 + 24}{2} = 22\\)\n(c) P(picking from the modal class) = \\(\\frac{25}{80}\\)\n= \\(\\frac{5}{16}\\).\n\n', '(a)\n\n\n\nClass Interval\nFreq (f)\nClassmark (x)\n\\(fx\\)\n\n\n10 - 14\n7\n12\n84\n\n\n15 - 19\n18\n17\n306\n\n\n20 - 24\n25\n22\n550\n\n\n25 - 29\n17\n27\n459\n\n\n30 - 34\n9\n32\n288\n\n\n35 - 39\n4\n37\n148\n\n\n\n80\n\n1835\n\n\n\n\\(\\bar{x} = \\frac{\\sum fx}{\\sum f}\\)\n= \\(\\frac{1835}{80}\\)\n\\(\\approxeq 22.9yrs\\)~(b)(i)\n\n\n\nClass Interval\nFreq (f)\nClass boundaries\n\n\n10 - 14\n7\n9.5 - 14.5\n\n\n15 - 19\n18\n14.5 - 19.5\n\n\n20 - 24\n25\n19.5 - 24.5\n\n\n25 - 29\n17\n24.5 - 29.5\n\n\n30 - 34\n9\n29.5 - 34.5\n\n\n35 - 39\n4\n34.5 - 39.5\n\n\n\n\n(ii) Modal age = \\(\\frac{20 + 24}{2} = 22\\)\n(c) P(picking from the modal class) = \\(\\frac{25}{80}\\)\n= \\(\\frac{5}{16}\\).\n~(c) P(picking from the modal class) = \\(\\frac{25}{80}\\)\n= \\(\\frac{5}{16}\\).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(497, '2014', '\n\nQuestion\n\n(a)In the diagram, \\(\\Delta\\) ABD isright-angled at B. |AB| = 3 cm, |AD| = 5 cm, \\(\\stackrel\\frown{ACB}\\) = 61° and \\(\\stackrel\\frown{DAC}\\) = x°. Calculate, correct to one decimal place, the value of x.\n(b)In the diagram, OABCD is a pyramid with a square base of side 2cm and a slant height of 4 cm. Calculate, correct to three significant figures : (i) the vertical height of the pyramid ; (ii) the volume of the pyramid.\n\n', '', '(a)In the diagram, \\(\\Delta\\) ABD isright-angled at B. |AB| = 3 cm, |AD| = 5 cm, \\(\\stackrel\\frown{ACB}\\) = 61° and \\(\\stackrel\\frown{DAC}\\) = x°. Calculate, correct to one decimal place, the value of x.\n(b)In the diagram, OABCD is a pyramid with a square base of side 2cm and a slant height of 4 cm. Calculate, correct to three significant figures : (i) the vertical height of the pyramid ; (ii) the volume of the pyramid.', '(a)In the diagram, \\(\\Delta\\) ABD isright-angled at B. |AB| = 3 cm, |AD| = 5 cm, \\(\\stackrel\\frown{ACB}\\) = 61° and \\(\\stackrel\\frown{DAC}\\) = x°. Calculate, correct to one decimal place, the value of x.~(b)In the diagram, OABCD is a pyramid with a square base of side 2cm and a slant height of 4 cm. Calculate, correct to three significant figures : (i) the vertical height of the pyramid ; (ii) the volume of the pyramid.', NULL, '\n\nExplanation\n(a) In \\(\\Delta ACB, < CAB = 180° - (90° + 61°)\\)\n= \\(180° - 151°\\)\n= \\(29°\\)\nIn \\(\\Delta ADB, < DAB = (29 + x)°\\)\n\\(\\cos (29 + x) = \\frac{3}{5} = 0.6\\)\n\\((29 + x)° = \\cos^{-1} (0.6)\\)\n\\((29 + x)° = 53.13°\\)\n\\(x = 53.13° - 29°\\)\n= \\(24.13°\\)\n\\(\\approxeq 24.1°\\) (1 decimal place).\n(b) (i)\\(DB = 2DN = 2NB\\)\n\\(OA = OB = OC = OD = 4cm\\)\n\\(AB = BC = CD = DA = 2cm\\)\n\\(DB^{2} = BC^{2} + DC^{2}\\)\n= \\(2^{2} + 2^{2}\\)\n= \\(\\sqrt{8}\\)\n= \\(2\\sqrt{2} cm\\)\n\\(2DN = DB \\implies DN = \\frac{2\\sqrt{2}}{2} = \\sqrt{2} cm\\)\nIn \\(\\Delta DON\\),\n\\(OD^{2} = ON^{2} + DN^{2}\\)\n\\(4^{2} = h^{2} + (\\sqrt{2})^{2} \\implies h^{2} = 16 - 2\\)\n\\(h^{2} = 14 \\implies h = \\sqrt{14}\\)\n(ii) \\(V = \\frac{1}{3} \\times \\base area \\times \\perp distance\\)\n\\(V = \\frac{1}{3} \\times 2^{2} \\times \\sqrt{14}\\)\n= \\(\\frac{4\\sqrt{14}}{3}\\)\n= \\(4.99 cm^{3}\\)\n\n', '', 'Explanation\n(a) In \\(\\Delta ACB, < CAB = 180° - (90° + 61°)\\)\n= \\(180° - 151°\\)\n= \\(29°\\)\nIn \\(\\Delta ADB, < DAB = (29 + x)°\\)\n\\(\\cos (29 + x) = \\frac{3}{5} = 0.6\\)\n\\((29 + x)° = \\cos^{-1} (0.6)\\)\n\\((29 + x)° = 53.13°\\)\n\\(x = 53.13° - 29°\\)\n= \\(24.13°\\)\n\\(\\approxeq 24.1°\\) (1 decimal place).\n(b) (i)\\(DB = 2DN = 2NB\\)\n\\(OA = OB = OC = OD = 4cm\\)\n\\(AB = BC = CD = DA = 2cm\\)\n\\(DB^{2} = BC^{2} + DC^{2}\\)\n= \\(2^{2} + 2^{2}\\)\n= \\(\\sqrt{8}\\)\n= \\(2\\sqrt{2} cm\\)\n\\(2DN = DB \\implies DN = \\frac{2\\sqrt{2}}{2} = \\sqrt{2} cm\\)\nIn \\(\\Delta DON\\),\n\\(OD^{2} = ON^{2} + DN^{2}\\)\n\\(4^{2} = h^{2} + (\\sqrt{2})^{2} \\implies h^{2} = 16 - 2\\)\n\\(h^{2} = 14 \\implies h = \\sqrt{14}\\)\n(ii) \\(V = \\frac{1}{3} \\times \\base area \\times \\perp distance\\)\n\\(V = \\frac{1}{3} \\times 2^{2} \\times \\sqrt{14}\\)\n= \\(\\frac{4\\sqrt{14}}{3}\\)\n= \\(4.99 cm^{3}\\)\n\n', '(a) In \\(\\Delta ACB, < CAB = 180° - (90° + 61°)\\)\n= \\(180° - 151°\\)\n= \\(29°\\)\nIn \\(\\Delta ADB, < DAB = (29 + x)°\\)\n\\(\\cos (29 + x) = \\frac{3}{5} = 0.6\\)\n\\((29 + x)° = \\cos^{-1} (0.6)\\)\n\\((29 + x)° = 53.13°\\)\n\\(x = 53.13° - 29°\\)\n= \\(24.13°\\)\n\\(\\approxeq 24.1°\\) (1 decimal place).~(b) (i)\\(DB = 2DN = 2NB\\)\n\\(OA = OB = OC = OD = 4cm\\)\n\\(AB = BC = CD = DA = 2cm\\)\n\\(DB^{2} = BC^{2} + DC^{2}\\)\n= \\(2^{2} + 2^{2}\\)\n= \\(\\sqrt{8}\\)\n= \\(2\\sqrt{2} cm\\)\n\\(2DN = DB \\implies DN = \\frac{2\\sqrt{2}}{2} = \\sqrt{2} cm\\)\nIn \\(\\Delta DON\\),\n\\(OD^{2} = ON^{2} + DN^{2}\\)\n\\(4^{2} = h^{2} + (\\sqrt{2})^{2} \\implies h^{2} = 16 - 2\\)\n\\(h^{2} = 14 \\implies h = \\sqrt{14}\\)\n(ii) \\(V = \\frac{1}{3} \\times \\base area \\times \\perp distance\\)\n\\(V = \\frac{1}{3} \\times 2^{2} \\times \\sqrt{14}\\)\n= \\(\\frac{4\\sqrt{14}}{3}\\)\n= \\(4.99 cm^{3}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(498, '2015', '\n\nQuestion\n\n(a) Copy and complete the table.\n\\(y = x^{2} - 2x - 2\\) for \\(-4 \\leq x \\leq 4\\)\n\n\n\nx\n-4\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\ny\n22\n\n\n\n-2\n\n\n1\n6\n\n\n\n(b) Using a scale of 2 cm to 1 unit on the x- axis and 2 cm to 5 units on the y- axis, draw the graph of \\(y = x^{2} - 2x - 2\\).\n(c) Use your graph to find : (i) the roots of the equation \\(x^{2} - 2x - 2 = 0\\) ; (ii) the values of x for which \\(x^{2} - 2x - 4\\frac{1}{2} = 0\\) ; (iii) the equation of the line of symmetry of the curve.\n\n\n(a) Copy and complete the table.\n\\(y = x^{2} - 2x - 2\\) for \\(-4 \\leq x \\leq 4\\)\n\n\n\nx\n-4\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\ny\n22\n\n\n\n-2\n\n\n1\n6\n\n\n', '', '(a) Copy and complete the table.\n\\(y = x^{2} - 2x - 2\\) for \\(-4 \\leq x \\leq 4\\)\n\n\n\nx\n-4\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\ny\n22\n\n\n\n-2\n\n\n1\n6\n\n\n\n(b) Using a scale of 2 cm to 1 unit on the x- axis and 2 cm to 5 units on the y- axis, draw the graph of \\(y = x^{2} - 2x - 2\\).\n(c) Use your graph to find : (i) the roots of the equation \\(x^{2} - 2x - 2 = 0\\) ; (ii) the values of x for which \\(x^{2} - 2x - 4\\frac{1}{2} = 0\\) ; (iii) the equation of the line of symmetry of the curve.\n\n\n(a) Copy and complete the table.\n\\(y = x^{2} - 2x - 2\\) for \\(-4 \\leq x \\leq 4\\)\n\n\n\nx\n-4\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\ny\n22\n\n\n\n-2\n\n\n1\n6\n', '(a) Copy and complete the table.\n\\(y = x^{2} - 2x - 2\\) for \\(-4 \\leq x \\leq 4\\)\n\n\n\nx\n-4\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\ny\n22\n\n\n\n-2\n\n\n1\n6\n\n\n~(b) Using a scale of 2 cm to 1 unit on the x- axis and 2 cm to 5 units on the y- axis, draw the graph of \\(y = x^{2} - 2x - 2\\).\n(c) Use your graph to find : (i) the roots of the equation \\(x^{2} - 2x - 2 = 0\\) ; (ii) the values of x for which \\(x^{2} - 2x - 4\\frac{1}{2} = 0\\) ; (iii) the equation of the line of symmetry of the curve.\n\n\n(a) Copy and complete the table.\n\\(y = x^{2} - 2x - 2\\) for \\(-4 \\leq x \\leq 4\\)\n\n\n\nx\n-4\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\ny\n22\n\n\n\n-2\n\n\n1\n6\n~(c) Use your graph to find : (i) the roots of the equation \\(x^{2} - 2x - 2 = 0\\) ; (ii) the values of x for which \\(x^{2} - 2x - 4\\frac{1}{2} = 0\\) ; (iii) the equation of the line of symmetry of the curve.\n\n\n(a) Copy and complete the table.\n\\(y = x^{2} - 2x - 2\\) for \\(-4 \\leq x \\leq 4\\)\n\n\n\nx\n-4\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\ny\n22\n\n\n\n-2\n\n\n1\n6\n', NULL, '\n\nExplanation\n(a) \\(y = x^{2} - 2x - 2\\)\n\n\n\nx\n-4\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\n\\(x^{2}\\)\n16\n9\n4\n1\n0\n1\n4\n9\n16\n\n\n\\(-2x\\)\n8\n6\n4\n2\n0\n-2\n-4\n-6\n-8\n\n\n\\(-2\\)\n-2\n-2\n-2\n-2\n-2\n-2\n-2\n-2\n-2\n\n\n\\(y\\)\n22\n13\n6\n1\n-2\n-3\n-2\n1\n6\n\n\n\n(b)\n(c)(i) \\(x^{2} - 2x - 2 = 0\\)\n\\(\\therefore y = 0 ; x = \\text{-0.7 or 2.7}\\)\n(ii) \\(x^{2} - 2x - 4\\frac{1}{2} = 0\\)\n\\(x^{2} - 2x - 4\\frac{1}{2} + 2\\frac{1}{2} = 0 + 2\\frac{1}{2}\\)\n\\(x^{2} - 2x - 2 = 2.5\\)\nWhen y = 2.5, \\(x = \\text{-1.3 or 3.3}\\).\n(iii) Line of symmetry at x = 1.\n\n\n', '', 'Explanation\n(a) \\(y = x^{2} - 2x - 2\\)\n\n\n\nx\n-4\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\n\\(x^{2}\\)\n16\n9\n4\n1\n0\n1\n4\n9\n16\n\n\n\\(-2x\\)\n8\n6\n4\n2\n0\n-2\n-4\n-6\n-8\n\n\n\\(-2\\)\n-2\n-2\n-2\n-2\n-2\n-2\n-2\n-2\n-2\n\n\n\\(y\\)\n22\n13\n6\n1\n-2\n-3\n-2\n1\n6\n\n\n\n(b)\n(c)(i) \\(x^{2} - 2x - 2 = 0\\)\n\\(\\therefore y = 0 ; x = \\text{-0.7 or 2.7}\\)\n(ii) \\(x^{2} - 2x - 4\\frac{1}{2} = 0\\)\n\\(x^{2} - 2x - 4\\frac{1}{2} + 2\\frac{1}{2} = 0 + 2\\frac{1}{2}\\)\n\\(x^{2} - 2x - 2 = 2.5\\)\nWhen y = 2.5, \\(x = \\text{-1.3 or 3.3}\\).\n(iii) Line of symmetry at x = 1.\n\n\n', '(a) \\(y = x^{2} - 2x - 2\\)\n\n\n\nx\n-4\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\n\\(x^{2}\\)\n16\n9\n4\n1\n0\n1\n4\n9\n16\n\n\n\\(-2x\\)\n8\n6\n4\n2\n0\n-2\n-4\n-6\n-8\n\n\n\\(-2\\)\n-2\n-2\n-2\n-2\n-2\n-2\n-2\n-2\n-2\n\n\n\\(y\\)\n22\n13\n6\n1\n-2\n-3\n-2\n1\n6\n\n\n~(b)\n(c)(i) \\(x^{2} - 2x - 2 = 0\\)\n\\(\\therefore y = 0 ; x = \\text{-0.7 or 2.7}\\)\n(ii) \\(x^{2} - 2x - 4\\frac{1}{2} = 0\\)\n\\(x^{2} - 2x - 4\\frac{1}{2} + 2\\frac{1}{2} = 0 + 2\\frac{1}{2}\\)\n\\(x^{2} - 2x - 2 = 2.5\\)\nWhen y = 2.5, \\(x = \\text{-1.3 or 3.3}\\).\n(iii) Line of symmetry at x = 1.\n\n~(c)(i) \\(x^{2} - 2x - 2 = 0\\)\n\\(\\therefore y = 0 ; x = \\text{-0.7 or 2.7}\\)\n(ii) \\(x^{2} - 2x - 4\\frac{1}{2} = 0\\)\n\\(x^{2} - 2x - 4\\frac{1}{2} + 2\\frac{1}{2} = 0 + 2\\frac{1}{2}\\)\n\\(x^{2} - 2x - 2 = 2.5\\)\nWhen y = 2.5, \\(x = \\text{-1.3 or 3.3}\\).\n(iii) Line of symmetry at x = 1.\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(499, '2015', '\n\nQuestion\n\n(a) Solve \\(\\frac{1}{81^{(x - 2)}} = 27^{(1 - x)}\\)\n(b) Simplify \\(\\frac{5}{\\sqrt{7} - \\sqrt{3}} + \\frac{1}{\\sqrt{7} + \\sqrt{3}}\\), leaving your answer in surd form.\n\n', '', '(a) Solve \\(\\frac{1}{81^{(x - 2)}} = 27^{(1 - x)}\\)\n(b) Simplify \\(\\frac{5}{\\sqrt{7} - \\sqrt{3}} + \\frac{1}{\\sqrt{7} + \\sqrt{3}}\\), leaving your answer in surd form.', '(a) Solve \\(\\frac{1}{81^{(x - 2)}} = 27^{(1 - x)}\\)~(b) Simplify \\(\\frac{5}{\\sqrt{7} - \\sqrt{3}} + \\frac{1}{\\sqrt{7} + \\sqrt{3}}\\), leaving your answer in surd form.', NULL, '\n\nExplanation\n(a) \\(\\frac{1}{81^{(x - 2)}} = 27^{(1 - x)}\\)\n\\(81^{-(x - 2)} = 27^{(1 - x)}\\)\n\\(3^{4[-(x - 2)]} = 3^{3(1 - x)}\\)\n\\(3^{-4(x - 2)} = 3^{3(1 - x)}\\)\n\\(-4(x - 2) = 3(1 - x)\\)\n\\(-4x + 8 = 3 - 3x\\)\n\\(8 - 3 = -3x + 4x \\implies x = 5\\)\n(b) \\(\\frac{5}{\\sqrt{7} - \\sqrt{3}} + \\frac{1}{\\sqrt{7} + \\sqrt{3}}\\)\n= \\(\\frac{5(\\sqrt{7} + \\sqrt{3}) + (\\sqrt{7} - \\sqrt{3})}{(\\sqrt{7} - \\sqrt{3})(\\sqrt{7} + \\sqrt{3})}\\)\n= \\(\\frac{5\\sqrt{7} + 5\\sqrt{3} + \\sqrt{7} - \\sqrt{3}}{7 - 3}\\)\n= \\(\\frac{6\\sqrt{7} + 4\\sqrt{3}}{4}\\)\n= \\(\\frac{3\\sqrt{7} + 2\\sqrt{3}}{2}\\).\n\n', '', 'Explanation\n(a) \\(\\frac{1}{81^{(x - 2)}} = 27^{(1 - x)}\\)\n\\(81^{-(x - 2)} = 27^{(1 - x)}\\)\n\\(3^{4[-(x - 2)]} = 3^{3(1 - x)}\\)\n\\(3^{-4(x - 2)} = 3^{3(1 - x)}\\)\n\\(-4(x - 2) = 3(1 - x)\\)\n\\(-4x + 8 = 3 - 3x\\)\n\\(8 - 3 = -3x + 4x \\implies x = 5\\)\n(b) \\(\\frac{5}{\\sqrt{7} - \\sqrt{3}} + \\frac{1}{\\sqrt{7} + \\sqrt{3}}\\)\n= \\(\\frac{5(\\sqrt{7} + \\sqrt{3}) + (\\sqrt{7} - \\sqrt{3})}{(\\sqrt{7} - \\sqrt{3})(\\sqrt{7} + \\sqrt{3})}\\)\n= \\(\\frac{5\\sqrt{7} + 5\\sqrt{3} + \\sqrt{7} - \\sqrt{3}}{7 - 3}\\)\n= \\(\\frac{6\\sqrt{7} + 4\\sqrt{3}}{4}\\)\n= \\(\\frac{3\\sqrt{7} + 2\\sqrt{3}}{2}\\).\n\n', '(a) \\(\\frac{1}{81^{(x - 2)}} = 27^{(1 - x)}\\)\n\\(81^{-(x - 2)} = 27^{(1 - x)}\\)\n\\(3^{4[-(x - 2)]} = 3^{3(1 - x)}\\)\n\\(3^{-4(x - 2)} = 3^{3(1 - x)}\\)\n\\(-4(x - 2) = 3(1 - x)\\)\n\\(-4x + 8 = 3 - 3x\\)\n\\(8 - 3 = -3x + 4x \\implies x = 5\\)~(b) \\(\\frac{5}{\\sqrt{7} - \\sqrt{3}} + \\frac{1}{\\sqrt{7} + \\sqrt{3}}\\)\n= \\(\\frac{5(\\sqrt{7} + \\sqrt{3}) + (\\sqrt{7} - \\sqrt{3})}{(\\sqrt{7} - \\sqrt{3})(\\sqrt{7} + \\sqrt{3})}\\)\n= \\(\\frac{5\\sqrt{7} + 5\\sqrt{3} + \\sqrt{7} - \\sqrt{3}}{7 - 3}\\)\n= \\(\\frac{6\\sqrt{7} + 4\\sqrt{3}}{4}\\)\n= \\(\\frac{3\\sqrt{7} + 2\\sqrt{3}}{2}\\).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(500, '2015', '\n\nQuestion\n\n(a) In the simultaneous equations : \\(px + qy = 5 ; qx + py = -10\\); p and q are constants. If x = 1 and y = -2 is a solution of the equations, find p and q.\n(b) Solve : \\(\\frac{4r - 3}{6r + 1} = \\frac{2r - 1}{3r + 4}\\).\n\n', '', '(a) In the simultaneous equations : \\(px + qy = 5 ; qx + py = -10\\); p and q are constants. If x = 1 and y = -2 is a solution of the equations, find p and q.\n(b) Solve : \\(\\frac{4r - 3}{6r + 1} = \\frac{2r - 1}{3r + 4}\\).', '(a) In the simultaneous equations : \\(px + qy = 5 ; qx + py = -10\\); p and q are constants. If x = 1 and y = -2 is a solution of the equations, find p and q.~(b) Solve : \\(\\frac{4r - 3}{6r + 1} = \\frac{2r - 1}{3r + 4}\\).', NULL, '\n\nExplanation\n(a) \\(px + qy = 5 \\)\nWhen x = 1 and y = -2, we have\n\\(p - 2q = 5 .... (1)\\)\n\\(qx + py = -10 \\)\nWhen x = 1 and y = -2, we have\n\\(q - 2p = -10 .... (2)\\)\nMultiplying (1) by -2, we have\n\\(-2p + 4q = -10 ... (1a)\\)\n\\(\\equiv 4q - 2p = -10\\)\n(2) - (1a) : \\(q - 4q = -10 + 10\\)\n\\(-3q = 0\\)\n\\(q = 0\\)\nPutting q = 0 in (1a), we have\n\\(-2p = -10 \\implies p = 5\\)\nHence, p = 5 and q = 0.\n(b) \\(\\frac{4r - 3}{6r + 1} = \\frac{2r - 1}{3r + 4}\\)\n\\((4r - 3)(3r + 4) = (2r - 1)(6r + 1)\\)\n\\(12r^{2} + 16r - 9r - 12 = 12r^{2} + 2r - 6r - 1\\)\n\\(12r^{2} - 12r^{2} + 7r + 4r = - 1 + 12\\)\n\\(11r = 11\\)\n\\(r = 1\\).\n\n', '', 'Explanation\n(a) \\(px + qy = 5 \\)\nWhen x = 1 and y = -2, we have\n\\(p - 2q = 5 .... (1)\\)\n\\(qx + py = -10 \\)\nWhen x = 1 and y = -2, we have\n\\(q - 2p = -10 .... (2)\\)\nMultiplying (1) by -2, we have\n\\(-2p + 4q = -10 ... (1a)\\)\n\\(\\equiv 4q - 2p = -10\\)\n(2) - (1a) : \\(q - 4q = -10 + 10\\)\n\\(-3q = 0\\)\n\\(q = 0\\)\nPutting q = 0 in (1a), we have\n\\(-2p = -10 \\implies p = 5\\)\nHence, p = 5 and q = 0.\n(b) \\(\\frac{4r - 3}{6r + 1} = \\frac{2r - 1}{3r + 4}\\)\n\\((4r - 3)(3r + 4) = (2r - 1)(6r + 1)\\)\n\\(12r^{2} + 16r - 9r - 12 = 12r^{2} + 2r - 6r - 1\\)\n\\(12r^{2} - 12r^{2} + 7r + 4r = - 1 + 12\\)\n\\(11r = 11\\)\n\\(r = 1\\).\n\n', '(a) \\(px + qy = 5 \\)\nWhen x = 1 and y = -2, we have\n\\(p - 2q = 5 .... (1)\\)\n\\(qx + py = -10 \\)\nWhen x = 1 and y = -2, we have\n\\(q - 2p = -10 .... (2)\\)\nMultiplying (1) by -2, we have\n\\(-2p + 4q = -10 ... (1a)\\)\n\\(\\equiv 4q - 2p = -10\\)\n(2) - (1a) : \\(q - 4q = -10 + 10\\)\n\\(-3q = 0\\)\n\\(q = 0\\)\nPutting q = 0 in (1a), we have\n\\(-2p = -10 \\implies p = 5\\)\nHence, p = 5 and q = 0.~(b) \\(\\frac{4r - 3}{6r + 1} = \\frac{2r - 1}{3r + 4}\\)\n\\((4r - 3)(3r + 4) = (2r - 1)(6r + 1)\\)\n\\(12r^{2} + 16r - 9r - 12 = 12r^{2} + 2r - 6r - 1\\)\n\\(12r^{2} - 12r^{2} + 7r + 4r = - 1 + 12\\)\n\\(11r = 11\\)\n\\(r = 1\\).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(501, '2015', '\n\nQuestion\n\n(a) AB is a chord of a circle centre O. If |AB| = 24.2 cm and the perimeter of \\(\\Delta\\) AOB is 52.2 cm, calculate < AOB, correct to the nearest degree.\n(b) A rectangular tank 60cm by 80cm by 100cm is half filled with water. How many litres of water is it holding?\n\n', '', '(a) AB is a chord of a circle centre O. If |AB| = 24.2 cm and the perimeter of \\(\\Delta\\) AOB is 52.2 cm, calculate < AOB, correct to the nearest degree.\n(b) A rectangular tank 60cm by 80cm by 100cm is half filled with water. How many litres of water is it holding?', '(a) AB is a chord of a circle centre O. If |AB| = 24.2 cm and the perimeter of \\(\\Delta\\) AOB is 52.2 cm, calculate < AOB, correct to the nearest degree.~(b) A rectangular tank 60cm by 80cm by 100cm is half filled with water. How many litres of water is it holding?', NULL, '\n\nExplanation\n(a)\n\\(2r + 24.2 = 52.2\\)\n\\(2r = 52.2 - 24.2 = 28\\)\n\\(r = 14 cm\\)\n\\(|AB|^{2} = OA^{2} + OB^{2} - 2(OA)(OB) \\cos <AOB\\)\n\\(24.2^{2} = 14^{2} + 14^{2} - 2(14)(14) \\cos < AOB\\)\n\\(\\cos < AOB = \\frac{14^{2} + 14^{2} - 24.2^{2}}{2(14)(14)}\\)\n\\(\\cos < AOB = \\frac{392 - 585.64}{392}\\)\n= \\(\\frac{-193.64}{392}\\)\n\\(\\cos < AOB = - 0.4940\\)\n\\(< AOB = \\cos^{-1} (0.4940)\\)\n= \\(119.6° \\approxeq 120°\\)\n(b) Volume of water in the tank = \\(\\frac{1}{2} \\times l \\times b \\times h\\)\n= \\(\\frac{1}{2} \\times 60 \\times 80 \\times 100\\)\n= \\(240000 cm^{3}\\)\nNumber of litres of water = \\(\\frac{240000}{1000}\\)\n= \\(240 litres\\)\n\n', '', 'Explanation\n(a)\n\\(2r + 24.2 = 52.2\\)\n\\(2r = 52.2 - 24.2 = 28\\)\n\\(r = 14 cm\\)\n\\(|AB|^{2} = OA^{2} + OB^{2} - 2(OA)(OB) \\cos <AOB\\)\n\\(24.2^{2} = 14^{2} + 14^{2} - 2(14)(14) \\cos < AOB\\)\n\\(\\cos < AOB = \\frac{14^{2} + 14^{2} - 24.2^{2}}{2(14)(14)}\\)\n\\(\\cos < AOB = \\frac{392 - 585.64}{392}\\)\n= \\(\\frac{-193.64}{392}\\)\n\\(\\cos < AOB = - 0.4940\\)\n\\(< AOB = \\cos^{-1} (0.4940)\\)\n= \\(119.6° \\approxeq 120°\\)\n(b) Volume of water in the tank = \\(\\frac{1}{2} \\times l \\times b \\times h\\)\n= \\(\\frac{1}{2} \\times 60 \\times 80 \\times 100\\)\n= \\(240000 cm^{3}\\)\nNumber of litres of water = \\(\\frac{240000}{1000}\\)\n= \\(240 litres\\)\n\n', '(a)\n\\(2r + 24.2 = 52.2\\)\n\\(2r = 52.2 - 24.2 = 28\\)\n\\(r = 14 cm\\)\n\\(|AB|^{2} = OA^{2} + OB^{2} - 2(OA)(OB) \\cos <AOB\\)\n\\(24.2^{2} = 14^{2} + 14^{2} - 2(14)(14) \\cos < AOB\\)\n\\(\\cos < AOB = \\frac{14^{2} + 14^{2} - 24.2^{2}}{2(14)(14)}\\)\n\\(\\cos < AOB = \\frac{392 - 585.64}{392}\\)\n= \\(\\frac{-193.64}{392}\\)\n\\(\\cos < AOB = - 0.4940\\)\n\\(< AOB = \\cos^{-1} (0.4940)\\)\n= \\(119.6° \\approxeq 120°\\)~(b) Volume of water in the tank = \\(\\frac{1}{2} \\times l \\times b \\times h\\)\n= \\(\\frac{1}{2} \\times 60 \\times 80 \\times 100\\)\n= \\(240000 cm^{3}\\)\nNumber of litres of water = \\(\\frac{240000}{1000}\\)\n= \\(240 litres\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(502, '2015', '\n\nQuestion\n\n(a) The sides of an isosceles triangle triangle are in the ratio \\(7 : 5 : 7\\). Calculate, correct to the nearest degree, the angle included between the equal sides.\n(b) The sum of the interior angles of a regular polygon is 1440°. Calculate : (i) the number of sides ; (ii) the size of one exterior angle of the polygon.\n\n', '', '(a) The sides of an isosceles triangle triangle are in the ratio \\(7 : 5 : 7\\). Calculate, correct to the nearest degree, the angle included between the equal sides.\n(b) The sum of the interior angles of a regular polygon is 1440°. Calculate : (i) the number of sides ; (ii) the size of one exterior angle of the polygon.', '(a) The sides of an isosceles triangle triangle are in the ratio \\(7 : 5 : 7\\). Calculate, correct to the nearest degree, the angle included between the equal sides.~(b) The sum of the interior angles of a regular polygon is 1440°. Calculate : (i) the number of sides ; (ii) the size of one exterior angle of the polygon.', NULL, '\n\nExplanation\n(a)\n\\(5^{2} = 7^{2} + 7^{2} - 2(7)(7) \\cos \\theta\\)\n\\(25 = 98 - 98 \\cos \\theta\\)\n\\(\\cos \\theta = \\frac{98 - 25}{98}\\)\n\\(\\cos \\theta = 0.7449\\)\n\\(\\theta = \\cos^{-1} (0.7449)\\)\n\\(\\approxeq 42°\\).\n(b) (i) \\((2n - 4) \\times 90° = 1440°\\)\n\\(2n - 4 = \\frac{1440}{90} = 16\\)\n\\(2n = 16 + 4 = 20\\)\n\\(n = 10\\)\n(ii) Size of exterior angle = \\(\\frac{360}{\\text{Number of sides}}\\)\n= \\(\\frac{360}{10}\\)\n= \\(36°\\)\n\n', '', 'Explanation\n(a)\n\\(5^{2} = 7^{2} + 7^{2} - 2(7)(7) \\cos \\theta\\)\n\\(25 = 98 - 98 \\cos \\theta\\)\n\\(\\cos \\theta = \\frac{98 - 25}{98}\\)\n\\(\\cos \\theta = 0.7449\\)\n\\(\\theta = \\cos^{-1} (0.7449)\\)\n\\(\\approxeq 42°\\).\n(b) (i) \\((2n - 4) \\times 90° = 1440°\\)\n\\(2n - 4 = \\frac{1440}{90} = 16\\)\n\\(2n = 16 + 4 = 20\\)\n\\(n = 10\\)\n(ii) Size of exterior angle = \\(\\frac{360}{\\text{Number of sides}}\\)\n= \\(\\frac{360}{10}\\)\n= \\(36°\\)\n\n', '(a)\n\\(5^{2} = 7^{2} + 7^{2} - 2(7)(7) \\cos \\theta\\)\n\\(25 = 98 - 98 \\cos \\theta\\)\n\\(\\cos \\theta = \\frac{98 - 25}{98}\\)\n\\(\\cos \\theta = 0.7449\\)\n\\(\\theta = \\cos^{-1} (0.7449)\\)\n\\(\\approxeq 42°\\).~(b) (i) \\((2n - 4) \\times 90° = 1440°\\)\n\\(2n - 4 = \\frac{1440}{90} = 16\\)\n\\(2n = 16 + 4 = 20\\)\n\\(n = 10\\)\n(ii) Size of exterior angle = \\(\\frac{360}{\\text{Number of sides}}\\)\n= \\(\\frac{360}{10}\\)\n= \\(36°\\)\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
(503, '2015', '\n\nQuestion\n\nThe table shows the marks scored by a group of students in a class test.\n\n\n\nMarks\n0\n1\n2\n3\n4\n5\n\n\nFrequency\n1\n4\n9\n8\n5\n3\n\n\n\n(a)(i) Calculate the mean mark ; (ii) Find the median.\n(b) If the information were to be represented in a pie chart, what would be the sectorial angle for the mark 2?\n\n\n\nThe table shows the marks scored by a group of students in a class test.\n\n\n\nMarks\n0\n1\n2\n3\n4\n5\n\n\nFrequency\n1\n4\n9\n8\n5\n3\n\n\n', '', 'The table shows the marks scored by a group of students in a class test.\n\n\n\nMarks\n0\n1\n2\n3\n4\n5\n\n\nFrequency\n1\n4\n9\n8\n5\n3\n\n\n\n(a)(i) Calculate the mean mark ; (ii) Find the median.\n(b) If the information were to be represented in a pie chart, what would be the sectorial angle for the mark 2?\n\n\n\nThe table shows the marks scored by a group of students in a class test.\n\n\n\nMarks\n0\n1\n2\n3\n4\n5\n\n\nFrequency\n1\n4\n9\n8\n5\n3\n', 'The table shows the marks scored by a group of students in a class test.\n\n\n\nMarks\n0\n1\n2\n3\n4\n5\n\n\nFrequency\n1\n4\n9\n8\n5\n3\n\n\n\n(a)(i) Calculate the mean mark ; (ii) Find the median.~(b) If the information were to be represented in a pie chart, what would be the sectorial angle for the mark 2?\n\n\n\nThe table shows the marks scored by a group of students in a class test.\n\n\n\nMarks\n0\n1\n2\n3\n4\n5\n\n\nFrequency\n1\n4\n9\n8\n5\n3\n', NULL, '\n\nExplanation\n(a)\n\n\n\nx\n0\n1\n2\n3\n4\n5\nTotal\n\n\nf\n1\n4\n9\n8\n5\n3\n30\n\n\nfx\n0\n4\n18\n24\n20\n15\n81\n\n\n\n(i) Mean(\\(\\bar{x}\\) = \\(\\frac{\\sum fx}{\\sum f}\\)\n= \\(\\frac{81}{30}\\)\n= 2.7\n(ii) \\(N = \\sum f = 30\\)\n\\(median = \\frac{1}{2}(N + 1)\\)\n= \\(\\frac{1}{2}(30 + 1)\\)\n=15.5\n= The 15th and 16th position.\n\\(\\therefore median = \\frac{3 + 3}{2} = 3\\)\n(b) Sectorial angle for the mark 2 = \\(\\frac{9}{30} \\times 360°\\)\n= 108°.\n\n', '', 'Explanation\n(a)\n\n\n\nx\n0\n1\n2\n3\n4\n5\nTotal\n\n\nf\n1\n4\n9\n8\n5\n3\n30\n\n\nfx\n0\n4\n18\n24\n20\n15\n81\n\n\n\n(i) Mean(\\(\\bar{x}\\) = \\(\\frac{\\sum fx}{\\sum f}\\)\n= \\(\\frac{81}{30}\\)\n= 2.7\n(ii) \\(N = \\sum f = 30\\)\n\\(median = \\frac{1}{2}(N + 1)\\)\n= \\(\\frac{1}{2}(30 + 1)\\)\n=15.5\n= The 15th and 16th position.\n\\(\\therefore median = \\frac{3 + 3}{2} = 3\\)\n(b) Sectorial angle for the mark 2 = \\(\\frac{9}{30} \\times 360°\\)\n= 108°.\n\n', '(a)\n\n\n\nx\n0\n1\n2\n3\n4\n5\nTotal\n\n\nf\n1\n4\n9\n8\n5\n3\n30\n\n\nfx\n0\n4\n18\n24\n20\n15\n81\n\n\n\n(i) Mean(\\(\\bar{x}\\) = \\(\\frac{\\sum fx}{\\sum f}\\)\n= \\(\\frac{81}{30}\\)\n= 2.7\n(ii) \\(N = \\sum f = 30\\)\n\\(median = \\frac{1}{2}(N + 1)\\)\n= \\(\\frac{1}{2}(30 + 1)\\)\n=15.5\n= The 15th and 16th position.\n\\(\\therefore median = \\frac{3 + 3}{2} = 3\\)~(b) Sectorial angle for the mark 2 = \\(\\frac{9}{30} \\times 360°\\)\n= 108°.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(504, '2015', '\n\nQuestion\n\n(a) Simplify : \\((\\frac{x^{2}}{2} - x + \\frac{1}{2})(\\frac{1}{x - 1})\\)\n(b) A point P is 40km from Q on a bearing 061°. Calculate, correct to one decimal place, the distance of P to (i) north of Q ; (ii) east of Q.\n(c) A man left N5,720 to be shared among his son and three daughters. Each daughter\'s share was \\(\\frac{3}{4}\\) of the son\'s share. How much did the son receive?\n\n', '', '(a) Simplify : \\((\\frac{x^{2}}{2} - x + \\frac{1}{2})(\\frac{1}{x - 1})\\)\n(b) A point P is 40km from Q on a bearing 061°. Calculate, correct to one decimal place, the distance of P to (i) north of Q ; (ii) east of Q.\n(c) A man left N5,720 to be shared among his son and three daughters. Each daughter\'s share was \\(\\frac{3}{4}\\) of the son\'s share. How much did the son receive?', '(a) Simplify : \\((\\frac{x^{2}}{2} - x + \\frac{1}{2})(\\frac{1}{x - 1})\\)~(b) A point P is 40km from Q on a bearing 061°. Calculate, correct to one decimal place, the distance of P to (i) north of Q ; (ii) east of Q.\n(c) A man left N5,720 to be shared among his son and three daughters. Each daughter\'s share was \\(\\frac{3}{4}\\) of the son\'s share. How much did the son receive?~(c) A man left N5,720 to be shared among his son and three daughters. Each daughter\'s share was \\(\\frac{3}{4}\\) of the son\'s share. How much did the son receive?', NULL, '\n\nExplanation\n(a) \\((\\frac{x^{2}}{2} - x + \\frac{1}{2})(\\frac{1}{x - 1})\\)\n\\(\\frac{x^{2}}{2} - x + \\frac{1}{2} = \\frac{x^{2} - 2x + 1}{2}\\)\n= \\(\\frac{(x - 1)^{2}}{2}\\)\n\\(\\therefore (\\frac{x^{2}}{2} - x + \\frac{1}{2})(\\frac{1}{x - 1}) = (\\frac{(x - 1)^{2}}{2})(\\frac{1}{x - 1})\\)\n= \\(\\frac{x - 1}{2}\\)\n(b)\n(i) TQ = PR (North of Q)\n\\(\\implies \\sin29 = \\frac{PR}{40}\\)\n\\(PR = 40 \\sin 29 = 19.39 km\\)\n(ii) QR = East of Q\n\\(\\frac{QR}{40} = \\cos 29\\)\n\\(QR = 40 \\cos 29\\)\n= 34.98km\n\\(\\approxeq\\) 35km.\n(c) Let the son\'s share = x.\nEach daughter\'s share = \\(\\frac{3}{4}x\\)\nFor the three daughters = \\(3 \\times \\frac{3}{4} = \\frac{9}{4}\\)\n\\(x + \\frac{9}{4}x = 5720 \\implies \\frac{13}{4}x = 5720\\)\n\\(x = \\frac{5720 \\times 4}{13} = N1760\\)\n\n', '', 'Explanation\n(a) \\((\\frac{x^{2}}{2} - x + \\frac{1}{2})(\\frac{1}{x - 1})\\)\n\\(\\frac{x^{2}}{2} - x + \\frac{1}{2} = \\frac{x^{2} - 2x + 1}{2}\\)\n= \\(\\frac{(x - 1)^{2}}{2}\\)\n\\(\\therefore (\\frac{x^{2}}{2} - x + \\frac{1}{2})(\\frac{1}{x - 1}) = (\\frac{(x - 1)^{2}}{2})(\\frac{1}{x - 1})\\)\n= \\(\\frac{x - 1}{2}\\)\n(b)\n(i) TQ = PR (North of Q)\n\\(\\implies \\sin29 = \\frac{PR}{40}\\)\n\\(PR = 40 \\sin 29 = 19.39 km\\)\n(ii) QR = East of Q\n\\(\\frac{QR}{40} = \\cos 29\\)\n\\(QR = 40 \\cos 29\\)\n= 34.98km\n\\(\\approxeq\\) 35km.\n(c) Let the son\'s share = x.\nEach daughter\'s share = \\(\\frac{3}{4}x\\)\nFor the three daughters = \\(3 \\times \\frac{3}{4} = \\frac{9}{4}\\)\n\\(x + \\frac{9}{4}x = 5720 \\implies \\frac{13}{4}x = 5720\\)\n\\(x = \\frac{5720 \\times 4}{13} = N1760\\)\n\n', '(a) \\((\\frac{x^{2}}{2} - x + \\frac{1}{2})(\\frac{1}{x - 1})\\)\n\\(\\frac{x^{2}}{2} - x + \\frac{1}{2} = \\frac{x^{2} - 2x + 1}{2}\\)\n= \\(\\frac{(x - 1)^{2}}{2}\\)\n\\(\\therefore (\\frac{x^{2}}{2} - x + \\frac{1}{2})(\\frac{1}{x - 1}) = (\\frac{(x - 1)^{2}}{2})(\\frac{1}{x - 1})\\)\n= \\(\\frac{x - 1}{2}\\)~(b)\n(i) TQ = PR (North of Q)\n\\(\\implies \\sin29 = \\frac{PR}{40}\\)\n\\(PR = 40 \\sin 29 = 19.39 km\\)\n(ii) QR = East of Q\n\\(\\frac{QR}{40} = \\cos 29\\)\n\\(QR = 40 \\cos 29\\)\n= 34.98km\n\\(\\approxeq\\) 35km.\n(c) Let the son\'s share = x.\nEach daughter\'s share = \\(\\frac{3}{4}x\\)\nFor the three daughters = \\(3 \\times \\frac{3}{4} = \\frac{9}{4}\\)\n\\(x + \\frac{9}{4}x = 5720 \\implies \\frac{13}{4}x = 5720\\)\n\\(x = \\frac{5720 \\times 4}{13} = N1760\\)\n~(c) Let the son\'s share = x.\nEach daughter\'s share = \\(\\frac{3}{4}x\\)\nFor the three daughters = \\(3 \\times \\frac{3}{4} = \\frac{9}{4}\\)\n\\(x + \\frac{9}{4}x = 5720 \\implies \\frac{13}{4}x = 5720\\)\n\\(x = \\frac{5720 \\times 4}{13} = N1760\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(505, '2015', '\n\nQuestion\n\n(a) The angles of depression of the top and bottom of a building are 51° and 62° respectively from the top of a tower 72m high. The base of the building is on the same horizontal level as the foot of the tower. Calculate the height of the building correct to 2 significant figures.\n(b)In the diagram, PR is a chord of the circle centre O and radius 30cm,< POR = 120°. Calculate correct to three significant figures : (i) the length of chord PR ; (ii) the length of arc PQR ; (iii) the perimeter of the shaded portion. (Take \\(\\pi = 3.142\\)).\n\n', '', '(a) The angles of depression of the top and bottom of a building are 51° and 62° respectively from the top of a tower 72m high. The base of the building is on the same horizontal level as the foot of the tower. Calculate the height of the building correct to 2 significant figures.\n(b)In the diagram, PR is a chord of the circle centre O and radius 30cm,< POR = 120°. Calculate correct to three significant figures : (i) the length of chord PR ; (ii) the length of arc PQR ; (iii) the perimeter of the shaded portion. (Take \\(\\pi = 3.142\\)).', '(a) The angles of depression of the top and bottom of a building are 51° and 62° respectively from the top of a tower 72m high. The base of the building is on the same horizontal level as the foot of the tower. Calculate the height of the building correct to 2 significant figures.~(b)In the diagram, PR is a chord of the circle centre O and radius 30cm,< POR = 120°. Calculate correct to three significant figures : (i) the length of chord PR ; (ii) the length of arc PQR ; (iii) the perimeter of the shaded portion. (Take \\(\\pi = 3.142\\)).', NULL, '\n\nExplanation\n(a)\nST = QR ; PS = height of tower.\nIn \\(\\Delta\\) PST,\n\\(\\frac{72}{ST} = \\tan 62°\\)\n\\(ST = \\frac{72}{\\tan 62}\\)\n= 38.283m \\(\\approxeq\\) 38.3m\nIn \\(\\Delta\\) PQR, \\(\\frac{h}{QR} = \\tan 51°\\)\n\\(h = QR \\times \\tan 51°\\)\n= \\(38.3 \\times 1.235\\)\n= \\(47.3m\\)\nQS = RT (height of the building)\nQS = PS - PQ\n= 72 - 47.3 = 24.7m\n(b)\n(i) \\(\\Delta POR = \\Delta QOR\\) (isosceles triangle)\n\\(\\frac{PT}{30} = \\sin 60\\)\n\\(PT = 30 \\sin 60 = 30 \\times \\frac{\\sqrt{3}}{2}\\)\n= \\(15\\sqrt{3} cm\\)\nPT = TR\nPR = 2PT\n= \\(2 \\times 15\\sqrt{3} = 30\\sqrt{3} cm\\)\n= \\(30 \\times 1.732 = 51.96 cm\\)\n\\(\\approxeq 52 cm\\)\n(ii) Length of minor arc, PQR = \\(\\frac{\\theta}{360} \\times 2\\pi r\\)\n= \\(\\frac{120}{360} \\times 2 \\times 3.142 \\times 30\\)\n= \\(62.84 cm\\)\n(iii) Perimeter of shaded portion = chord PR + Arc PQR\n= 51.96 + 62.84\n= 114.8 cm \\(\\approxeq\\) 115cm.\n\n', '', 'Explanation\n(a)\nST = QR ; PS = height of tower.\nIn \\(\\Delta\\) PST,\n\\(\\frac{72}{ST} = \\tan 62°\\)\n\\(ST = \\frac{72}{\\tan 62}\\)\n= 38.283m \\(\\approxeq\\) 38.3m\nIn \\(\\Delta\\) PQR, \\(\\frac{h}{QR} = \\tan 51°\\)\n\\(h = QR \\times \\tan 51°\\)\n= \\(38.3 \\times 1.235\\)\n= \\(47.3m\\)\nQS = RT (height of the building)\nQS = PS - PQ\n= 72 - 47.3 = 24.7m\n(b)\n(i) \\(\\Delta POR = \\Delta QOR\\) (isosceles triangle)\n\\(\\frac{PT}{30} = \\sin 60\\)\n\\(PT = 30 \\sin 60 = 30 \\times \\frac{\\sqrt{3}}{2}\\)\n= \\(15\\sqrt{3} cm\\)\nPT = TR\nPR = 2PT\n= \\(2 \\times 15\\sqrt{3} = 30\\sqrt{3} cm\\)\n= \\(30 \\times 1.732 = 51.96 cm\\)\n\\(\\approxeq 52 cm\\)\n(ii) Length of minor arc, PQR = \\(\\frac{\\theta}{360} \\times 2\\pi r\\)\n= \\(\\frac{120}{360} \\times 2 \\times 3.142 \\times 30\\)\n= \\(62.84 cm\\)\n(iii) Perimeter of shaded portion = chord PR + Arc PQR\n= 51.96 + 62.84\n= 114.8 cm \\(\\approxeq\\) 115cm.\n\n', '(a)\nST = QR ; PS = height of tower.\nIn \\(\\Delta\\) PST,\n\\(\\frac{72}{ST} = \\tan 62°\\)\n\\(ST = \\frac{72}{\\tan 62}\\)\n= 38.283m \\(\\approxeq\\) 38.3m\nIn \\(\\Delta\\) PQR, \\(\\frac{h}{QR} = \\tan 51°\\)\n\\(h = QR \\times \\tan 51°\\)\n= \\(38.3 \\times 1.235\\)\n= \\(47.3m\\)\nQS = RT (height of the building)\nQS = PS - PQ\n= 72 - 47.3 = 24.7m~(b)\n(i) \\(\\Delta POR = \\Delta QOR\\) (isosceles triangle)\n\\(\\frac{PT}{30} = \\sin 60\\)\n\\(PT = 30 \\sin 60 = 30 \\times \\frac{\\sqrt{3}}{2}\\)\n= \\(15\\sqrt{3} cm\\)\nPT = TR\nPR = 2PT\n= \\(2 \\times 15\\sqrt{3} = 30\\sqrt{3} cm\\)\n= \\(30 \\times 1.732 = 51.96 cm\\)\n\\(\\approxeq 52 cm\\)\n(ii) Length of minor arc, PQR = \\(\\frac{\\theta}{360} \\times 2\\pi r\\)\n= \\(\\frac{120}{360} \\times 2 \\times 3.142 \\times 30\\)\n= \\(62.84 cm\\)\n(iii) Perimeter of shaded portion = chord PR + Arc PQR\n= 51.96 + 62.84\n= 114.8 cm \\(\\approxeq\\) 115cm.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(506, '2015', '\n\nQuestion\n\n(a)\nIn the diagram, XY is a chord of a circle of radius 5cm. The chord subtends an angle 96° at the centre. Calculate, correct to three significant figures, the area of the minor segment cut-off. (Take \\(\\pi = \\frac{22}{7}\\)).\n(b)The figure shows a circle inscribed in a square. If a portion of the circle is shaded with some portions of the square, calculate the total area of the shaded portions. [Take \\(\\pi = \\frac{22}{7}\\)].\n\n', '', '(a)\nIn the diagram, XY is a chord of a circle of radius 5cm. The chord subtends an angle 96° at the centre. Calculate, correct to three significant figures, the area of the minor segment cut-off. (Take \\(\\pi = \\frac{22}{7}\\)).\n(b)The figure shows a circle inscribed in a square. If a portion of the circle is shaded with some portions of the square, calculate the total area of the shaded portions. [Take \\(\\pi = \\frac{22}{7}\\)].', '(a)\nIn the diagram, XY is a chord of a circle of radius 5cm. The chord subtends an angle 96° at the centre. Calculate, correct to three significant figures, the area of the minor segment cut-off. (Take \\(\\pi = \\frac{22}{7}\\)).~(b)The figure shows a circle inscribed in a square. If a portion of the circle is shaded with some portions of the square, calculate the total area of the shaded portions. [Take \\(\\pi = \\frac{22}{7}\\)].', NULL, '\n\nExplanation\n(a) Area of minor sector = \\(\\frac{\\theta}{360} \\times \\pi r^{2}\\)\n= \\(\\frac{96}{360} \\times \\frac{22}{7} \\times 5^{2}\\)\n= \\(20.95 cm^{2}\\)\nArea of triangle formed from the sector = \\(\\frac{1}{2} r^{2} \\sin \\theta\\)\n= \\(\\frac{1}{2} \\times 5^{2} \\times \\sin 96\\)\n= \\(12.43 cm^{2}\\)\n\\(\\therefore \\text{The area of minor segment} = 20.95 - 12.43\\)\n= \\(8.52 cm^{2}\\)\n(b) Area of minor sector = \\(\\frac{\\theta}{360} \\times \\pi r^{2}\\)\n= \\(\\frac{80}{360} \\times \\frac{22}{7} \\times 7^{2}\\)\n= \\(34.22 cm^{2}\\)\nArea of the square = \\((14)^{2} = 196 cm^{2}\\)\nArea of the circle = \\(\\pi r^{2} = \\frac{22}{7} \\times 7^{2} = 154 cm^{2}\\)\nArea of the shaded portion in the square = \\(196 - 154 = 42 cm^{2}\\)\nTotal area of the shaded portions = \\(42 + 34.22\\)\n= \\(76.22 cm^{2} \\approxeq 76.2 cm^{2}\\)\n\n', '', 'Explanation\n(a) Area of minor sector = \\(\\frac{\\theta}{360} \\times \\pi r^{2}\\)\n= \\(\\frac{96}{360} \\times \\frac{22}{7} \\times 5^{2}\\)\n= \\(20.95 cm^{2}\\)\nArea of triangle formed from the sector = \\(\\frac{1}{2} r^{2} \\sin \\theta\\)\n= \\(\\frac{1}{2} \\times 5^{2} \\times \\sin 96\\)\n= \\(12.43 cm^{2}\\)\n\\(\\therefore \\text{The area of minor segment} = 20.95 - 12.43\\)\n= \\(8.52 cm^{2}\\)\n(b) Area of minor sector = \\(\\frac{\\theta}{360} \\times \\pi r^{2}\\)\n= \\(\\frac{80}{360} \\times \\frac{22}{7} \\times 7^{2}\\)\n= \\(34.22 cm^{2}\\)\nArea of the square = \\((14)^{2} = 196 cm^{2}\\)\nArea of the circle = \\(\\pi r^{2} = \\frac{22}{7} \\times 7^{2} = 154 cm^{2}\\)\nArea of the shaded portion in the square = \\(196 - 154 = 42 cm^{2}\\)\nTotal area of the shaded portions = \\(42 + 34.22\\)\n= \\(76.22 cm^{2} \\approxeq 76.2 cm^{2}\\)\n\n', '(a) Area of minor sector = \\(\\frac{\\theta}{360} \\times \\pi r^{2}\\)\n= \\(\\frac{96}{360} \\times \\frac{22}{7} \\times 5^{2}\\)\n= \\(20.95 cm^{2}\\)\nArea of triangle formed from the sector = \\(\\frac{1}{2} r^{2} \\sin \\theta\\)\n= \\(\\frac{1}{2} \\times 5^{2} \\times \\sin 96\\)\n= \\(12.43 cm^{2}\\)\n\\(\\therefore \\text{The area of minor segment} = 20.95 - 12.43\\)\n= \\(8.52 cm^{2}\\)~(b) Area of minor sector = \\(\\frac{\\theta}{360} \\times \\pi r^{2}\\)\n= \\(\\frac{80}{360} \\times \\frac{22}{7} \\times 7^{2}\\)\n= \\(34.22 cm^{2}\\)\nArea of the square = \\((14)^{2} = 196 cm^{2}\\)\nArea of the circle = \\(\\pi r^{2} = \\frac{22}{7} \\times 7^{2} = 154 cm^{2}\\)\nArea of the shaded portion in the square = \\(196 - 154 = 42 cm^{2}\\)\nTotal area of the shaded portions = \\(42 + 34.22\\)\n= \\(76.22 cm^{2} \\approxeq 76.2 cm^{2}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(507, '2015', '\n\nQuestion\n\nUsing a ruler and a pair of compasses only,\n(a) Construct : (i) \\(\\Delta PQR\\) such that /PQ/ = 8cm, /PR/ = 7cm and < QPR = 105°. (ii) locus \\(L_{1}\\) of points equidistant from P and Q. (iii) locus \\(l_{2}\\) of points equidistant Q and R.\n(b)(i) Label the point T where \\(l_{1}\\) and \\(l_{2}\\) intersect ; (ii) With centre T and radius /TQ/, construct a circle \\(l_{3}\\). (iii) Complete quadrilateral PQSR such that /RS/ = /QS/ and /RQ/ = /TS/.\n\n', '', 'Using a ruler and a pair of compasses only,\n(a) Construct : (i) \\(\\Delta PQR\\) such that /PQ/ = 8cm, /PR/ = 7cm and < QPR = 105°. (ii) locus \\(L_{1}\\) of points equidistant from P and Q. (iii) locus \\(l_{2}\\) of points equidistant Q and R.\n(b)(i) Label the point T where \\(l_{1}\\) and \\(l_{2}\\) intersect ; (ii) With centre T and radius /TQ/, construct a circle \\(l_{3}\\). (iii) Complete quadrilateral PQSR such that /RS/ = /QS/ and /RQ/ = /TS/.', 'Using a ruler and a pair of compasses only,\n(a) Construct : (i) \\(\\Delta PQR\\) such that /PQ/ = 8cm, /PR/ = 7cm and < QPR = 105°. (ii) locus \\(L_{1}\\) of points equidistant from P and Q. (iii) locus \\(l_{2}\\) of points equidistant Q and R.~(b)(i) Label the point T where \\(l_{1}\\) and \\(l_{2}\\) intersect ; (ii) With centre T and radius /TQ/, construct a circle \\(l_{3}\\). (iii) Complete quadrilateral PQSR such that /RS/ = /QS/ and /RQ/ = /TS/.', NULL, '\n\nExplanation\n', '', '', '', 'WAEC-PK-1', 'WAEC-MATH', 1),
(508, '2015', '\n\nQuestion\n\nK(lat. 60°N, long. 50°W) is a point on the eart\'s surface. L is another point due East of K and the third point N is due North of K. The distance KL is 3520km and KN is 10951km.\n(a) Calculate: (i) The longitude of L ; (ii) The latitude of N. (Take \\(\\pi = \\frac{22}{7}\\) and the radius of the earth = 6400km).\n(b) A man was allowed 20% of his income as tax free. He then paid 25 kobo in the naira on the remainder. If he paid N1,200.00 as tax, calculate his total income.\n\n', '', 'K(lat. 60°N, long. 50°W) is a point on the eart\'s surface. L is another point due East of K and the third point N is due North of K. The distance KL is 3520km and KN is 10951km.\n(a) Calculate: (i) The longitude of L ; (ii) The latitude of N. (Take \\(\\pi = \\frac{22}{7}\\) and the radius of the earth = 6400km).\n(b) A man was allowed 20% of his income as tax free. He then paid 25 kobo in the naira on the remainder. If he paid N1,200.00 as tax, calculate his total income.', 'K(lat. 60°N, long. 50°W) is a point on the eart\'s surface. L is another point due East of K and the third point N is due North of K. The distance KL is 3520km and KN is 10951km.\n(a) Calculate: (i) The longitude of L ; (ii) The latitude of N. (Take \\(\\pi = \\frac{22}{7}\\) and the radius of the earth = 6400km).~(b) A man was allowed 20% of his income as tax free. He then paid 25 kobo in the naira on the remainder. If he paid N1,200.00 as tax, calculate his total income.', NULL, '\n\nExplanation\n(a)\n(i) Let the length of the longitude be \\(\\theta\\).\nLongitude difference = \\(50 + \\theta\\)\nDistance along the parallel of latitude (KL) = \\(\\frac{50 + \\theta}{360} \\times 2\\pi R \\cos 60\\)\n\\(3520 = \\frac{50 + \\theta}{360} \\times 2 \\times \\frac{22}{7} \\times 6400 \\cos 60\\)\n\\(3520 = \\frac{50 + \\theta}{360} \\times 2 \\times \\frac{22}{7} \\times 3200\\)\n\\(50 + \\theta = \\frac{3520 \\times 360 \\times 7}{22 times 2 \\times 3200}\\)\n\\(50 + \\theta = 63°\\)\n\\(\\theta = 63° - 50° = 13°\\)\n(ii) Let the latitude of N be \\(\\alpha\\).\nThe distance along the parallel of longitude (KN) = \\(\\frac{60 + \\alpha}{360} \\times 2 \\times \\frac{22}{7} \\times 6400\\)\n\\(10951 = \\frac{60 + \\alpha}{360} \\times 2 \\times \\frac{22}{7} \\times 6400\\)\n\\(60 + \\alpha = \\frac{10951 \\times 360 \\times 7}{2 \\times 22 \\times 6400}\\)\n\\(60 + \\alpha = 98°\\)\n\\(\\alpha = 98° - 60° = 38°\\)\nThe latitude of N = 38°S.\n(b) Let the man\'s income tax be x.\n80% of x = \\(\\frac{1200}{0.25} = N4800\\)\n\\(0.8x = N4,800\\)\n\\(x = \\frac{4800}{0.8} = N6000\\)\n\n', '', 'Explanation\n(a)\n(i) Let the length of the longitude be \\(\\theta\\).\nLongitude difference = \\(50 + \\theta\\)\nDistance along the parallel of latitude (KL) = \\(\\frac{50 + \\theta}{360} \\times 2\\pi R \\cos 60\\)\n\\(3520 = \\frac{50 + \\theta}{360} \\times 2 \\times \\frac{22}{7} \\times 6400 \\cos 60\\)\n\\(3520 = \\frac{50 + \\theta}{360} \\times 2 \\times \\frac{22}{7} \\times 3200\\)\n\\(50 + \\theta = \\frac{3520 \\times 360 \\times 7}{22 times 2 \\times 3200}\\)\n\\(50 + \\theta = 63°\\)\n\\(\\theta = 63° - 50° = 13°\\)\n(ii) Let the latitude of N be \\(\\alpha\\).\nThe distance along the parallel of longitude (KN) = \\(\\frac{60 + \\alpha}{360} \\times 2 \\times \\frac{22}{7} \\times 6400\\)\n\\(10951 = \\frac{60 + \\alpha}{360} \\times 2 \\times \\frac{22}{7} \\times 6400\\)\n\\(60 + \\alpha = \\frac{10951 \\times 360 \\times 7}{2 \\times 22 \\times 6400}\\)\n\\(60 + \\alpha = 98°\\)\n\\(\\alpha = 98° - 60° = 38°\\)\nThe latitude of N = 38°S.\n(b) Let the man\'s income tax be x.\n80% of x = \\(\\frac{1200}{0.25} = N4800\\)\n\\(0.8x = N4,800\\)\n\\(x = \\frac{4800}{0.8} = N6000\\)\n\n', '(a)\n(i) Let the length of the longitude be \\(\\theta\\).\nLongitude difference = \\(50 + \\theta\\)\nDistance along the parallel of latitude (KL) = \\(\\frac{50 + \\theta}{360} \\times 2\\pi R \\cos 60\\)\n\\(3520 = \\frac{50 + \\theta}{360} \\times 2 \\times \\frac{22}{7} \\times 6400 \\cos 60\\)\n\\(3520 = \\frac{50 + \\theta}{360} \\times 2 \\times \\frac{22}{7} \\times 3200\\)\n\\(50 + \\theta = \\frac{3520 \\times 360 \\times 7}{22 times 2 \\times 3200}\\)\n\\(50 + \\theta = 63°\\)\n\\(\\theta = 63° - 50° = 13°\\)\n(ii) Let the latitude of N be \\(\\alpha\\).\nThe distance along the parallel of longitude (KN) = \\(\\frac{60 + \\alpha}{360} \\times 2 \\times \\frac{22}{7} \\times 6400\\)\n\\(10951 = \\frac{60 + \\alpha}{360} \\times 2 \\times \\frac{22}{7} \\times 6400\\)\n\\(60 + \\alpha = \\frac{10951 \\times 360 \\times 7}{2 \\times 22 \\times 6400}\\)\n\\(60 + \\alpha = 98°\\)\n\\(\\alpha = 98° - 60° = 38°\\)\nThe latitude of N = 38°S.~(b) Let the man\'s income tax be x.\n80% of x = \\(\\frac{1200}{0.25} = N4800\\)\n\\(0.8x = N4,800\\)\n\\(x = \\frac{4800}{0.8} = N6000\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(509, '2015', '\n\nQuestion\n\nThe frequency distribution shows tha marks of 100 students in a Mathematics test.\n\n\n\nMarks\n1-10\n11-20\n21-30\n31-40\n41-50\n51-60\n61-70\n71-80\n81-90\n91-100\n\n\n\nNo. of\nStudents\n\n2\n4\n9\n13\n18\n32\n13\n5\n3\n1\n\n\n\n(a) Draw cumulative frequency curve for the distribution .\n(b) Use your curve to estimate : (i) the median ; (ii) the lower quartile ; (iii) the 60th percentile.\n\n\n\nThe frequency distribution shows tha marks of 100 students in a Mathematics test.\n\n\n\nMarks\n1-10\n11-20\n21-30\n31-40\n41-50\n51-60\n61-70\n71-80\n81-90\n91-100\n\n\n\nNo. of\nStudents\n\n2\n4\n9\n13\n18\n32\n13\n5\n3\n1\n\n\n', '', 'The frequency distribution shows tha marks of 100 students in a Mathematics test.\n\n\n\nMarks\n1-10\n11-20\n21-30\n31-40\n41-50\n51-60\n61-70\n71-80\n81-90\n91-100\n\n\n\nNo. of\nStudents\n\n2\n4\n9\n13\n18\n32\n13\n5\n3\n1\n\n\n\n(a) Draw cumulative frequency curve for the distribution .\n(b) Use your curve to estimate : (i) the median ; (ii) the lower quartile ; (iii) the 60th percentile.\n\n\n\nThe frequency distribution shows tha marks of 100 students in a Mathematics test.\n\n\n\nMarks\n1-10\n11-20\n21-30\n31-40\n41-50\n51-60\n61-70\n71-80\n81-90\n91-100\n\n\n\nNo. of\nStudents\n\n2\n4\n9\n13\n18\n32\n13\n5\n3\n1\n', 'The frequency distribution shows tha marks of 100 students in a Mathematics test.\n\n\n\nMarks\n1-10\n11-20\n21-30\n31-40\n41-50\n51-60\n61-70\n71-80\n81-90\n91-100\n\n\n\nNo. of\nStudents\n\n2\n4\n9\n13\n18\n32\n13\n5\n3\n1\n\n\n\n(a) Draw cumulative frequency curve for the distribution .~(b) Use your curve to estimate : (i) the median ; (ii) the lower quartile ; (iii) the 60th percentile.\n\n\n\nThe frequency distribution shows tha marks of 100 students in a Mathematics test.\n\n\n\nMarks\n1-10\n11-20\n21-30\n31-40\n41-50\n51-60\n61-70\n71-80\n81-90\n91-100\n\n\n\nNo. of\nStudents\n\n2\n4\n9\n13\n18\n32\n13\n5\n3\n1\n', NULL, '\n\nExplanation\n(a)\n\n\n\nMarks\nFrequency\nCum. freq\n\n\n1 - 10\n2\n2\n\n\n11 - 20\n4\n6\n\n\n21 - 30\n9\n15\n\n\n31 - 40\n13\n28\n\n\n41 - 50\n18\n46\n\n\n51 - 60\n32\n78\n\n\n61 - 70\n13\n91\n\n\n71 - 80\n5\n96\n\n\n81 - 90\n3\n99\n\n\n91 - 100\n1\n100\n\n\n\n\n(b)(i) \\(N = \\sum f = 100\\)\n\\(Median = \\frac{1}{2}N = \\frac{1}{2}(100)\\)\n= 50th percentile.\n\\(\\therefore Median = 52\\).\n(ii) \\(Q_{1} = \\frac{1}{4} N = \\frac{1}{4}(100)\\)\n= 25th percentile.\n\\(\\therefore Q_{1} = 39\\)\n(iii) 60th percentile = 55.\n\n\n', '', 'Explanation\n(a)\n\n\n\nMarks\nFrequency\nCum. freq\n\n\n1 - 10\n2\n2\n\n\n11 - 20\n4\n6\n\n\n21 - 30\n9\n15\n\n\n31 - 40\n13\n28\n\n\n41 - 50\n18\n46\n\n\n51 - 60\n32\n78\n\n\n61 - 70\n13\n91\n\n\n71 - 80\n5\n96\n\n\n81 - 90\n3\n99\n\n\n91 - 100\n1\n100\n\n\n\n\n(b)(i) \\(N = \\sum f = 100\\)\n\\(Median = \\frac{1}{2}N = \\frac{1}{2}(100)\\)\n= 50th percentile.\n\\(\\therefore Median = 52\\).\n(ii) \\(Q_{1} = \\frac{1}{4} N = \\frac{1}{4}(100)\\)\n= 25th percentile.\n\\(\\therefore Q_{1} = 39\\)\n(iii) 60th percentile = 55.\n\n\n', '(a)\n\n\n\nMarks\nFrequency\nCum. freq\n\n\n1 - 10\n2\n2\n\n\n11 - 20\n4\n6\n\n\n21 - 30\n9\n15\n\n\n31 - 40\n13\n28\n\n\n41 - 50\n18\n46\n\n\n51 - 60\n32\n78\n\n\n61 - 70\n13\n91\n\n\n71 - 80\n5\n96\n\n\n81 - 90\n3\n99\n\n\n91 - 100\n1\n100\n\n\n\n~(b)(i) \\(N = \\sum f = 100\\)\n\\(Median = \\frac{1}{2}N = \\frac{1}{2}(100)\\)\n= 50th percentile.\n\\(\\therefore Median = 52\\).\n(ii) \\(Q_{1} = \\frac{1}{4} N = \\frac{1}{4}(100)\\)\n= 25th percentile.\n\\(\\therefore Q_{1} = 39\\)\n(iii) 60th percentile = 55.\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(510, '2015', '\n\nQuestion\n\n(a) Simplify : \\(\\sqrt{1001_{two}}\\), leaving your answer in base two.\n(b)\nIn the diagram, O is the centre of the circle radius x. /PQ/ = z, /OK/ = y and < OKP = 90°. Find the value of z in terms of x and y.\n(c)\nIn the diagram, P, Q, R and S are points of the circle centre O. \\(\\stackrel\\frown{POQ} = 160°\\), \\(\\stackrel\\frown{QSR} = 45°\\) and \\(\\stackrel\\frown{PQS} = 40°\\). Calculate, (i) < QPS ; (ii) < RQS.\n\n', '', '(a) Simplify : \\(\\sqrt{1001_{two}}\\), leaving your answer in base two.\n(b)\nIn the diagram, O is the centre of the circle radius x. /PQ/ = z, /OK/ = y and < OKP = 90°. Find the value of z in terms of x and y.\n(c)\nIn the diagram, P, Q, R and S are points of the circle centre O. \\(\\stackrel\\frown{POQ} = 160°\\), \\(\\stackrel\\frown{QSR} = 45°\\) and \\(\\stackrel\\frown{PQS} = 40°\\). Calculate, (i) < QPS ; (ii) < RQS.', '(a) Simplify : \\(\\sqrt{1001_{two}}\\), leaving your answer in base two.~(b)\nIn the diagram, O is the centre of the circle radius x. /PQ/ = z, /OK/ = y and < OKP = 90°. Find the value of z in terms of x and y.\n(c)\nIn the diagram, P, Q, R and S are points of the circle centre O. \\(\\stackrel\\frown{POQ} = 160°\\), \\(\\stackrel\\frown{QSR} = 45°\\) and \\(\\stackrel\\frown{PQS} = 40°\\). Calculate, (i) < QPS ; (ii) < RQS.~(c)\nIn the diagram, P, Q, R and S are points of the circle centre O. \\(\\stackrel\\frown{POQ} = 160°\\), \\(\\stackrel\\frown{QSR} = 45°\\) and \\(\\stackrel\\frown{PQS} = 40°\\). Calculate, (i) < QPS ; (ii) < RQS.', NULL, '\n\nExplanation\n(a) \\(1001_{2} = 1 \\times 2^{3} + 0 \\times 2^{2} + 0 \\times 2^{1} + 1 \\times 2^{0}\\)\n= \\(8 + 0 + 0 + 1\\)\n= 9\n\\(\\sqrt{9} = 3_{10}\\)\n\n\n\n2\n3\n\n\n2\n1 r 1\n\n\n\n0 r 1\n\n\n\n\\(\\therefore \\sqrt{1001_{2}} = 11_{2}\\)\n(b) In \\(\\Delta QOP, OQ = OP\\)\n\\(\\therefore QK = KP\\)\n\\(\\Delta QOK = \\Delta POK\\) (right- angled triangle)\nIn \\(\\Delta POK\\),\n\\(OP^{2} = PK^{2} + OK^{2}\\)\n\\(x^{2} = y^{2} + (\\frac{z}{2})^{2}\\)\n\\(\\frac{z^{2}}{4} = x^{2} - y^{2}\\)\n\\(z^{2} = 4x^{2} - 4y^{2}\\)\n\\(z = \\sqrt{4(x^{2} - y^{2})}\\)\n\\(z = 2\\sqrt{x^{2} - y^{2}}\\)\n(c)(i) \\(< PSQ = \\frac{1}{2}(< POQ) = \\frac{1}{2}(160°)\\)\n= 80°\nIn \\(\\Delta PSQ\\),\n\\(< PQS+ < QPS + < PSQ = 180°\\)\n\\(40° + < QPS + 80° = 180°\\)\n\\(< QPS = 180° - 120° = 60°\\)\n(ii) \\(< PSR = < PSQ + < QSR\\)\n= \\(80° + 45°\\)\n= \\(125°\\)\n\\(< PQS = < PQO + < RQS \\)\n= \\(40° + < RQS\\)\n\\(< PSR + < PQS = 180°\\)\n\\(\\therefore 125° + 40° + < RQS = 180°\\)\n\\(< RQS = 180° - 165°\\)\n= \\(15°\\)\n\n', '', 'Explanation\n(a) \\(1001_{2} = 1 \\times 2^{3} + 0 \\times 2^{2} + 0 \\times 2^{1} + 1 \\times 2^{0}\\)\n= \\(8 + 0 + 0 + 1\\)\n= 9\n\\(\\sqrt{9} = 3_{10}\\)\n\n\n\n2\n3\n\n\n2\n1 r 1\n\n\n\n0 r 1\n\n\n\n\\(\\therefore \\sqrt{1001_{2}} = 11_{2}\\)\n(b) In \\(\\Delta QOP, OQ = OP\\)\n\\(\\therefore QK = KP\\)\n\\(\\Delta QOK = \\Delta POK\\) (right- angled triangle)\nIn \\(\\Delta POK\\),\n\\(OP^{2} = PK^{2} + OK^{2}\\)\n\\(x^{2} = y^{2} + (\\frac{z}{2})^{2}\\)\n\\(\\frac{z^{2}}{4} = x^{2} - y^{2}\\)\n\\(z^{2} = 4x^{2} - 4y^{2}\\)\n\\(z = \\sqrt{4(x^{2} - y^{2})}\\)\n\\(z = 2\\sqrt{x^{2} - y^{2}}\\)\n(c)(i) \\(< PSQ = \\frac{1}{2}(< POQ) = \\frac{1}{2}(160°)\\)\n= 80°\nIn \\(\\Delta PSQ\\),\n\\(< PQS+ < QPS + < PSQ = 180°\\)\n\\(40° + < QPS + 80° = 180°\\)\n\\(< QPS = 180° - 120° = 60°\\)\n(ii) \\(< PSR = < PSQ + < QSR\\)\n= \\(80° + 45°\\)\n= \\(125°\\)\n\\(< PQS = < PQO + < RQS \\)\n= \\(40° + < RQS\\)\n\\(< PSR + < PQS = 180°\\)\n\\(\\therefore 125° + 40° + < RQS = 180°\\)\n\\(< RQS = 180° - 165°\\)\n= \\(15°\\)\n\n', '(a) \\(1001_{2} = 1 \\times 2^{3} + 0 \\times 2^{2} + 0 \\times 2^{1} + 1 \\times 2^{0}\\)\n= \\(8 + 0 + 0 + 1\\)\n= 9\n\\(\\sqrt{9} = 3_{10}\\)\n\n\n\n2\n3\n\n\n2\n1 r 1\n\n\n\n0 r 1\n\n\n\n\\(\\therefore \\sqrt{1001_{2}} = 11_{2}\\)~(b) In \\(\\Delta QOP, OQ = OP\\)\n\\(\\therefore QK = KP\\)\n\\(\\Delta QOK = \\Delta POK\\) (right- angled triangle)\nIn \\(\\Delta POK\\),\n\\(OP^{2} = PK^{2} + OK^{2}\\)\n\\(x^{2} = y^{2} + (\\frac{z}{2})^{2}\\)\n\\(\\frac{z^{2}}{4} = x^{2} - y^{2}\\)\n\\(z^{2} = 4x^{2} - 4y^{2}\\)\n\\(z = \\sqrt{4(x^{2} - y^{2})}\\)\n\\(z = 2\\sqrt{x^{2} - y^{2}}\\)\n(c)(i) \\(< PSQ = \\frac{1}{2}(< POQ) = \\frac{1}{2}(160°)\\)\n= 80°\nIn \\(\\Delta PSQ\\),\n\\(< PQS+ < QPS + < PSQ = 180°\\)\n\\(40° + < QPS + 80° = 180°\\)\n\\(< QPS = 180° - 120° = 60°\\)\n(ii) \\(< PSR = < PSQ + < QSR\\)\n= \\(80° + 45°\\)\n= \\(125°\\)\n\\(< PQS = < PQO + < RQS \\)\n= \\(40° + < RQS\\)\n\\(< PSR + < PQS = 180°\\)\n\\(\\therefore 125° + 40° + < RQS = 180°\\)\n\\(< RQS = 180° - 165°\\)\n= \\(15°\\)\n~(c)(i) \\(< PSQ = \\frac{1}{2}(< POQ) = \\frac{1}{2}(160°)\\)\n= 80°\nIn \\(\\Delta PSQ\\),\n\\(< PQS+ < QPS + < PSQ = 180°\\)\n\\(40° + < QPS + 80° = 180°\\)\n\\(< QPS = 180° - 120° = 60°\\)\n(ii) \\(< PSR = < PSQ + < QSR\\)\n= \\(80° + 45°\\)\n= \\(125°\\)\n\\(< PQS = < PQO + < RQS \\)\n= \\(40° + < RQS\\)\n\\(< PSR + < PQS = 180°\\)\n\\(\\therefore 125° + 40° + < RQS = 180°\\)\n\\(< RQS = 180° - 165°\\)\n= \\(15°\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(511, '2015', '\n\nQuestion\n\n(a) Copy and complete the following table of values for the relation \\(y = 2x^{2} - 7x - 3\\).\n\n\n\nx\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n19\n\n-3\n\n-9\n\n\n\n\n\n\n(b) Using 2 cm to 1 unit on the x- axis and 2 cm to 5 units on the y- axis, draw the graph of \\(y = 2x^{2} - 7x - 3\\) for \\(-2 \\leq x \\leq 5\\).\n(c) From your graph, find the : (i) minimum value of y ;\n                        (ii) gradient of the curve at x = 1.\n(d) By drawing a suitable straight line, find the values of x for which \\(2x^{2} - 7x - 5 = x + 4\\).\n\n\n(a) Copy and complete the following table of values for the relation \\(y = 2x^{2} - 7x - 3\\).\n\n\n\nx\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n19\n\n-3\n\n-9\n\n\n\n\n\n', '', '(a) Copy and complete the following table of values for the relation \\(y = 2x^{2} - 7x - 3\\).\n\n\n\nx\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n19\n\n-3\n\n-9\n\n\n\n\n\n\n(b) Using 2 cm to 1 unit on the x- axis and 2 cm to 5 units on the y- axis, draw the graph of \\(y = 2x^{2} - 7x - 3\\) for \\(-2 \\leq x \\leq 5\\).\n(c) From your graph, find the : (i) minimum value of y ;\n                        (ii) gradient of the curve at x = 1.\n(d) By drawing a suitable straight line, find the values of x for which \\(2x^{2} - 7x - 5 = x + 4\\).\n\n\n(a) Copy and complete the following table of values for the relation \\(y = 2x^{2} - 7x - 3\\).\n\n\n\nx\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n19\n\n-3\n\n-9\n\n\n\n', '(a) Copy and complete the following table of values for the relation \\(y = 2x^{2} - 7x - 3\\).\n\n\n\nx\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n19\n\n-3\n\n-9\n\n\n\n\n\n~(b) Using 2 cm to 1 unit on the x- axis and 2 cm to 5 units on the y- axis, draw the graph of \\(y = 2x^{2} - 7x - 3\\) for \\(-2 \\leq x \\leq 5\\).\n(c) From your graph, find the : (i) minimum value of y ;\n                        (ii) gradient of the curve at x = 1.\n(d) By drawing a suitable straight line, find the values of x for which \\(2x^{2} - 7x - 5 = x + 4\\).\n\n\n(a) Copy and complete the following table of values for the relation \\(y = 2x^{2} - 7x - 3\\).\n\n\n\nx\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n19\n\n-3\n\n-9\n\n\n\n~(c) From your graph, find the : (i) minimum value of y ;\n                        (ii) gradient of the curve at x = 1.\n(d) By drawing a suitable straight line, find the values of x for which \\(2x^{2} - 7x - 5 = x + 4\\).\n\n\n(a) Copy and complete the following table of values for the relation \\(y = 2x^{2} - 7x - 3\\).\n\n\n\nx\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n19\n\n-3\n\n-9\n\n\n\n~(d) By drawing a suitable straight line, find the values of x for which \\(2x^{2} - 7x - 5 = x + 4\\).\n\n\n(a) Copy and complete the following table of values for the relation \\(y = 2x^{2} - 7x - 3\\).\n\n\n\nx\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n19\n\n-3\n\n-9\n\n\n\n', NULL, '\n\nExplanation\n(a) \\(y = 2x^{2} - 7x - 3\\)\n\n\n\nx\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\n\\(2x^{2}\\)\n8\n2\n0\n2\n8\n18\n32\n50\n\n\n\\(-7x\\)\n14\n7\n0\n-7\n-14\n-21\n-28\n-35\n\n\n\\(-3\\)\n-3\n-3\n-3\n-3\n-3\n-3\n-3\n-3\n\n\n\\(y\\)\n19\n6\n-3\n-8\n-9\n-6\n1\n12\n\n\n\n(b)\n(c)(i) Minimum value of y = -9\n(ii) \\(y = 2x^{2} - 7x - 3\\)\nGradient = \\(\\frac{\\mathrm d y}{\\mathrm d x} = 4x - 7\\)\nGradient at x = 1 : \\(4(1) - 7 = 4 - 7 = -3\\).\n(d) \\(2x^{2} - 7x - 5 = x + 4\\)\n\\(2x^{2} - 7x - 5 + 2 = x + 4 + 2\\)\n\\(2x^{2} - 7x - 3 = x + 6\\)\n\\(\\implies y = x + 6\\)\n\n\n\nx\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\n6\n6\n6\n6\n6\n6\n6\n6\n6\n\n\ny\n4\n5\n6\n7\n8\n9\n10\n11\n\n\n\nFrom the graph, x = -0.9 or 5.\n\n\n', '', 'Explanation\n(a) \\(y = 2x^{2} - 7x - 3\\)\n\n\n\nx\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\n\\(2x^{2}\\)\n8\n2\n0\n2\n8\n18\n32\n50\n\n\n\\(-7x\\)\n14\n7\n0\n-7\n-14\n-21\n-28\n-35\n\n\n\\(-3\\)\n-3\n-3\n-3\n-3\n-3\n-3\n-3\n-3\n\n\n\\(y\\)\n19\n6\n-3\n-8\n-9\n-6\n1\n12\n\n\n\n(b)\n(c)(i) Minimum value of y = -9\n(ii) \\(y = 2x^{2} - 7x - 3\\)\nGradient = \\(\\frac{\\mathrm d y}{\\mathrm d x} = 4x - 7\\)\nGradient at x = 1 : \\(4(1) - 7 = 4 - 7 = -3\\).\n(d) \\(2x^{2} - 7x - 5 = x + 4\\)\n\\(2x^{2} - 7x - 5 + 2 = x + 4 + 2\\)\n\\(2x^{2} - 7x - 3 = x + 6\\)\n\\(\\implies y = x + 6\\)\n\n\n\nx\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\n6\n6\n6\n6\n6\n6\n6\n6\n6\n\n\ny\n4\n5\n6\n7\n8\n9\n10\n11\n\n\n\nFrom the graph, x = -0.9 or 5.\n\n\n', '(a) \\(y = 2x^{2} - 7x - 3\\)\n\n\n\nx\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\n\\(2x^{2}\\)\n8\n2\n0\n2\n8\n18\n32\n50\n\n\n\\(-7x\\)\n14\n7\n0\n-7\n-14\n-21\n-28\n-35\n\n\n\\(-3\\)\n-3\n-3\n-3\n-3\n-3\n-3\n-3\n-3\n\n\n\\(y\\)\n19\n6\n-3\n-8\n-9\n-6\n1\n12\n\n\n~(b)\n(c)(i) Minimum value of y = -9\n(ii) \\(y = 2x^{2} - 7x - 3\\)\nGradient = \\(\\frac{\\mathrm d y}{\\mathrm d x} = 4x - 7\\)\nGradient at x = 1 : \\(4(1) - 7 = 4 - 7 = -3\\).\n(d) \\(2x^{2} - 7x - 5 = x + 4\\)\n\\(2x^{2} - 7x - 5 + 2 = x + 4 + 2\\)\n\\(2x^{2} - 7x - 3 = x + 6\\)\n\\(\\implies y = x + 6\\)\n\n\n\nx\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\n6\n6\n6\n6\n6\n6\n6\n6\n6\n\n\ny\n4\n5\n6\n7\n8\n9\n10\n11\n\n\n\nFrom the graph, x = -0.9 or 5.\n\n~(c)(i) Minimum value of y = -9\n(ii) \\(y = 2x^{2} - 7x - 3\\)\nGradient = \\(\\frac{\\mathrm d y}{\\mathrm d x} = 4x - 7\\)\nGradient at x = 1 : \\(4(1) - 7 = 4 - 7 = -3\\).\n(d) \\(2x^{2} - 7x - 5 = x + 4\\)\n\\(2x^{2} - 7x - 5 + 2 = x + 4 + 2\\)\n\\(2x^{2} - 7x - 3 = x + 6\\)\n\\(\\implies y = x + 6\\)\n\n\n\nx\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\n6\n6\n6\n6\n6\n6\n6\n6\n6\n\n\ny\n4\n5\n6\n7\n8\n9\n10\n11\n\n\n\nFrom the graph, x = -0.9 or 5.\n\n~(d) \\(2x^{2} - 7x - 5 = x + 4\\)\n\\(2x^{2} - 7x - 5 + 2 = x + 4 + 2\\)\n\\(2x^{2} - 7x - 3 = x + 6\\)\n\\(\\implies y = x + 6\\)\n\n\n\nx\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\n6\n6\n6\n6\n6\n6\n6\n6\n6\n\n\ny\n4\n5\n6\n7\n8\n9\n10\n11\n\n\n\nFrom the graph, x = -0.9 or 5.\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(512, '2015', '\n\nQuestion\n\n(a) If p varies directly as \\(r^{2}\\) and p = 3.2 whenr = 4, find the value of p when r = 6.5.\n(b) Solve the simultaneous equations :\n\\(\\frac{x}{2} + \\frac{y}{4} = 1 ; \\frac{x}{3} - \\frac{y}{4} = \\frac{-1}{6}\\)\n\n', '', '(a) If p varies directly as \\(r^{2}\\) and p = 3.2 whenr = 4, find the value of p when r = 6.5.\n(b) Solve the simultaneous equations :\n\\(\\frac{x}{2} + \\frac{y}{4} = 1 ; \\frac{x}{3} - \\frac{y}{4} = \\frac{-1}{6}\\)', '(a) If p varies directly as \\(r^{2}\\) and p = 3.2 whenr = 4, find the value of p when r = 6.5.~(b) Solve the simultaneous equations :\n\\(\\frac{x}{2} + \\frac{y}{4} = 1 ; \\frac{x}{3} - \\frac{y}{4} = \\frac{-1}{6}\\)', NULL, '\n\nExplanation\n(a) \\(p \\propto r^{2}\\)\n\\(\\implies p = kr^{2}\\)\n\\(3.2 = k(4^{2}) \\implies 3.2 = 16k\\)\n\\(k = \\frac{3.2}{16} = 0.2\\)\n\\(\\therefore p = 0.2r^{2}\\)\nWhen r = 6.5,\n\\(p = 0.2 (6.5^{2}) \\)\n= \\(0.2(42.25)\\)\n\\(p = 8.45\\) \n(b) \\(\\frac{x}{2} + \\frac{y}{4} = 1 ... (1)\\)\n\\(\\frac{x}{3} - \\frac{y}{4} = -\\frac{1}{6} .... (2)\\)\nAdding equation (1) and (2), we have\n\\(\\frac{x}{2} + \\frac{x}{3} = 1 - \\frac{1}{6}\\)\n\\(\\frac{5x}{6} = \\frac{5}{6} \\implies x = 1\\)\nPut x = 1 in (1),\n\\(\\frac{1}{2} + \\frac{y}{4} = 1 \\implies \\frac{y}{4} = 1 - \\frac{1}{2}\\)\n\\(\\frac{y}{4} = \\frac{1}{2} \\implies y = 2\\)\n\\((x, y) = (1, 2)\\).\n\n', '', 'Explanation\n(a) \\(p \\propto r^{2}\\)\n\\(\\implies p = kr^{2}\\)\n\\(3.2 = k(4^{2}) \\implies 3.2 = 16k\\)\n\\(k = \\frac{3.2}{16} = 0.2\\)\n\\(\\therefore p = 0.2r^{2}\\)\nWhen r = 6.5,\n\\(p = 0.2 (6.5^{2}) \\)\n= \\(0.2(42.25)\\)\n\\(p = 8.45\\) \n(b) \\(\\frac{x}{2} + \\frac{y}{4} = 1 ... (1)\\)\n\\(\\frac{x}{3} - \\frac{y}{4} = -\\frac{1}{6} .... (2)\\)\nAdding equation (1) and (2), we have\n\\(\\frac{x}{2} + \\frac{x}{3} = 1 - \\frac{1}{6}\\)\n\\(\\frac{5x}{6} = \\frac{5}{6} \\implies x = 1\\)\nPut x = 1 in (1),\n\\(\\frac{1}{2} + \\frac{y}{4} = 1 \\implies \\frac{y}{4} = 1 - \\frac{1}{2}\\)\n\\(\\frac{y}{4} = \\frac{1}{2} \\implies y = 2\\)\n\\((x, y) = (1, 2)\\).\n\n', '(a) \\(p \\propto r^{2}\\)\n\\(\\implies p = kr^{2}\\)\n\\(3.2 = k(4^{2}) \\implies 3.2 = 16k\\)\n\\(k = \\frac{3.2}{16} = 0.2\\)\n\\(\\therefore p = 0.2r^{2}\\)\nWhen r = 6.5,\n\\(p = 0.2 (6.5^{2}) \\)\n= \\(0.2(42.25)\\)\n\\(p = 8.45\\) ~(b) \\(\\frac{x}{2} + \\frac{y}{4} = 1 ... (1)\\)\n\\(\\frac{x}{3} - \\frac{y}{4} = -\\frac{1}{6} .... (2)\\)\nAdding equation (1) and (2), we have\n\\(\\frac{x}{2} + \\frac{x}{3} = 1 - \\frac{1}{6}\\)\n\\(\\frac{5x}{6} = \\frac{5}{6} \\implies x = 1\\)\nPut x = 1 in (1),\n\\(\\frac{1}{2} + \\frac{y}{4} = 1 \\implies \\frac{y}{4} = 1 - \\frac{1}{2}\\)\n\\(\\frac{y}{4} = \\frac{1}{2} \\implies y = 2\\)\n\\((x, y) = (1, 2)\\).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(513, '2015', '\n\nQuestion\n\nWithout using Mathematical tables or a calculator, simplify :\n(a) \\(\\sqrt{50} - 3\\sqrt{2}(2\\sqrt{2} - 5) - 5\\sqrt{32}\\)\n(b) \\(\\frac{1}{2} \\log_{10} \\frac{25}{4} - 2 \\log_{10} \\frac{4}{5} + \\log_{10} \\frac{320}{125}\\).\n\n', '', 'Without using Mathematical tables or a calculator, simplify :\n(a) \\(\\sqrt{50} - 3\\sqrt{2}(2\\sqrt{2} - 5) - 5\\sqrt{32}\\)\n(b) \\(\\frac{1}{2} \\log_{10} \\frac{25}{4} - 2 \\log_{10} \\frac{4}{5} + \\log_{10} \\frac{320}{125}\\).', 'Without using Mathematical tables or a calculator, simplify :\n(a) \\(\\sqrt{50} - 3\\sqrt{2}(2\\sqrt{2} - 5) - 5\\sqrt{32}\\)~(b) \\(\\frac{1}{2} \\log_{10} \\frac{25}{4} - 2 \\log_{10} \\frac{4}{5} + \\log_{10} \\frac{320}{125}\\).', NULL, '\n\nExplanation\n(a) \\(\\sqrt{50} - 3\\sqrt{2}(2\\sqrt{2} - 5) - 5\\sqrt{32}\\)\n\\(\\sqrt{25 \\times 2} - (3 \\times 2)\\sqrt{2 \\times 2} +(3 \\times 5)\\sqrt{2} - 5\\sqrt{16 \\times 2}\\)\n= \\(5\\sqrt{2} - 6(2) +15\\sqrt{2} - 5(4\\sqrt{2})\\)\n= \\(5\\sqrt{2} - 12 +15\\sqrt{2} - 20\\sqrt{2}\\)\n= \\((5 + 15 - 20)\\sqrt{2} - 12\\)\n= -12.\n(b) \\(\\frac{1}{2} \\log_{10} (\\frac{25}{4}) - 2 \\log_{10} (\\frac{4}{5}) + \\log_{10} (\\frac{320}{125})\\)\n= \\(\\log_{10} (\\frac{25}{4})^{\\frac{1}{2}} - \\log_{10} (\\frac{4}{5})^{2} + \\log_{10} (\\frac{320}{125})\\)\n= \\(\\log_{10} (\\frac{5}{2}) - \\log_{10} (\\frac{16}{25}) + \\log_{10} (\\frac{320}{125})\\)\n= \\(\\log_{10} (\\frac{\\frac{5}{2} \\times \\frac{320}{125}}{\\frac{16}{25}}\\)\n= \\(\\log_{10} (\\frac{\\frac{160}{25}}{\\frac{16}{25}}\\)\n= \\(\\log_{10} 10= 1\\)\n\n', '', 'Explanation\n(a) \\(\\sqrt{50} - 3\\sqrt{2}(2\\sqrt{2} - 5) - 5\\sqrt{32}\\)\n\\(\\sqrt{25 \\times 2} - (3 \\times 2)\\sqrt{2 \\times 2} +(3 \\times 5)\\sqrt{2} - 5\\sqrt{16 \\times 2}\\)\n= \\(5\\sqrt{2} - 6(2) +15\\sqrt{2} - 5(4\\sqrt{2})\\)\n= \\(5\\sqrt{2} - 12 +15\\sqrt{2} - 20\\sqrt{2}\\)\n= \\((5 + 15 - 20)\\sqrt{2} - 12\\)\n= -12.\n(b) \\(\\frac{1}{2} \\log_{10} (\\frac{25}{4}) - 2 \\log_{10} (\\frac{4}{5}) + \\log_{10} (\\frac{320}{125})\\)\n= \\(\\log_{10} (\\frac{25}{4})^{\\frac{1}{2}} - \\log_{10} (\\frac{4}{5})^{2} + \\log_{10} (\\frac{320}{125})\\)\n= \\(\\log_{10} (\\frac{5}{2}) - \\log_{10} (\\frac{16}{25}) + \\log_{10} (\\frac{320}{125})\\)\n= \\(\\log_{10} (\\frac{\\frac{5}{2} \\times \\frac{320}{125}}{\\frac{16}{25}}\\)\n= \\(\\log_{10} (\\frac{\\frac{160}{25}}{\\frac{16}{25}}\\)\n= \\(\\log_{10} 10= 1\\)\n\n', '(a) \\(\\sqrt{50} - 3\\sqrt{2}(2\\sqrt{2} - 5) - 5\\sqrt{32}\\)\n\\(\\sqrt{25 \\times 2} - (3 \\times 2)\\sqrt{2 \\times 2} +(3 \\times 5)\\sqrt{2} - 5\\sqrt{16 \\times 2}\\)\n= \\(5\\sqrt{2} - 6(2) +15\\sqrt{2} - 5(4\\sqrt{2})\\)\n= \\(5\\sqrt{2} - 12 +15\\sqrt{2} - 20\\sqrt{2}\\)\n= \\((5 + 15 - 20)\\sqrt{2} - 12\\)\n= -12.~(b) \\(\\frac{1}{2} \\log_{10} (\\frac{25}{4}) - 2 \\log_{10} (\\frac{4}{5}) + \\log_{10} (\\frac{320}{125})\\)\n= \\(\\log_{10} (\\frac{25}{4})^{\\frac{1}{2}} - \\log_{10} (\\frac{4}{5})^{2} + \\log_{10} (\\frac{320}{125})\\)\n= \\(\\log_{10} (\\frac{5}{2}) - \\log_{10} (\\frac{16}{25}) + \\log_{10} (\\frac{320}{125})\\)\n= \\(\\log_{10} (\\frac{\\frac{5}{2} \\times \\frac{320}{125}}{\\frac{16}{25}}\\)\n= \\(\\log_{10} (\\frac{\\frac{160}{25}}{\\frac{16}{25}}\\)\n= \\(\\log_{10} 10= 1\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(514, '2015', '\n\nQuestion\n\n(a) Simplify : \\(625^{\\frac{3}{8}} \\times 5^{\\frac{1}{2}} \\div 25\\)\n(b) Solve the following equations correct to one decimal place.\n(i) \\(\\tan (\\theta + 25)° = 5.145\\)\n(ii) \\(5\\cos \\theta - 1 = 0\\), where \\(0° \\leq \\theta \\leq 90°\\).\n\n', '', '(a) Simplify : \\(625^{\\frac{3}{8}} \\times 5^{\\frac{1}{2}} \\div 25\\)\n(b) Solve the following equations correct to one decimal place.\n(i) \\(\\tan (\\theta + 25)° = 5.145\\)\n(ii) \\(5\\cos \\theta - 1 = 0\\), where \\(0° \\leq \\theta \\leq 90°\\).', '(a) Simplify : \\(625^{\\frac{3}{8}} \\times 5^{\\frac{1}{2}} \\div 25\\)~(b) Solve the following equations correct to one decimal place.\n(i) \\(\\tan (\\theta + 25)° = 5.145\\)\n(ii) \\(5\\cos \\theta - 1 = 0\\), where \\(0° \\leq \\theta \\leq 90°\\).', NULL, '\n\nExplanation\n(a) \\(625^{\\frac{3}{8}} \\times 5^{\\frac{1}{2}} \\div 25\\)\n= \\((5^{4})^{\\frac{3}{8}} \\times 5^{\\frac{1}{2}} \\div 5^{2}\\)\n= \\(5^{\\frac{3}{2}} \\times 5^{\\frac{1}{2}} \\div 5^{2}\\)\n= \\(5^{\\frac{3}{2} + \\frac{1}{2} - 2}\\)\n= \\(5^{0} = 1\\)\n(b)(i) \\(\\tan (\\theta + 25)° = 5.145\\)\n\\(\\theta + 25= \\tan^{-1} (5.145)\\)\n\\(\\implies \\theta + 25= 79°\\)\n\\(\\theta = 79° - 25° = 54°\\)\n(ii) \\(5 \\cos \\theta - 1 = 0\\)\n\\(\\implies 5 \\cos \\theta = 1\\)\n\\(\\cos \\theta = \\frac{1}{5} = 0.2\\)\n\\(\\theta = \\cos^{-1} (0.2) = 78.463°\\)\n\\(\\approxeq 78.5°\\) (to one d.p)\n\n', '', 'Explanation\n(a) \\(625^{\\frac{3}{8}} \\times 5^{\\frac{1}{2}} \\div 25\\)\n= \\((5^{4})^{\\frac{3}{8}} \\times 5^{\\frac{1}{2}} \\div 5^{2}\\)\n= \\(5^{\\frac{3}{2}} \\times 5^{\\frac{1}{2}} \\div 5^{2}\\)\n= \\(5^{\\frac{3}{2} + \\frac{1}{2} - 2}\\)\n= \\(5^{0} = 1\\)\n(b)(i) \\(\\tan (\\theta + 25)° = 5.145\\)\n\\(\\theta + 25= \\tan^{-1} (5.145)\\)\n\\(\\implies \\theta + 25= 79°\\)\n\\(\\theta = 79° - 25° = 54°\\)\n(ii) \\(5 \\cos \\theta - 1 = 0\\)\n\\(\\implies 5 \\cos \\theta = 1\\)\n\\(\\cos \\theta = \\frac{1}{5} = 0.2\\)\n\\(\\theta = \\cos^{-1} (0.2) = 78.463°\\)\n\\(\\approxeq 78.5°\\) (to one d.p)\n\n', '(a) \\(625^{\\frac{3}{8}} \\times 5^{\\frac{1}{2}} \\div 25\\)\n= \\((5^{4})^{\\frac{3}{8}} \\times 5^{\\frac{1}{2}} \\div 5^{2}\\)\n= \\(5^{\\frac{3}{2}} \\times 5^{\\frac{1}{2}} \\div 5^{2}\\)\n= \\(5^{\\frac{3}{2} + \\frac{1}{2} - 2}\\)\n= \\(5^{0} = 1\\)~(b)(i) \\(\\tan (\\theta + 25)° = 5.145\\)\n\\(\\theta + 25= \\tan^{-1} (5.145)\\)\n\\(\\implies \\theta + 25= 79°\\)\n\\(\\theta = 79° - 25° = 54°\\)\n(ii) \\(5 \\cos \\theta - 1 = 0\\)\n\\(\\implies 5 \\cos \\theta = 1\\)\n\\(\\cos \\theta = \\frac{1}{5} = 0.2\\)\n\\(\\theta = \\cos^{-1} (0.2) = 78.463°\\)\n\\(\\approxeq 78.5°\\) (to one d.p)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(515, '2015', '\n\nQuestion\n\n In the diagram, PQRS is a circle with centre O and radius 7cm. SQ and PR intersect at K and < SKR = 90°. If the length of the arc SR is four times that of arc PQ, find the length of the arc SR. [Take \\(\\pi = \\frac{22}{7}\\)].\n\n', ' In the diagram, PQRS is a circle with centre O and radius 7cm. SQ and PR intersect at K and < SKR = 90°. If the length of the arc SR is four times that of arc PQ, find the length of the arc SR. [Take \\(\\pi = \\frac{22}{7}\\)].', '', '', NULL, '\n\nExplanation\nLet < POQ = \\(\\theta\\)\n\\(\\implies\\) SR = \\(4\\theta\\).\n\\(< PSQ = \\frac{\\theta}{2} ; < SPR = 2\\theta\\)\nIn \\(\\Delta PSK\\),\n\\(2\\theta + \\frac{\\theta}{2} = 90°\\)\n\\(\\frac{5\\theta}{2} = 90° \\implies 5\\theta = 180°\\)\n\\(\\theta = \\frac{180°}{5} = 36°\\)\n\\(\\therefore < SOR = 4\\theta = 4 \\times 36° = 144°\\)\nLength of arc = \\(\\frac{\\theta}{360°} \\times 2\\pi r\\)\n\\(\\frac{144}{360} \\times 2 \\times \\frac{22}{7} \\times 7 \\)\n= \\(17.6 cm\\).\n\n', 'Let < POQ = \\(\\theta\\)\n\\(\\implies\\) SR = \\(4\\theta\\).\n\\(< PSQ = \\frac{\\theta}{2} ; < SPR = 2\\theta\\)\nIn \\(\\Delta PSK\\),\n\\(2\\theta + \\frac{\\theta}{2} = 90°\\)\n\\(\\frac{5\\theta}{2} = 90° \\implies 5\\theta = 180°\\)\n\\(\\theta = \\frac{180°}{5} = 36°\\)\n\\(\\therefore < SOR = 4\\theta = 4 \\times 36° = 144°\\)\nLength of arc = \\(\\frac{\\theta}{360°} \\times 2\\pi r\\)\n\\(\\frac{144}{360} \\times 2 \\times \\frac{22}{7} \\times 7 \\)\n= \\(17.6 cm\\).\n\n', '', '', 'WAEC-PK-1', 'WAEC-MATH', 1),
(516, '2015', '\n\nQuestion\n\nThe table shows the number of suitcases possessed by a group of travellers.\n\n\n\nNo. of suitcases\n0\n1\n2\n3\n4\n5\n\n\nTravellers\n2\n7\n7\n2\n3\n9\n\n\n\n(a) Calculate the (i) median (ii) mean, correct to the nearest whole number.\n(b) Draw a bar chart to represent the information.\n\n\n\nThe table shows the number of suitcases possessed by a group of travellers.\n\n\n\nNo. of suitcases\n0\n1\n2\n3\n4\n5\n\n\nTravellers\n2\n7\n7\n2\n3\n9\n\n\n', '', 'The table shows the number of suitcases possessed by a group of travellers.\n\n\n\nNo. of suitcases\n0\n1\n2\n3\n4\n5\n\n\nTravellers\n2\n7\n7\n2\n3\n9\n\n\n\n(a) Calculate the (i) median (ii) mean, correct to the nearest whole number.\n(b) Draw a bar chart to represent the information.\n\n\n\nThe table shows the number of suitcases possessed by a group of travellers.\n\n\n\nNo. of suitcases\n0\n1\n2\n3\n4\n5\n\n\nTravellers\n2\n7\n7\n2\n3\n9\n', 'The table shows the number of suitcases possessed by a group of travellers.\n\n\n\nNo. of suitcases\n0\n1\n2\n3\n4\n5\n\n\nTravellers\n2\n7\n7\n2\n3\n9\n\n\n\n(a) Calculate the (i) median (ii) mean, correct to the nearest whole number.~(b) Draw a bar chart to represent the information.\n\n\n\nThe table shows the number of suitcases possessed by a group of travellers.\n\n\n\nNo. of suitcases\n0\n1\n2\n3\n4\n5\n\n\nTravellers\n2\n7\n7\n2\n3\n9\n', NULL, '\n\nExplanation\nx\n0\n1\n2\n3\n4\n5\n\n\n7\n2\n7\n7\n2\n3\n9\n\n\nfx\n0\n7\n14\n6\n12\n45\n\n\n\n(a)(i) Median = \\(\\frac{1}{2} (N + 1)\\)\n= \\(\\frac{1}{2}(30 + 1) = 15.5\\)\n= 15th and 16th position\n= \\(\\frac{2 + 2}{2} = 2\\)\n(ii) Mean = \\(\\frac{\\sum fx}{\\sum f}\\)\n= \\(\\frac{84}{30}\\)\n= \\(2.8 \\approxeq 3\\) (nearest whole no).\n(b) \n\n', '', 'Explanation\nx\n0\n1\n2\n3\n4\n5\n\n\n7\n2\n7\n7\n2\n3\n9\n\n\nfx\n0\n7\n14\n6\n12\n45\n\n\n\n(a)(i) Median = \\(\\frac{1}{2} (N + 1)\\)\n= \\(\\frac{1}{2}(30 + 1) = 15.5\\)\n= 15th and 16th position\n= \\(\\frac{2 + 2}{2} = 2\\)\n(ii) Mean = \\(\\frac{\\sum fx}{\\sum f}\\)\n= \\(\\frac{84}{30}\\)\n= \\(2.8 \\approxeq 3\\) (nearest whole no).\n(b) \n\n', '(a)(i) Median = \\(\\frac{1}{2} (N + 1)\\)\n= \\(\\frac{1}{2}(30 + 1) = 15.5\\)\n= 15th and 16th position\n= \\(\\frac{2 + 2}{2} = 2\\)\n(ii) Mean = \\(\\frac{\\sum fx}{\\sum f}\\)\n= \\(\\frac{84}{30}\\)\n= \\(2.8 \\approxeq 3\\) (nearest whole no).~(b) \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
(517, '2015', '\n\nQuestion\n\n(a) Simplify : \\(\\frac{\\frac{1}{3}c^{2} - \\frac{2}{3}cd}{\\frac{1}{2}d^{2} - \\frac{1}{4}cd}\\)\n(b)\nIn the diagram, YPF is a straight line. < XPY = 44°, < MPF = 46°, < XYP = < MFP = 90°, /XY/ = 7cm and /MP/ = 9 cm.\n(i) Calculate, correct to 3 significant figures, /XM/ and /YF/ ; (ii) Find < XMP.\n\n', '', '(a) Simplify : \\(\\frac{\\frac{1}{3}c^{2} - \\frac{2}{3}cd}{\\frac{1}{2}d^{2} - \\frac{1}{4}cd}\\)\n(b)\nIn the diagram, YPF is a straight line. < XPY = 44°, < MPF = 46°, < XYP = < MFP = 90°, /XY/ = 7cm and /MP/ = 9 cm.\n(i) Calculate, correct to 3 significant figures, /XM/ and /YF/ ; (ii) Find < XMP.', '(a) Simplify : \\(\\frac{\\frac{1}{3}c^{2} - \\frac{2}{3}cd}{\\frac{1}{2}d^{2} - \\frac{1}{4}cd}\\)~(b)\nIn the diagram, YPF is a straight line. < XPY = 44°, < MPF = 46°, < XYP = < MFP = 90°, /XY/ = 7cm and /MP/ = 9 cm.\n(i) Calculate, correct to 3 significant figures, /XM/ and /YF/ ; (ii) Find < XMP.', NULL, '\n\nExplanation\n(a) \\(\\frac{\\frac{1}{3}c^{2} - \\frac{2}{3}cd}{\\frac{1}{2}d^{2} - \\frac{1}{4}cd}\\)\n= \\(\\frac{\\frac{1}{3}c(c - 2d)}{\\frac{1}{4}d(2d - c)}\\)\n= \\(\\frac{\\frac{1}{3}c (c - 2d)}{-\\frac{1}{4}d (c - 2d)}\\)\n= \\(-\\frac{4c}{3d}\\)\n(b)(i) In \\(\\Delta XYP\\),\n\\(\\frac{7}{XP} = \\sin 44°\\)\n\\(XP = \\frac{7}{\\sin 44}\\)\n= \\(10.08 cm\\)\n< XPM + 44° + 46° = 180°\n< XPM = 90°.\nIn \\(\\Delta XPM\\),\n\\(XM^{2} = XP^{2} + PM^{2}\\)\n\\(XM^{2} = (10.08)^{2} + 9^{2}\\)\n= \\(101.6064 + 81 = 182.6064\\)\n\\(XM = \\sqrt{182.6064}\\)\n= \\(15.513 cm \\approxeq 13.5 cm\\)\nIn \\(\\Delta XYP\\),\n\\(\\frac{7}{YP} = \\tan 44\\)\n\\(YP = \\frac{7}{\\tan 44}\\)\n\\(YP = 7.249 cm \\approxeq 7.2 cm\\)\nIn \\(\\Delta MPF\\),\n\\(\\frac{PF}{9} = \\cos 46\\)\n\\(PF = 9\\cos 46 = 6.2519 cm\\)\n\\(YF = YP + PF = 7.249 + 6.2519\\)\n= \\(13.5009 cm \\approxeq 13.5 cm\\)\n(ii)\n\\(\\frac{10.1}{9} = \\tan \\theta\\)\n\\(\\tan \\theta = 1.122\\)\n\\(\\theta = \\tan^{-1} (1.122)\\)\n= 48.296°\n\n', '', 'Explanation\n(a) \\(\\frac{\\frac{1}{3}c^{2} - \\frac{2}{3}cd}{\\frac{1}{2}d^{2} - \\frac{1}{4}cd}\\)\n= \\(\\frac{\\frac{1}{3}c(c - 2d)}{\\frac{1}{4}d(2d - c)}\\)\n= \\(\\frac{\\frac{1}{3}c (c - 2d)}{-\\frac{1}{4}d (c - 2d)}\\)\n= \\(-\\frac{4c}{3d}\\)\n(b)(i) In \\(\\Delta XYP\\),\n\\(\\frac{7}{XP} = \\sin 44°\\)\n\\(XP = \\frac{7}{\\sin 44}\\)\n= \\(10.08 cm\\)\n< XPM + 44° + 46° = 180°\n< XPM = 90°.\nIn \\(\\Delta XPM\\),\n\\(XM^{2} = XP^{2} + PM^{2}\\)\n\\(XM^{2} = (10.08)^{2} + 9^{2}\\)\n= \\(101.6064 + 81 = 182.6064\\)\n\\(XM = \\sqrt{182.6064}\\)\n= \\(15.513 cm \\approxeq 13.5 cm\\)\nIn \\(\\Delta XYP\\),\n\\(\\frac{7}{YP} = \\tan 44\\)\n\\(YP = \\frac{7}{\\tan 44}\\)\n\\(YP = 7.249 cm \\approxeq 7.2 cm\\)\nIn \\(\\Delta MPF\\),\n\\(\\frac{PF}{9} = \\cos 46\\)\n\\(PF = 9\\cos 46 = 6.2519 cm\\)\n\\(YF = YP + PF = 7.249 + 6.2519\\)\n= \\(13.5009 cm \\approxeq 13.5 cm\\)\n(ii)\n\\(\\frac{10.1}{9} = \\tan \\theta\\)\n\\(\\tan \\theta = 1.122\\)\n\\(\\theta = \\tan^{-1} (1.122)\\)\n= 48.296°\n\n', '(a) \\(\\frac{\\frac{1}{3}c^{2} - \\frac{2}{3}cd}{\\frac{1}{2}d^{2} - \\frac{1}{4}cd}\\)\n= \\(\\frac{\\frac{1}{3}c(c - 2d)}{\\frac{1}{4}d(2d - c)}\\)\n= \\(\\frac{\\frac{1}{3}c (c - 2d)}{-\\frac{1}{4}d (c - 2d)}\\)\n= \\(-\\frac{4c}{3d}\\)~(b)(i) In \\(\\Delta XYP\\),\n\\(\\frac{7}{XP} = \\sin 44°\\)\n\\(XP = \\frac{7}{\\sin 44}\\)\n= \\(10.08 cm\\)\n< XPM + 44° + 46° = 180°\n< XPM = 90°.\nIn \\(\\Delta XPM\\),\n\\(XM^{2} = XP^{2} + PM^{2}\\)\n\\(XM^{2} = (10.08)^{2} + 9^{2}\\)\n= \\(101.6064 + 81 = 182.6064\\)\n\\(XM = \\sqrt{182.6064}\\)\n= \\(15.513 cm \\approxeq 13.5 cm\\)\nIn \\(\\Delta XYP\\),\n\\(\\frac{7}{YP} = \\tan 44\\)\n\\(YP = \\frac{7}{\\tan 44}\\)\n\\(YP = 7.249 cm \\approxeq 7.2 cm\\)\nIn \\(\\Delta MPF\\),\n\\(\\frac{PF}{9} = \\cos 46\\)\n\\(PF = 9\\cos 46 = 6.2519 cm\\)\n\\(YF = YP + PF = 7.249 + 6.2519\\)\n= \\(13.5009 cm \\approxeq 13.5 cm\\)\n(ii)\n\\(\\frac{10.1}{9} = \\tan \\theta\\)\n\\(\\tan \\theta = 1.122\\)\n\\(\\theta = \\tan^{-1} (1.122)\\)\n= 48.296°\n~c)}\\)\n= \\(\\frac{\\frac{1}{3}c (c - 2d)}{-\\frac{1}{4}d (c - 2d)}\\)\n= \\(-\\frac{4c}{3d}\\)\n(b)(i) In \\(\\Delta XYP\\),\n\\(\\frac{7}{XP} = \\sin 44°\\)\n\\(XP = \\frac{7}{\\sin 44}\\)\n= \\(10.08 cm\\)\n< XPM + 44° + 46° = 180°\n< XPM = 90°.\nIn \\(\\Delta XPM\\),\n\\(XM^{2} = XP^{2} + PM^{2}\\)\n\\(XM^{2} = (10.08)^{2} + 9^{2}\\)\n= \\(101.6064 + 81 = 182.6064\\)\n\\(XM = \\sqrt{182.6064}\\)\n= \\(15.513 cm \\approxeq 13.5 cm\\)\nIn \\(\\Delta XYP\\),\n\\(\\frac{7}{YP} = \\tan 44\\)\n\\(YP = \\frac{7}{\\tan 44}\\)\n\\(YP = 7.249 cm \\approxeq 7.2 cm\\)\nIn \\(\\Delta MPF\\),\n\\(\\frac{PF}{9} = \\cos 46\\)\n\\(PF = 9\\cos 46 = 6.2519 cm\\)\n\\(YF = YP + PF = 7.249 + 6.2519\\)\n= \\(13.5009 cm \\approxeq 13.5 cm\\)\n(ii)\n\\(\\frac{10.1}{9} = \\tan \\theta\\)\n\\(\\tan \\theta = 1.122\\)\n\\(\\theta = \\tan^{-1} (1.122)\\)\n= 48.296°\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(518, '2015', '\n\nQuestion\n\nThe table shows the monthly contributions and expenditure pattern of an employee in 1999.\n\n\n\nItem\nPercentage\n\n\nPension\n5\n\n\nIncome Tax\n25\n\n\nFood\n40\n\n\nTransport\n10\n\n\nRent\n12.5\n\n\nOthers\n7.5\n\n\n\n(a) Draw a pie chart to illustrate the data.\n(b) If the employee\'s gross monthly salary was N10,800.00, calculate (i) the pension contribution of the employee ; (ii) the income tax paid by the employee.\n(c) If the pension contribution and income tax were deducted from the gross monthly salary, before payment, calculate the take- home pay of the employee.\n\n\n\nThe table shows the monthly contributions and expenditure pattern of an employee in 1999.\n\n\n\nItem\nPercentage\n\n\nPension\n5\n\n\nIncome Tax\n25\n\n\nFood\n40\n\n\nTransport\n10\n\n\nRent\n12.5\n\n\nOthers\n7.5\n\n\n', '', 'The table shows the monthly contributions and expenditure pattern of an employee in 1999.\n\n\n\nItem\nPercentage\n\n\nPension\n5\n\n\nIncome Tax\n25\n\n\nFood\n40\n\n\nTransport\n10\n\n\nRent\n12.5\n\n\nOthers\n7.5\n\n\n\n(a) Draw a pie chart to illustrate the data.\n(b) If the employee\'s gross monthly salary was N10,800.00, calculate (i) the pension contribution of the employee ; (ii) the income tax paid by the employee.\n(c) If the pension contribution and income tax were deducted from the gross monthly salary, before payment, calculate the take- home pay of the employee.\n\n\n\nThe table shows the monthly contributions and expenditure pattern of an employee in 1999.\n\n\n\nItem\nPercentage\n\n\nPension\n5\n\n\nIncome Tax\n25\n\n\nFood\n40\n\n\nTransport\n10\n\n\nRent\n12.5\n\n\nOthers\n7.5\n', 'The table shows the monthly contributions and expenditure pattern of an employee in 1999.\n\n\n\nItem\nPercentage\n\n\nPension\n5\n\n\nIncome Tax\n25\n\n\nFood\n40\n\n\nTransport\n10\n\n\nRent\n12.5\n\n\nOthers\n7.5\n\n\n\n(a) Draw a pie chart to illustrate the data.~(b) If the employee\'s gross monthly salary was N10,800.00, calculate (i) the pension contribution of the employee ; (ii) the income tax paid by the employee.\n(c) If the pension contribution and income tax were deducted from the gross monthly salary, before payment, calculate the take- home pay of the employee.\n\n\n\nThe table shows the monthly contributions and expenditure pattern of an employee in 1999.\n\n\n\nItem\nPercentage\n\n\nPension\n5\n\n\nIncome Tax\n25\n\n\nFood\n40\n\n\nTransport\n10\n\n\nRent\n12.5\n\n\nOthers\n7.5\n~(c) If the pension contribution and income tax were deducted from the gross monthly salary, before payment, calculate the take- home pay of the employee.\n\n\n\nThe table shows the monthly contributions and expenditure pattern of an employee in 1999.\n\n\n\nItem\nPercentage\n\n\nPension\n5\n\n\nIncome Tax\n25\n\n\nFood\n40\n\n\nTransport\n10\n\n\nRent\n12.5\n\n\nOthers\n7.5\n', NULL, '\n\nExplanation\n(a)\n\n\n\nItems\nPercentage(%)\nDegree in pie chart\n\n\nPension\n5\n\\(\\frac{5}{100} \\times 360° = 18°\\)\n\n\nIncome tax\n25\n\\(\\frac{25}{100} \\times 360° = 90°\\)\n\n\nFood\n40\n\\(\\frac{40}{100} \\times 360° = 144°\\)\n\n\nTransport\n10\n\\(\\frac{10}{100} \\times 360° = 36°\\)\n\n\nRent\n12.5\n\\(\\frac{12.5}{100} \\times 360° = 45°\\)\n\n\nOthers\n7.5\n\\(\\frac{7.5}{100} \\times 360° = 27°\\)\n\n\n\n100\n360°\n\n\n\nA pie- chart showing the monthly contributions and expenditure pattern of an employee in 1999.\n\n(b)(i) \\(360° = N10,800.00\\)\n\\(\\text{Pension} - 18° = \\frac{18}{360} \\times N10,800\\)\n= \\(N540\\)\n(ii) \\(360° = N10,800.00\\)\n\\(\\text{Income tax} - 90° = \\frac{90}{360} \\times N10,800\\)\n= \\(N2700\\)\n(c) The percentage sum of pension and income tax = (5 + 25)% = 30%\nTake home pay = 100% - 30% = 70%\n= \\(\\frac{70}{100} \\times N10,800\\)\n= \\(N7560\\)\n\n', '', 'Explanation\n(a)\n\n\n\nItems\nPercentage(%)\nDegree in pie chart\n\n\nPension\n5\n\\(\\frac{5}{100} \\times 360° = 18°\\)\n\n\nIncome tax\n25\n\\(\\frac{25}{100} \\times 360° = 90°\\)\n\n\nFood\n40\n\\(\\frac{40}{100} \\times 360° = 144°\\)\n\n\nTransport\n10\n\\(\\frac{10}{100} \\times 360° = 36°\\)\n\n\nRent\n12.5\n\\(\\frac{12.5}{100} \\times 360° = 45°\\)\n\n\nOthers\n7.5\n\\(\\frac{7.5}{100} \\times 360° = 27°\\)\n\n\n\n100\n360°\n\n\n\nA pie- chart showing the monthly contributions and expenditure pattern of an employee in 1999.\n\n(b)(i) \\(360° = N10,800.00\\)\n\\(\\text{Pension} - 18° = \\frac{18}{360} \\times N10,800\\)\n= \\(N540\\)\n(ii) \\(360° = N10,800.00\\)\n\\(\\text{Income tax} - 90° = \\frac{90}{360} \\times N10,800\\)\n= \\(N2700\\)\n(c) The percentage sum of pension and income tax = (5 + 25)% = 30%\nTake home pay = 100% - 30% = 70%\n= \\(\\frac{70}{100} \\times N10,800\\)\n= \\(N7560\\)\n\n', '(a)\n\n\n\nItems\nPercentage(%)\nDegree in pie chart\n\n\nPension\n5\n\\(\\frac{5}{100} \\times 360° = 18°\\)\n\n\nIncome tax\n25\n\\(\\frac{25}{100} \\times 360° = 90°\\)\n\n\nFood\n40\n\\(\\frac{40}{100} \\times 360° = 144°\\)\n\n\nTransport\n10\n\\(\\frac{10}{100} \\times 360° = 36°\\)\n\n\nRent\n12.5\n\\(\\frac{12.5}{100} \\times 360° = 45°\\)\n\n\nOthers\n7.5\n\\(\\frac{7.5}{100} \\times 360° = 27°\\)\n\n\n\n100\n360°\n\n\n\nA pie- chart showing the monthly contributions and expenditure pattern of an employee in 1999.\n~(b)(i) \\(360° = N10,800.00\\)\n\\(\\text{Pension} - 18° = \\frac{18}{360} \\times N10,800\\)\n= \\(N540\\)\n(ii) \\(360° = N10,800.00\\)\n\\(\\text{Income tax} - 90° = \\frac{90}{360} \\times N10,800\\)\n= \\(N2700\\)\n(c) The percentage sum of pension and income tax = (5 + 25)% = 30%\nTake home pay = 100% - 30% = 70%\n= \\(\\frac{70}{100} \\times N10,800\\)\n= \\(N7560\\)\n~(c) The percentage sum of pension and income tax = (5 + 25)% = 30%\nTake home pay = 100% - 30% = 70%\n= \\(\\frac{70}{100} \\times N10,800\\)\n= \\(N7560\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(519, '2015', '\n\nQuestion\n\n(a) A = {1, 2, 5, 7} and B = {1, 3, 6, 7} are subsets of the universal set U = {1, 2, 3,...., 10}. Find (i) \\(A\'\\) ; (ii) \\((A \\cap B)\'\\) ; (iii) \\((A \\cup B)\'\\) ; (iv) the subsets of B each of which has three elements.\n(b) Write down the 15th term of the sequence, \\(\\frac{2}{1 \\times 3}, \\frac{2}{2 \\times 4}, \\frac{4}{3 \\times 5}, \\frac{5}{4 \\times 6},...\\).\n(c) An Arithmetic Progression (A.P) has 3 as its first term and 4 as the common difference, (i) write an expression in its simplest form for the nth term ; (ii) find the least term of the A.P that is greater than 100.\n\n', '', '(a) A = {1, 2, 5, 7} and B = {1, 3, 6, 7} are subsets of the universal set U = {1, 2, 3,...., 10}. Find (i) \\(A\'\\) ; (ii) \\((A \\cap B)\'\\) ; (iii) \\((A \\cup B)\'\\) ; (iv) the subsets of B each of which has three elements.\n(b) Write down the 15th term of the sequence, \\(\\frac{2}{1 \\times 3}, \\frac{2}{2 \\times 4}, \\frac{4}{3 \\times 5}, \\frac{5}{4 \\times 6},...\\).\n(c) An Arithmetic Progression (A.P) has 3 as its first term and 4 as the common difference, (i) write an expression in its simplest form for the nth term ; (ii) find the least term of the A.P that is greater than 100.', '(a) A = {1, 2, 5, 7} and B = {1, 3, 6, 7} are subsets of the universal set U = {1, 2, 3,...., 10}. Find (i) \\(A\'\\) ; (ii) \\((A \\cap B)\'\\) ; (iii) \\((A \\cup B)\'\\) ; (iv) the subsets of B each of which has three elements.~(b) Write down the 15th term of the sequence, \\(\\frac{2}{1 \\times 3}, \\frac{2}{2 \\times 4}, \\frac{4}{3 \\times 5}, \\frac{5}{4 \\times 6},...\\).\n(c) An Arithmetic Progression (A.P) has 3 as its first term and 4 as the common difference, (i) write an expression in its simplest form for the nth term ; (ii) find the least term of the A.P that is greater than 100.~(c) An Arithmetic Progression (A.P) has 3 as its first term and 4 as the common difference, (i) write an expression in its simplest form for the nth term ; (ii) find the least term of the A.P that is greater than 100.', NULL, '\n\nExplanation\n(a)(i) U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}\nA = {1, 2, 5, 7}\nA\' = U - A = {3, 4, 6, 8, 9, 10}\n(ii) \\(A \\cap B = {1, 2, 5, 7} \\cap {1, 3, 6, 7} = {1, 7}\\)\n\\((A \\cap B)\' = {2, 3, 4, 5, 6, 8, 9, 10}\\)\n(iii) \\(A \\cup B = {1, 2, 5, 7} \\cup {1, 3, 6, 7}\\)\n= \\({1, 2, 3, 5, 6, 7}\\)\n\\((A \\cup B)\' = {4, 8, 9,10}\\)\n(iv) {1, 3, 6}, {1, 3, 7}, {1, 6, 7} and {3, 6, 7}.\n(b) The nth term of the sequence is given as\n\\(U_{n} = \\frac{n + 1}{n(n + 2)}, n = 1, 2,3\\)\nFor the 15th term,\n\\(U_{15} = \\frac{16}{15 \\times 17}\\)\n(c)(i) \\(T_{n} = a + (n - 1)d\\)\n\\(T_{n} = 3 + (n - 1)4 = 3 + 4n - 4\\)\n= \\(4n - 1\\)\n(ii) \\(4n - 1 > 100\\)\n\\(4n > 100 + 1 \\implies 4n > 101\\)\n\\(n > 25.25\\)\nThe least term greater than 100 = 26th term.\n\n', '', 'Explanation\n(a)(i) U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}\nA = {1, 2, 5, 7}\nA\' = U - A = {3, 4, 6, 8, 9, 10}\n(ii) \\(A \\cap B = {1, 2, 5, 7} \\cap {1, 3, 6, 7} = {1, 7}\\)\n\\((A \\cap B)\' = {2, 3, 4, 5, 6, 8, 9, 10}\\)\n(iii) \\(A \\cup B = {1, 2, 5, 7} \\cup {1, 3, 6, 7}\\)\n= \\({1, 2, 3, 5, 6, 7}\\)\n\\((A \\cup B)\' = {4, 8, 9,10}\\)\n(iv) {1, 3, 6}, {1, 3, 7}, {1, 6, 7} and {3, 6, 7}.\n(b) The nth term of the sequence is given as\n\\(U_{n} = \\frac{n + 1}{n(n + 2)}, n = 1, 2,3\\)\nFor the 15th term,\n\\(U_{15} = \\frac{16}{15 \\times 17}\\)\n(c)(i) \\(T_{n} = a + (n - 1)d\\)\n\\(T_{n} = 3 + (n - 1)4 = 3 + 4n - 4\\)\n= \\(4n - 1\\)\n(ii) \\(4n - 1 > 100\\)\n\\(4n > 100 + 1 \\implies 4n > 101\\)\n\\(n > 25.25\\)\nThe least term greater than 100 = 26th term.\n\n', '(a)(i) U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}\nA = {1, 2, 5, 7}\nA\' = U - A = {3, 4, 6, 8, 9, 10}\n(ii) \\(A \\cap B = {1, 2, 5, 7} \\cap {1, 3, 6, 7} = {1, 7}\\)\n\\((A \\cap B)\' = {2, 3, 4, 5, 6, 8, 9, 10}\\)\n(iii) \\(A \\cup B = {1, 2, 5, 7} \\cup {1, 3, 6, 7}\\)\n= \\({1, 2, 3, 5, 6, 7}\\)\n\\((A \\cup B)\' = {4, 8, 9,10}\\)\n(iv) {1, 3, 6}, {1, 3, 7}, {1, 6, 7} and {3, 6, 7}.~(b) The nth term of the sequence is given as\n\\(U_{n} = \\frac{n + 1}{n(n + 2)}, n = 1, 2,3\\)\nFor the 15th term,\n\\(U_{15} = \\frac{16}{15 \\times 17}\\)\n(c)(i) \\(T_{n} = a + (n - 1)d\\)\n\\(T_{n} = 3 + (n - 1)4 = 3 + 4n - 4\\)\n= \\(4n - 1\\)\n(ii) \\(4n - 1 > 100\\)\n\\(4n > 100 + 1 \\implies 4n > 101\\)\n\\(n > 25.25\\)\nThe least term greater than 100 = 26th term.\n~(c)(i) \\(T_{n} = a + (n - 1)d\\)\n\\(T_{n} = 3 + (n - 1)4 = 3 + 4n - 4\\)\n= \\(4n - 1\\)\n(ii) \\(4n - 1 > 100\\)\n\\(4n > 100 + 1 \\implies 4n > 101\\)\n\\(n > 25.25\\)\nThe least term greater than 100 = 26th term.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(520, '2015', '\n\nQuestion\n\nThe marks obtained by 40 students in an examination are as follows :\n85 77 87 74 77 78 79 89 95 90 78 73 86 83 91 74 84 81 83 75 77 70 81 69 75 63 76 87 61 78 69 96 65 80 84 80 77 74 88 72.\n(a) Copy and complete the table for the distribution using the above data.\n\n\n\nClass Boundaries\nTally\nFrequency\n\n\n59.5 - 64.5\n\n\n\n\n64.5 - 69.5\n\n\n\n\n69.5 - 74.5\n\n\n\n\n74.5 - 79.5\n\n\n\n\n79.5 - 84.5\n\n\n\n\n84.5 - 89.5\n\n\n\n\n89.5 - 94.5\n\n\n\n\n94.5 - 99.5\n\n\n\n\n\n(b) Draw a histogram to represent the distribution.\n(c) Using your histogram, estimate the modal mark.\n(d) If a student is chosen at random, find the probability that the student obtains a mark greater than 79.\n\n\nThe marks obtained by 40 students in an examination are as follows :\n85 77 87 74 77 78 79 89 95 90 78 73 86 83 91 74 84 81 83 75 77 70 81 69 75 63 76 87 61 78 69 96 65 80 84 80 77 74 88 72.\n(a) Copy and complete the table for the distribution using the above data.\n\n\n\nClass Boundaries\nTally\nFrequency\n\n\n59.5 - 64.5\n\n\n\n\n64.5 - 69.5\n\n\n\n\n69.5 - 74.5\n\n\n\n\n74.5 - 79.5\n\n\n\n\n79.5 - 84.5\n\n\n\n\n84.5 - 89.5\n\n\n\n\n89.5 - 94.5\n\n\n\n\n94.5 - 99.5\n\n\n\n\n', '', 'The marks obtained by 40 students in an examination are as follows :\n85 77 87 74 77 78 79 89 95 90 78 73 86 83 91 74 84 81 83 75 77 70 81 69 75 63 76 87 61 78 69 96 65 80 84 80 77 74 88 72.\n(a) Copy and complete the table for the distribution using the above data.\n\n\n\nClass Boundaries\nTally\nFrequency\n\n\n59.5 - 64.5\n\n\n\n\n64.5 - 69.5\n\n\n\n\n69.5 - 74.5\n\n\n\n\n74.5 - 79.5\n\n\n\n\n79.5 - 84.5\n\n\n\n\n84.5 - 89.5\n\n\n\n\n89.5 - 94.5\n\n\n\n\n94.5 - 99.5\n\n\n\n\n\n(b) Draw a histogram to represent the distribution.\n(c) Using your histogram, estimate the modal mark.\n(d) If a student is chosen at random, find the probability that the student obtains a mark greater than 79.\n\n\nThe marks obtained by 40 students in an examination are as follows :\n85 77 87 74 77 78 79 89 95 90 78 73 86 83 91 74 84 81 83 75 77 70 81 69 75 63 76 87 61 78 69 96 65 80 84 80 77 74 88 72.\n(a) Copy and complete the table for the distribution using the above data.\n\n\n\nClass Boundaries\nTally\nFrequency\n\n\n59.5 - 64.5\n\n\n\n\n64.5 - 69.5\n\n\n\n\n69.5 - 74.5\n\n\n\n\n74.5 - 79.5\n\n\n\n\n79.5 - 84.5\n\n\n\n\n84.5 - 89.5\n\n\n\n\n89.5 - 94.5\n\n\n\n\n94.5 - 99.5\n\n\n', 'The marks obtained by 40 students in an examination are as follows :\n85 77 87 74 77 78 79 89 95 90 78 73 86 83 91 74 84 81 83 75 77 70 81 69 75 63 76 87 61 78 69 96 65 80 84 80 77 74 88 72.\n(a) Copy and complete the table for the distribution using the above data.\n\n\n\nClass Boundaries\nTally\nFrequency\n\n\n59.5 - 64.5\n\n\n\n\n64.5 - 69.5\n\n\n\n\n69.5 - 74.5\n\n\n\n\n74.5 - 79.5\n\n\n\n\n79.5 - 84.5\n\n\n\n\n84.5 - 89.5\n\n\n\n\n89.5 - 94.5\n\n\n\n\n94.5 - 99.5\n\n\n\n\n~(b) Draw a histogram to represent the distribution.\n(c) Using your histogram, estimate the modal mark.\n(d) If a student is chosen at random, find the probability that the student obtains a mark greater than 79.\n\n\nThe marks obtained by 40 students in an examination are as follows :\n85 77 87 74 77 78 79 89 95 90 78 73 86 83 91 74 84 81 83 75 77 70 81 69 75 63 76 87 61 78 69 96 65 80 84 80 77 74 88 72.\n(a) Copy and complete the table for the distribution using the above data.\n\n\n\nClass Boundaries\nTally\nFrequency\n\n\n59.5 - 64.5\n\n\n\n\n64.5 - 69.5\n\n\n\n\n69.5 - 74.5\n\n\n\n\n74.5 - 79.5\n\n\n\n\n79.5 - 84.5\n\n\n\n\n84.5 - 89.5\n\n\n\n\n89.5 - 94.5\n\n\n\n\n94.5 - 99.5\n\n\n~(c) Using your histogram, estimate the modal mark.\n(d) If a student is chosen at random, find the probability that the student obtains a mark greater than 79.\n\n\nThe marks obtained by 40 students in an examination are as follows :\n85 77 87 74 77 78 79 89 95 90 78 73 86 83 91 74 84 81 83 75 77 70 81 69 75 63 76 87 61 78 69 96 65 80 84 80 77 74 88 72.\n(a) Copy and complete the table for the distribution using the above data.\n\n\n\nClass Boundaries\nTally\nFrequency\n\n\n59.5 - 64.5\n\n\n\n\n64.5 - 69.5\n\n\n\n\n69.5 - 74.5\n\n\n\n\n74.5 - 79.5\n\n\n\n\n79.5 - 84.5\n\n\n\n\n84.5 - 89.5\n\n\n\n\n89.5 - 94.5\n\n\n\n\n94.5 - 99.5\n\n\n~(d) If a student is chosen at random, find the probability that the student obtains a mark greater than 79.\n\n\nThe marks obtained by 40 students in an examination are as follows :\n85 77 87 74 77 78 79 89 95 90 78 73 86 83 91 74 84 81 83 75 77 70 81 69 75 63 76 87 61 78 69 96 65 80 84 80 77 74 88 72.\n(a) Copy and complete the table for the distribution using the above data.\n\n\n\nClass Boundaries\nTally\nFrequency\n\n\n59.5 - 64.5\n\n\n\n\n64.5 - 69.5\n\n\n\n\n69.5 - 74.5\n\n\n\n\n74.5 - 79.5\n\n\n\n\n79.5 - 84.5\n\n\n\n\n84.5 - 89.5\n\n\n\n\n89.5 - 94.5\n\n\n\n\n94.5 - 99.5\n\n\n', NULL, '\n\nExplanation\n(a)\n\n\n\nClass Boundaries\nTally\nFrequency\n\n\n59.5 - 64.5\n||\n2\n\n\n64.5 - 69.5\n|||\n3\n\n\n69.5 - 74.5\n|||||\n6\n\n\n74.5 - 79.5\n|||||||||\n11\n\n\n79.5 - 84.5\n|||||||\n8\n\n\n84.5 - 89.5\n||||||\n7\n\n\n89.5 - 94.5\n||\n2\n\n\n94.5 - 99.5\n|\n1\n\n\n\n(b)\nAn histogram to represent the distribution of 40 students in an examination.\n(c) Modal mark : \\(\\frac{74.5 + 79.5}{2} = \\frac{154}{2} = 77\\)\n(d) Number of students who obtained marks greater than 79 : 8 + 7 + 2 + 1 = 18.\nTotal number of students = 40\n\\(\\therefore\\) P(mark greater than 79) = \\(\\frac{18}{40} = \\frac{9}{20}\\)\n\n', '', 'Explanation\n(a)\n\n\n\nClass Boundaries\nTally\nFrequency\n\n\n59.5 - 64.5\n||\n2\n\n\n64.5 - 69.5\n|||\n3\n\n\n69.5 - 74.5\n|||||\n6\n\n\n74.5 - 79.5\n|||||||||\n11\n\n\n79.5 - 84.5\n|||||||\n8\n\n\n84.5 - 89.5\n||||||\n7\n\n\n89.5 - 94.5\n||\n2\n\n\n94.5 - 99.5\n|\n1\n\n\n\n(b)\nAn histogram to represent the distribution of 40 students in an examination.\n(c) Modal mark : \\(\\frac{74.5 + 79.5}{2} = \\frac{154}{2} = 77\\)\n(d) Number of students who obtained marks greater than 79 : 8 + 7 + 2 + 1 = 18.\nTotal number of students = 40\n\\(\\therefore\\) P(mark greater than 79) = \\(\\frac{18}{40} = \\frac{9}{20}\\)\n\n', '(a)\n\n\n\nClass Boundaries\nTally\nFrequency\n\n\n59.5 - 64.5\n||\n2\n\n\n64.5 - 69.5\n|||\n3\n\n\n69.5 - 74.5\n|||||\n6\n\n\n74.5 - 79.5\n|||||||||\n11\n\n\n79.5 - 84.5\n|||||||\n8\n\n\n84.5 - 89.5\n||||||\n7\n\n\n89.5 - 94.5\n||\n2\n\n\n94.5 - 99.5\n|\n1\n\n\n~(b)\nAn histogram to represent the distribution of 40 students in an examination.\n(c) Modal mark : \\(\\frac{74.5 + 79.5}{2} = \\frac{154}{2} = 77\\)\n(d) Number of students who obtained marks greater than 79 : 8 + 7 + 2 + 1 = 18.\nTotal number of students = 40\n\\(\\therefore\\) P(mark greater than 79) = \\(\\frac{18}{40} = \\frac{9}{20}\\)\n~(c) Modal mark : \\(\\frac{74.5 + 79.5}{2} = \\frac{154}{2} = 77\\)\n(d) Number of students who obtained marks greater than 79 : 8 + 7 + 2 + 1 = 18.\nTotal number of students = 40\n\\(\\therefore\\) P(mark greater than 79) = \\(\\frac{18}{40} = \\frac{9}{20}\\)\n~(d) Number of students who obtained marks greater than 79 : 8 + 7 + 2 + 1 = 18.\nTotal number of students = 40\n\\(\\therefore\\) P(mark greater than 79) = \\(\\frac{18}{40} = \\frac{9}{20}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(521, '2015', '\n\nQuestion\n\nIn the diagram, PQT is a straight line and SQ // RT.\n\n(a) Join QR and show that : (i) < RPS = < QRT ; (ii) < PRS = < QTR.\n(b) ABC is a triangle. The sides AB and AC are produced to D and E respectively such that < DBC = 132° and < ECD = 96°. Show that \\(\\Delta\\) ABC is isosceles.\n\n', '', 'In the diagram, PQT is a straight line and SQ // RT.\n\n(a) Join QR and show that : (i) < RPS = < QRT ; (ii) < PRS = < QTR.\n(b) ABC is a triangle. The sides AB and AC are produced to D and E respectively such that < DBC = 132° and < ECD = 96°. Show that \\(\\Delta\\) ABC is isosceles.', 'In the diagram, PQT is a straight line and SQ // RT.\n\n(a) Join QR and show that : (i) < RPS = < QRT ; (ii) < PRS = < QTR.~(b) ABC is a triangle. The sides AB and AC are produced to D and E respectively such that < DBC = 132° and < ECD = 96°. Show that \\(\\Delta\\) ABC is isosceles.', NULL, '\n\nExplanation\n(a) < QRT = < RQS (alternate angles)\n< RPS = < RQS (angles in the same segment)\n\\(\\therefore\\) < RPS = < QRT\n\n(ii) < QRT = < PQS (corresponding angles)\n< PQS = < PRS (angles in the same segment)\n\\(\\therefore\\) < QTR = < PRS.\n(b) \\(< ABC = \\hat{B} = 180° - 132° = 48°\\)\n\\(< ACB = \\hat{C} = 180° - 96° = 84°\\)\n\\(< BAC = \\hat{A} = 180° - (48° + 84°) = 48°\\)\n\\(< BAC = < ABC = 48°\\)\n\\(\\therefore \\Delta ABC = isosceles\\).\n\n\n', '', 'Explanation\n(a) < QRT = < RQS (alternate angles)\n< RPS = < RQS (angles in the same segment)\n\\(\\therefore\\) < RPS = < QRT\n\n(ii) < QRT = < PQS (corresponding angles)\n< PQS = < PRS (angles in the same segment)\n\\(\\therefore\\) < QTR = < PRS.\n(b) \\(< ABC = \\hat{B} = 180° - 132° = 48°\\)\n\\(< ACB = \\hat{C} = 180° - 96° = 84°\\)\n\\(< BAC = \\hat{A} = 180° - (48° + 84°) = 48°\\)\n\\(< BAC = < ABC = 48°\\)\n\\(\\therefore \\Delta ABC = isosceles\\).\n\n\n', '(a) < QRT = < RQS (alternate angles)\n< RPS = < RQS (angles in the same segment)\n\\(\\therefore\\) < RPS = < QRT\n\n(ii) < QRT = < PQS (corresponding angles)\n< PQS = < PRS (angles in the same segment)\n\\(\\therefore\\) < QTR = < PRS.~(b) \\(< ABC = \\hat{B} = 180° - 132° = 48°\\)\n\\(< ACB = \\hat{C} = 180° - 96° = 84°\\)\n\\(< BAC = \\hat{A} = 180° - (48° + 84°) = 48°\\)\n\\(< BAC = < ABC = 48°\\)\n\\(\\therefore \\Delta ABC = isosceles\\).\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(522, '2015', '\n\nQuestion\n\nAn aeroplane flies due west for 3 hours from P (lat. 50°N, long. 60°W) to a point Q at an average speed of 600km/h. The aeroplane then flies due south from Q to a point Y 500km away. Calculate, correct to 3 significant figures,\n(a) the longitude of Q ;\n(b) the latitude of Y . [Take the radius of the earth = 6400km and \\(\\pi = \\frac{22}{7}\\)].\n\n', '', 'An aeroplane flies due west for 3 hours from P (lat. 50°N, long. 60°W) to a point Q at an average speed of 600km/h. The aeroplane then flies due south from Q to a point Y 500km away. Calculate, correct to 3 significant figures,\n(a) the longitude of Q ;\n(b) the latitude of Y . [Take the radius of the earth = 6400km and \\(\\pi = \\frac{22}{7}\\)].', 'An aeroplane flies due west for 3 hours from P (lat. 50°N, long. 60°W) to a point Q at an average speed of 600km/h. The aeroplane then flies due south from Q to a point Y 500km away. Calculate, correct to 3 significant figures,\n(a) the longitude of Q ;~(b) the latitude of Y . [Take the radius of the earth = 6400km and \\(\\pi = \\frac{22}{7}\\)].', NULL, '\n\nExplanation\n(a)\n\\(\\frac{r}{R} = \\cos 50\\)\n\\(r = 6400 \\cos 50 = 4,113.84 km\\)\n\\(\\approxeq 4,114km\\)\nThe distance PQ along the parallel of latitude = speed (km/h) \\(\\times\\) time (h)\n= 600km/h x 3h\n= 1800km\n\\(\\frac{\\theta}{360} \\times 2 \\times \\frac{22}{7} \\times r = 1800\\)\n\\(\\theta = \\frac{1800 \\times 7 \\times 360}{44 \\times 4113.84}\\)\n\\(\\theta = 25.059° \\approxeq 25.1°\\)\n\\(\\therefore\\) The longitude of Q = \\(60 + \\theta\\)\n= \\(60 + 25.1 = 85.1°W\\)\n(b) The distance (Q) along the meridian\n\\(\\frac{\\theta}{360} \\times 2\\pi R\\)\n\\(500 = \\frac{\\theta}{360} \\times 2 \\times \\frac{22}{7} \\times 6400\\)\n\\(\\theta = \\frac{500 \\times 360 \\times 7}{44 \\times 6400}\\)\n\\(\\theta = 4.47°\\)\n\\(\\therefore \\text{The latitude of Y} = 50° - 4.47° = 45.53°\\)\n\\(\\approxeq 45.5°N\\)\n\n', '', 'Explanation\n(a)\n\\(\\frac{r}{R} = \\cos 50\\)\n\\(r = 6400 \\cos 50 = 4,113.84 km\\)\n\\(\\approxeq 4,114km\\)\nThe distance PQ along the parallel of latitude = speed (km/h) \\(\\times\\) time (h)\n= 600km/h x 3h\n= 1800km\n\\(\\frac{\\theta}{360} \\times 2 \\times \\frac{22}{7} \\times r = 1800\\)\n\\(\\theta = \\frac{1800 \\times 7 \\times 360}{44 \\times 4113.84}\\)\n\\(\\theta = 25.059° \\approxeq 25.1°\\)\n\\(\\therefore\\) The longitude of Q = \\(60 + \\theta\\)\n= \\(60 + 25.1 = 85.1°W\\)\n(b) The distance (Q) along the meridian\n\\(\\frac{\\theta}{360} \\times 2\\pi R\\)\n\\(500 = \\frac{\\theta}{360} \\times 2 \\times \\frac{22}{7} \\times 6400\\)\n\\(\\theta = \\frac{500 \\times 360 \\times 7}{44 \\times 6400}\\)\n\\(\\theta = 4.47°\\)\n\\(\\therefore \\text{The latitude of Y} = 50° - 4.47° = 45.53°\\)\n\\(\\approxeq 45.5°N\\)\n\n', '(a)\n\\(\\frac{r}{R} = \\cos 50\\)\n\\(r = 6400 \\cos 50 = 4,113.84 km\\)\n\\(\\approxeq 4,114km\\)\nThe distance PQ along the parallel of latitude = speed (km/h) \\(\\times\\) time (h)\n= 600km/h x 3h\n= 1800km\n\\(\\frac{\\theta}{360} \\times 2 \\times \\frac{22}{7} \\times r = 1800\\)\n\\(\\theta = \\frac{1800 \\times 7 \\times 360}{44 \\times 4113.84}\\)\n\\(\\theta = 25.059° \\approxeq 25.1°\\)\n\\(\\therefore\\) The longitude of Q = \\(60 + \\theta\\)\n= \\(60 + 25.1 = 85.1°W\\)~(b) The distance (Q) along the meridian\n\\(\\frac{\\theta}{360} \\times 2\\pi R\\)\n\\(500 = \\frac{\\theta}{360} \\times 2 \\times \\frac{22}{7} \\times 6400\\)\n\\(\\theta = \\frac{500 \\times 360 \\times 7}{44 \\times 6400}\\)\n\\(\\theta = 4.47°\\)\n\\(\\therefore \\text{The latitude of Y} = 50° - 4.47° = 45.53°\\)\n\\(\\approxeq 45.5°N\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(523, '2015', '\n\nQuestion\n\nUsing ruler and a pair of compasses only,\n(a) construct a quadrilateral PXYQ such that /PX/ = 9.9 cm, /QX/ = 10.2 cm, < QPZ = 75°, /QY/ = 10.4 cm and PQ // XY.\n(b) Construct the (i) locus \\(l_{1}\\) of points equidistant from X and Y ; (ii) locus \\(l_{2}\\) of points equidistant from QY and YX.\n(c) Locate M, the point of intersection of \\(l_{1}\\) and \\(l_{2}\\).\n(d) Measure /PM/.\n\n', '', 'Using ruler and a pair of compasses only,\n(a) construct a quadrilateral PXYQ such that /PX/ = 9.9 cm, /QX/ = 10.2 cm, < QPZ = 75°, /QY/ = 10.4 cm and PQ // XY.\n(b) Construct the (i) locus \\(l_{1}\\) of points equidistant from X and Y ; (ii) locus \\(l_{2}\\) of points equidistant from QY and YX.\n(c) Locate M, the point of intersection of \\(l_{1}\\) and \\(l_{2}\\).\n(d) Measure /PM/.', 'Using ruler and a pair of compasses only,\n(a) construct a quadrilateral PXYQ such that /PX/ = 9.9 cm, /QX/ = 10.2 cm, < QPZ = 75°, /QY/ = 10.4 cm and PQ // XY.~(b) Construct the (i) locus \\(l_{1}\\) of points equidistant from X and Y ; (ii) locus \\(l_{2}\\) of points equidistant from QY and YX.\n(c) Locate M, the point of intersection of \\(l_{1}\\) and \\(l_{2}\\).\n(d) Measure /PM/.~(c) Locate M, the point of intersection of \\(l_{1}\\) and \\(l_{2}\\).\n(d) Measure /PM/.~(d) Measure /PM/.', NULL, '\n\nExplanation\n(d) /PM/ = 9.7 cm.\n\n', '(d) /PM/ = 9.7 cm.\n\n', '', '', 'WAEC-PK-1', 'WAEC-MATH', 1),
(524, '2015', '\n\nQuestion\n\nThe table below shows the values of the relation \\(y = 11 - 2x - 2x^{2}\\) for \\(-4 \\leq x \\leq 3\\).\n\n\n\nx\n-4\n-3\n-2\n-1\n0\n1\n2\n3\n\n\ny\n-13\n\n\n\n11\n\n\n\n\n\n\n(a) Copy and complete the table.\n(b) Using a scale of 2 cm to 1 unit on the x- axis and 2 cm to 5 units on the y- axis, draw the graph of \\(y = 11 - 2x - 2x^{2}\\).\n(c) Use your graph to find : (i) the roots of the equation \\(11 - 2x - 2x^{2} = 0\\) ; (ii) the values of x for which \\(3 - 2x - 2x^{2} = 0\\) ; (iii) the gradient of the curve at x = 1.\n\n\n\nThe table below shows the values of the relation \\(y = 11 - 2x - 2x^{2}\\) for \\(-4 \\leq x \\leq 3\\).\n\n\n\nx\n-4\n-3\n-2\n-1\n0\n1\n2\n3\n\n\ny\n-13\n\n\n\n11\n\n\n\n\n\n', '', 'The table below shows the values of the relation \\(y = 11 - 2x - 2x^{2}\\) for \\(-4 \\leq x \\leq 3\\).\n\n\n\nx\n-4\n-3\n-2\n-1\n0\n1\n2\n3\n\n\ny\n-13\n\n\n\n11\n\n\n\n\n\n\n(a) Copy and complete the table.\n(b) Using a scale of 2 cm to 1 unit on the x- axis and 2 cm to 5 units on the y- axis, draw the graph of \\(y = 11 - 2x - 2x^{2}\\).\n(c) Use your graph to find : (i) the roots of the equation \\(11 - 2x - 2x^{2} = 0\\) ; (ii) the values of x for which \\(3 - 2x - 2x^{2} = 0\\) ; (iii) the gradient of the curve at x = 1.\n\n\n\nThe table below shows the values of the relation \\(y = 11 - 2x - 2x^{2}\\) for \\(-4 \\leq x \\leq 3\\).\n\n\n\nx\n-4\n-3\n-2\n-1\n0\n1\n2\n3\n\n\ny\n-13\n\n\n\n11\n\n\n\n', 'The table below shows the values of the relation \\(y = 11 - 2x - 2x^{2}\\) for \\(-4 \\leq x \\leq 3\\).\n\n\n\nx\n-4\n-3\n-2\n-1\n0\n1\n2\n3\n\n\ny\n-13\n\n\n\n11\n\n\n\n\n\n\n(a) Copy and complete the table.~(b) Using a scale of 2 cm to 1 unit on the x- axis and 2 cm to 5 units on the y- axis, draw the graph of \\(y = 11 - 2x - 2x^{2}\\).\n(c) Use your graph to find : (i) the roots of the equation \\(11 - 2x - 2x^{2} = 0\\) ; (ii) the values of x for which \\(3 - 2x - 2x^{2} = 0\\) ; (iii) the gradient of the curve at x = 1.\n\n\n\nThe table below shows the values of the relation \\(y = 11 - 2x - 2x^{2}\\) for \\(-4 \\leq x \\leq 3\\).\n\n\n\nx\n-4\n-3\n-2\n-1\n0\n1\n2\n3\n\n\ny\n-13\n\n\n\n11\n\n\n\n~(c) Use your graph to find : (i) the roots of the equation \\(11 - 2x - 2x^{2} = 0\\) ; (ii) the values of x for which \\(3 - 2x - 2x^{2} = 0\\) ; (iii) the gradient of the curve at x = 1.\n\n\n\nThe table below shows the values of the relation \\(y = 11 - 2x - 2x^{2}\\) for \\(-4 \\leq x \\leq 3\\).\n\n\n\nx\n-4\n-3\n-2\n-1\n0\n1\n2\n3\n\n\ny\n-13\n\n\n\n11\n\n\n\n', NULL, '\n\nExplanation\n(a)\n\n\n\nx\n-4\n-3\n-2\n-1\n0\n1\n2\n3\n\n\n\\(x^{2}\\)\n16\n9\n4\n1\n0\n1\n4\n9\n\n\n\\(11\\)\n11\n11\n11\n11\n11\n11\n11\n11\n\n\n\\(-2x\\)\n8\n6\n4\n2\n0\n-2\n-4\n-6\n\n\n\\(-2x^{2}\\)\n-32\n-18\n-8\n-2\n0\n-2\n-8\n-18\n\n\ny\n-13\n-1\n7\n11\n11\n7\n-1\n-13\n\n\n\n(b)\n(c)(i) x = -2.9 or 1.9\n(ii) \\(y + (3 - 2x - 2x^{2}) = 11 - 2x - 2x^{2}\\)\n\\(y = 11 - 2x - 2x^{2} - 3 + 2x + 2x^{2}\\)\n\\(y = 8\\)\n\\(\\therefore x = \\text{-1.8 or 0.8}\\)\n(iii) \\(y = 11 - 2x - 2x^{2}\\)\n\\(\\frac{\\mathrm d y}{\\mathrm d x} = -2 - 4x\\)\nGradient at x = 1 : \\(-2 - 4(1) = -2 - 4 = -6\\)\n\n', '', 'Explanation\n(a)\n\n\n\nx\n-4\n-3\n-2\n-1\n0\n1\n2\n3\n\n\n\\(x^{2}\\)\n16\n9\n4\n1\n0\n1\n4\n9\n\n\n\\(11\\)\n11\n11\n11\n11\n11\n11\n11\n11\n\n\n\\(-2x\\)\n8\n6\n4\n2\n0\n-2\n-4\n-6\n\n\n\\(-2x^{2}\\)\n-32\n-18\n-8\n-2\n0\n-2\n-8\n-18\n\n\ny\n-13\n-1\n7\n11\n11\n7\n-1\n-13\n\n\n\n(b)\n(c)(i) x = -2.9 or 1.9\n(ii) \\(y + (3 - 2x - 2x^{2}) = 11 - 2x - 2x^{2}\\)\n\\(y = 11 - 2x - 2x^{2} - 3 + 2x + 2x^{2}\\)\n\\(y = 8\\)\n\\(\\therefore x = \\text{-1.8 or 0.8}\\)\n(iii) \\(y = 11 - 2x - 2x^{2}\\)\n\\(\\frac{\\mathrm d y}{\\mathrm d x} = -2 - 4x\\)\nGradient at x = 1 : \\(-2 - 4(1) = -2 - 4 = -6\\)\n\n', '(a)\n\n\n\nx\n-4\n-3\n-2\n-1\n0\n1\n2\n3\n\n\n\\(x^{2}\\)\n16\n9\n4\n1\n0\n1\n4\n9\n\n\n\\(11\\)\n11\n11\n11\n11\n11\n11\n11\n11\n\n\n\\(-2x\\)\n8\n6\n4\n2\n0\n-2\n-4\n-6\n\n\n\\(-2x^{2}\\)\n-32\n-18\n-8\n-2\n0\n-2\n-8\n-18\n\n\ny\n-13\n-1\n7\n11\n11\n7\n-1\n-13\n\n\n~(b)\n(c)(i) x = -2.9 or 1.9\n(ii) \\(y + (3 - 2x - 2x^{2}) = 11 - 2x - 2x^{2}\\)\n\\(y = 11 - 2x - 2x^{2} - 3 + 2x + 2x^{2}\\)\n\\(y = 8\\)\n\\(\\therefore x = \\text{-1.8 or 0.8}\\)\n(iii) \\(y = 11 - 2x - 2x^{2}\\)\n\\(\\frac{\\mathrm d y}{\\mathrm d x} = -2 - 4x\\)\nGradient at x = 1 : \\(-2 - 4(1) = -2 - 4 = -6\\)\n~(c)(i) x = -2.9 or 1.9\n(ii) \\(y + (3 - 2x - 2x^{2}) = 11 - 2x - 2x^{2}\\)\n\\(y = 11 - 2x - 2x^{2} - 3 + 2x + 2x^{2}\\)\n\\(y = 8\\)\n\\(\\therefore x = \\text{-1.8 or 0.8}\\)\n(iii) \\(y = 11 - 2x - 2x^{2}\\)\n\\(\\frac{\\mathrm d y}{\\mathrm d x} = -2 - 4x\\)\nGradient at x = 1 : \\(-2 - 4(1) = -2 - 4 = -6\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(525, '2015', '\n\nQuestion\n\n(a) Simplify : \\((2a + b)^{2} - (b - 2a)^{2}\\)\n(b) Given that \\(S = K\\sqrt{m^{2} + n^{2}}\\); (i) make m the subject of the relations ; (ii) if S = 12.2, K = 0.02 and n = 1.1, find, correct to the nearest whole number, the positive value of m.\n\n', '', '(a) Simplify : \\((2a + b)^{2} - (b - 2a)^{2}\\)\n(b) Given that \\(S = K\\sqrt{m^{2} + n^{2}}\\); (i) make m the subject of the relations ; (ii) if S = 12.2, K = 0.02 and n = 1.1, find, correct to the nearest whole number, the positive value of m.', '(a) Simplify : \\((2a + b)^{2} - (b - 2a)^{2}\\)~(b) Given that \\(S = K\\sqrt{m^{2} + n^{2}}\\); (i) make m the subject of the relations ; (ii) if S = 12.2, K = 0.02 and n = 1.1, find, correct to the nearest whole number, the positive value of m.', NULL, '\n\nExplanation\n(a) \\((2a + b)^{2} - (b - 2a)^{2}\\)\nUsing the method of difference of two squares,\n\\((2a + b)^{2} - (b - 2a)^{2} = ((2a + b) + (b - 2a))((2a + b) - (b - 2a))\\)\n= \\((2b)(4a)\\)\n= \\(8ab\\).\n(b)(i) \\(S = K\\sqrt{m^{2} + n^{2}}\\)\n\\(S^{2} = K^{2}[m^{2} + n^{2}]\\)\n\\(\\frac{S^{2}}{K^{2}} = m^{2} + n^{2}\\)\n\\(m^{2} = \\frac{S^{2}}{K^{2}} - n^{2}\\)\n\\(m = \\sqrt{\\frac{S^{2}}{K^{2}} - n^{2}}\\)\n(ii) When S = 12.2, K = 0.02, n = 1.1\n\\(m = \\sqrt{\\frac{12.2^{2}}{0.02^{2}} - (1.1^{2})}\\)\n\\(m = \\sqrt{\\frac{148.84}{0.0004} - 1.21}\\)\n\\(m = \\sqrt{372,098.79}\\)\n= \\(609.999 \\approxeq 610\\)\n\n', '', 'Explanation\n(a) \\((2a + b)^{2} - (b - 2a)^{2}\\)\nUsing the method of difference of two squares,\n\\((2a + b)^{2} - (b - 2a)^{2} = ((2a + b) + (b - 2a))((2a + b) - (b - 2a))\\)\n= \\((2b)(4a)\\)\n= \\(8ab\\).\n(b)(i) \\(S = K\\sqrt{m^{2} + n^{2}}\\)\n\\(S^{2} = K^{2}[m^{2} + n^{2}]\\)\n\\(\\frac{S^{2}}{K^{2}} = m^{2} + n^{2}\\)\n\\(m^{2} = \\frac{S^{2}}{K^{2}} - n^{2}\\)\n\\(m = \\sqrt{\\frac{S^{2}}{K^{2}} - n^{2}}\\)\n(ii) When S = 12.2, K = 0.02, n = 1.1\n\\(m = \\sqrt{\\frac{12.2^{2}}{0.02^{2}} - (1.1^{2})}\\)\n\\(m = \\sqrt{\\frac{148.84}{0.0004} - 1.21}\\)\n\\(m = \\sqrt{372,098.79}\\)\n= \\(609.999 \\approxeq 610\\)\n\n', '(a) \\((2a + b)^{2} - (b - 2a)^{2}\\)\nUsing the method of difference of two squares,\n\\((2a + b)^{2} - (b - 2a)^{2} = ((2a + b) + (b - 2a))((2a + b) - (b - 2a))\\)\n= \\((2b)(4a)\\)\n= \\(8ab\\).~b)^{2} - (b - 2a)^{2}\\)\nUsing the method of difference of two squares,\n\\((2a + b)^{2} - (b - 2a)^{2} = ((2a + b) + (b - 2a))((2a + b) - (b - 2a))\\)\n= \\((2b)(4a)\\)\n= \\(8ab\\).\n(b)(i) \\(S = K\\sqrt{m^{2} + n^{2}}\\)\n\\(S^{2} = K^{2}[m^{2} + n^{2}]\\)\n\\(\\frac{S^{2}}{K^{2}} = m^{2} + n^{2}\\)\n\\(m^{2} = \\frac{S^{2}}{K^{2}} - n^{2}\\)\n\\(m = \\sqrt{\\frac{S^{2}}{K^{2}} - n^{2}}\\)\n(ii) When S = 12.2, K = 0.02, n = 1.1\n\\(m = \\sqrt{\\frac{12.2^{2}}{0.02^{2}} - (1.1^{2})}\\)\n\\(m = \\sqrt{\\frac{148.84}{0.0004} - 1.21}\\)\n\\(m = \\sqrt{372,098.79}\\)\n= \\(609.999 \\approxeq 610\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(526, '2015', '\n\nQuestion\n\nThe sets A = {1, 3, 5, 7, 9, 11}, B = {2, 3, 5, 7, 11, 15} and C = {3, 6, 9, 12, 15} are subsets of \\(\\varepsilon\\) = {1, 2, 3, ..., 15}.\n(a) Draw a Venn diagram to illustrate the given information.\n(b) Use your diagram to find : (i) \\(C \\cap A\'\\) ; (ii) \\(A\' \\cap (B \\cup C)\\).\n\n', '', 'The sets A = {1, 3, 5, 7, 9, 11}, B = {2, 3, 5, 7, 11, 15} and C = {3, 6, 9, 12, 15} are subsets of \\(\\varepsilon\\) = {1, 2, 3, ..., 15}.\n(a) Draw a Venn diagram to illustrate the given information.\n(b) Use your diagram to find : (i) \\(C \\cap A\'\\) ; (ii) \\(A\' \\cap (B \\cup C)\\).', 'The sets A = {1, 3, 5, 7, 9, 11}, B = {2, 3, 5, 7, 11, 15} and C = {3, 6, 9, 12, 15} are subsets of \\(\\varepsilon\\) = {1, 2, 3, ..., 15}.\n(a) Draw a Venn diagram to illustrate the given information.~(b) Use your diagram to find : (i) \\(C \\cap A\'\\) ; (ii) \\(A\' \\cap (B \\cup C)\\).', NULL, '\n\nExplanation\n(a)\n(b)(i) \\(C \\cap A\' = {15}\\)\n(ii) \\(A\' \\cap (B \\cup C) = {2, 6, 12, 15}\\)\n\n', '', 'Explanation\n(a)\n(b)(i) \\(C \\cap A\' = {15}\\)\n(ii) \\(A\' \\cap (B \\cup C) = {2, 6, 12, 15}\\)\n\n', '~(b)(i) \\(C \\cap A\' = {15}\\)\n(ii) \\(A\' \\cap (B \\cup C) = {2, 6, 12, 15}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(527, '2015', '\n\nQuestion\n\n(a) A manufacturer offers distributors a discount of \\(20%\\) on any article bought and a further discount of \\(2\\frac{1}{2}%\\) for prompt payment.\n(i) if the marked price of an article is N25,000, find the total amount saved by adistributor for paying promptly. (ii) if a distributor pays N11,700 promptly for an article marked Nx, find the value of x.\n(b) Factorize \\(6y^{2} - 149y - 102\\), hence solve the equation \\(6y^{2} - 149y - 102 = 0\\).\n\n', '', '(a) A manufacturer offers distributors a discount of \\(20%\\) on any article bought and a further discount of \\(2\\frac{1}{2}%\\) for prompt payment.\n(i) if the marked price of an article is N25,000, find the total amount saved by adistributor for paying promptly. (ii) if a distributor pays N11,700 promptly for an article marked Nx, find the value of x.\n(b) Factorize \\(6y^{2} - 149y - 102\\), hence solve the equation \\(6y^{2} - 149y - 102 = 0\\).', '(a) A manufacturer offers distributors a discount of \\(20%\\) on any article bought and a further discount of \\(2\\frac{1}{2}%\\) for prompt payment.\n(i) if the marked price of an article is N25,000, find the total amount saved by adistributor for paying promptly. (ii) if a distributor pays N11,700 promptly for an article marked Nx, find the value of x.~(b) Factorize \\(6y^{2} - 149y - 102\\), hence solve the equation \\(6y^{2} - 149y - 102 = 0\\).', NULL, '\n\nExplanation\n(a) (i) First discount = 20% of N25,000\n= \\(\\frac{20}{100} \\times N25,000 = N5,000\\)\nFurther discount for prompt payment = \\(2\\frac{1}{2}%\\) of N25,000.\n= \\(\\frac{5}{200} \\times N25,000 = N625\\)\n\\(\\therefore \\text{Total amount saved by distributor} = N(5,000 + 625)\\)\n= \\(N5625\\)\n(ii) Total %age discount for prompt payment = \\((20 + 2\\frac{1}{2})% = 22\\frac{1}{2}%\\)\nTherefore, the distributor paid \\((100% - 22\\frac{1}{2}%) \\times Nx\\)\n= \\(77\\frac{1}{2}% \\times Nx\\)\n\\(\\implies 77\\frac{1}{2}% \\times x = 11,700\\)\n\\(x = \\frac{11,700 \\times 200}{155}\\)\n= \\(N15,096.77\\)\n(b) \\(6y^{2} - 149y - 102 \\)\n= \\(6y^{2} - 153y + 4y - 102\\)\n= \\(3y(2y - 51) + 2(2y - 51)\\)\n= \\((3y + 2)(2y - 51)\\)\n\\(\\therefore 6y^{2} - 149y - 102 = 0 \\implies (3y + 2)(2y - 51) = 0\\)\n\\(3y + 2 = 0 \\implies 3y = -2 \\)\n\\(y = -\\frac{2}{3}\\)\n\\(2y - 51 = 0 \\implies 2y = 51\\)\n\\(y = 25\\frac{1}{2}\\)\n\\(\\therefore y = -\\frac{2}{3} ; y = 25\\frac{1}{2}\\)\n\n', '', 'Explanation\n(a) (i) First discount = 20% of N25,000\n= \\(\\frac{20}{100} \\times N25,000 = N5,000\\)\nFurther discount for prompt payment = \\(2\\frac{1}{2}%\\) of N25,000.\n= \\(\\frac{5}{200} \\times N25,000 = N625\\)\n\\(\\therefore \\text{Total amount saved by distributor} = N(5,000 + 625)\\)\n= \\(N5625\\)\n(ii) Total %age discount for prompt payment = \\((20 + 2\\frac{1}{2})% = 22\\frac{1}{2}%\\)\nTherefore, the distributor paid \\((100% - 22\\frac{1}{2}%) \\times Nx\\)\n= \\(77\\frac{1}{2}% \\times Nx\\)\n\\(\\implies 77\\frac{1}{2}% \\times x = 11,700\\)\n\\(x = \\frac{11,700 \\times 200}{155}\\)\n= \\(N15,096.77\\)\n(b) \\(6y^{2} - 149y - 102 \\)\n= \\(6y^{2} - 153y + 4y - 102\\)\n= \\(3y(2y - 51) + 2(2y - 51)\\)\n= \\((3y + 2)(2y - 51)\\)\n\\(\\therefore 6y^{2} - 149y - 102 = 0 \\implies (3y + 2)(2y - 51) = 0\\)\n\\(3y + 2 = 0 \\implies 3y = -2 \\)\n\\(y = -\\frac{2}{3}\\)\n\\(2y - 51 = 0 \\implies 2y = 51\\)\n\\(y = 25\\frac{1}{2}\\)\n\\(\\therefore y = -\\frac{2}{3} ; y = 25\\frac{1}{2}\\)\n\n', '(a) (i) First discount = 20% of N25,000\n= \\(\\frac{20}{100} \\times N25,000 = N5,000\\)\nFurther discount for prompt payment = \\(2\\frac{1}{2}%\\) of N25,000.\n= \\(\\frac{5}{200} \\times N25,000 = N625\\)\n\\(\\therefore \\text{Total amount saved by distributor} = N(5,000 + 625)\\)\n= \\(N5625\\)\n(ii) Total %age discount for prompt payment = \\((20 + 2\\frac{1}{2})% = 22\\frac{1}{2}%\\)\nTherefore, the distributor paid \\((100% - 22\\frac{1}{2}%) \\times Nx\\)\n= \\(77\\frac{1}{2}% \\times Nx\\)\n\\(\\implies 77\\frac{1}{2}% \\times x = 11,700\\)\n\\(x = \\frac{11,700 \\times 200}{155}\\)\n= \\(N15,096.77\\)~(b) \\(6y^{2} - 149y - 102 \\)\n= \\(6y^{2} - 153y + 4y - 102\\)\n= \\(3y(2y - 51) + 2(2y - 51)\\)\n= \\((3y + 2)(2y - 51)\\)\n\\(\\therefore 6y^{2} - 149y - 102 = 0 \\implies (3y + 2)(2y - 51) = 0\\)\n\\(3y + 2 = 0 \\implies 3y = -2 \\)\n\\(y = -\\frac{2}{3}\\)\n\\(2y - 51 = 0 \\implies 2y = 51\\)\n\\(y = 25\\frac{1}{2}\\)\n\\(\\therefore y = -\\frac{2}{3} ; y = 25\\frac{1}{2}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(528, '2015', '\n\nQuestion\n\n(a) Without using calculator or mathematical tables, evaluate \\(\\frac{3}{\\sqrt{3}}(\\frac{2}{\\sqrt{3}} - \\frac{\\sqrt{12}}{6})\\)\n(b)In the diagram, O is the centre of the circle. The side AB is produced to E, < ACB = 49° and < CBE = 68°. Calculate,\n(i) the interior angle AOC ; (ii) < BOC.\n\n', '', '(a) Without using calculator or mathematical tables, evaluate \\(\\frac{3}{\\sqrt{3}}(\\frac{2}{\\sqrt{3}} - \\frac{\\sqrt{12}}{6})\\)\n(b)In the diagram, O is the centre of the circle. The side AB is produced to E, < ACB = 49° and < CBE = 68°. Calculate,\n(i) the interior angle AOC ; (ii) < BOC.', '(a) Without using calculator or mathematical tables, evaluate \\(\\frac{3}{\\sqrt{3}}(\\frac{2}{\\sqrt{3}} - \\frac{\\sqrt{12}}{6})\\)~(b)In the diagram, O is the centre of the circle. The side AB is produced to E, < ACB = 49° and < CBE = 68°. Calculate,\n(i) the interior angle AOC ; (ii) < BOC.', NULL, '\n\nExplanation\n(a) \\(\\frac{3}{\\sqrt{3}}(\\frac{2}{\\sqrt{3}} - \\frac{\\sqrt{12}}{6})\\)\n= \\(\\frac{6}{\\sqrt{9}} - \\frac{3\\sqrt{12}}{6\\sqrt{3}}\\)\n= \\(2 - \\frac{2\\sqrt{3}}{2\\sqrt{3}}\\)\n= \\(2 - 1 = 1\\)\n(b)(i) \\(< ABC = 180° - 68° = 112°\\)\n\\(< AOC = 2 \\times 112° = 224°\\)\n\\(\\hat{O} = < AOC (obtuse)\\)\n= \\(360° - 224° = 136°\\)\n(ii) In \\(\\Delta AOC\\), OA= OC (radii)\n\\(\\therefore \\Delta AOC = isosceles\\)\n\\(\\hat{A} = < OAC ; \\hat{C} = < ACO ; \\hat{A} = \\hat{C} = x\\)\n\\(\\hat{O} + \\hat{A} + \\hat{C} = 180°\\)\n\\(136° + x + x = 180° \\implies 2x = 44°\\)\n\\(x = 22°\\)\nIn \\(\\Delta BOC\\), OB = OC (radii)\n\\(\\therefore \\Delta BOC = isosceles\\)\n\\(< OCB = 49° + 22°= 71°\\)\n\\(< OBC = < OCB = 71°\\)\n\\(< OBC + < OCB + < BOC = 180°\\)\n\\(71° + 71° + < BOC = 180°\\)\n\\(< BOC = 180° - 142° = 38°\\)\n\n', '', 'Explanation\n(a) \\(\\frac{3}{\\sqrt{3}}(\\frac{2}{\\sqrt{3}} - \\frac{\\sqrt{12}}{6})\\)\n= \\(\\frac{6}{\\sqrt{9}} - \\frac{3\\sqrt{12}}{6\\sqrt{3}}\\)\n= \\(2 - \\frac{2\\sqrt{3}}{2\\sqrt{3}}\\)\n= \\(2 - 1 = 1\\)\n(b)(i) \\(< ABC = 180° - 68° = 112°\\)\n\\(< AOC = 2 \\times 112° = 224°\\)\n\\(\\hat{O} = < AOC (obtuse)\\)\n= \\(360° - 224° = 136°\\)\n(ii) In \\(\\Delta AOC\\), OA= OC (radii)\n\\(\\therefore \\Delta AOC = isosceles\\)\n\\(\\hat{A} = < OAC ; \\hat{C} = < ACO ; \\hat{A} = \\hat{C} = x\\)\n\\(\\hat{O} + \\hat{A} + \\hat{C} = 180°\\)\n\\(136° + x + x = 180° \\implies 2x = 44°\\)\n\\(x = 22°\\)\nIn \\(\\Delta BOC\\), OB = OC (radii)\n\\(\\therefore \\Delta BOC = isosceles\\)\n\\(< OCB = 49° + 22°= 71°\\)\n\\(< OBC = < OCB = 71°\\)\n\\(< OBC + < OCB + < BOC = 180°\\)\n\\(71° + 71° + < BOC = 180°\\)\n\\(< BOC = 180° - 142° = 38°\\)\n\n', '(a) \\(\\frac{3}{\\sqrt{3}}(\\frac{2}{\\sqrt{3}} - \\frac{\\sqrt{12}}{6})\\)\n= \\(\\frac{6}{\\sqrt{9}} - \\frac{3\\sqrt{12}}{6\\sqrt{3}}\\)\n= \\(2 - \\frac{2\\sqrt{3}}{2\\sqrt{3}}\\)\n= \\(2 - 1 = 1\\)~(b)(i) \\(< ABC = 180° - 68° = 112°\\)\n\\(< AOC = 2 \\times 112° = 224°\\)\n\\(\\hat{O} = < AOC (obtuse)\\)\n= \\(360° - 224° = 136°\\)\n(ii) In \\(\\Delta AOC\\), OA= OC (radii)\n\\(\\therefore \\Delta AOC = isosceles\\)\n\\(\\hat{A} = < OAC ; \\hat{C} = < ACO ; \\hat{A} = \\hat{C} = x\\)\n\\(\\hat{O} + \\hat{A} + \\hat{C} = 180°\\)\n\\(136° + x + x = 180° \\implies 2x = 44°\\)\n\\(x = 22°\\)\nIn \\(\\Delta BOC\\), OB = OC (radii)\n\\(\\therefore \\Delta BOC = isosceles\\)\n\\(< OCB = 49° + 22°= 71°\\)\n\\(< OBC = < OCB = 71°\\)\n\\(< OBC + < OCB + < BOC = 180°\\)\n\\(71° + 71° + < BOC = 180°\\)\n\\(< BOC = 180° - 142° = 38°\\)\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
(529, '2015', '\n\nQuestion\n\nThe probabilities that Ade, Kujo and Fati will pass an examination are \\(\\frac{2}{3}, \\frac{5}{8}\\) and \\(\\frac{3}{4}\\) respectively. Find the probability that\n(a) the three ;\n(b) none of them ;\n(c) Ade and Kujo only ; will pass the examination.\n\n', '', 'The probabilities that Ade, Kujo and Fati will pass an examination are \\(\\frac{2}{3}, \\frac{5}{8}\\) and \\(\\frac{3}{4}\\) respectively. Find the probability that\n(a) the three ;\n(b) none of them ;\n(c) Ade and Kujo only ; will pass the examination.', 'The probabilities that Ade, Kujo and Fati will pass an examination are \\(\\frac{2}{3}, \\frac{5}{8}\\) and \\(\\frac{3}{4}\\) respectively. Find the probability that\n(a) the three ;~(b) none of them ;\n(c) Ade and Kujo only ; will pass the examination.~(c) Ade and Kujo only ; will pass the examination.', NULL, '\n\nExplanation\n(a) \\(P(Ade) = \\frac{2}{3} ; P(Kujo) = \\frac{5}{8} ; P(Fati) = \\frac{3}{4}\\)\n\\(P(\\text{all three pass the exam}) = \\frac{2}{3} \\times \\frac{5}{8} \\times \\frac{3}{4}\\)\n= \\(\\frac{5}{16}\\)\n(b) P(Ade fails) = \\(1 - \\frac{2}{3} = \\frac{1}{3}\\)\nP(Kujo fails) = \\(1 - \\frac{5}{8} = \\frac{3}{8}\\)\nP(Fati fails) = \\(1 - \\frac{3}{4} = \\frac{1}{4}\\)\nP(none passes) = \\(\\frac{1}{3} \\times \\frac{3}{8} \\times \\frac{1}{4}\\)\n= \\(\\frac{1}{32}\\).\n(c) P(Ade and Kujo only pass) = \\(\\frac{2}{3} \\times \\frac{5}{8} \\times \\frac{1}{4}\\)\n= \\(\\frac{5}{48}\\)\n\n', '', 'Explanation\n(a) \\(P(Ade) = \\frac{2}{3} ; P(Kujo) = \\frac{5}{8} ; P(Fati) = \\frac{3}{4}\\)\n\\(P(\\text{all three pass the exam}) = \\frac{2}{3} \\times \\frac{5}{8} \\times \\frac{3}{4}\\)\n= \\(\\frac{5}{16}\\)\n(b) P(Ade fails) = \\(1 - \\frac{2}{3} = \\frac{1}{3}\\)\nP(Kujo fails) = \\(1 - \\frac{5}{8} = \\frac{3}{8}\\)\nP(Fati fails) = \\(1 - \\frac{3}{4} = \\frac{1}{4}\\)\nP(none passes) = \\(\\frac{1}{3} \\times \\frac{3}{8} \\times \\frac{1}{4}\\)\n= \\(\\frac{1}{32}\\).\n(c) P(Ade and Kujo only pass) = \\(\\frac{2}{3} \\times \\frac{5}{8} \\times \\frac{1}{4}\\)\n= \\(\\frac{5}{48}\\)\n\n', '(a) \\(P(Ade) = \\frac{2}{3} ; P(Kujo) = \\frac{5}{8} ; P(Fati) = \\frac{3}{4}\\)\n\\(P(\\text{all three pass the exam}) = \\frac{2}{3} \\times \\frac{5}{8} \\times \\frac{3}{4}\\)\n= \\(\\frac{5}{16}\\)~(b) P(Ade fails) = \\(1 - \\frac{2}{3} = \\frac{1}{3}\\)\nP(Kujo fails) = \\(1 - \\frac{5}{8} = \\frac{3}{8}\\)\nP(Fati fails) = \\(1 - \\frac{3}{4} = \\frac{1}{4}\\)\nP(none passes) = \\(\\frac{1}{3} \\times \\frac{3}{8} \\times \\frac{1}{4}\\)\n= \\(\\frac{1}{32}\\).\n(c) P(Ade and Kujo only pass) = \\(\\frac{2}{3} \\times \\frac{5}{8} \\times \\frac{1}{4}\\)\n= \\(\\frac{5}{48}\\)\n~(c) P(Ade and Kujo only pass) = \\(\\frac{2}{3} \\times \\frac{5}{8} \\times \\frac{1}{4}\\)\n= \\(\\frac{5}{48}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(530, '2015', '\n\nQuestion\n\n(a) Simplify : \\(\\frac{1}{2}\\log_{10} 25 - 2\\log_{10} 3 + \\log_{10} 18\\)\n(b) If \\(123_{y} = 83_{10}\\), obtain an equation in y, hence find the value of y.\n(c) Solve the equation \\(\\frac{9^{2x - 3}}{3^{x + 3}} = 1\\)\n\n', '', '(a) Simplify : \\(\\frac{1}{2}\\log_{10} 25 - 2\\log_{10} 3 + \\log_{10} 18\\)\n(b) If \\(123_{y} = 83_{10}\\), obtain an equation in y, hence find the value of y.\n(c) Solve the equation \\(\\frac{9^{2x - 3}}{3^{x + 3}} = 1\\)', '(a) Simplify : \\(\\frac{1}{2}\\log_{10} 25 - 2\\log_{10} 3 + \\log_{10} 18\\)~(b) If \\(123_{y} = 83_{10}\\), obtain an equation in y, hence find the value of y.\n(c) Solve the equation \\(\\frac{9^{2x - 3}}{3^{x + 3}} = 1\\)~(c) Solve the equation \\(\\frac{9^{2x - 3}}{3^{x + 3}} = 1\\)', NULL, '\n\nExplanation\n(a) \\(\\frac{1}{2}\\log_{10} 25 - 2\\log_{10} 3 + \\log_{10} 18\\)\n= \\(\\log_{10} (25^{\\frac{1}{2}}) - \\log_{10} (3^{2}) + \\log_{10} 18\\)\n= \\(\\log_{10} (\\frac{5 \\times 18}{9})\\)\n= \\(\\log_{10} 10\\)\n= 1\n(b) \\(123_{y} = (1 \\times y^{2}) + (2 \\times y^{1}) + (3 \\times y^{0})\\)\n\\(y^{2} + 2y + 3 = 83\\)\n\\(y^{2} + 2y + 3 - 83 = 0 \\implies y^{2} + 2y - 80 = 0\\)\nEquation : \\(y^{2} + 2y - 80 = 0\\)\n\\(y^{2} - 8y + 10y - 80 = 0\\)\n\\(y(y - 8) + 10(y - 8) = 0\\)\n\\((y + 10)(y - 8) = 0 \\implies y = -10 ; 8\\)\ny = 8 since y cannot be negative.\n(c) \\(\\frac{9^{2x - 3}}{3^{x + 3}} = 1\\)\n\\(\\implies 9^{2x - 3} = 3^{x + 3}\\)\n\\((3^{2})^{2x - 3} = 3^{x + 3}\\)\n\\(3^{4x - 6} = 3^{x + 3}\\)\n\\(\\implies 4x - 6 = x + 3\\)\n\\(4x - x = 3 + 6 \\implies 3x = 9\\)\n\\(x = 3\\)\n\n', '', 'Explanation\n(a) \\(\\frac{1}{2}\\log_{10} 25 - 2\\log_{10} 3 + \\log_{10} 18\\)\n= \\(\\log_{10} (25^{\\frac{1}{2}}) - \\log_{10} (3^{2}) + \\log_{10} 18\\)\n= \\(\\log_{10} (\\frac{5 \\times 18}{9})\\)\n= \\(\\log_{10} 10\\)\n= 1\n(b) \\(123_{y} = (1 \\times y^{2}) + (2 \\times y^{1}) + (3 \\times y^{0})\\)\n\\(y^{2} + 2y + 3 = 83\\)\n\\(y^{2} + 2y + 3 - 83 = 0 \\implies y^{2} + 2y - 80 = 0\\)\nEquation : \\(y^{2} + 2y - 80 = 0\\)\n\\(y^{2} - 8y + 10y - 80 = 0\\)\n\\(y(y - 8) + 10(y - 8) = 0\\)\n\\((y + 10)(y - 8) = 0 \\implies y = -10 ; 8\\)\ny = 8 since y cannot be negative.\n(c) \\(\\frac{9^{2x - 3}}{3^{x + 3}} = 1\\)\n\\(\\implies 9^{2x - 3} = 3^{x + 3}\\)\n\\((3^{2})^{2x - 3} = 3^{x + 3}\\)\n\\(3^{4x - 6} = 3^{x + 3}\\)\n\\(\\implies 4x - 6 = x + 3\\)\n\\(4x - x = 3 + 6 \\implies 3x = 9\\)\n\\(x = 3\\)\n\n', '(a) \\(\\frac{1}{2}\\log_{10} 25 - 2\\log_{10} 3 + \\log_{10} 18\\)\n= \\(\\log_{10} (25^{\\frac{1}{2}}) - \\log_{10} (3^{2}) + \\log_{10} 18\\)\n= \\(\\log_{10} (\\frac{5 \\times 18}{9})\\)\n= \\(\\log_{10} 10\\)\n= 1~(b) \\(123_{y} = (1 \\times y^{2}) + (2 \\times y^{1}) + (3 \\times y^{0})\\)\n\\(y^{2} + 2y + 3 = 83\\)\n\\(y^{2} + 2y + 3 - 83 = 0 \\implies y^{2} + 2y - 80 = 0\\)\nEquation : \\(y^{2} + 2y - 80 = 0\\)\n\\(y^{2} - 8y + 10y - 80 = 0\\)\n\\(y(y - 8) + 10(y - 8) = 0\\)\n\\((y + 10)(y - 8) = 0 \\implies y = -10 ; 8\\)\ny = 8 since y cannot be negative.\n(c) \\(\\frac{9^{2x - 3}}{3^{x + 3}} = 1\\)\n\\(\\implies 9^{2x - 3} = 3^{x + 3}\\)\n\\((3^{2})^{2x - 3} = 3^{x + 3}\\)\n\\(3^{4x - 6} = 3^{x + 3}\\)\n\\(\\implies 4x - 6 = x + 3\\)\n\\(4x - x = 3 + 6 \\implies 3x = 9\\)\n\\(x = 3\\)\n~(c) \\(\\frac{9^{2x - 3}}{3^{x + 3}} = 1\\)\n\\(\\implies 9^{2x - 3} = 3^{x + 3}\\)\n\\((3^{2})^{2x - 3} = 3^{x + 3}\\)\n\\(3^{4x - 6} = 3^{x + 3}\\)\n\\(\\implies 4x - 6 = x + 3\\)\n\\(4x - x = 3 + 6 \\implies 3x = 9\\)\n\\(x = 3\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(531, '2015', '\n\nQuestion\n\n(a) The first term of an Arithmetic Progression(AP) is 3 and the common difference is 4. Find the sum of the first 28 terms.\n(b) Given that \\(x = \\frac{2m}{1 - m^{2}}\\) and \\(y = \\frac{2m}{1 + m}\\), express 2x - y in terms of m in the simplest form.\n(c) The angles of pentagon are x°, 2x°, 3x°, 2x° and (3x - 10)°. Find the value of x.\n\n', '', '(a) The first term of an Arithmetic Progression(AP) is 3 and the common difference is 4. Find the sum of the first 28 terms.\n(b) Given that \\(x = \\frac{2m}{1 - m^{2}}\\) and \\(y = \\frac{2m}{1 + m}\\), express 2x - y in terms of m in the simplest form.\n(c) The angles of pentagon are x°, 2x°, 3x°, 2x° and (3x - 10)°. Find the value of x.', '(a) The first term of an Arithmetic Progression(AP) is 3 and the common difference is 4. Find the sum of the first 28 terms.~(b) Given that \\(x = \\frac{2m}{1 - m^{2}}\\) and \\(y = \\frac{2m}{1 + m}\\), express 2x - y in terms of m in the simplest form.\n(c) The angles of pentagon are x°, 2x°, 3x°, 2x° and (3x - 10)°. Find the value of x.~(c) The angles of pentagon are x°, 2x°, 3x°, 2x° and (3x - 10)°. Find the value of x.', NULL, '\n\nExplanation\n(a) \\(a = 3; d = 4 ; n = 28\\)\n\\(S_{n} = \\frac{n}{2}(2a + (n - 1)d)\\)\n\\(S_{28} = \\frac{28}{2}(2(3) + (28 - 1) 4)\\)\n= \\(14(6 + 108)\\)\n= \\(14(114)\\)\n= \\(1596\\)\n(b) \\(x = \\frac{2m}{1 - m^{2}} ; y = \\frac{2m}{1 + m}\\)\n\\(2x - y = 2(\\frac{2m}{1 - m^{2}}) - (\\frac{2m}{1 + m})\\)\n= \\(\\frac{4m}{1 - m^{2}} - \\frac{2m}{1 + m}\\)\n= \\(\\frac{4m - 2m(1 - m)}{1 - m^{2}}\\)\n= \\(\\frac{4m - 2m + 2m^{2}}{1 - m^{2}}\\)\n= \\(\\frac{2m^{2} + 2m}{1 - m^{2}}\\)\n= \\(\\frac{2m(m + 1)}{(1 + m)(1 - m)}\\)\n= \\(\\frac{2m}{1 - m}\\)\n(c) Sum of interior angles = (2n - 4)90°\nWhen n = 5 (pentagon),\n= \\((2(5) - 4) \\times 90°\\)\n= \\(540°\\)\n\\(x + 2x + 3x + 2x + (3x - 10) = 11x - 10\\)\n= \\(11x - 10 = 540°\\)\n= \\(11x = 550°\\)\n= \\(x = 50°\\)\n\n', '', 'Explanation\n(a) \\(a = 3; d = 4 ; n = 28\\)\n\\(S_{n} = \\frac{n}{2}(2a + (n - 1)d)\\)\n\\(S_{28} = \\frac{28}{2}(2(3) + (28 - 1) 4)\\)\n= \\(14(6 + 108)\\)\n= \\(14(114)\\)\n= \\(1596\\)\n(b) \\(x = \\frac{2m}{1 - m^{2}} ; y = \\frac{2m}{1 + m}\\)\n\\(2x - y = 2(\\frac{2m}{1 - m^{2}}) - (\\frac{2m}{1 + m})\\)\n= \\(\\frac{4m}{1 - m^{2}} - \\frac{2m}{1 + m}\\)\n= \\(\\frac{4m - 2m(1 - m)}{1 - m^{2}}\\)\n= \\(\\frac{4m - 2m + 2m^{2}}{1 - m^{2}}\\)\n= \\(\\frac{2m^{2} + 2m}{1 - m^{2}}\\)\n= \\(\\frac{2m(m + 1)}{(1 + m)(1 - m)}\\)\n= \\(\\frac{2m}{1 - m}\\)\n(c) Sum of interior angles = (2n - 4)90°\nWhen n = 5 (pentagon),\n= \\((2(5) - 4) \\times 90°\\)\n= \\(540°\\)\n\\(x + 2x + 3x + 2x + (3x - 10) = 11x - 10\\)\n= \\(11x - 10 = 540°\\)\n= \\(11x = 550°\\)\n= \\(x = 50°\\)\n\n', '(a) \\(a = 3; d = 4 ; n = 28\\)\n\\(S_{n} = \\frac{n}{2}(2a + (n - 1)d)\\)\n\\(S_{28} = \\frac{28}{2}(2(3) + (28 - 1) 4)\\)\n= \\(14(6 + 108)\\)\n= \\(14(114)\\)\n= \\(1596\\)~(b) \\(x = \\frac{2m}{1 - m^{2}} ; y = \\frac{2m}{1 + m}\\)\n\\(2x - y = 2(\\frac{2m}{1 - m^{2}}) - (\\frac{2m}{1 + m})\\)\n= \\(\\frac{4m}{1 - m^{2}} - \\frac{2m}{1 + m}\\)\n= \\(\\frac{4m - 2m(1 - m)}{1 - m^{2}}\\)\n= \\(\\frac{4m - 2m + 2m^{2}}{1 - m^{2}}\\)\n= \\(\\frac{2m^{2} + 2m}{1 - m^{2}}\\)\n= \\(\\frac{2m(m + 1)}{(1 + m)(1 - m)}\\)\n= \\(\\frac{2m}{1 - m}\\)\n(c) Sum of interior angles = (2n - 4)90°\nWhen n = 5 (pentagon),\n= \\((2(5) - 4) \\times 90°\\)\n= \\(540°\\)\n\\(x + 2x + 3x + 2x + (3x - 10) = 11x - 10\\)\n= \\(11x - 10 = 540°\\)\n= \\(11x = 550°\\)\n= \\(x = 50°\\)\n~(c) Sum of interior angles = (2n - 4)90°\nWhen n = 5 (pentagon),\n= \\((2(5) - 4) \\times 90°\\)\n= \\(540°\\)\n\\(x + 2x + 3x + 2x + (3x - 10) = 11x - 10\\)\n= \\(11x - 10 = 540°\\)\n= \\(11x = 550°\\)\n= \\(x = 50°\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(532, '2015', '\n\nQuestion\n\n(a) An open rectangular tank is made of a steel plate of area 1440\\(m^{2}\\). Its length is twice its width . If the depth of the tank is 4m less than its width, find its length.\n(b) A man saved N3,000 in a bank P, whose interest rate was x% per annum and N2,000 in another bank Q whose interest rate was y% per annum. His total interest in one year was N640. If he had saved N2,000 in P and N3,000 in Q for the same period, he would have gained N20 as additional interest. Find the values of x and y.\n\n', '', '(a) An open rectangular tank is made of a steel plate of area 1440\\(m^{2}\\). Its length is twice its width . If the depth of the tank is 4m less than its width, find its length.\n(b) A man saved N3,000 in a bank P, whose interest rate was x% per annum and N2,000 in another bank Q whose interest rate was y% per annum. His total interest in one year was N640. If he had saved N2,000 in P and N3,000 in Q for the same period, he would have gained N20 as additional interest. Find the values of x and y.', '(a) An open rectangular tank is made of a steel plate of area 1440\\(m^{2}\\). Its length is twice its width . If the depth of the tank is 4m less than its width, find its length.~(b) A man saved N3,000 in a bank P, whose interest rate was x% per annum and N2,000 in another bank Q whose interest rate was y% per annum. His total interest in one year was N640. If he had saved N2,000 in P and N3,000 in Q for the same period, he would have gained N20 as additional interest. Find the values of x and y.', NULL, '\n\nExplanation\n(a) Width = x m; Length = 2x m; Height = (x - 4) m.\nTotal surface area = 2Lh + WL + 2Wh\n= \\(2(2x)(x - 4) + x(2x) + 2(x)(x - 4) = 1440 m^{2}\\)\n= \\(4x^{2} - 16x + 2x^{2} + 2x^{2} - 8x = 1440\\)\n= \\(8x^{2} - 24x = 1440\\)\n= \\(x^{2} - 3x - 180 = 0\\)\n= \\((x - 15)(x + 12) = 0\\)\n\\(x = 15m\\)\nLength = 2(15m) = 30m.\n(b) \\(I = \\frac{PRT}{100}\\)\n\\(I_{P} + I_{Q} = I_{T}\\)\n\\(\\frac{3000 \\times x \\times 1}{100} + \\frac{2000 \\times y \\times 1}{100} = 640\\)\n\\(30x + 20y = 640 \\implies 3x + 2y = 64 .... (1)\\)\n\\(\\frac{2000 \\times x \\times 1}{100} + \\frac{3000 \\times y \\times 1}{100} = 660\\)\n\\(20x + 30y = 660 \\implies 2x + 3y = 66 .... (2)\\)\nTo eliminate x , multiply (1) by 2 and (2) by 3.\n\\(6x + 4y = 128 ... (1)\\)\n\\(6x + 9y = 198 .... (2)\\)\n(2) - (1) : \\(9y - 4y = 198 - 128 \\implies 5y = 70\\)\n\\(y = 14\\)\n\\(2x + 3y = 66 \\implies 2x + 3(14) = 66\\)\n\\(2x + 42 = 66 \\implies 2x = 24\\)\n\\(x = 12\\)\n\\(\\therefore x = 12 ; y = 14\\).\n\n', '', 'Explanation\n(a) Width = x m; Length = 2x m; Height = (x - 4) m.\nTotal surface area = 2Lh + WL + 2Wh\n= \\(2(2x)(x - 4) + x(2x) + 2(x)(x - 4) = 1440 m^{2}\\)\n= \\(4x^{2} - 16x + 2x^{2} + 2x^{2} - 8x = 1440\\)\n= \\(8x^{2} - 24x = 1440\\)\n= \\(x^{2} - 3x - 180 = 0\\)\n= \\((x - 15)(x + 12) = 0\\)\n\\(x = 15m\\)\nLength = 2(15m) = 30m.\n(b) \\(I = \\frac{PRT}{100}\\)\n\\(I_{P} + I_{Q} = I_{T}\\)\n\\(\\frac{3000 \\times x \\times 1}{100} + \\frac{2000 \\times y \\times 1}{100} = 640\\)\n\\(30x + 20y = 640 \\implies 3x + 2y = 64 .... (1)\\)\n\\(\\frac{2000 \\times x \\times 1}{100} + \\frac{3000 \\times y \\times 1}{100} = 660\\)\n\\(20x + 30y = 660 \\implies 2x + 3y = 66 .... (2)\\)\nTo eliminate x , multiply (1) by 2 and (2) by 3.\n\\(6x + 4y = 128 ... (1)\\)\n\\(6x + 9y = 198 .... (2)\\)\n(2) - (1) : \\(9y - 4y = 198 - 128 \\implies 5y = 70\\)\n\\(y = 14\\)\n\\(2x + 3y = 66 \\implies 2x + 3(14) = 66\\)\n\\(2x + 42 = 66 \\implies 2x = 24\\)\n\\(x = 12\\)\n\\(\\therefore x = 12 ; y = 14\\).\n\n', '(a) Width = x m; Length = 2x m; Height = (x - 4) m.\nTotal surface area = 2Lh + WL + 2Wh\n= \\(2(2x)(x - 4) + x(2x) + 2(x)(x - 4) = 1440 m^{2}\\)\n= \\(4x^{2} - 16x + 2x^{2} + 2x^{2} - 8x = 1440\\)\n= \\(8x^{2} - 24x = 1440\\)\n= \\(x^{2} - 3x - 180 = 0\\)\n= \\((x - 15)(x + 12) = 0\\)\n\\(x = 15m\\)\nLength = 2(15m) = 30m.~(b) \\(I = \\frac{PRT}{100}\\)\n\\(I_{P} + I_{Q} = I_{T}\\)\n\\(\\frac{3000 \\times x \\times 1}{100} + \\frac{2000 \\times y \\times 1}{100} = 640\\)\n\\(30x + 20y = 640 \\implies 3x + 2y = 64 .... (1)\\)\n\\(\\frac{2000 \\times x \\times 1}{100} + \\frac{3000 \\times y \\times 1}{100} = 660\\)\n\\(20x + 30y = 660 \\implies 2x + 3y = 66 .... (2)\\)\nTo eliminate x , multiply (1) by 2 and (2) by 3.\n\\(6x + 4y = 128 ... (1)\\)\n\\(6x + 9y = 198 .... (2)\\)\n(2) - (1) : \\(9y - 4y = 198 - 128 \\implies 5y = 70\\)\n\\(y = 14\\)\n\\(2x + 3y = 66 \\implies 2x + 3(14) = 66\\)\n\\(2x + 42 = 66 \\implies 2x = 24\\)\n\\(x = 12\\)\n\\(\\therefore x = 12 ; y = 14\\).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(533, '2015', '\n\nQuestion\n\nThe diagram is a portion of a right circular solid cylinder of radius 7 cm and height 15 cm. The centre of the base of the cylinder is Q, while that of the top is B, where \\(\\stackrel\\frown{ABC} = \\stackrel\\frown{PQR} = 120°\\). Calculate, correct to one decimal place:\n(a) The volume\n(b) the total surface area of the solid. [Take \\(\\pi = \\frac{22}{7}\\)].\n\n', '', 'The diagram is a portion of a right circular solid cylinder of radius 7 cm and height 15 cm. The centre of the base of the cylinder is Q, while that of the top is B, where \\(\\stackrel\\frown{ABC} = \\stackrel\\frown{PQR} = 120°\\). Calculate, correct to one decimal place:\n(a) The volume\n(b) the total surface area of the solid. [Take \\(\\pi = \\frac{22}{7}\\)].', 'The diagram is a portion of a right circular solid cylinder of radius 7 cm and height 15 cm. The centre of the base of the cylinder is Q, while that of the top is B, where \\(\\stackrel\\frown{ABC} = \\stackrel\\frown{PQR} = 120°\\). Calculate, correct to one decimal place:\n(a) The volume~(b) the total surface area of the solid. [Take \\(\\pi = \\frac{22}{7}\\)].', NULL, '\n\nExplanation\n(a) Area of the major sector = \\(\\frac{\\theta}{360°} \\times \\pi r^{2}\\)\n= \\(\\frac{360 - 120}{360} \\times \\frac{22}{7} \\times 7 \\times 7\\)\n= \\((\\frac{2}{3} \\times 154) cm^{2}\\)\nVolume = \\(\\text{Area of the major sector} \\times \\text{height (cross section)}\\)\n= \\(\\frac{2}{3} \\times 154 \\times 15 = 1540 cm^{3}\\)\n(b) Curved surface area = \\(\\frac{240}{360} \\times 2\\pi rh\\)\n= \\(\\frac{240}{360} \\times 2 \\times \\frac{22}{7} \\times 7 \\times 15\\)\n= \\(440 cm^{2}\\)\nArea of rectangular surface = \\(2 \\times l \\times b\\)\n= \\(2 \\times 7 \\times 15 = 210 cm^{2}\\)\nArea of top and bottom of cylinder (major sector)\n= \\(2 \\times \\frac{240}{360} \\times \\frac{22}{7} \\times 7 \\times 7\\)\n= \\(205.33 cm^{2}\\)\n\\(\\therefore \\text{Total surface area} = 440 + 205.33 + 210 = 855.33 cm^{2}\\)\n\n', '', 'Explanation\n(a) Area of the major sector = \\(\\frac{\\theta}{360°} \\times \\pi r^{2}\\)\n= \\(\\frac{360 - 120}{360} \\times \\frac{22}{7} \\times 7 \\times 7\\)\n= \\((\\frac{2}{3} \\times 154) cm^{2}\\)\nVolume = \\(\\text{Area of the major sector} \\times \\text{height (cross section)}\\)\n= \\(\\frac{2}{3} \\times 154 \\times 15 = 1540 cm^{3}\\)\n(b) Curved surface area = \\(\\frac{240}{360} \\times 2\\pi rh\\)\n= \\(\\frac{240}{360} \\times 2 \\times \\frac{22}{7} \\times 7 \\times 15\\)\n= \\(440 cm^{2}\\)\nArea of rectangular surface = \\(2 \\times l \\times b\\)\n= \\(2 \\times 7 \\times 15 = 210 cm^{2}\\)\nArea of top and bottom of cylinder (major sector)\n= \\(2 \\times \\frac{240}{360} \\times \\frac{22}{7} \\times 7 \\times 7\\)\n= \\(205.33 cm^{2}\\)\n\\(\\therefore \\text{Total surface area} = 440 + 205.33 + 210 = 855.33 cm^{2}\\)\n\n', '(a) Area of the major sector = \\(\\frac{\\theta}{360°} \\times \\pi r^{2}\\)\n= \\(\\frac{360 - 120}{360} \\times \\frac{22}{7} \\times 7 \\times 7\\)\n= \\((\\frac{2}{3} \\times 154) cm^{2}\\)\nVolume = \\(\\text{Area of the major sector} \\times \\text{height (cross section)}\\)\n= \\(\\frac{2}{3} \\times 154 \\times 15 = 1540 cm^{3}\\)~(b) Curved surface area = \\(\\frac{240}{360} \\times 2\\pi rh\\)\n= \\(\\frac{240}{360} \\times 2 \\times \\frac{22}{7} \\times 7 \\times 15\\)\n= \\(440 cm^{2}\\)\nArea of rectangular surface = \\(2 \\times l \\times b\\)\n= \\(2 \\times 7 \\times 15 = 210 cm^{2}\\)\nArea of top and bottom of cylinder (major sector)\n= \\(2 \\times \\frac{240}{360} \\times \\frac{22}{7} \\times 7 \\times 7\\)\n= \\(205.33 cm^{2}\\)\n\\(\\therefore \\text{Total surface area} = 440 + 205.33 + 210 = 855.33 cm^{2}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(534, '2015', '\n\nQuestion\n\nUsing ruler and a pair of compasses only,\n(a) construct, (i) triangle XYZ with |XY| = 8cm, < YXZ = 60° and < XYZ = 30° ; (ii) the perpendicular ZT to meet XY in T ; (iii) the locus \\(l_{1}\\) of points equidistant from ZY and XY.\n(b) If \\(l_{1}\\) and ZT intersect at S, measure |ST|.\n\n', '', 'Using ruler and a pair of compasses only,\n(a) construct, (i) triangle XYZ with |XY| = 8cm, < YXZ = 60° and < XYZ = 30° ; (ii) the perpendicular ZT to meet XY in T ; (iii) the locus \\(l_{1}\\) of points equidistant from ZY and XY.\n(b) If \\(l_{1}\\) and ZT intersect at S, measure |ST|.', 'Using ruler and a pair of compasses only,\n(a) construct, (i) triangle XYZ with |XY| = 8cm, < YXZ = 60° and < XYZ = 30° ; (ii) the perpendicular ZT to meet XY in T ; (iii) the locus \\(l_{1}\\) of points equidistant from ZY and XY.~(b) If \\(l_{1}\\) and ZT intersect at S, measure |ST|.', NULL, '\n\nExplanation\n', '', '', '', 'WAEC-PK-1', 'WAEC-MATH', 1),
(535, '2015', '\n\nQuestion\n\nIn the diagram, /PQ/ = 8m, /QR/ = 13m, the bearing of Q from P is 050° and the bearing of R from Q is 130°.\n(a) Calculate, correct to 3 significant figures, (i) /PR/ ; (ii) the bearing of R from P.\n(b) Calculate the shortest distance between Q and PR, hence the area of triangle PQR.\n\n', '', 'In the diagram, /PQ/ = 8m, /QR/ = 13m, the bearing of Q from P is 050° and the bearing of R from Q is 130°.\n(a) Calculate, correct to 3 significant figures, (i) /PR/ ; (ii) the bearing of R from P.\n(b) Calculate the shortest distance between Q and PR, hence the area of triangle PQR.', 'In the diagram, /PQ/ = 8m, /QR/ = 13m, the bearing of Q from P is 050° and the bearing of R from Q is 130°.\n(a) Calculate, correct to 3 significant figures, (i) /PR/ ; (ii) the bearing of R from P.~(b) Calculate the shortest distance between Q and PR, hence the area of triangle PQR.', NULL, '\n\nExplanation\n(a)(i) < PQR = (180° - 130°) + 50° = 100°\n\\(PR^{2} = PQ^{2} + QR^{2} - 2(PQ)(QR) \\cos < PQR\\)\n= \\(8^{2} + 13^{2} - 2(8)(13) \\cos 100\\)\n= \\(233 + 36.118\\)\n\\(PR^{2} = 269.118\\)\n\\(PR = 16.405m \\approxeq 16.4 m\\)\n(ii) \\(\\frac{QR}{\\sin < QPR} = \\frac{PR}{\\sin < PQR}\\)\n\\(\\frac{13}{\\sin < QPR} = \\frac{16.4}{\\sin 100}\\)\n\\(\\sin < QPR = \\frac{13 \\times \\sin 100}{16.4}\\)\n\\(\\sin < QPR = 0.7806\\)\n\\(< QPR = 51.32° \\approxeq 51.3°\\)\nThe bearing of R from P = 50° + 51.3° = 101.3°\n\\(\\approxeq\\) 101°.\n(b)\nIn \\(\\Delta PQD\\),\n\\(\\frac{h}{8} = \\sin 51.3\\)\n\\(h = 8 \\sin 51.3\\)\n= \\(6.243 m \\)\n\\(\\approxeq 6.24 m\\)\nArea of \\(\\Delta PQR = \\frac{1}{2} \\times PR \\times h\\)\n\\(\\frac{1}{2} \\times 6.24 \\times 16.4 = 51.166 m^{2}\\)\n\\(\\approxeq 51.2 m^{2}\\)\n\n', '', 'Explanation\n(a)(i) < PQR = (180° - 130°) + 50° = 100°\n\\(PR^{2} = PQ^{2} + QR^{2} - 2(PQ)(QR) \\cos < PQR\\)\n= \\(8^{2} + 13^{2} - 2(8)(13) \\cos 100\\)\n= \\(233 + 36.118\\)\n\\(PR^{2} = 269.118\\)\n\\(PR = 16.405m \\approxeq 16.4 m\\)\n(ii) \\(\\frac{QR}{\\sin < QPR} = \\frac{PR}{\\sin < PQR}\\)\n\\(\\frac{13}{\\sin < QPR} = \\frac{16.4}{\\sin 100}\\)\n\\(\\sin < QPR = \\frac{13 \\times \\sin 100}{16.4}\\)\n\\(\\sin < QPR = 0.7806\\)\n\\(< QPR = 51.32° \\approxeq 51.3°\\)\nThe bearing of R from P = 50° + 51.3° = 101.3°\n\\(\\approxeq\\) 101°.\n(b)\nIn \\(\\Delta PQD\\),\n\\(\\frac{h}{8} = \\sin 51.3\\)\n\\(h = 8 \\sin 51.3\\)\n= \\(6.243 m \\)\n\\(\\approxeq 6.24 m\\)\nArea of \\(\\Delta PQR = \\frac{1}{2} \\times PR \\times h\\)\n\\(\\frac{1}{2} \\times 6.24 \\times 16.4 = 51.166 m^{2}\\)\n\\(\\approxeq 51.2 m^{2}\\)\n\n', '(a)(i) < PQR = (180° - 130°) + 50° = 100°\n\\(PR^{2} = PQ^{2} + QR^{2} - 2(PQ)(QR) \\cos < PQR\\)\n= \\(8^{2} + 13^{2} - 2(8)(13) \\cos 100\\)\n= \\(233 + 36.118\\)\n\\(PR^{2} = 269.118\\)\n\\(PR = 16.405m \\approxeq 16.4 m\\)\n(ii) \\(\\frac{QR}{\\sin < QPR} = \\frac{PR}{\\sin < PQR}\\)\n\\(\\frac{13}{\\sin < QPR} = \\frac{16.4}{\\sin 100}\\)\n\\(\\sin < QPR = \\frac{13 \\times \\sin 100}{16.4}\\)\n\\(\\sin < QPR = 0.7806\\)\n\\(< QPR = 51.32° \\approxeq 51.3°\\)\nThe bearing of R from P = 50° + 51.3° = 101.3°\n\\(\\approxeq\\) 101°.~(b)\nIn \\(\\Delta PQD\\),\n\\(\\frac{h}{8} = \\sin 51.3\\)\n\\(h = 8 \\sin 51.3\\)\n= \\(6.243 m \\)\n\\(\\approxeq 6.24 m\\)\nArea of \\(\\Delta PQR = \\frac{1}{2} \\times PR \\times h\\)\n\\(\\frac{1}{2} \\times 6.24 \\times 16.4 = 51.166 m^{2}\\)\n\\(\\approxeq 51.2 m^{2}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(536, '2015', '\n\nQuestion\n\nThe following data gives the lengths, in cm, of 30 pieces of iron rods :\n45 55 65 60 61 68 59 54 64 76 50 68 72 68 80 67 70 62 79 67 64 63 71 59 64 53 57 74 55 57\n(a) Using class intervals of 45 - 49, 50 - 54, 55 - 59, ... construct a frequency table of the data.\n(b) Draw the histogram for the distribution\n(c) Calculate the mean of the distribution\n(d) What is the probability of selecting an iron rod whose length is in the modal class?\n\n', '', 'The following data gives the lengths, in cm, of 30 pieces of iron rods :\n45 55 65 60 61 68 59 54 64 76 50 68 72 68 80 67 70 62 79 67 64 63 71 59 64 53 57 74 55 57\n(a) Using class intervals of 45 - 49, 50 - 54, 55 - 59, ... construct a frequency table of the data.\n(b) Draw the histogram for the distribution\n(c) Calculate the mean of the distribution\n(d) What is the probability of selecting an iron rod whose length is in the modal class?', 'The following data gives the lengths, in cm, of 30 pieces of iron rods :\n45 55 65 60 61 68 59 54 64 76 50 68 72 68 80 67 70 62 79 67 64 63 71 59 64 53 57 74 55 57\n(a) Using class intervals of 45 - 49, 50 - 54, 55 - 59, ... construct a frequency table of the data.~(b) Draw the histogram for the distribution\n(c) Calculate the mean of the distribution\n(d) What is the probability of selecting an iron rod whose length is in the modal class?~(c) Calculate the mean of the distribution\n(d) What is the probability of selecting an iron rod whose length is in the modal class?~(d) What is the probability of selecting an iron rod whose length is in the modal class?', NULL, '\n\nExplanation\n(a)\n\n\n\n\nClass\nInterval\n\nFreq (f)\nTally\n\nMid-point\n(x)\n\n\\(fx\\)\n\n\n45 - 49\n1\n|\n47\n47\n\n\n50 - 54\n3\n|||\n52\n156\n\n\n55 - 59\n6\n|||||\n57\n342\n\n\n60 - 64\n7\n||||||\n62\n434\n\n\n65 - 69\n6\n|||||\n67\n402\n\n\n70 - 74\n4\n||||\n72\n288\n\n\n75 - 79\n2\n||\n77\n154\n\n\n80 - 84\n1\n|\n82\n82\n\n\n\\(\\sum\\)\n30\n\n1905\n\n\n\n\n(b)\n(c) Mean \\(\\bar{x} = \\frac{\\sum fx}{\\sum f}\\)\n= \\(\\frac{1905}{30}\\)\n= \\(63.5\\).\n(d) P(iron rod in modal class) = \\(\\frac{7}{30}\\)\n\n\n', '', 'Explanation\n(a)\n\n\n\n\nClass\nInterval\n\nFreq (f)\nTally\n\nMid-point\n(x)\n\n\\(fx\\)\n\n\n45 - 49\n1\n|\n47\n47\n\n\n50 - 54\n3\n|||\n52\n156\n\n\n55 - 59\n6\n|||||\n57\n342\n\n\n60 - 64\n7\n||||||\n62\n434\n\n\n65 - 69\n6\n|||||\n67\n402\n\n\n70 - 74\n4\n||||\n72\n288\n\n\n75 - 79\n2\n||\n77\n154\n\n\n80 - 84\n1\n|\n82\n82\n\n\n\\(\\sum\\)\n30\n\n1905\n\n\n\n\n(b)\n(c) Mean \\(\\bar{x} = \\frac{\\sum fx}{\\sum f}\\)\n= \\(\\frac{1905}{30}\\)\n= \\(63.5\\).\n(d) P(iron rod in modal class) = \\(\\frac{7}{30}\\)\n\n\n', '(a)\n\n\n\n\nClass\nInterval\n\nFreq (f)\nTally\n\nMid-point\n(x)\n\n\\(fx\\)\n\n\n45 - 49\n1\n|\n47\n47\n\n\n50 - 54\n3\n|||\n52\n156\n\n\n55 - 59\n6\n|||||\n57\n342\n\n\n60 - 64\n7\n||||||\n62\n434\n\n\n65 - 69\n6\n|||||\n67\n402\n\n\n70 - 74\n4\n||||\n72\n288\n\n\n75 - 79\n2\n||\n77\n154\n\n\n80 - 84\n1\n|\n82\n82\n\n\n\\(\\sum\\)\n30\n\n1905\n\n\n\n~(b)\n(c) Mean \\(\\bar{x} = \\frac{\\sum fx}{\\sum f}\\)\n= \\(\\frac{1905}{30}\\)\n= \\(63.5\\).\n(d) P(iron rod in modal class) = \\(\\frac{7}{30}\\)\n\n~(c) Mean \\(\\bar{x} = \\frac{\\sum fx}{\\sum f}\\)\n= \\(\\frac{1905}{30}\\)\n= \\(63.5\\).\n(d) P(iron rod in modal class) = \\(\\frac{7}{30}\\)\n\n~(d) P(iron rod in modal class) = \\(\\frac{7}{30}\\)\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(537, '2015', '\n\nQuestion\n\nGiven is the graph of the relation \\(y = ax^{2} + bx + c\\) where a, b and c are constants. Use the graph to :\n(a) find the roots of the equation \\(ax^{2} + bx + c = 0\\);\n(b) determine the values of constants a, b and c in the relation using the values of the coordinates P and Q and hence write down the relation illustrated in the graph\n(c) find the maximum value of y and the corresponding value of x at this point.\n(d) find the values of x when y = 2.\n\n\n', '', 'Given is the graph of the relation \\(y = ax^{2} + bx + c\\) where a, b and c are constants. Use the graph to :\n(a) find the roots of the equation \\(ax^{2} + bx + c = 0\\);\n(b) determine the values of constants a, b and c in the relation using the values of the coordinates P and Q and hence write down the relation illustrated in the graph\n(c) find the maximum value of y and the corresponding value of x at this point.\n(d) find the values of x when y = 2.\n', 'Given is the graph of the relation \\(y = ax^{2} + bx + c\\) where a, b and c are constants. Use the graph to :\n(a) find the roots of the equation \\(ax^{2} + bx + c = 0\\);~(b) determine the values of constants a, b and c in the relation using the values of the coordinates P and Q and hence write down the relation illustrated in the graph\n(c) find the maximum value of y and the corresponding value of x at this point.\n(d) find the values of x when y = 2.\n~(c) find the maximum value of y and the corresponding value of x at this point.\n(d) find the values of x when y = 2.\n~(d) find the values of x when y = 2.\n', NULL, '\n\nExplanation\n(a) The roots of the equation are -1.5 or 2.\n(b) C = 6 (intercept on the positive vertical axis)\n\\(ax^{2} + bx + c = 0\\)\nWhen x = 2, c = 6\n\\(4a + 2b + 6 = 0 \\implies 4a + 2b = -6 ....(1)\\)\nWhen x = -1.5, c = 6\n\\(a(-1.5)^{2} + b(-1.5) = -6\\)\n\\(\\frac{9a}{4} - \\frac{3b}{2} = -6 \\implies 9a - 6b = -24 ... (2)\\)\nMultiply (1) by 3 : \\(12a + 6b = -18 ... (1a)\\)\n(1a) + (2) : \\(12a + 9a = -18 - 24 \\implies 21a = -42\\)\n\\(a = -2\\)\n\\(4a + 2b = -6 \\implies 4(-2) + 2b = -6\\)\n\\(2b = -6 + 8 = 2 \\implies b = 1\\)\n(a, b, c) = (-2, 1, 6)\nHence the relation illustrated in the graph is \\(y = -2x^{2} + x + 6\\).\n(c) When \\(x = 0.2, y_{max} = 6.2\\)\n(d) When \\(y = 2, x = \\text{ -1.2 or 1.7}\\)\n\n', '', 'Explanation\n(a) The roots of the equation are -1.5 or 2.\n(b) C = 6 (intercept on the positive vertical axis)\n\\(ax^{2} + bx + c = 0\\)\nWhen x = 2, c = 6\n\\(4a + 2b + 6 = 0 \\implies 4a + 2b = -6 ....(1)\\)\nWhen x = -1.5, c = 6\n\\(a(-1.5)^{2} + b(-1.5) = -6\\)\n\\(\\frac{9a}{4} - \\frac{3b}{2} = -6 \\implies 9a - 6b = -24 ... (2)\\)\nMultiply (1) by 3 : \\(12a + 6b = -18 ... (1a)\\)\n(1a) + (2) : \\(12a + 9a = -18 - 24 \\implies 21a = -42\\)\n\\(a = -2\\)\n\\(4a + 2b = -6 \\implies 4(-2) + 2b = -6\\)\n\\(2b = -6 + 8 = 2 \\implies b = 1\\)\n(a, b, c) = (-2, 1, 6)\nHence the relation illustrated in the graph is \\(y = -2x^{2} + x + 6\\).\n(c) When \\(x = 0.2, y_{max} = 6.2\\)\n(d) When \\(y = 2, x = \\text{ -1.2 or 1.7}\\)\n\n', '(a) The roots of the equation are -1.5 or 2.~(b) C = 6 (intercept on the positive vertical axis)\n\\(ax^{2} + bx + c = 0\\)\nWhen x = 2, c = 6\n\\(4a + 2b + 6 = 0 \\implies 4a + 2b = -6 ....(1)\\)\nWhen x = -1.5, c = 6\n\\(a(-1.5)^{2} + b(-1.5) = -6\\)\n\\(\\frac{9a}{4} - \\frac{3b}{2} = -6 \\implies 9a - 6b = -24 ... (2)\\)\nMultiply (1) by 3 : \\(12a + 6b = -18 ... (1a)\\)\n(1a) + (2) : \\(12a + 9a = -18 - 24 \\implies 21a = -42\\)\n\\(a = -2\\)\n\\(4a + 2b = -6 \\implies 4(-2) + 2b = -6\\)\n\\(2b = -6 + 8 = 2 \\implies b = 1\\)\n(a, b, c) = (-2, 1, 6)\nHence the relation illustrated in the graph is \\(y = -2x^{2} + x + 6\\).\n(c) When \\(x = 0.2, y_{max} = 6.2\\)\n(d) When \\(y = 2, x = \\text{ -1.2 or 1.7}\\)\n~c) = (-2, 1, 6)\nHence the relation illustrated in the graph is \\(y = -2x^{2} + x + 6\\).\n(c) When \\(x = 0.2, y_{max} = 6.2\\)\n(d) When \\(y = 2, x = \\text{ -1.2 or 1.7}\\)\n~(d) When \\(y = 2, x = \\text{ -1.2 or 1.7}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(538, '2015', '\n\nQuestion\n\nGiven that \\(\\log_{10} 2 = 0.3010\\) and \\(\\log_{10} 3 = 0.4771\\), calculate without using mathematical tables or calculator, the value of :\n(a) \\(\\log_{10} 54\\) ;\n(b) \\(\\log_{10} 0.24\\).\n\n', '', 'Given that \\(\\log_{10} 2 = 0.3010\\) and \\(\\log_{10} 3 = 0.4771\\), calculate without using mathematical tables or calculator, the value of :\n(a) \\(\\log_{10} 54\\) ;\n(b) \\(\\log_{10} 0.24\\).', 'Given that \\(\\log_{10} 2 = 0.3010\\) and \\(\\log_{10} 3 = 0.4771\\), calculate without using mathematical tables or calculator, the value of :\n(a) \\(\\log_{10} 54\\) ;~(b) \\(\\log_{10} 0.24\\).', NULL, '\n\nExplanation\n(a) \\(\\log_{10} 54 \\)\n\\(54 = 2 \\times 3^{3}\\)\n\\(\\log_{10} 54 = \\log_{10} (2 \\times 3^{3})\\)\n= \\(\\log_{10} 2 + \\log_{10} 3^{3}\\)\n= \\(\\log_{10} 2 + 3\\log_{10} 3\\)\n= \\(0.3010 + (3 \\times 0.4771)\\)\n= \\(0.3010 + 1.4313\\)\n= \\(1.7323\\)\n(b) \\(\\log_{10} 0.24\\)\n\\(0.24 = \\frac{24}{100} = \\frac{2^{3} \\times 3}{100}\\)\n\\(\\log_{10} 0.24 = \\log_{10} 24 - \\log_{10} 100\\)\n\\(\\log_{10} (2^{3} \\times 3) - \\log_{10} (10^{2})\\)\n\\(\\log_{10} 2^{3} + \\log_{10} 3 - 2\\log_{10} 10\\)\n= \\(3\\log_{10} 2 + \\log_{10} 3 - 2\\)\n= \\((3 \\times 0.3010) + 0.4771 - 2\\)\n= \\(0.9030 + 0.4771 - 2\\)\n= \\(-0.6199\\)\n\n', '', 'Explanation\n(a) \\(\\log_{10} 54 \\)\n\\(54 = 2 \\times 3^{3}\\)\n\\(\\log_{10} 54 = \\log_{10} (2 \\times 3^{3})\\)\n= \\(\\log_{10} 2 + \\log_{10} 3^{3}\\)\n= \\(\\log_{10} 2 + 3\\log_{10} 3\\)\n= \\(0.3010 + (3 \\times 0.4771)\\)\n= \\(0.3010 + 1.4313\\)\n= \\(1.7323\\)\n(b) \\(\\log_{10} 0.24\\)\n\\(0.24 = \\frac{24}{100} = \\frac{2^{3} \\times 3}{100}\\)\n\\(\\log_{10} 0.24 = \\log_{10} 24 - \\log_{10} 100\\)\n\\(\\log_{10} (2^{3} \\times 3) - \\log_{10} (10^{2})\\)\n\\(\\log_{10} 2^{3} + \\log_{10} 3 - 2\\log_{10} 10\\)\n= \\(3\\log_{10} 2 + \\log_{10} 3 - 2\\)\n= \\((3 \\times 0.3010) + 0.4771 - 2\\)\n= \\(0.9030 + 0.4771 - 2\\)\n= \\(-0.6199\\)\n\n', '(a) \\(\\log_{10} 54 \\)\n\\(54 = 2 \\times 3^{3}\\)\n\\(\\log_{10} 54 = \\log_{10} (2 \\times 3^{3})\\)\n= \\(\\log_{10} 2 + \\log_{10} 3^{3}\\)\n= \\(\\log_{10} 2 + 3\\log_{10} 3\\)\n= \\(0.3010 + (3 \\times 0.4771)\\)\n= \\(0.3010 + 1.4313\\)\n= \\(1.7323\\)~(b) \\(\\log_{10} 0.24\\)\n\\(0.24 = \\frac{24}{100} = \\frac{2^{3} \\times 3}{100}\\)\n\\(\\log_{10} 0.24 = \\log_{10} 24 - \\log_{10} 100\\)\n\\(\\log_{10} (2^{3} \\times 3) - \\log_{10} (10^{2})\\)\n\\(\\log_{10} 2^{3} + \\log_{10} 3 - 2\\log_{10} 10\\)\n= \\(3\\log_{10} 2 + \\log_{10} 3 - 2\\)\n= \\((3 \\times 0.3010) + 0.4771 - 2\\)\n= \\(0.9030 + 0.4771 - 2\\)\n= \\(-0.6199\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(539, '2015', '\n\nQuestion\n\n(a) Simplify : \\(\\frac{1}{3^{5n}} \\times 9^{n - 1} \\times 27^{n +1}\\)\n(b) The sum of the ages of a woman and her daughter is 46 years. In 4 years\' time, the ratio of their ages will be 7 : 2. Find their present ages.\n\n', '', '(a) Simplify : \\(\\frac{1}{3^{5n}} \\times 9^{n - 1} \\times 27^{n +1}\\)\n(b) The sum of the ages of a woman and her daughter is 46 years. In 4 years\' time, the ratio of their ages will be 7 : 2. Find their present ages.', '(a) Simplify : \\(\\frac{1}{3^{5n}} \\times 9^{n - 1} \\times 27^{n +1}\\)~(b) The sum of the ages of a woman and her daughter is 46 years. In 4 years\' time, the ratio of their ages will be 7 : 2. Find their present ages.', NULL, '\n\nExplanation\n(a) \\(\\frac{1}{3^{5n}} \\times 9^{n - 1} \\times 27^{n + 1}\\)\n\\(3^{-5n} \\times (3^{2})^{n - 1} \\times (3^{3})^{n + 1}\\)\n\\(3^{-5n} \\times 3^{2n - 2} \\times 3^{3n + 3}\\)\n\\(3^{-5n + 2n - 2 + 3n + 3}\\)\n= \\(3^{1} = 3\\)\n(b) Let the daughter\'s age be c and the woman\'s age be d.\n\\(c + d = 46 .... (1)\\)\nIn 4 years time, the daughter\'s age = c + 4\nThe woman\'s age = d + 4\n\\(\\frac{d + 4}{c + 4} = \\frac{7}{2}\\)\n\\(2(d + 4) = 7(c + 4) \\implies 2d + 8 = 7c + 28\\)\n\\(2d - 7c = 28 - 8 \\implies 2d - 7c = 20 ... (2)\\)\n\\(c + d = 46 \\implies d = 46 - c\\)\n\\(\\therefore 2(46 - c) - 7c = 20\\)\n\\(92 - 2c - 7c = 20 \\implies 92 - 20 = 9c\\)\n\\(72 = 9c \\implies c = 8\\)\n\\(d = 46 - c \\implies d = 46 - 8 = 38\\)\nTherefore, the daughter is 8 years old and the woman is 38 years.\n\n', '', 'Explanation\n(a) \\(\\frac{1}{3^{5n}} \\times 9^{n - 1} \\times 27^{n + 1}\\)\n\\(3^{-5n} \\times (3^{2})^{n - 1} \\times (3^{3})^{n + 1}\\)\n\\(3^{-5n} \\times 3^{2n - 2} \\times 3^{3n + 3}\\)\n\\(3^{-5n + 2n - 2 + 3n + 3}\\)\n= \\(3^{1} = 3\\)\n(b) Let the daughter\'s age be c and the woman\'s age be d.\n\\(c + d = 46 .... (1)\\)\nIn 4 years time, the daughter\'s age = c + 4\nThe woman\'s age = d + 4\n\\(\\frac{d + 4}{c + 4} = \\frac{7}{2}\\)\n\\(2(d + 4) = 7(c + 4) \\implies 2d + 8 = 7c + 28\\)\n\\(2d - 7c = 28 - 8 \\implies 2d - 7c = 20 ... (2)\\)\n\\(c + d = 46 \\implies d = 46 - c\\)\n\\(\\therefore 2(46 - c) - 7c = 20\\)\n\\(92 - 2c - 7c = 20 \\implies 92 - 20 = 9c\\)\n\\(72 = 9c \\implies c = 8\\)\n\\(d = 46 - c \\implies d = 46 - 8 = 38\\)\nTherefore, the daughter is 8 years old and the woman is 38 years.\n\n', '(a) \\(\\frac{1}{3^{5n}} \\times 9^{n - 1} \\times 27^{n + 1}\\)\n\\(3^{-5n} \\times (3^{2})^{n - 1} \\times (3^{3})^{n + 1}\\)\n\\(3^{-5n} \\times 3^{2n - 2} \\times 3^{3n + 3}\\)\n\\(3^{-5n + 2n - 2 + 3n + 3}\\)\n= \\(3^{1} = 3\\)~(b) Let the daughter\'s age be c and the woman\'s age be d.\n\\(c + d = 46 .... (1)\\)\nIn 4 years time, the daughter\'s age = c + 4\nThe woman\'s age = d + 4\n\\(\\frac{d + 4}{c + 4} = \\frac{7}{2}\\)\n\\(2(d + 4) = 7(c + 4) \\implies 2d + 8 = 7c + 28\\)\n\\(2d - 7c = 28 - 8 \\implies 2d - 7c = 20 ... (2)\\)\n\\(c + d = 46 \\implies d = 46 - c\\)\n\\(\\therefore 2(46 - c) - 7c = 20\\)\n\\(92 - 2c - 7c = 20 \\implies 92 - 20 = 9c\\)\n\\(72 = 9c \\implies c = 8\\)\n\\(d = 46 - c \\implies d = 46 - 8 = 38\\)\nTherefore, the daughter is 8 years old and the woman is 38 years.\n~c) - 7c = 20\\)\n\\(92 - 2c - 7c = 20 \\implies 92 - 20 = 9c\\)\n\\(72 = 9c \\implies c = 8\\)\n\\(d = 46 - c \\implies d = 46 - 8 = 38\\)\nTherefore, the daughter is 8 years old and the woman is 38 years.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(540, '2015', '\n\nQuestion\n\nThe sides of a rectangular floor are xm and (x + 7)m. The diagonal is (x + 8)m. Calculate, in metres :\n(a) the value of x ;\n(b) the area of the floor.\n\n', '', 'The sides of a rectangular floor are xm and (x + 7)m. The diagonal is (x + 8)m. Calculate, in metres :\n(a) the value of x ;\n(b) the area of the floor.', 'The sides of a rectangular floor are xm and (x + 7)m. The diagonal is (x + 8)m. Calculate, in metres :\n(a) the value of x ;~(b) the area of the floor.', NULL, '\n\nExplanation\n(a)\n\\((x + 8)^{2} = x^{2} + (x + 7)^{2}\\)\n\\(x^{2} + 16x + 64 = x^{2} + x^{2} + 14x + 49\\)\n\\(x^{2} + 16x + 64 - 2x^{2} - 14x - 49 = 0\\)\n\\(15 + 2x - x^{2} = 0\\)\n\\(x^{2} - 2x - 15 = 0\\)\n\\((x + 3)(x - 5) = 0\\)\nx = 5m since measurements cannot be negative.\n(b) \\(Area = length \\times breadth\\)\nLength = (x + 7) = (5 + 7) = 12 m\nBreadth = x m = 5 m\n\\(Area = 12 \\times 5 = 60 m^{2}\\)\n\n', '', 'Explanation\n(a)\n\\((x + 8)^{2} = x^{2} + (x + 7)^{2}\\)\n\\(x^{2} + 16x + 64 = x^{2} + x^{2} + 14x + 49\\)\n\\(x^{2} + 16x + 64 - 2x^{2} - 14x - 49 = 0\\)\n\\(15 + 2x - x^{2} = 0\\)\n\\(x^{2} - 2x - 15 = 0\\)\n\\((x + 3)(x - 5) = 0\\)\nx = 5m since measurements cannot be negative.\n(b) \\(Area = length \\times breadth\\)\nLength = (x + 7) = (5 + 7) = 12 m\nBreadth = x m = 5 m\n\\(Area = 12 \\times 5 = 60 m^{2}\\)\n\n', '(a)\n\\((x + 8)^{2} = x^{2} + (x + 7)^{2}\\)\n\\(x^{2} + 16x + 64 = x^{2} + x^{2} + 14x + 49\\)\n\\(x^{2} + 16x + 64 - 2x^{2} - 14x - 49 = 0\\)\n\\(15 + 2x - x^{2} = 0\\)\n\\(x^{2} - 2x - 15 = 0\\)\n\\((x + 3)(x - 5) = 0\\)\nx = 5m since measurements cannot be negative.~(b) \\(Area = length \\times breadth\\)\nLength = (x + 7) = (5 + 7) = 12 m\nBreadth = x m = 5 m\n\\(Area = 12 \\times 5 = 60 m^{2}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(541, '2015', '\n\nQuestion\n\nThe diagram above shows the bar charts representing the number of vehicles manufactured by a company in January, February and March, 1992.\n(a) How many vehicles were produced in February?\n(b) What fraction of the vehicles manufactured in February were cars?\n(c) How many buses were produced altogether from January to March, 1992?\n(d) What is the ratio in the lowest term of the number of lorries produced in February to that in March?\n\n\n', '', 'The diagram above shows the bar charts representing the number of vehicles manufactured by a company in January, February and March, 1992.\n(a) How many vehicles were produced in February?\n(b) What fraction of the vehicles manufactured in February were cars?\n(c) How many buses were produced altogether from January to March, 1992?\n(d) What is the ratio in the lowest term of the number of lorries produced in February to that in March?\n', 'The diagram above shows the bar charts representing the number of vehicles manufactured by a company in January, February and March, 1992.\n(a) How many vehicles were produced in February?~(b) What fraction of the vehicles manufactured in February were cars?\n(c) How many buses were produced altogether from January to March, 1992?\n(d) What is the ratio in the lowest term of the number of lorries produced in February to that in March?\n~(c) How many buses were produced altogether from January to March, 1992?\n(d) What is the ratio in the lowest term of the number of lorries produced in February to that in March?\n~(d) What is the ratio in the lowest term of the number of lorries produced in February to that in March?\n', NULL, '\n\nExplanation\nVEHICLES\nJan\nFeb\nMar\n\n\nCAR\n10\n40\n5\n\n\nLORRY\n15\n25\n35\n\n\nBUS\n20\n30\n50\n\n\nTOTAL\n45\n95\n90\n\n\n\n(a) Vehicles produced in February = 40 + 25 + 30 = 95\n(b) Fraction of the vehicles in February that were cars = \\(\\frac{40}{95} = \\frac{8}{19}\\)\n(c) Total number of buses from January to March = 20 + 30 + 50 = 100\n(d) Ratio of lorries in February to that in March = \\(25 : 35 \\equiv 5 : 7\\).\n\n\n', '', 'Explanation\nVEHICLES\nJan\nFeb\nMar\n\n\nCAR\n10\n40\n5\n\n\nLORRY\n15\n25\n35\n\n\nBUS\n20\n30\n50\n\n\nTOTAL\n45\n95\n90\n\n\n\n(a) Vehicles produced in February = 40 + 25 + 30 = 95\n(b) Fraction of the vehicles in February that were cars = \\(\\frac{40}{95} = \\frac{8}{19}\\)\n(c) Total number of buses from January to March = 20 + 30 + 50 = 100\n(d) Ratio of lorries in February to that in March = \\(25 : 35 \\equiv 5 : 7\\).\n\n\n', '(a) Vehicles produced in February = 40 + 25 + 30 = 95~(b) Fraction of the vehicles in February that were cars = \\(\\frac{40}{95} = \\frac{8}{19}\\)\n(c) Total number of buses from January to March = 20 + 30 + 50 = 100\n(d) Ratio of lorries in February to that in March = \\(25 : 35 \\equiv 5 : 7\\).\n\n~(c) Total number of buses from January to March = 20 + 30 + 50 = 100\n(d) Ratio of lorries in February to that in March = \\(25 : 35 \\equiv 5 : 7\\).\n\n~(d) Ratio of lorries in February to that in March = \\(25 : 35 \\equiv 5 : 7\\).\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(542, '2015', '\n\nQuestion\n\n(a)In the diagram, MN || ST, NP || QT and < STQ = 70°. Find x.\n(b)In the diagram above, AC is a straight line, |BC| = |BD|, \\(\\stackrel\\frown{BCD} = 50°\\) and \\(\\stackrel\\frown{BAD} = 55°\\). Find \\(\\stackrel\\frown{BDA}\\).\n\n', '', '(a)In the diagram, MN || ST, NP || QT and < STQ = 70°. Find x.\n(b)In the diagram above, AC is a straight line, |BC| = |BD|, \\(\\stackrel\\frown{BCD} = 50°\\) and \\(\\stackrel\\frown{BAD} = 55°\\). Find \\(\\stackrel\\frown{BDA}\\).', '(a)In the diagram, MN || ST, NP || QT and < STQ = 70°. Find x.~(b)In the diagram above, AC is a straight line, |BC| = |BD|, \\(\\stackrel\\frown{BCD} = 50°\\) and \\(\\stackrel\\frown{BAD} = 55°\\). Find \\(\\stackrel\\frown{BDA}\\).', NULL, '\n\nExplanation\n(a)\\(< QTR = < TRP = 70°\\) (alternate angle)\n\\(< MNR = < TRP = 70°\\) (alternate angle; MN || ST)\n\\(< MNR + < PNM = 180°\\)\n\\(70° + < PNM = 180° \\implies < PNM = 110°\\)\n\\(\\therefore x = 110°\\)\n(b) \\(< BDC = < BCD = 50°\\) (BD = BC)\n\\(\\therefore \\Delta BDC = isosceles\\)\n\\(< ABD = 2(50°) = 100°\\) (exterior angle)\n\\(55° + < BDA + 100° = 180°\\)\n\\(< BDA = 180° - 155° = 25°\\)\n\n\n', '', 'Explanation\n(a)\\(< QTR = < TRP = 70°\\) (alternate angle)\n\\(< MNR = < TRP = 70°\\) (alternate angle; MN || ST)\n\\(< MNR + < PNM = 180°\\)\n\\(70° + < PNM = 180° \\implies < PNM = 110°\\)\n\\(\\therefore x = 110°\\)\n(b) \\(< BDC = < BCD = 50°\\) (BD = BC)\n\\(\\therefore \\Delta BDC = isosceles\\)\n\\(< ABD = 2(50°) = 100°\\) (exterior angle)\n\\(55° + < BDA + 100° = 180°\\)\n\\(< BDA = 180° - 155° = 25°\\)\n\n\n', '(a)\\(< QTR = < TRP = 70°\\) (alternate angle)\n\\(< MNR = < TRP = 70°\\) (alternate angle; MN || ST)\n\\(< MNR + < PNM = 180°\\)\n\\(70° + < PNM = 180° \\implies < PNM = 110°\\)\n\\(\\therefore x = 110°\\)~(b) \\(< BDC = < BCD = 50°\\) (BD = BC)\n\\(\\therefore \\Delta BDC = isosceles\\)\n\\(< ABD = 2(50°) = 100°\\) (exterior angle)\n\\(55° + < BDA + 100° = 180°\\)\n\\(< BDA = 180° - 155° = 25°\\)\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(543, '2015', '\n\nQuestion\n\n(a) Draw the table of values for the relation \\(y = x^{2}\\) for the interval \\(-3 \\leq x \\leq 4\\).\n(b) Using a scale of 2 cm to 1 unit on the x- axis and 2 cm to 2 units on the y- axis, draw the graphs of : (i) \\(y = x^{2}\\) ; (ii) \\(y = 2x + 3\\) for \\(-3 \\leq x \\leq 4\\).\n(c) Use your graph to find : (i) the roots of the equation \\(x^{2} = 2x + 3\\) ; (ii) the gradient of \\(y = x^{2}\\) at x = -2.\n\n', '', '(a) Draw the table of values for the relation \\(y = x^{2}\\) for the interval \\(-3 \\leq x \\leq 4\\).\n(b) Using a scale of 2 cm to 1 unit on the x- axis and 2 cm to 2 units on the y- axis, draw the graphs of : (i) \\(y = x^{2}\\) ; (ii) \\(y = 2x + 3\\) for \\(-3 \\leq x \\leq 4\\).\n(c) Use your graph to find : (i) the roots of the equation \\(x^{2} = 2x + 3\\) ; (ii) the gradient of \\(y = x^{2}\\) at x = -2.', '(a) Draw the table of values for the relation \\(y = x^{2}\\) for the interval \\(-3 \\leq x \\leq 4\\).~(b) Using a scale of 2 cm to 1 unit on the x- axis and 2 cm to 2 units on the y- axis, draw the graphs of : (i) \\(y = x^{2}\\) ; (ii) \\(y = 2x + 3\\) for \\(-3 \\leq x \\leq 4\\).\n(c) Use your graph to find : (i) the roots of the equation \\(x^{2} = 2x + 3\\) ; (ii) the gradient of \\(y = x^{2}\\) at x = -2.~(c) Use your graph to find : (i) the roots of the equation \\(x^{2} = 2x + 3\\) ; (ii) the gradient of \\(y = x^{2}\\) at x = -2.', NULL, '\n\nExplanation\n(a)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\n\\(y = x^{2}\\)\n9\n4\n1\n0\n1\n4\n9\n16\n\n\n\n(b) \\(y = 2x + 3\\)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\n2x\n-6\n-4\n-2\n0\n2\n4\n6\n8\n\n\n3\n3\n3\n3\n3\n3\n3\n3\n3\n\n\ny\n-3\n-1\n1\n3\n5\n7\n9\n11\n\n\n\n\n(c)(i) The roots of the equation are -1 and 3, from the graph.\n(ii) The gradient of \\(y = x^{2}\\) at x = -2 : \\(\\frac{8}{-2.5} = -3.2\\)\n\n', '', 'Explanation\n(a)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\n\\(y = x^{2}\\)\n9\n4\n1\n0\n1\n4\n9\n16\n\n\n\n(b) \\(y = 2x + 3\\)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\n2x\n-6\n-4\n-2\n0\n2\n4\n6\n8\n\n\n3\n3\n3\n3\n3\n3\n3\n3\n3\n\n\ny\n-3\n-1\n1\n3\n5\n7\n9\n11\n\n\n\n\n(c)(i) The roots of the equation are -1 and 3, from the graph.\n(ii) The gradient of \\(y = x^{2}\\) at x = -2 : \\(\\frac{8}{-2.5} = -3.2\\)\n\n', '(a)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\n\\(y = x^{2}\\)\n9\n4\n1\n0\n1\n4\n9\n16\n\n\n~(b) \\(y = 2x + 3\\)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\n2x\n-6\n-4\n-2\n0\n2\n4\n6\n8\n\n\n3\n3\n3\n3\n3\n3\n3\n3\n3\n\n\ny\n-3\n-1\n1\n3\n5\n7\n9\n11\n\n\n\n\n(c)(i) The roots of the equation are -1 and 3, from the graph.\n(ii) The gradient of \\(y = x^{2}\\) at x = -2 : \\(\\frac{8}{-2.5} = -3.2\\)\n~(c)(i) The roots of the equation are -1 and 3, from the graph.\n(ii) The gradient of \\(y = x^{2}\\) at x = -2 : \\(\\frac{8}{-2.5} = -3.2\\)\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
(544, '2015', '\n\nQuestion\n\nThe number of child births recorded in 50 maternity centres of a local government in August 1993 are as follows :\n50 99 81 86 69 85 93 63 92 65 77 74 76 71 90 74 81 94 67 75 95 81 68 105 99 68 75 75 76 73 79 74 80 69 74 62 74 80 79 68 79 75 75 71 83 75 80 85 81 82\n(a) Construct a frequency distribution table, using class intervals 45 - 54, 55 - 64, etc.\n(b) Draw the histogram for the distribution\n(c) Use your histogram to estimate the mode.\n(d) Calculate the mean number of births.\n\n', '', 'The number of child births recorded in 50 maternity centres of a local government in August 1993 are as follows :\n50 99 81 86 69 85 93 63 92 65 77 74 76 71 90 74 81 94 67 75 95 81 68 105 99 68 75 75 76 73 79 74 80 69 74 62 74 80 79 68 79 75 75 71 83 75 80 85 81 82\n(a) Construct a frequency distribution table, using class intervals 45 - 54, 55 - 64, etc.\n(b) Draw the histogram for the distribution\n(c) Use your histogram to estimate the mode.\n(d) Calculate the mean number of births.', 'The number of child births recorded in 50 maternity centres of a local government in August 1993 are as follows :\n50 99 81 86 69 85 93 63 92 65 77 74 76 71 90 74 81 94 67 75 95 81 68 105 99 68 75 75 76 73 79 74 80 69 74 62 74 80 79 68 79 75 75 71 83 75 80 85 81 82\n(a) Construct a frequency distribution table, using class intervals 45 - 54, 55 - 64, etc.~(b) Draw the histogram for the distribution\n(c) Use your histogram to estimate the mode.\n(d) Calculate the mean number of births.~(c) Use your histogram to estimate the mode.\n(d) Calculate the mean number of births.~(d) Calculate the mean number of births.', NULL, '\n\nExplanation\n (a)\n\nFrequency Distribution Table\n\n\nBirth\n45 - 54\n55 - 64\n65 - 74\n75 - 84\n85 - 94\n95 - 104\n105 - 114\n\n\nFreq(f)\n1\n2\n15\n21\n7\n3\n1\n\n\n\n(b)\n(c) Modal class is 75 - 84\nMode = \\(\\frac{75 + 84}{2} = 79.5\\)\n(d)\n\n\n\nBirth\nfreq (f)\n\nclass midpoint\n(x)\n\n\\(fx\\)\n\n\n45 - 54\n1\n49.5\n49.5\n\n\n55 - 64\n2\n59.5\n119\n\n\n65 - 74\n15\n69.5\n1042.5\n\n\n75 - 84\n21\n79.5\n1669.5\n\n\n85 - 94\n7\n89.5\n626.5\n\n\n95 - 104\n3\n99.5\n298.5\n\n\n105 - 114\n1\n109.5\n109.5\n\n\nTotal\n50\n\n3915\n\n\n\nMean \\(\\bar{x} = \\frac{\\sum fx}{\\sum f}\\)\n= \\(\\frac{3915}{50}\\)\n= \\(78.3\\).\n\n\n', '', 'Explanation\n (a)\n\nFrequency Distribution Table\n\n\nBirth\n45 - 54\n55 - 64\n65 - 74\n75 - 84\n85 - 94\n95 - 104\n105 - 114\n\n\nFreq(f)\n1\n2\n15\n21\n7\n3\n1\n\n\n\n(b)\n(c) Modal class is 75 - 84\nMode = \\(\\frac{75 + 84}{2} = 79.5\\)\n(d)\n\n\n\nBirth\nfreq (f)\n\nclass midpoint\n(x)\n\n\\(fx\\)\n\n\n45 - 54\n1\n49.5\n49.5\n\n\n55 - 64\n2\n59.5\n119\n\n\n65 - 74\n15\n69.5\n1042.5\n\n\n75 - 84\n21\n79.5\n1669.5\n\n\n85 - 94\n7\n89.5\n626.5\n\n\n95 - 104\n3\n99.5\n298.5\n\n\n105 - 114\n1\n109.5\n109.5\n\n\nTotal\n50\n\n3915\n\n\n\nMean \\(\\bar{x} = \\frac{\\sum fx}{\\sum f}\\)\n= \\(\\frac{3915}{50}\\)\n= \\(78.3\\).\n\n\n', '(a)\n\nFrequency Distribution Table\n\n\nBirth\n45 - 54\n55 - 64\n65 - 74\n75 - 84\n85 - 94\n95 - 104\n105 - 114\n\n\nFreq(f)\n1\n2\n15\n21\n7\n3\n1\n\n\n~(b)\n(c) Modal class is 75 - 84\nMode = \\(\\frac{75 + 84}{2} = 79.5\\)\n(d)\n\n\n\nBirth\nfreq (f)\n\nclass midpoint\n(x)\n\n\\(fx\\)\n\n\n45 - 54\n1\n49.5\n49.5\n\n\n55 - 64\n2\n59.5\n119\n\n\n65 - 74\n15\n69.5\n1042.5\n\n\n75 - 84\n21\n79.5\n1669.5\n\n\n85 - 94\n7\n89.5\n626.5\n\n\n95 - 104\n3\n99.5\n298.5\n\n\n105 - 114\n1\n109.5\n109.5\n\n\nTotal\n50\n\n3915\n\n\n\nMean \\(\\bar{x} = \\frac{\\sum fx}{\\sum f}\\)\n= \\(\\frac{3915}{50}\\)\n= \\(78.3\\).\n\n~(c) Modal class is 75 - 84\nMode = \\(\\frac{75 + 84}{2} = 79.5\\)\n(d)\n\n\n\nBirth\nfreq (f)\n\nclass midpoint\n(x)\n\n\\(fx\\)\n\n\n45 - 54\n1\n49.5\n49.5\n\n\n55 - 64\n2\n59.5\n119\n\n\n65 - 74\n15\n69.5\n1042.5\n\n\n75 - 84\n21\n79.5\n1669.5\n\n\n85 - 94\n7\n89.5\n626.5\n\n\n95 - 104\n3\n99.5\n298.5\n\n\n105 - 114\n1\n109.5\n109.5\n\n\nTotal\n50\n\n3915\n\n\n\nMean \\(\\bar{x} = \\frac{\\sum fx}{\\sum f}\\)\n= \\(\\frac{3915}{50}\\)\n= \\(78.3\\).\n\n~(d)\n\n\n\nBirth\nfreq (f)\n\nclass midpoint\n(x)\n\n\\(fx\\)\n\n\n45 - 54\n1\n49.5\n49.5\n\n\n55 - 64\n2\n59.5\n119\n\n\n65 - 74\n15\n69.5\n1042.5\n\n\n75 - 84\n21\n79.5\n1669.5\n\n\n85 - 94\n7\n89.5\n626.5\n\n\n95 - 104\n3\n99.5\n298.5\n\n\n105 - 114\n1\n109.5\n109.5\n\n\nTotal\n50\n\n3915\n\n\n\nMean \\(\\bar{x} = \\frac{\\sum fx}{\\sum f}\\)\n= \\(\\frac{3915}{50}\\)\n= \\(78.3\\).\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(545, '2015', '\n\nQuestion\n\n(a) In a market survey, 100 traders sell fruits, 40 sell apples, 46 oranges, 50 mangoes, 14 apples and oranges, 15 apples and mangoes and 10 sell the three types of fruits. Each of the 100 traders sells at least one of the three fruits.\n(i) Represent the information in a Venn diagram ; (ii) Find the number that sell oranges and mangoes only.\n(b) Find the value of x for which \\(312_{four} + 52_{x} = 96_{ten}\\)\n\n', '', '(a) In a market survey, 100 traders sell fruits, 40 sell apples, 46 oranges, 50 mangoes, 14 apples and oranges, 15 apples and mangoes and 10 sell the three types of fruits. Each of the 100 traders sells at least one of the three fruits.\n(i) Represent the information in a Venn diagram ; (ii) Find the number that sell oranges and mangoes only.\n(b) Find the value of x for which \\(312_{four} + 52_{x} = 96_{ten}\\)', '(a) In a market survey, 100 traders sell fruits, 40 sell apples, 46 oranges, 50 mangoes, 14 apples and oranges, 15 apples and mangoes and 10 sell the three types of fruits. Each of the 100 traders sells at least one of the three fruits.\n(i) Represent the information in a Venn diagram ; (ii) Find the number that sell oranges and mangoes only.~(b) Find the value of x for which \\(312_{four} + 52_{x} = 96_{ten}\\)', NULL, '\n\nExplanation\n(a)(i)\n(ii) \\(21 + 4 + 5 + 10 + x + 35 - x + 32 - x = 100\\)\n\\(107 - x = 100\\)\n\\(x = 107 - 100 = 7\\)\nTherefore, 7 traders sell oranges and mangoes alone.\n(b) \\(312_{four} + 52_{x} = 96_{ten}\\)\n\\(312_{4} = 3 \\times 4^{2} + 1 \\times 4^{1} + 2 \\times 4^{0}\\)\n= \\(48 + 4 + 2 = 54_{10}\\)\n\\(52_{x} = 5 \\times x^{1} + 2\\times x^{0}\\)\n= \\((5x + 2)_{10}\\)\n\\(\\therefore 54 + (5x + 2) = 96\\)\n\\(56 + 5x = 96\\implies 5x = 96 - 56 = 40\\)\n\\(x = 8\\)\n\\(\\therefore 312_{4} + 52_{8} = 96_{10}\\)\n\n', '', 'Explanation\n(a)(i)\n(ii) \\(21 + 4 + 5 + 10 + x + 35 - x + 32 - x = 100\\)\n\\(107 - x = 100\\)\n\\(x = 107 - 100 = 7\\)\nTherefore, 7 traders sell oranges and mangoes alone.\n(b) \\(312_{four} + 52_{x} = 96_{ten}\\)\n\\(312_{4} = 3 \\times 4^{2} + 1 \\times 4^{1} + 2 \\times 4^{0}\\)\n= \\(48 + 4 + 2 = 54_{10}\\)\n\\(52_{x} = 5 \\times x^{1} + 2\\times x^{0}\\)\n= \\((5x + 2)_{10}\\)\n\\(\\therefore 54 + (5x + 2) = 96\\)\n\\(56 + 5x = 96\\implies 5x = 96 - 56 = 40\\)\n\\(x = 8\\)\n\\(\\therefore 312_{4} + 52_{8} = 96_{10}\\)\n\n', '(a)(i)\n(ii) \\(21 + 4 + 5 + 10 + x + 35 - x + 32 - x = 100\\)\n\\(107 - x = 100\\)\n\\(x = 107 - 100 = 7\\)\nTherefore, 7 traders sell oranges and mangoes alone.~(b) \\(312_{four} + 52_{x} = 96_{ten}\\)\n\\(312_{4} = 3 \\times 4^{2} + 1 \\times 4^{1} + 2 \\times 4^{0}\\)\n= \\(48 + 4 + 2 = 54_{10}\\)\n\\(52_{x} = 5 \\times x^{1} + 2\\times x^{0}\\)\n= \\((5x + 2)_{10}\\)\n\\(\\therefore 54 + (5x + 2) = 96\\)\n\\(56 + 5x = 96\\implies 5x = 96 - 56 = 40\\)\n\\(x = 8\\)\n\\(\\therefore 312_{4} + 52_{8} = 96_{10}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(546, '2015', '\n\nQuestion\n\n(a) The probabilities that three boys pass an examination are \\(\\frac{2}{3}, \\frac{5}{8}\\) and \\(\\frac{3}{4}\\) respectively. Find the probability that :\n(i) all three boys pass ; (ii) none of the boyspass ; (iii) only two of the boys pass.\n(b) A shop-keeper marks a television set for sale at N36,000 so as to make a profit of 20% on the cost price. When he sells it, he allows a discount of 5% of the marked price. Calculate the actual percentage profit.\n\n', '', '(a) The probabilities that three boys pass an examination are \\(\\frac{2}{3}, \\frac{5}{8}\\) and \\(\\frac{3}{4}\\) respectively. Find the probability that :\n(i) all three boys pass ; (ii) none of the boyspass ; (iii) only two of the boys pass.\n(b) A shop-keeper marks a television set for sale at N36,000 so as to make a profit of 20% on the cost price. When he sells it, he allows a discount of 5% of the marked price. Calculate the actual percentage profit.', '(a) The probabilities that three boys pass an examination are \\(\\frac{2}{3}, \\frac{5}{8}\\) and \\(\\frac{3}{4}\\) respectively. Find the probability that :\n(i) all three boys pass ; (ii) none of the boyspass ; (iii) only two of the boys pass.~(b) A shop-keeper marks a television set for sale at N36,000 so as to make a profit of 20% on the cost price. When he sells it, he allows a discount of 5% of the marked price. Calculate the actual percentage profit.', NULL, '\n\nExplanation\n(a)(i) P(A) = \\(\\frac{2}{3}\\) ; P(B) = \\(\\frac{5}{8}\\) ; P(C) = \\(\\frac{3}{4}\\).\n\\(P(A \\cap B \\cap C) = P(A) \\times P(B) \\times P(C)\\)\n= \\(\\frac{2}{3} \\times \\frac{5}{8} \\times \\frac{3}{4} = \\frac{5}{16}\\)\n(ii) P(A\') = \\(1 - \\frac{2}{3} = \\frac{1}{3}\\)\nP(B\') = \\(1 - \\frac{5}{8} = \\frac{3}{8}\\)\nP(C\') = \\(1 - \\frac{3}{4} = \\frac{1}{4}\\)\n\\(P(A\' \\cap B\' \\cap C\') = P(A\') \\times P(B\') \\times P(C\')\\)\n= \\(\\frac{1}{3} \\times \\frac{3}{8} \\times \\frac{1}{4} \\)\n= \\(\\frac{1}{32}\\)\n(iii) P(A and B pass, C fails) = \\(\\frac{2}{3} \\times \\frac{5}{8} \\times \\frac{1}{4} = \\frac{5}{48}\\)\nP( A and C pass, B fails) = \\(\\frac{2}{3} \\times \\frac{3}{4} \\times \\frac{3}{8} = \\frac{3}{16}\\)\nP(B and C pass, A fails) = \\(\\frac{5}{8} \\times \\frac{3}{4} \\times \\frac{1}{3} = \\frac{5}{32}\\)\nP(only two of the boys pass) = \\(\\frac{5}{48} + \\frac{3}{16} + \\frac{5}{32} = \\frac{43}{96}\\)\n(b) Let the cost price be x\n% gain = \\(\\frac{\\text{actual gain}}{\\text{cost price}} \\times 100%\\)\n\\(20 = \\frac{36,000 - x}{x} \\times 100%\\)\n\\(\\frac{36,000 - x}{x} = \\frac{1}{5}\\)\n\\(5(36,000 - x) = x \\implies 180,000 - 5x = x\\)\n\\(180,000 = x + 5x = 6x\\)\n\\(x = N30,000\\)\nDiscount (5%) : \\(\\frac{5}{100} \\times N36,000 = N1,800\\)\nSelling price : \\(N(36,000 - 1,800) = N34,200\\)\nActual gain : \\(\\frac{34,200 - 30,000}{30,000} \\times 100% \\)\n= 14%\n\n', '', 'Explanation\n(a)(i) P(A) = \\(\\frac{2}{3}\\) ; P(B) = \\(\\frac{5}{8}\\) ; P(C) = \\(\\frac{3}{4}\\).\n\\(P(A \\cap B \\cap C) = P(A) \\times P(B) \\times P(C)\\)\n= \\(\\frac{2}{3} \\times \\frac{5}{8} \\times \\frac{3}{4} = \\frac{5}{16}\\)\n(ii) P(A\') = \\(1 - \\frac{2}{3} = \\frac{1}{3}\\)\nP(B\') = \\(1 - \\frac{5}{8} = \\frac{3}{8}\\)\nP(C\') = \\(1 - \\frac{3}{4} = \\frac{1}{4}\\)\n\\(P(A\' \\cap B\' \\cap C\') = P(A\') \\times P(B\') \\times P(C\')\\)\n= \\(\\frac{1}{3} \\times \\frac{3}{8} \\times \\frac{1}{4} \\)\n= \\(\\frac{1}{32}\\)\n(iii) P(A and B pass, C fails) = \\(\\frac{2}{3} \\times \\frac{5}{8} \\times \\frac{1}{4} = \\frac{5}{48}\\)\nP( A and C pass, B fails) = \\(\\frac{2}{3} \\times \\frac{3}{4} \\times \\frac{3}{8} = \\frac{3}{16}\\)\nP(B and C pass, A fails) = \\(\\frac{5}{8} \\times \\frac{3}{4} \\times \\frac{1}{3} = \\frac{5}{32}\\)\nP(only two of the boys pass) = \\(\\frac{5}{48} + \\frac{3}{16} + \\frac{5}{32} = \\frac{43}{96}\\)\n(b) Let the cost price be x\n% gain = \\(\\frac{\\text{actual gain}}{\\text{cost price}} \\times 100%\\)\n\\(20 = \\frac{36,000 - x}{x} \\times 100%\\)\n\\(\\frac{36,000 - x}{x} = \\frac{1}{5}\\)\n\\(5(36,000 - x) = x \\implies 180,000 - 5x = x\\)\n\\(180,000 = x + 5x = 6x\\)\n\\(x = N30,000\\)\nDiscount (5%) : \\(\\frac{5}{100} \\times N36,000 = N1,800\\)\nSelling price : \\(N(36,000 - 1,800) = N34,200\\)\nActual gain : \\(\\frac{34,200 - 30,000}{30,000} \\times 100% \\)\n= 14%\n\n', '(a)(i) P(A) = \\(\\frac{2}{3}\\) ; P(B) = \\(\\frac{5}{8}\\) ; P(C) = \\(\\frac{3}{4}\\).\n\\(P(A \\cap B \\cap C) = P(A) \\times P(B) \\times P(C)\\)\n= \\(\\frac{2}{3} \\times \\frac{5}{8} \\times \\frac{3}{4} = \\frac{5}{16}\\)\n(ii) P(A\') = \\(1 - \\frac{2}{3} = \\frac{1}{3}\\)\nP(B\') = \\(1 - \\frac{5}{8} = \\frac{3}{8}\\)\nP(C\') = \\(1 - \\frac{3}{4} = \\frac{1}{4}\\)\n\\(P(A\' \\cap B\' \\cap C\') = P(A\') \\times P(B\') \\times P(C\')\\)\n= \\(\\frac{1}{3} \\times \\frac{3}{8} \\times \\frac{1}{4} \\)\n= \\(\\frac{1}{32}\\)\n(iii) P(A and B pass, C fails) = \\(\\frac{2}{3} \\times \\frac{5}{8} \\times \\frac{1}{4} = \\frac{5}{48}\\)\nP( A and C pass, B fails) = \\(\\frac{2}{3} \\times \\frac{3}{4} \\times \\frac{3}{8} = \\frac{3}{16}\\)\nP(B and C pass, A fails) = \\(\\frac{5}{8} \\times \\frac{3}{4} \\times \\frac{1}{3} = \\frac{5}{32}\\)\nP(only two of the boys pass) = \\(\\frac{5}{48} + \\frac{3}{16} + \\frac{5}{32} = \\frac{43}{96}\\)~(b) Let the cost price be x\n% gain = \\(\\frac{\\text{actual gain}}{\\text{cost price}} \\times 100%\\)\n\\(20 = \\frac{36,000 - x}{x} \\times 100%\\)\n\\(\\frac{36,000 - x}{x} = \\frac{1}{5}\\)\n\\(5(36,000 - x) = x \\implies 180,000 - 5x = x\\)\n\\(180,000 = x + 5x = 6x\\)\n\\(x = N30,000\\)\nDiscount (5%) : \\(\\frac{5}{100} \\times N36,000 = N1,800\\)\nSelling price : \\(N(36,000 - 1,800) = N34,200\\)\nActual gain : \\(\\frac{34,200 - 30,000}{30,000} \\times 100% \\)\n= 14%\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(547, '2015', '\n\nQuestion\n\n(a) The roots of the equation \\(2x^{2} + (p + 1)x + 9 = 0\\), are 1 and 3, where p and q are constants. Find the values of p and q.\n(b) The weight of an object varies inversely as the square of its distance from the centre of the earth. A small satellite weighs 80kg on the earth\'s surface. Calculate, correct to the nearest whole number, the weight of the satellite when it is 800km above the surface of the earth. [Take the radius of the earth as 6,400km].\n\n', '', '(a) The roots of the equation \\(2x^{2} + (p + 1)x + 9 = 0\\), are 1 and 3, where p and q are constants. Find the values of p and q.\n(b) The weight of an object varies inversely as the square of its distance from the centre of the earth. A small satellite weighs 80kg on the earth\'s surface. Calculate, correct to the nearest whole number, the weight of the satellite when it is 800km above the surface of the earth. [Take the radius of the earth as 6,400km].', '(a) The roots of the equation \\(2x^{2} + (p + 1)x + 9 = 0\\), are 1 and 3, where p and q are constants. Find the values of p and q.~(b) The weight of an object varies inversely as the square of its distance from the centre of the earth. A small satellite weighs 80kg on the earth\'s surface. Calculate, correct to the nearest whole number, the weight of the satellite when it is 800km above the surface of the earth. [Take the radius of the earth as 6,400km].', NULL, '\n\nExplanation\n(a) \\(2x^{2} + (p + 1)x + q = 0\\)\nThe roots of the equation are 1 and 3, therefore (x - 1)(x - 3) = 0.\nEquation : \\(x^{2} - 3x - x + 3 = 0\\)\n\\(x^{2} - 4x + 3 = 0\\)\n\\(\\therefore 2x^{2} + (p + 1)x + q \\equiv x^{2} - 4x + 3\\)\n\\(\\frac{2x^{2} + (p + 1)x + q}{2} = x^{2} + \\frac{(p + 1)}{2} x + \\frac{q}{2}\\)\n\\(\\implies \\frac{q}{2} = 3\\)\n\\(q = 6\\)\n\\(\\frac{p + 1}{2} = -4 \\implies p + 1 = -8\\)\n\\(p = -9\\)\n(p, q) = (-9, 6).\n(b) \\(W \\propto \\frac{1}{d^{2}} \\)\n\\(W = \\frac{k}{d^{2}}\\)\n\\(k = Wd^{2}\\)\n\\(W = \\frac{80 \\times (6400)^{2}}{d^{2}}\\) (equation of variation).\n\\(W = \\frac{80 \\times (6400)^{2}}{(6400 + 800)^{2}}\\)\n\\(W = \\frac{80 \\times (6400)^{2}}{(7200)^{2}}\\)\n= 63.21 kg\n\n', '', 'Explanation\n(a) \\(2x^{2} + (p + 1)x + q = 0\\)\nThe roots of the equation are 1 and 3, therefore (x - 1)(x - 3) = 0.\nEquation : \\(x^{2} - 3x - x + 3 = 0\\)\n\\(x^{2} - 4x + 3 = 0\\)\n\\(\\therefore 2x^{2} + (p + 1)x + q \\equiv x^{2} - 4x + 3\\)\n\\(\\frac{2x^{2} + (p + 1)x + q}{2} = x^{2} + \\frac{(p + 1)}{2} x + \\frac{q}{2}\\)\n\\(\\implies \\frac{q}{2} = 3\\)\n\\(q = 6\\)\n\\(\\frac{p + 1}{2} = -4 \\implies p + 1 = -8\\)\n\\(p = -9\\)\n(p, q) = (-9, 6).\n(b) \\(W \\propto \\frac{1}{d^{2}} \\)\n\\(W = \\frac{k}{d^{2}}\\)\n\\(k = Wd^{2}\\)\n\\(W = \\frac{80 \\times (6400)^{2}}{d^{2}}\\) (equation of variation).\n\\(W = \\frac{80 \\times (6400)^{2}}{(6400 + 800)^{2}}\\)\n\\(W = \\frac{80 \\times (6400)^{2}}{(7200)^{2}}\\)\n= 63.21 kg\n\n', '(a) \\(2x^{2} + (p + 1)x + q = 0\\)\nThe roots of the equation are 1 and 3, therefore (x - 1)(x - 3) = 0.\nEquation : \\(x^{2} - 3x - x + 3 = 0\\)\n\\(x^{2} - 4x + 3 = 0\\)\n\\(\\therefore 2x^{2} + (p + 1)x + q \\equiv x^{2} - 4x + 3\\)\n\\(\\frac{2x^{2} + (p + 1)x + q}{2} = x^{2} + \\frac{(p + 1)}{2} x + \\frac{q}{2}\\)\n\\(\\implies \\frac{q}{2} = 3\\)\n\\(q = 6\\)\n\\(\\frac{p + 1}{2} = -4 \\implies p + 1 = -8\\)\n\\(p = -9\\)\n(p, q) = (-9, 6).~(b) \\(W \\propto \\frac{1}{d^{2}} \\)\n\\(W = \\frac{k}{d^{2}}\\)\n\\(k = Wd^{2}\\)\n\\(W = \\frac{80 \\times (6400)^{2}}{d^{2}}\\) (equation of variation).\n\\(W = \\frac{80 \\times (6400)^{2}}{(6400 + 800)^{2}}\\)\n\\(W = \\frac{80 \\times (6400)^{2}}{(7200)^{2}}\\)\n= 63.21 kg\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(548, '2015', '\n\nQuestion\n\nIn the diagram, three points A, B and C are on the same horizontal ground. B is 15m from A, on a bearing of 053°, C is 18m from B on a bearing of 161°. A vertical pole with top T is erected at B such that < ATB = 58°. Calculate, correct to three significant figures,\n(a) the length of AC.\n(b) the bearing of C from A ;\n(c) the height of the pole BT.\n\n', '', 'In the diagram, three points A, B and C are on the same horizontal ground. B is 15m from A, on a bearing of 053°, C is 18m from B on a bearing of 161°. A vertical pole with top T is erected at B such that < ATB = 58°. Calculate, correct to three significant figures,\n(a) the length of AC.\n(b) the bearing of C from A ;\n(c) the height of the pole BT.', 'In the diagram, three points A, B and C are on the same horizontal ground. B is 15m from A, on a bearing of 053°, C is 18m from B on a bearing of 161°. A vertical pole with top T is erected at B such that < ATB = 58°. Calculate, correct to three significant figures,\n(a) the length of AC.~(b) the bearing of C from A ;\n(c) the height of the pole BT.~(c) the height of the pole BT.', NULL, '\n\nExplanation\n< PAB = < ABE = 53° (alternate angles)\n< CBE = 180° - < DBC = 180° - 161° = 19°\n< ABC = < ABE + < CBE\n= 53° + 19° = 72°\n(a) In \\(\\Delta ABC\\),\n\\(AC^{2} = AB^{2} + BC^{2} - 2(AB)(BC) \\cos < ABC\\)\n\\(AC^{2} = 15^{2} + 18^{2} - 2(15)(18) \\cos 72\\)\n= \\(225 + 324 - 540 \\cos 72\\)\n= \\(549 - 166.869\\)\n\\(AC^{2}= 382.131\\)\n\\(AC = \\sqrt{382.131} = 19.548m\\)\n\\(\\approxeq 19.5m\\)\n(b) \\(\\frac{\\sin A}{18} = \\frac{\\sin 72}{19.548}\\)\n\\(\\sin A = \\frac{18 \\times \\sin 72}{19.548}\\)\n\\(\\sin A = 0.8757\\)\n\\(A = \\sin^{-1} (0.8757) = 61.13°\\)\nThe bearing of C from A = 61.13° + 53° = 114.13° \\(\\approxeq\\) 114°.\n(c) In \\(\\Delta ATB\\),\n\\(\\frac{15}{BT} = \\tan 58\\)\n\\(BT = \\frac{15}{\\tan 58}\\)\n\\(9.373 m \\approxeq 9.37m\\)\n\n', '', 'Explanation\n< PAB = < ABE = 53° (alternate angles)\n< CBE = 180° - < DBC = 180° - 161° = 19°\n< ABC = < ABE + < CBE\n= 53° + 19° = 72°\n(a) In \\(\\Delta ABC\\),\n\\(AC^{2} = AB^{2} + BC^{2} - 2(AB)(BC) \\cos < ABC\\)\n\\(AC^{2} = 15^{2} + 18^{2} - 2(15)(18) \\cos 72\\)\n= \\(225 + 324 - 540 \\cos 72\\)\n= \\(549 - 166.869\\)\n\\(AC^{2}= 382.131\\)\n\\(AC = \\sqrt{382.131} = 19.548m\\)\n\\(\\approxeq 19.5m\\)\n(b) \\(\\frac{\\sin A}{18} = \\frac{\\sin 72}{19.548}\\)\n\\(\\sin A = \\frac{18 \\times \\sin 72}{19.548}\\)\n\\(\\sin A = 0.8757\\)\n\\(A = \\sin^{-1} (0.8757) = 61.13°\\)\nThe bearing of C from A = 61.13° + 53° = 114.13° \\(\\approxeq\\) 114°.\n(c) In \\(\\Delta ATB\\),\n\\(\\frac{15}{BT} = \\tan 58\\)\n\\(BT = \\frac{15}{\\tan 58}\\)\n\\(9.373 m \\approxeq 9.37m\\)\n\n', '(a) In \\(\\Delta ABC\\),\n\\(AC^{2} = AB^{2} + BC^{2} - 2(AB)(BC) \\cos < ABC\\)\n\\(AC^{2} = 15^{2} + 18^{2} - 2(15)(18) \\cos 72\\)\n= \\(225 + 324 - 540 \\cos 72\\)\n= \\(549 - 166.869\\)\n\\(AC^{2}= 382.131\\)\n\\(AC = \\sqrt{382.131} = 19.548m\\)\n\\(\\approxeq 19.5m\\)~(b) \\(\\frac{\\sin A}{18} = \\frac{\\sin 72}{19.548}\\)\n\\(\\sin A = \\frac{18 \\times \\sin 72}{19.548}\\)\n\\(\\sin A = 0.8757\\)\n\\(A = \\sin^{-1} (0.8757) = 61.13°\\)\nThe bearing of C from A = 61.13° + 53° = 114.13° \\(\\approxeq\\) 114°.\n(c) In \\(\\Delta ATB\\),\n\\(\\frac{15}{BT} = \\tan 58\\)\n\\(BT = \\frac{15}{\\tan 58}\\)\n\\(9.373 m \\approxeq 9.37m\\)\n~(c) In \\(\\Delta ATB\\),\n\\(\\frac{15}{BT} = \\tan 58\\)\n\\(BT = \\frac{15}{\\tan 58}\\)\n\\(9.373 m \\approxeq 9.37m\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(549, '2015', '\n\nQuestion\n\n(a) A boy blew his rubber balloon to a spherical shape. The balloon burst when its diameter was 15 cm. Calculate, correct to the nearest whole number, the volume of air in the balloon at the point of bursting. [Take \\(\\pi = \\frac{22}{7}\\)]\n(b) A point X is on latitude 28°N and longitude 105°W. Y is another point on the same latitude as X but on longitude 35°E. (i) Calculate, correct to three significant figures, the distance between X and Y along latitude 28°N ; (ii) How far is X from the equator? [Take \\(\\pi = \\frac{22}{7}\\) and radius of the earth = 6,400km].\n\n', '', '(a) A boy blew his rubber balloon to a spherical shape. The balloon burst when its diameter was 15 cm. Calculate, correct to the nearest whole number, the volume of air in the balloon at the point of bursting. [Take \\(\\pi = \\frac{22}{7}\\)]\n(b) A point X is on latitude 28°N and longitude 105°W. Y is another point on the same latitude as X but on longitude 35°E. (i) Calculate, correct to three significant figures, the distance between X and Y along latitude 28°N ; (ii) How far is X from the equator? [Take \\(\\pi = \\frac{22}{7}\\) and radius of the earth = 6,400km].', '(a) A boy blew his rubber balloon to a spherical shape. The balloon burst when its diameter was 15 cm. Calculate, correct to the nearest whole number, the volume of air in the balloon at the point of bursting. [Take \\(\\pi = \\frac{22}{7}\\)]~(b) A point X is on latitude 28°N and longitude 105°W. Y is another point on the same latitude as X but on longitude 35°E. (i) Calculate, correct to three significant figures, the distance between X and Y along latitude 28°N ; (ii) How far is X from the equator? [Take \\(\\pi = \\frac{22}{7}\\) and radius of the earth = 6,400km].', NULL, '\n\nExplanation\n(a) \\(d = 15 cm \\implies r = \\frac{15}{2} cm\\)\n\\(V = \\frac{4\\pi r^{3}}{3}\\)\n\\(V = \\frac{4}{3} \\times \\frac{22}{7} \\times (\\frac{15}{2})^{3}\\)\n\\(V = \\frac{1125 \\times 22}{7 \\times 2}\\)\n\\(V = 1,767.857 cm^{3}\\)\n(b)(i) Longitude difference : 105° + 35° = 140°\nDistance along the parallel of latitude = \\(\\frac{140}{360} \\times 2 \\times \\frac{22}{7} \\times 6400 \\cos 28\\)\n= \\(\\frac{140 \\times 44 \\times 6400 \\times 0.8829}{360 \\times 7}\\)\n= \\(13812.48 km\\)\n\\(\\approxeq 13800 km\\)\n(ii) Distance from the equator : \\(\\frac{28}{360} \\times 2 \\times \\frac{22}{7} \\times 6400\\)\n= \\(3128.89 km\\)\n\\(\\approxeq 3130 km\\)\n\n', '', 'Explanation\n(a) \\(d = 15 cm \\implies r = \\frac{15}{2} cm\\)\n\\(V = \\frac{4\\pi r^{3}}{3}\\)\n\\(V = \\frac{4}{3} \\times \\frac{22}{7} \\times (\\frac{15}{2})^{3}\\)\n\\(V = \\frac{1125 \\times 22}{7 \\times 2}\\)\n\\(V = 1,767.857 cm^{3}\\)\n(b)(i) Longitude difference : 105° + 35° = 140°\nDistance along the parallel of latitude = \\(\\frac{140}{360} \\times 2 \\times \\frac{22}{7} \\times 6400 \\cos 28\\)\n= \\(\\frac{140 \\times 44 \\times 6400 \\times 0.8829}{360 \\times 7}\\)\n= \\(13812.48 km\\)\n\\(\\approxeq 13800 km\\)\n(ii) Distance from the equator : \\(\\frac{28}{360} \\times 2 \\times \\frac{22}{7} \\times 6400\\)\n= \\(3128.89 km\\)\n\\(\\approxeq 3130 km\\)\n\n', '(a) \\(d = 15 cm \\implies r = \\frac{15}{2} cm\\)\n\\(V = \\frac{4\\pi r^{3}}{3}\\)\n\\(V = \\frac{4}{3} \\times \\frac{22}{7} \\times (\\frac{15}{2})^{3}\\)\n\\(V = \\frac{1125 \\times 22}{7 \\times 2}\\)\n\\(V = 1,767.857 cm^{3}\\)~(b)(i) Longitude difference : 105° + 35° = 140°\nDistance along the parallel of latitude = \\(\\frac{140}{360} \\times 2 \\times \\frac{22}{7} \\times 6400 \\cos 28\\)\n= \\(\\frac{140 \\times 44 \\times 6400 \\times 0.8829}{360 \\times 7}\\)\n= \\(13812.48 km\\)\n\\(\\approxeq 13800 km\\)\n(ii) Distance from the equator : \\(\\frac{28}{360} \\times 2 \\times \\frac{22}{7} \\times 6400\\)\n= \\(3128.89 km\\)\n\\(\\approxeq 3130 km\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(550, '2015', '\n\nQuestion\n\n(a) Evaluate and express your answer in standard form : \\(\\frac{4.56 \\times 3.6}{0.12}\\)\n(b) Without using mathematical tables or calculator, evaluate \\((73.8)^{2} - (26.2)^{2}\\).\n(c) Simplify \\(\\sqrt{1\\frac{19}{81}}\\), expressing your answer in the form \\(\\frac{a}{b}\\) where a and b are positive integers.\n\n', '', '(a) Evaluate and express your answer in standard form : \\(\\frac{4.56 \\times 3.6}{0.12}\\)\n(b) Without using mathematical tables or calculator, evaluate \\((73.8)^{2} - (26.2)^{2}\\).\n(c) Simplify \\(\\sqrt{1\\frac{19}{81}}\\), expressing your answer in the form \\(\\frac{a}{b}\\) where a and b are positive integers.', '(a) Evaluate and express your answer in standard form : \\(\\frac{4.56 \\times 3.6}{0.12}\\)~(b) Without using mathematical tables or calculator, evaluate \\((73.8)^{2} - (26.2)^{2}\\).\n(c) Simplify \\(\\sqrt{1\\frac{19}{81}}\\), expressing your answer in the form \\(\\frac{a}{b}\\) where a and b are positive integers.~(c) Simplify \\(\\sqrt{1\\frac{19}{81}}\\), expressing your answer in the form \\(\\frac{a}{b}\\) where a and b are positive integers.', NULL, '\n\nExplanation\n(a) \\(\\frac{4.56 \\times 3.6}{0.12} \\equiv \\frac{4.56 \\times 360}{12}\\)\n= \\(4.56 \\times 30 \\)\n= \\(136.8 \\)\n= \\(1.368 \\times 10^{2}\\)\n(b) \\((73.8)^{2} - (26.2)^{2}\\)\nUsing the method of difference of two squares,\n= \\((73.8 + 26.2) \\times (73.8 - 26.2)\\)\n= \\(100 \\times 47.6\\)\n= \\(4760\\)\n(c) \\(\\sqrt{1\\frac{19}{81}}\\)\n= \\(\\sqrt{\\frac{100}{81}}\\)\n= \\(\\frac{10}{9}\\)\n\n', '', 'Explanation\n(a) \\(\\frac{4.56 \\times 3.6}{0.12} \\equiv \\frac{4.56 \\times 360}{12}\\)\n= \\(4.56 \\times 30 \\)\n= \\(136.8 \\)\n= \\(1.368 \\times 10^{2}\\)\n(b) \\((73.8)^{2} - (26.2)^{2}\\)\nUsing the method of difference of two squares,\n= \\((73.8 + 26.2) \\times (73.8 - 26.2)\\)\n= \\(100 \\times 47.6\\)\n= \\(4760\\)\n(c) \\(\\sqrt{1\\frac{19}{81}}\\)\n= \\(\\sqrt{\\frac{100}{81}}\\)\n= \\(\\frac{10}{9}\\)\n\n', '(a) \\(\\frac{4.56 \\times 3.6}{0.12} \\equiv \\frac{4.56 \\times 360}{12}\\)\n= \\(4.56 \\times 30 \\)\n= \\(136.8 \\)\n= \\(1.368 \\times 10^{2}\\)~(b) \\((73.8)^{2} - (26.2)^{2}\\)\nUsing the method of difference of two squares,\n= \\((73.8 + 26.2) \\times (73.8 - 26.2)\\)\n= \\(100 \\times 47.6\\)\n= \\(4760\\)\n(c) \\(\\sqrt{1\\frac{19}{81}}\\)\n= \\(\\sqrt{\\frac{100}{81}}\\)\n= \\(\\frac{10}{9}\\)\n~(c) \\(\\sqrt{1\\frac{19}{81}}\\)\n= \\(\\sqrt{\\frac{100}{81}}\\)\n= \\(\\frac{10}{9}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(551, '2015', '\n\nQuestion\n\n(a) Given that \\(\\cos x = 0.7431, 0° < x < 90°\\), use tables to find the values of : (i) \\(2 \\sin x\\) ; (ii) \\(\\tan \\frac{x}{2}\\).\n(b) The interior angles of a pentagon are in ratio 2 : 3 : 4 : 4 : 5. Find the value of the largest angle.\n\n', '', '(a) Given that \\(\\cos x = 0.7431, 0° < x < 90°\\), use tables to find the values of : (i) \\(2 \\sin x\\) ; (ii) \\(\\tan \\frac{x}{2}\\).\n(b) The interior angles of a pentagon are in ratio 2 : 3 : 4 : 4 : 5. Find the value of the largest angle.', '(a) Given that \\(\\cos x = 0.7431, 0° < x < 90°\\), use tables to find the values of : (i) \\(2 \\sin x\\) ; (ii) \\(\\tan \\frac{x}{2}\\).~(b) The interior angles of a pentagon are in ratio 2 : 3 : 4 : 4 : 5. Find the value of the largest angle.', NULL, '\n\nExplanation\n(a) \\(\\cos x = 0.7431\\)\n\\(x = \\cos^{-1} (0.7431)\\)\n\\(x = 42°\\)\n(i) \\(2 \\sin x = 2 \\sin 42\\)\n= \\(2 \\times 0.6692\\)\n= \\(1.3384\\)\n(ii) \\(\\tan \\frac{x}{2} = \\tan \\frac{42}{2}\\)\n= \\(\\tan 21°\\)\n= \\(0.3839\\)\n(b) Sum of the interior angles of a polygon = \\((2n - 4) \\times 90°\\)\nFor a pentagon, n = 5\n\\((2(5) - 4) \\times 90° = 6 \\times 90°\\)\n= \\(540°\\)\nRatio of sides = 2:3:4:4:5\nTotal = 2 + 3 + 4 + 4 + 5 = 18\nLargest angle = \\(\\frac{5}{18} \\times 540° = 150°\\)\n\n', '', 'Explanation\n(a) \\(\\cos x = 0.7431\\)\n\\(x = \\cos^{-1} (0.7431)\\)\n\\(x = 42°\\)\n(i) \\(2 \\sin x = 2 \\sin 42\\)\n= \\(2 \\times 0.6692\\)\n= \\(1.3384\\)\n(ii) \\(\\tan \\frac{x}{2} = \\tan \\frac{42}{2}\\)\n= \\(\\tan 21°\\)\n= \\(0.3839\\)\n(b) Sum of the interior angles of a polygon = \\((2n - 4) \\times 90°\\)\nFor a pentagon, n = 5\n\\((2(5) - 4) \\times 90° = 6 \\times 90°\\)\n= \\(540°\\)\nRatio of sides = 2:3:4:4:5\nTotal = 2 + 3 + 4 + 4 + 5 = 18\nLargest angle = \\(\\frac{5}{18} \\times 540° = 150°\\)\n\n', '(a) \\(\\cos x = 0.7431\\)\n\\(x = \\cos^{-1} (0.7431)\\)\n\\(x = 42°\\)\n(i) \\(2 \\sin x = 2 \\sin 42\\)\n= \\(2 \\times 0.6692\\)\n= \\(1.3384\\)\n(ii) \\(\\tan \\frac{x}{2} = \\tan \\frac{42}{2}\\)\n= \\(\\tan 21°\\)\n= \\(0.3839\\)~(b) Sum of the interior angles of a polygon = \\((2n - 4) \\times 90°\\)\nFor a pentagon, n = 5\n\\((2(5) - 4) \\times 90° = 6 \\times 90°\\)\n= \\(540°\\)\nRatio of sides = 2:3:4:4:5\nTotal = 2 + 3 + 4 + 4 + 5 = 18\nLargest angle = \\(\\frac{5}{18} \\times 540° = 150°\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(552, '2015', '\n\nQuestion\n\n(a) Given the expression \\(y = ax^{2} - bx - 12\\) , find the values of x when a = 1, b = 2 and y = 3.\n(b) If \\(\\sqrt{x^{2} + 1} = \\frac{5}{4}\\), find the positive value of x.\n\n', '', '(a) Given the expression \\(y = ax^{2} - bx - 12\\) , find the values of x when a = 1, b = 2 and y = 3.\n(b) If \\(\\sqrt{x^{2} + 1} = \\frac{5}{4}\\), find the positive value of x.', '(a) Given the expression \\(y = ax^{2} - bx - 12\\) , find the values of x when a = 1, b = 2 and y = 3.~(b) If \\(\\sqrt{x^{2} + 1} = \\frac{5}{4}\\), find the positive value of x.', NULL, '\n\nExplanation\n(a) \\(y = ax^{2} - bx - 12\\)\nWhen a = 1, b = 2 and y = 3.\n\\(3 = x^{2} - 2x - 12\\)\n\\(x^{2} - 2x - 12 - 3 = 0 \\implies x^{2} - 2x - 15 = 0\\)\n\\(x^{2} - 5x + 3x - 15 = 0\\)\n\\((x - 5)(x + 3) = 0\\)\n\\(\\text{x = 5 or -3}\\)\n(b) \\(\\sqrt{x^{2} + 1} = \\frac{5}{4}\\)\nSquaring both sides,\n\\(x^{2} + 1 = \\frac{25}{16}\\)\n\\(x^{2} = \\frac{25}{16} - 1 = \\frac{9}{16}\\)\n\\(x = \\sqrt{\\frac{9}{16}} = \\pm \\frac{3}{4}\\)\nThe positive value of x = \\(\\frac{3}{4}\\).\n\n', '', 'Explanation\n(a) \\(y = ax^{2} - bx - 12\\)\nWhen a = 1, b = 2 and y = 3.\n\\(3 = x^{2} - 2x - 12\\)\n\\(x^{2} - 2x - 12 - 3 = 0 \\implies x^{2} - 2x - 15 = 0\\)\n\\(x^{2} - 5x + 3x - 15 = 0\\)\n\\((x - 5)(x + 3) = 0\\)\n\\(\\text{x = 5 or -3}\\)\n(b) \\(\\sqrt{x^{2} + 1} = \\frac{5}{4}\\)\nSquaring both sides,\n\\(x^{2} + 1 = \\frac{25}{16}\\)\n\\(x^{2} = \\frac{25}{16} - 1 = \\frac{9}{16}\\)\n\\(x = \\sqrt{\\frac{9}{16}} = \\pm \\frac{3}{4}\\)\nThe positive value of x = \\(\\frac{3}{4}\\).\n\n', '(a) \\(y = ax^{2} - bx - 12\\)\nWhen a = 1, b = 2 and y = 3.\n\\(3 = x^{2} - 2x - 12\\)\n\\(x^{2} - 2x - 12 - 3 = 0 \\implies x^{2} - 2x - 15 = 0\\)\n\\(x^{2} - 5x + 3x - 15 = 0\\)\n\\((x - 5)(x + 3) = 0\\)\n\\(\\text{x = 5 or -3}\\)~(b) \\(\\sqrt{x^{2} + 1} = \\frac{5}{4}\\)\nSquaring both sides,\n\\(x^{2} + 1 = \\frac{25}{16}\\)\n\\(x^{2} = \\frac{25}{16} - 1 = \\frac{9}{16}\\)\n\\(x = \\sqrt{\\frac{9}{16}} = \\pm \\frac{3}{4}\\)\nThe positive value of x = \\(\\frac{3}{4}\\).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(553, '2015', '\n\nQuestion\n\nUsing ruler and a pair of compasses only,\n(a) construct \\(\\Delta PQR\\) such that |PQ| = 7 cm, |PR| = 6 cm and < PQR = 60°.\n(b) locate point M, the mid-point of PQ.\n(c) Measure < RMQ.\n\n', '', 'Using ruler and a pair of compasses only,\n(a) construct \\(\\Delta PQR\\) such that |PQ| = 7 cm, |PR| = 6 cm and < PQR = 60°.\n(b) locate point M, the mid-point of PQ.\n(c) Measure < RMQ.', 'Using ruler and a pair of compasses only,\n(a) construct \\(\\Delta PQR\\) such that |PQ| = 7 cm, |PR| = 6 cm and < PQR = 60°.~(b) locate point M, the mid-point of PQ.\n(c) Measure < RMQ.~(c) Measure < RMQ.', NULL, '\n\nExplanation\n < RMQ = 60°.\n\n', ' < RMQ = 60°.\n\n', '', '', 'WAEC-PK-1', 'WAEC-MATH', 1),
(554, '2015', '\n\nQuestion\n\n\nThe pie chart shows the distribution of marks scored by 200 pupils in a test.\n(a) How many pupils scored : (i) between 41 and 50 marks? ; (ii) above 80 marks ?\n(b) What fraction of the pupils scored at most 50 marks?\n(c) What is the modal class?\n\n\n', '', 'The pie chart shows the distribution of marks scored by 200 pupils in a test.\n(a) How many pupils scored : (i) between 41 and 50 marks? ; (ii) above 80 marks ?\n(b) What fraction of the pupils scored at most 50 marks?\n(c) What is the modal class?\n', 'The pie chart shows the distribution of marks scored by 200 pupils in a test.\n(a) How many pupils scored : (i) between 41 and 50 marks? ; (ii) above 80 marks ?~(b) What fraction of the pupils scored at most 50 marks?\n(c) What is the modal class?\n~(c) What is the modal class?\n', NULL, '\n\nExplanation\n(a)(i) The number of pupils who scored between 41 and 50 marks = \\(\\frac{90}{360} \\times 200 = 50\\)\n(ii) The angles for pupils above 60 marks = 54° + 36° + 18° = 108°\n\\(\\frac{108}{360} \\times 200 = 60\\)\n(b) The fraction of the pupils who scored at least 50 marks : \\(\\frac{54 + 90}{360} = \\frac{2}{5}\\)\n(c) The modal class = 51 - 60; It has the largest angle in the pie- chart.\n\n', '', 'Explanation\n(a)(i) The number of pupils who scored between 41 and 50 marks = \\(\\frac{90}{360} \\times 200 = 50\\)\n(ii) The angles for pupils above 60 marks = 54° + 36° + 18° = 108°\n\\(\\frac{108}{360} \\times 200 = 60\\)\n(b) The fraction of the pupils who scored at least 50 marks : \\(\\frac{54 + 90}{360} = \\frac{2}{5}\\)\n(c) The modal class = 51 - 60; It has the largest angle in the pie- chart.\n\n', '(a)(i) The number of pupils who scored between 41 and 50 marks = \\(\\frac{90}{360} \\times 200 = 50\\)\n(ii) The angles for pupils above 60 marks = 54° + 36° + 18° = 108°\n\\(\\frac{108}{360} \\times 200 = 60\\)~(b) The fraction of the pupils who scored at least 50 marks : \\(\\frac{54 + 90}{360} = \\frac{2}{5}\\)\n(c) The modal class = 51 - 60; It has the largest angle in the pie- chart.\n~(c) The modal class = 51 - 60; It has the largest angle in the pie- chart.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(555, '2015', '\n\nQuestion\n\n(a)\n\n\n\n\nLimes\nApples\n\n\nGood\n10\n8\n\n\nBad\n6\n6\n\n\n\nThe table shows the number of limes and apples of the same size in a bag. If two of the fruits are picked at random, one at a time, without replacement, find the probability that : (i) both are good limes ; (ii) both are bad fruits ; (iii) one is a good apple and the other a bad lime.\n(b) Solve the equation \\(\\log_{3} (4x + 1) - \\log_{3} (3x - 5) = 2\\).\n\n\n\n(a)\n\n\n\n\nLimes\nApples\n\n\nGood\n10\n8\n\n\nBad\n6\n6\n\n\n', '', '(a)\n\n\n\n\nLimes\nApples\n\n\nGood\n10\n8\n\n\nBad\n6\n6\n\n\n\nThe table shows the number of limes and apples of the same size in a bag. If two of the fruits are picked at random, one at a time, without replacement, find the probability that : (i) both are good limes ; (ii) both are bad fruits ; (iii) one is a good apple and the other a bad lime.\n(b) Solve the equation \\(\\log_{3} (4x + 1) - \\log_{3} (3x - 5) = 2\\).\n\n\n\n(a)\n\n\n\n\nLimes\nApples\n\n\nGood\n10\n8\n\n\nBad\n6\n6\n', '(a)\n\n\n\n\nLimes\nApples\n\n\nGood\n10\n8\n\n\nBad\n6\n6\n\n\n\nThe table shows the number of limes and apples of the same size in a bag. If two of the fruits are picked at random, one at a time, without replacement, find the probability that : (i) both are good limes ; (ii) both are bad fruits ; (iii) one is a good apple and the other a bad lime.~(b) Solve the equation \\(\\log_{3} (4x + 1) - \\log_{3} (3x - 5) = 2\\).\n\n\n\n(a)\n\n\n\n\nLimes\nApples\n\n\nGood\n10\n8\n\n\nBad\n6\n6\n', NULL, '\n\nExplanation\n(a) Total number of fruits = 16 limes + 14 apples = 30\n(i) P(both are good limes) = \\(\\frac{10}{30} \\times \\frac{9}{29} = \\frac{3}{29}\\)\n(ii) P(both are bad fruits) = \\(\\frac{12}{30} \\times \\frac{11}{29} = \\frac{22}{145}\\)\n(iii) P(one is a good apple and the other a bad lime) = \\((\\frac{8}{30} \\times \\frac{6}{29}) + (\\frac{6}{30} \\times \\frac{8}{29})\\)\n= \\(\\frac{8}{145} + \\frac{8}{145}\\)\n= \\(\\frac{16}{145}\\)\nNote : The arrangement is not specific. Hence, it can be a good apple first and then a bad lime or a bad lime first and then a good apple.\n(b) \\(\\log_{3} (4x + 1) - \\log_{3} (3x - 5) = 2\\)\n\\(\\log_{3} (\\frac{4x + 1}{3x - 5}) = 2\\)\n\\((\\frac{4x + 1}{3x - 5}) = 3^{2}\\)\n\\(\\frac{4x + 1}{3x - 5} = 9 \\implies 4x + 1 = 9(3x - 5)\\)\n\\(4x + 1 = 27x - 45 \\implies 23x = 46\\)\n\\(x = 2\\)\n\n', '', 'Explanation\n(a) Total number of fruits = 16 limes + 14 apples = 30\n(i) P(both are good limes) = \\(\\frac{10}{30} \\times \\frac{9}{29} = \\frac{3}{29}\\)\n(ii) P(both are bad fruits) = \\(\\frac{12}{30} \\times \\frac{11}{29} = \\frac{22}{145}\\)\n(iii) P(one is a good apple and the other a bad lime) = \\((\\frac{8}{30} \\times \\frac{6}{29}) + (\\frac{6}{30} \\times \\frac{8}{29})\\)\n= \\(\\frac{8}{145} + \\frac{8}{145}\\)\n= \\(\\frac{16}{145}\\)\nNote : The arrangement is not specific. Hence, it can be a good apple first and then a bad lime or a bad lime first and then a good apple.\n(b) \\(\\log_{3} (4x + 1) - \\log_{3} (3x - 5) = 2\\)\n\\(\\log_{3} (\\frac{4x + 1}{3x - 5}) = 2\\)\n\\((\\frac{4x + 1}{3x - 5}) = 3^{2}\\)\n\\(\\frac{4x + 1}{3x - 5} = 9 \\implies 4x + 1 = 9(3x - 5)\\)\n\\(4x + 1 = 27x - 45 \\implies 23x = 46\\)\n\\(x = 2\\)\n\n', '(a) Total number of fruits = 16 limes + 14 apples = 30\n(i) P(both are good limes) = \\(\\frac{10}{30} \\times \\frac{9}{29} = \\frac{3}{29}\\)\n(ii) P(both are bad fruits) = \\(\\frac{12}{30} \\times \\frac{11}{29} = \\frac{22}{145}\\)\n(iii) P(one is a good apple and the other a bad lime) = \\((\\frac{8}{30} \\times \\frac{6}{29}) + (\\frac{6}{30} \\times \\frac{8}{29})\\)\n= \\(\\frac{8}{145} + \\frac{8}{145}\\)\n= \\(\\frac{16}{145}\\)\nNote : The arrangement is not specific. Hence, it can be a good apple first and then a bad lime or a bad lime first and then a good apple.~(b) \\(\\log_{3} (4x + 1) - \\log_{3} (3x - 5) = 2\\)\n\\(\\log_{3} (\\frac{4x + 1}{3x - 5}) = 2\\)\n\\((\\frac{4x + 1}{3x - 5}) = 3^{2}\\)\n\\(\\frac{4x + 1}{3x - 5} = 9 \\implies 4x + 1 = 9(3x - 5)\\)\n\\(4x + 1 = 27x - 45 \\implies 23x = 46\\)\n\\(x = 2\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(556, '2015', '\n\nQuestion\n\n(a) A man earns N150,000 per annum. He is allowed a tax free pay on N40,000. If he pays 25 kobo in the naira as tax on his taxable income, how much has he left?\n(b) A bookshop has 650 copies of a book for sale. The books were marked at N75 per copy in order to make a profit of 30%. A bookseller bought 300 copies at 5% discount. If the remaining copies are sold at N75 each, calculate the percentage profit the bookshop would make on the whole.\n\n', '', '(a) A man earns N150,000 per annum. He is allowed a tax free pay on N40,000. If he pays 25 kobo in the naira as tax on his taxable income, how much has he left?\n(b) A bookshop has 650 copies of a book for sale. The books were marked at N75 per copy in order to make a profit of 30%. A bookseller bought 300 copies at 5% discount. If the remaining copies are sold at N75 each, calculate the percentage profit the bookshop would make on the whole.', '(a) A man earns N150,000 per annum. He is allowed a tax free pay on N40,000. If he pays 25 kobo in the naira as tax on his taxable income, how much has he left?~(b) A bookshop has 650 copies of a book for sale. The books were marked at N75 per copy in order to make a profit of 30%. A bookseller bought 300 copies at 5% discount. If the remaining copies are sold at N75 each, calculate the percentage profit the bookshop would make on the whole.', NULL, '\n\nExplanation\n(a) Taxable income = \\(N150,000 - N40,000 = N110,000\\)\nTax payment = \\(N110,000 \\times 0.25 = N27,500\\)\nIncome left = \\(N(150,000 - 27,500) = N122,500\\)\n(b) \\(% gain = \\frac{\\text{actual profit}}{\\text{cost price}} \\times 100%\\)\n= \\(\\frac{SP - CP}{CP} \\times 100%\\)\n\\(\\therefore \\frac{75 - CP}{CP} = \\frac{30}{100} \\implies 0.3CP = 75 - CP\\)\n\\(1.3CP = 75 \\implies CP = N57.69\\)\nTotal cost price = \\(650 \\times \\frac{750}{13} = N37,500\\)\n350 copies sold at N75 ==> \\(SP = 350 \\times N75 = N26,250\\)\n300 copies sold at 5% discouunt ==> \\((75 - \\frac{5}{100} \\times 75) \\times 300\\)\n= \\((75 - 3.75) \\times 300 = N21,375\\)\nTotal selling price = \\(N(26,250 + 21,375) = N47, 625\\)\nProfit = \\(N(47,625 - 37,500) = N10,125\\)\n\\(% = \\frac{10,125}{37,500} \\times 100% = 27%\\)\n\n', '', 'Explanation\n(a) Taxable income = \\(N150,000 - N40,000 = N110,000\\)\nTax payment = \\(N110,000 \\times 0.25 = N27,500\\)\nIncome left = \\(N(150,000 - 27,500) = N122,500\\)\n(b) \\(% gain = \\frac{\\text{actual profit}}{\\text{cost price}} \\times 100%\\)\n= \\(\\frac{SP - CP}{CP} \\times 100%\\)\n\\(\\therefore \\frac{75 - CP}{CP} = \\frac{30}{100} \\implies 0.3CP = 75 - CP\\)\n\\(1.3CP = 75 \\implies CP = N57.69\\)\nTotal cost price = \\(650 \\times \\frac{750}{13} = N37,500\\)\n350 copies sold at N75 ==> \\(SP = 350 \\times N75 = N26,250\\)\n300 copies sold at 5% discouunt ==> \\((75 - \\frac{5}{100} \\times 75) \\times 300\\)\n= \\((75 - 3.75) \\times 300 = N21,375\\)\nTotal selling price = \\(N(26,250 + 21,375) = N47, 625\\)\nProfit = \\(N(47,625 - 37,500) = N10,125\\)\n\\(% = \\frac{10,125}{37,500} \\times 100% = 27%\\)\n\n', '(a) Taxable income = \\(N150,000 - N40,000 = N110,000\\)\nTax payment = \\(N110,000 \\times 0.25 = N27,500\\)\nIncome left = \\(N(150,000 - 27,500) = N122,500\\)~(b) \\(% gain = \\frac{\\text{actual profit}}{\\text{cost price}} \\times 100%\\)\n= \\(\\frac{SP - CP}{CP} \\times 100%\\)\n\\(\\therefore \\frac{75 - CP}{CP} = \\frac{30}{100} \\implies 0.3CP = 75 - CP\\)\n\\(1.3CP = 75 \\implies CP = N57.69\\)\nTotal cost price = \\(650 \\times \\frac{750}{13} = N37,500\\)\n350 copies sold at N75 ==> \\(SP = 350 \\times N75 = N26,250\\)\n300 copies sold at 5% discouunt ==> \\((75 - \\frac{5}{100} \\times 75) \\times 300\\)\n= \\((75 - 3.75) \\times 300 = N21,375\\)\nTotal selling price = \\(N(26,250 + 21,375) = N47, 625\\)\nProfit = \\(N(47,625 - 37,500) = N10,125\\)\n\\(% = \\frac{10,125}{37,500} \\times 100% = 27%\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(557, '2015', '\n\nQuestion\n\n(a) Copy and complete the following table of values for the relation \\(y = x^{2} - 2x - 5\\)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n\n\n-2\n\n-6\n-2\n3\n10\n\n\n\n\n(b) Draw the graph of the relation \\(y = x^{2} - 2x - 5\\); using a scale of 2 cm to 1 unit on the x- axis, and 2 cm to 2 units on the y- axis.\n(c) Using the same axes, draw the graph of \\(y = 2x + 3\\).\n(d) Obtain in the form \\(ax^{2} + bx + c = 0\\) where a, b and c are integers, the equation which is satisfied by the x- coordinate of the points of intersection of the two graphs.\n(e) From your graphs, determine the roots of the equation obtained in (d) above.\n\n\n\n(a) Copy and complete the following table of values for the relation \\(y = x^{2} - 2x - 5\\)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n\n\n-2\n\n-6\n-2\n3\n10\n\n\n\n', '', '(a) Copy and complete the following table of values for the relation \\(y = x^{2} - 2x - 5\\)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n\n\n-2\n\n-6\n-2\n3\n10\n\n\n\n\n(b) Draw the graph of the relation \\(y = x^{2} - 2x - 5\\); using a scale of 2 cm to 1 unit on the x- axis, and 2 cm to 2 units on the y- axis.\n(c) Using the same axes, draw the graph of \\(y = 2x + 3\\).\n(d) Obtain in the form \\(ax^{2} + bx + c = 0\\) where a, b and c are integers, the equation which is satisfied by the x- coordinate of the points of intersection of the two graphs.\n(e) From your graphs, determine the roots of the equation obtained in (d) above.\n\n\n\n(a) Copy and complete the following table of values for the relation \\(y = x^{2} - 2x - 5\\)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n\n\n-2\n\n-6\n-2\n3\n10\n\n', '(a) Copy and complete the following table of values for the relation \\(y = x^{2} - 2x - 5\\)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n\n\n-2\n\n-6\n-2\n3\n10\n\n\n\n~(b) Draw the graph of the relation \\(y = x^{2} - 2x - 5\\); using a scale of 2 cm to 1 unit on the x- axis, and 2 cm to 2 units on the y- axis.\n(c) Using the same axes, draw the graph of \\(y = 2x + 3\\).\n(d) Obtain in the form \\(ax^{2} + bx + c = 0\\) where a, b and c are integers, the equation which is satisfied by the x- coordinate of the points of intersection of the two graphs.\n(e) From your graphs, determine the roots of the equation obtained in (d) above.\n\n\n\n(a) Copy and complete the following table of values for the relation \\(y = x^{2} - 2x - 5\\)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n\n\n-2\n\n-6\n-2\n3\n10\n\n~(c) Using the same axes, draw the graph of \\(y = 2x + 3\\).\n(d) Obtain in the form \\(ax^{2} + bx + c = 0\\) where a, b and c are integers, the equation which is satisfied by the x- coordinate of the points of intersection of the two graphs.\n(e) From your graphs, determine the roots of the equation obtained in (d) above.\n\n\n\n(a) Copy and complete the following table of values for the relation \\(y = x^{2} - 2x - 5\\)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n\n\n-2\n\n-6\n-2\n3\n10\n\n~(d) Obtain in the form \\(ax^{2} + bx + c = 0\\) where a, b and c are integers, the equation which is satisfied by the x- coordinate of the points of intersection of the two graphs.\n(e) From your graphs, determine the roots of the equation obtained in (d) above.\n\n\n\n(a) Copy and complete the following table of values for the relation \\(y = x^{2} - 2x - 5\\)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n\n\n-2\n\n-6\n-2\n3\n10\n\n~(e) From your graphs, determine the roots of the equation obtained in (d) above.\n\n\n\n(a) Copy and complete the following table of values for the relation \\(y = x^{2} - 2x - 5\\)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n\n\n-2\n\n-6\n-2\n3\n10\n\n', NULL, '\n\nExplanation\n(a)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\n\\(x^{2}\\)\n9\n4\n1\n0\n1\n4\n9\n16\n25\n\n\n\\(-2x\\)\n6\n4\n2\n0\n-2\n-4\n-6\n-8\n-10\n\n\n-5\n-5\n-5\n-5\n-5\n-5\n-5\n-5\n-5\n-5\n\n\ny\n10\n3\n-2\n-5\n-6\n-5\n-2\n3\n10\n\n\n\n(b)\n(c)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\n\\(2x\\)\n-6\n-4\n-2\n0\n2\n4\n6\n8\n10\n\n\n-3\n-3\n-3\n-3\n-3\n-3\n-3\n-3\n-3\n-3\n\n\ny\n-9\n-7\n-5\n-3\n-1\n1\n3\n5\n7\n\n\n\n(d) \\(x^{2} - 2x - 5 = 2x - 3\\)\n\\(x^{2} - 2x - 2x - 5 + 3 = 0\\)\n\\(x^{2} - 4x - 2 = 0\\)\n(e) x = -0.8 or 4.3\n\n', '', 'Explanation\n(a)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\n\\(x^{2}\\)\n9\n4\n1\n0\n1\n4\n9\n16\n25\n\n\n\\(-2x\\)\n6\n4\n2\n0\n-2\n-4\n-6\n-8\n-10\n\n\n-5\n-5\n-5\n-5\n-5\n-5\n-5\n-5\n-5\n-5\n\n\ny\n10\n3\n-2\n-5\n-6\n-5\n-2\n3\n10\n\n\n\n(b)\n(c)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\n\\(2x\\)\n-6\n-4\n-2\n0\n2\n4\n6\n8\n10\n\n\n-3\n-3\n-3\n-3\n-3\n-3\n-3\n-3\n-3\n-3\n\n\ny\n-9\n-7\n-5\n-3\n-1\n1\n3\n5\n7\n\n\n\n(d) \\(x^{2} - 2x - 5 = 2x - 3\\)\n\\(x^{2} - 2x - 2x - 5 + 3 = 0\\)\n\\(x^{2} - 4x - 2 = 0\\)\n(e) x = -0.8 or 4.3\n\n', '(a)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\n\\(x^{2}\\)\n9\n4\n1\n0\n1\n4\n9\n16\n25\n\n\n\\(-2x\\)\n6\n4\n2\n0\n-2\n-4\n-6\n-8\n-10\n\n\n-5\n-5\n-5\n-5\n-5\n-5\n-5\n-5\n-5\n-5\n\n\ny\n10\n3\n-2\n-5\n-6\n-5\n-2\n3\n10\n\n\n~(b)\n(c)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\n\\(2x\\)\n-6\n-4\n-2\n0\n2\n4\n6\n8\n10\n\n\n-3\n-3\n-3\n-3\n-3\n-3\n-3\n-3\n-3\n-3\n\n\ny\n-9\n-7\n-5\n-3\n-1\n1\n3\n5\n7\n\n\n\n(d) \\(x^{2} - 2x - 5 = 2x - 3\\)\n\\(x^{2} - 2x - 2x - 5 + 3 = 0\\)\n\\(x^{2} - 4x - 2 = 0\\)\n(e) x = -0.8 or 4.3\n~(c)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\n\\(2x\\)\n-6\n-4\n-2\n0\n2\n4\n6\n8\n10\n\n\n-3\n-3\n-3\n-3\n-3\n-3\n-3\n-3\n-3\n-3\n\n\ny\n-9\n-7\n-5\n-3\n-1\n1\n3\n5\n7\n\n\n\n(d) \\(x^{2} - 2x - 5 = 2x - 3\\)\n\\(x^{2} - 2x - 2x - 5 + 3 = 0\\)\n\\(x^{2} - 4x - 2 = 0\\)\n(e) x = -0.8 or 4.3\n~(d) \\(x^{2} - 2x - 5 = 2x - 3\\)\n\\(x^{2} - 2x - 2x - 5 + 3 = 0\\)\n\\(x^{2} - 4x - 2 = 0\\)\n(e) x = -0.8 or 4.3\n~(e) x = -0.8 or 4.3\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
(558, '2016', '\n\nQuestion\n\n(a) The mean of 1, 2, x, 11, y, 14, arranged in ascending order, is 8 and the median is 9. Find the values of x and y.\n(b)\nIn the diagram, MN || PQ, |LM| = 3cm and |LP| = 4cm. If the area of \\(\\Delta\\) LMN is 18\\(cm^{2}\\), find the area of the quadrilateral MPQN.\n\n', '', '(a) The mean of 1, 2, x, 11, y, 14, arranged in ascending order, is 8 and the median is 9. Find the values of x and y.\n(b)\nIn the diagram, MN || PQ, |LM| = 3cm and |LP| = 4cm. If the area of \\(\\Delta\\) LMN is 18\\(cm^{2}\\), find the area of the quadrilateral MPQN.', '(a) The mean of 1, 2, x, 11, y, 14, arranged in ascending order, is 8 and the median is 9. Find the values of x and y.~(b)\nIn the diagram, MN || PQ, |LM| = 3cm and |LP| = 4cm. If the area of \\(\\Delta\\) LMN is 18\\(cm^{2}\\), find the area of the quadrilateral MPQN.', NULL, '\n\nExplanation\n(a) Mean \\(\\bar{x} = \\frac{\\sum x}{n} = 8\\)\n\\(\\frac{1 + 2 + x + 11 + y + 14}{6} = 8\\)\n\\(28 + x + y = 48 \\implies x + y = 20 .... (1) \\)\nThe middle numbers are x and 11 therefore, \\(\\frac{x + 11}{2} = 9\\)\n\\(x + 11 = 18 \\implies x = 18 - 11 = 7\\)\nPutting x = 7 in (1),\n\\(7 + y = 20 \\implies y = 20 - 7 = 13\\)\n(x, y) = (7, 13).\n(b) \\(\\frac{LM}{LP} = \\frac{3}{4}\\)\n\\(\\frac{\\text{Area of LMN}}{\\text{Area of LPQ}} = \\frac{3^{2}}{4^{2}}\\)\nArea of \\(\\Delta\\) LPQ = \\(\\frac{16}{9} \\times 18 = 32 cm^{2}\\)\n\\(\\therefore \\text{Area of quadrilateral MPQN} = 32 - 18 = 14 cm^{2}\\)\n\n', '', 'Explanation\n(a) Mean \\(\\bar{x} = \\frac{\\sum x}{n} = 8\\)\n\\(\\frac{1 + 2 + x + 11 + y + 14}{6} = 8\\)\n\\(28 + x + y = 48 \\implies x + y = 20 .... (1) \\)\nThe middle numbers are x and 11 therefore, \\(\\frac{x + 11}{2} = 9\\)\n\\(x + 11 = 18 \\implies x = 18 - 11 = 7\\)\nPutting x = 7 in (1),\n\\(7 + y = 20 \\implies y = 20 - 7 = 13\\)\n(x, y) = (7, 13).\n(b) \\(\\frac{LM}{LP} = \\frac{3}{4}\\)\n\\(\\frac{\\text{Area of LMN}}{\\text{Area of LPQ}} = \\frac{3^{2}}{4^{2}}\\)\nArea of \\(\\Delta\\) LPQ = \\(\\frac{16}{9} \\times 18 = 32 cm^{2}\\)\n\\(\\therefore \\text{Area of quadrilateral MPQN} = 32 - 18 = 14 cm^{2}\\)\n\n', '(a) Mean \\(\\bar{x} = \\frac{\\sum x}{n} = 8\\)\n\\(\\frac{1 + 2 + x + 11 + y + 14}{6} = 8\\)\n\\(28 + x + y = 48 \\implies x + y = 20 .... (1) \\)\nThe middle numbers are x and 11 therefore, \\(\\frac{x + 11}{2} = 9\\)\n\\(x + 11 = 18 \\implies x = 18 - 11 = 7\\)\nPutting x = 7 in (1),\n\\(7 + y = 20 \\implies y = 20 - 7 = 13\\)\n(x, y) = (7, 13).~(b) \\(\\frac{LM}{LP} = \\frac{3}{4}\\)\n\\(\\frac{\\text{Area of LMN}}{\\text{Area of LPQ}} = \\frac{3^{2}}{4^{2}}\\)\nArea of \\(\\Delta\\) LPQ = \\(\\frac{16}{9} \\times 18 = 32 cm^{2}\\)\n\\(\\therefore \\text{Area of quadrilateral MPQN} = 32 - 18 = 14 cm^{2}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(559, '2016', '\n\nQuestion\n\n(a) A surveyor walks 100m up a hill which slopes at an angle of 24° to the horizontal. Calculate, correct to the nearest metre, the height through which he rises.\n(b)\nIn the diagram, ABC is an isosceles triangle. |AB| = |AC| = 5 cm, and |BC| = 8 cm. Calculate, correct to the nearest degree, < BAC.\n(c) Two boats, 70 metres apart and on opposite sides of a light-house, are in a straight line with the light-house. The angles of elevation of the top of the light-house from the two boats are 71.6° and 45°. Find the height of the light-house. [Take \\(\\tan 71.6° = 3\\)].\n\n', '', '(a) A surveyor walks 100m up a hill which slopes at an angle of 24° to the horizontal. Calculate, correct to the nearest metre, the height through which he rises.\n(b)\nIn the diagram, ABC is an isosceles triangle. |AB| = |AC| = 5 cm, and |BC| = 8 cm. Calculate, correct to the nearest degree, < BAC.\n(c) Two boats, 70 metres apart and on opposite sides of a light-house, are in a straight line with the light-house. The angles of elevation of the top of the light-house from the two boats are 71.6° and 45°. Find the height of the light-house. [Take \\(\\tan 71.6° = 3\\)].', '(a) A surveyor walks 100m up a hill which slopes at an angle of 24° to the horizontal. Calculate, correct to the nearest metre, the height through which he rises.~(b)\nIn the diagram, ABC is an isosceles triangle. |AB| = |AC| = 5 cm, and |BC| = 8 cm. Calculate, correct to the nearest degree, < BAC.\n(c) Two boats, 70 metres apart and on opposite sides of a light-house, are in a straight line with the light-house. The angles of elevation of the top of the light-house from the two boats are 71.6° and 45°. Find the height of the light-house. [Take \\(\\tan 71.6° = 3\\)].~(c) Two boats, 70 metres apart and on opposite sides of a light-house, are in a straight line with the light-house. The angles of elevation of the top of the light-house from the two boats are 71.6° and 45°. Find the height of the light-house. [Take \\(\\tan 71.6° = 3\\)].', NULL, '\n\nExplanation\n(a)\\(\\frac{h}{100} = \\sin 24\\)\n\\(h = 100 \\sin 24\\)\n= \\(100 \\times 0.4067\\)\n= \\(40.67m\\)\n(b)\\(\\sin \\theta = \\frac{4}{5} = 0.8\\)\n\\(\\theta = \\sin^{-1} (0.8) = 53.13°\\)\n\\(< BAC = 2 \\theta = 2(53.13°)\\)\n= \\(106.26°\\)\n\\(\\approxeq 106°\\) (to the nearest degree).\n(c)\nIn \\(\\Delta PSQ\\),\n\\(\\frac{h}{PQ} = \\tan 71.6\\)\n\\(PQ = \\frac{h}{\\tan 71.6}\\)\nIn \\(\\Delta QSR\\),\n\\(\\frac{h}{QR} = \\tan 45\\)\n\\(QR = \\frac{h}{\\tan 45}\\)\n\\(PR = PQ + QR\\)\n\\(\\frac{h}{3} + h = 70 \\implies \\frac{4}{3}h = 70\\)\n\\(h = \\frac{70 \\times 3}{4} = 52.5m\\)\n\n', '', 'Explanation\n(a)\\(\\frac{h}{100} = \\sin 24\\)\n\\(h = 100 \\sin 24\\)\n= \\(100 \\times 0.4067\\)\n= \\(40.67m\\)\n(b)\\(\\sin \\theta = \\frac{4}{5} = 0.8\\)\n\\(\\theta = \\sin^{-1} (0.8) = 53.13°\\)\n\\(< BAC = 2 \\theta = 2(53.13°)\\)\n= \\(106.26°\\)\n\\(\\approxeq 106°\\) (to the nearest degree).\n(c)\nIn \\(\\Delta PSQ\\),\n\\(\\frac{h}{PQ} = \\tan 71.6\\)\n\\(PQ = \\frac{h}{\\tan 71.6}\\)\nIn \\(\\Delta QSR\\),\n\\(\\frac{h}{QR} = \\tan 45\\)\n\\(QR = \\frac{h}{\\tan 45}\\)\n\\(PR = PQ + QR\\)\n\\(\\frac{h}{3} + h = 70 \\implies \\frac{4}{3}h = 70\\)\n\\(h = \\frac{70 \\times 3}{4} = 52.5m\\)\n\n', '(a)\\(\\frac{h}{100} = \\sin 24\\)\n\\(h = 100 \\sin 24\\)\n= \\(100 \\times 0.4067\\)\n= \\(40.67m\\)~(b)\\(\\sin \\theta = \\frac{4}{5} = 0.8\\)\n\\(\\theta = \\sin^{-1} (0.8) = 53.13°\\)\n\\(< BAC = 2 \\theta = 2(53.13°)\\)\n= \\(106.26°\\)\n\\(\\approxeq 106°\\) (to the nearest degree).\n(c)\nIn \\(\\Delta PSQ\\),\n\\(\\frac{h}{PQ} = \\tan 71.6\\)\n\\(PQ = \\frac{h}{\\tan 71.6}\\)\nIn \\(\\Delta QSR\\),\n\\(\\frac{h}{QR} = \\tan 45\\)\n\\(QR = \\frac{h}{\\tan 45}\\)\n\\(PR = PQ + QR\\)\n\\(\\frac{h}{3} + h = 70 \\implies \\frac{4}{3}h = 70\\)\n\\(h = \\frac{70 \\times 3}{4} = 52.5m\\)\n~(c)\nIn \\(\\Delta PSQ\\),\n\\(\\frac{h}{PQ} = \\tan 71.6\\)\n\\(PQ = \\frac{h}{\\tan 71.6}\\)\nIn \\(\\Delta QSR\\),\n\\(\\frac{h}{QR} = \\tan 45\\)\n\\(QR = \\frac{h}{\\tan 45}\\)\n\\(PR = PQ + QR\\)\n\\(\\frac{h}{3} + h = 70 \\implies \\frac{4}{3}h = 70\\)\n\\(h = \\frac{70 \\times 3}{4} = 52.5m\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(560, '2016', '\n\nQuestion\n\n(a) A cylindrical well of radius 1 metre is dug out to a depth of 8 metres. (i) calculate, in m\\(^{3}\\), the volume of soil dug out ; (ii) if the soil is used to raise the level of rectangular floor of a room 4m by 12m, calculate, correct to the nearest cm, the thickness of the new layer of soil. [Take \\(\\pi = \\frac{22}{7}\\)].\n(b)\nThe diagram shows a quadrilateral ABCD in which < DAB is a right- angle. |AB| = 3.3 cm, |BC| = 3.9 cm, |CD| = 5.6 cm. (i) find the length of BD. (ii) show that < BCD = 90°.\n\n', '', '(a) A cylindrical well of radius 1 metre is dug out to a depth of 8 metres. (i) calculate, in m\\(^{3}\\), the volume of soil dug out ; (ii) if the soil is used to raise the level of rectangular floor of a room 4m by 12m, calculate, correct to the nearest cm, the thickness of the new layer of soil. [Take \\(\\pi = \\frac{22}{7}\\)].\n(b)\nThe diagram shows a quadrilateral ABCD in which < DAB is a right- angle. |AB| = 3.3 cm, |BC| = 3.9 cm, |CD| = 5.6 cm. (i) find the length of BD. (ii) show that < BCD = 90°.', '(a) A cylindrical well of radius 1 metre is dug out to a depth of 8 metres. (i) calculate, in m\\(^{3}\\), the volume of soil dug out ; (ii) if the soil is used to raise the level of rectangular floor of a room 4m by 12m, calculate, correct to the nearest cm, the thickness of the new layer of soil. [Take \\(\\pi = \\frac{22}{7}\\)].~(b)\nThe diagram shows a quadrilateral ABCD in which < DAB is a right- angle. |AB| = 3.3 cm, |BC| = 3.9 cm, |CD| = 5.6 cm. (i) find the length of BD. (ii) show that < BCD = 90°.', NULL, '\n\nExplanation\n(a)(i) r = 1m, h = 8m.\n\\(V = \\pi r^{2} h = \\frac{22}{7} \\times 1^{2} \\times 8\\)\n= \\(\\frac{176}{7}\\)\n= \\(25.14 m^{3}\\)\n(ii) \\(V = L \\times B \\times H\\)\n\\(\\frac{176}{7} = 4 \\times 12 \\times h\\)\n\\(h = \\frac{176}{7} \\div 48\\)\n\\(h = \\frac{11}{21}m\\)\n= \\(\\frac{1100}{21} cm = 52.38 cm\\)\n\\(\\approxeq 52 cm\\) (nearest whole number)\n(b)(i)\\(3.3^{2} + 5.6^{2} = BD^{2}\\)\n\\(BD = \\sqrt{3.3^{2} + 5.6^{2}}\\)\n= \\(\\sqrt{10.89 + 31.36}\\)\n= \\(\\sqrt{42.25} = 6.5 cm\\)\n(ii)\nIf < BCD is right- angled, then Pythagoras rule holds true.\n\\(BD^{2} = 5.2^{2} + 3.9^{2} = 27.04 + 15.21\\)\n= \\(42.25\\)\n\\(BD = \\sqrt{42.25} = 6.5 cm\\)\nHence, \\(\\Delta BCD\\) is right-angled.\n\n', '', 'Explanation\n(a)(i) r = 1m, h = 8m.\n\\(V = \\pi r^{2} h = \\frac{22}{7} \\times 1^{2} \\times 8\\)\n= \\(\\frac{176}{7}\\)\n= \\(25.14 m^{3}\\)\n(ii) \\(V = L \\times B \\times H\\)\n\\(\\frac{176}{7} = 4 \\times 12 \\times h\\)\n\\(h = \\frac{176}{7} \\div 48\\)\n\\(h = \\frac{11}{21}m\\)\n= \\(\\frac{1100}{21} cm = 52.38 cm\\)\n\\(\\approxeq 52 cm\\) (nearest whole number)\n(b)(i)\\(3.3^{2} + 5.6^{2} = BD^{2}\\)\n\\(BD = \\sqrt{3.3^{2} + 5.6^{2}}\\)\n= \\(\\sqrt{10.89 + 31.36}\\)\n= \\(\\sqrt{42.25} = 6.5 cm\\)\n(ii)\nIf < BCD is right- angled, then Pythagoras rule holds true.\n\\(BD^{2} = 5.2^{2} + 3.9^{2} = 27.04 + 15.21\\)\n= \\(42.25\\)\n\\(BD = \\sqrt{42.25} = 6.5 cm\\)\nHence, \\(\\Delta BCD\\) is right-angled.\n\n', '(a)(i) r = 1m, h = 8m.\n\\(V = \\pi r^{2} h = \\frac{22}{7} \\times 1^{2} \\times 8\\)\n= \\(\\frac{176}{7}\\)\n= \\(25.14 m^{3}\\)\n(ii) \\(V = L \\times B \\times H\\)\n\\(\\frac{176}{7} = 4 \\times 12 \\times h\\)\n\\(h = \\frac{176}{7} \\div 48\\)\n\\(h = \\frac{11}{21}m\\)\n= \\(\\frac{1100}{21} cm = 52.38 cm\\)\n\\(\\approxeq 52 cm\\) (nearest whole number)~(b)(i)\\(3.3^{2} + 5.6^{2} = BD^{2}\\)\n\\(BD = \\sqrt{3.3^{2} + 5.6^{2}}\\)\n= \\(\\sqrt{10.89 + 31.36}\\)\n= \\(\\sqrt{42.25} = 6.5 cm\\)\n(ii)\nIf < BCD is right- angled, then Pythagoras rule holds true.\n\\(BD^{2} = 5.2^{2} + 3.9^{2} = 27.04 + 15.21\\)\n= \\(42.25\\)\n\\(BD = \\sqrt{42.25} = 6.5 cm\\)\nHence, \\(\\Delta BCD\\) is right-angled.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(561, '2016', '\n\nQuestion\n\n(a) The first term of an Arithmetic Progression (A.P) is 8. The ratio of the 7th term to the 9th term is 5 : 8. Calculate the common difference of the progression.\n(b) A sphere of radius 2 cm is of mass 11.2g. Find (i) the volume of the sphere ; (ii) the density of the sphere ; (iii) the mass of a sphere of the same material but with radius 3cm. [Take \\(\\pi = \\frac{22}{7}\\)].\n\n', '', '(a) The first term of an Arithmetic Progression (A.P) is 8. The ratio of the 7th term to the 9th term is 5 : 8. Calculate the common difference of the progression.\n(b) A sphere of radius 2 cm is of mass 11.2g. Find (i) the volume of the sphere ; (ii) the density of the sphere ; (iii) the mass of a sphere of the same material but with radius 3cm. [Take \\(\\pi = \\frac{22}{7}\\)].', '(a) The first term of an Arithmetic Progression (A.P) is 8. The ratio of the 7th term to the 9th term is 5 : 8. Calculate the common difference of the progression.~(b) A sphere of radius 2 cm is of mass 11.2g. Find (i) the volume of the sphere ; (ii) the density of the sphere ; (iii) the mass of a sphere of the same material but with radius 3cm. [Take \\(\\pi = \\frac{22}{7}\\)].', NULL, '\n\nExplanation\n(a) \\(T_{n} = a + (n - 1)d\\) (terms of an AP)\nGiven a = -8;\n\\(T_{7} = a + 6d = -8 + 6d\\)\n\\(T_{9} = a + 8d = -8 + 8d\\)\n\\(\\frac{-8 + 6d}{-8 + 8d} = \\frac{5}{8}\\)\n\\(5(-8 + 8d) = 8(-8 + 6d)\\)\n\\(-40 + 40d = -64 + 48d\\)\n\\(-40 + 64 = 48d - 40d \\times 24 = 8d\\)\n\\( d = 3\\)\n(b) Given r = 2 cm, m = 11.2g\n(i) \\(V = \\frac{4}{3} \\pi r^{3}\\)\n= \\(\\frac{4}{3} \\times \\frac{22}{7} \\times 2^{3}\\)\n= \\(\\frac{704}{21}\\)\n= \\(33.52 cm^{3} = 33.52 \\times 10^{-6} m^{3}\\)\n= \\(3.352 \\times 10^{-5} m^{3}\\)\n(ii) \\(Density = \\frac{mass}{volume}\\)\n= \\(\\frac{11.2}{33.52}\\)\n= \\(0.334 g/cm^{3}\\)\n(iii) \\(V = \\frac{4}{3} \\pi r^{3}\\)\n= \\(\\frac{4}{3} \\times \\frac{22}{7} \\times 3^{3}\\)\n= \\(\\frac{2376}{21}\\)\n= \\(113.14 cm^{3}\\)\n\\(mass = density \\times volume\\)\n\\(0.334 \\times 113.14 = 37.789g\\)\n\n\n', '', 'Explanation\n(a) \\(T_{n} = a + (n - 1)d\\) (terms of an AP)\nGiven a = -8;\n\\(T_{7} = a + 6d = -8 + 6d\\)\n\\(T_{9} = a + 8d = -8 + 8d\\)\n\\(\\frac{-8 + 6d}{-8 + 8d} = \\frac{5}{8}\\)\n\\(5(-8 + 8d) = 8(-8 + 6d)\\)\n\\(-40 + 40d = -64 + 48d\\)\n\\(-40 + 64 = 48d - 40d \\times 24 = 8d\\)\n\\( d = 3\\)\n(b) Given r = 2 cm, m = 11.2g\n(i) \\(V = \\frac{4}{3} \\pi r^{3}\\)\n= \\(\\frac{4}{3} \\times \\frac{22}{7} \\times 2^{3}\\)\n= \\(\\frac{704}{21}\\)\n= \\(33.52 cm^{3} = 33.52 \\times 10^{-6} m^{3}\\)\n= \\(3.352 \\times 10^{-5} m^{3}\\)\n(ii) \\(Density = \\frac{mass}{volume}\\)\n= \\(\\frac{11.2}{33.52}\\)\n= \\(0.334 g/cm^{3}\\)\n(iii) \\(V = \\frac{4}{3} \\pi r^{3}\\)\n= \\(\\frac{4}{3} \\times \\frac{22}{7} \\times 3^{3}\\)\n= \\(\\frac{2376}{21}\\)\n= \\(113.14 cm^{3}\\)\n\\(mass = density \\times volume\\)\n\\(0.334 \\times 113.14 = 37.789g\\)\n\n\n', '(a) \\(T_{n} = a + (n - 1)d\\) (terms of an AP)\nGiven a = -8;\n\\(T_{7} = a + 6d = -8 + 6d\\)\n\\(T_{9} = a + 8d = -8 + 8d\\)\n\\(\\frac{-8 + 6d}{-8 + 8d} = \\frac{5}{8}\\)\n\\(5(-8 + 8d) = 8(-8 + 6d)\\)\n\\(-40 + 40d = -64 + 48d\\)\n\\(-40 + 64 = 48d - 40d \\times 24 = 8d\\)\n\\( d = 3\\)~(b) Given r = 2 cm, m = 11.2g\n(i) \\(V = \\frac{4}{3} \\pi r^{3}\\)\n= \\(\\frac{4}{3} \\times \\frac{22}{7} \\times 2^{3}\\)\n= \\(\\frac{704}{21}\\)\n= \\(33.52 cm^{3} = 33.52 \\times 10^{-6} m^{3}\\)\n= \\(3.352 \\times 10^{-5} m^{3}\\)\n(ii) \\(Density = \\frac{mass}{volume}\\)\n= \\(\\frac{11.2}{33.52}\\)\n= \\(0.334 g/cm^{3}\\)\n(iii) \\(V = \\frac{4}{3} \\pi r^{3}\\)\n= \\(\\frac{4}{3} \\times \\frac{22}{7} \\times 3^{3}\\)\n= \\(\\frac{2376}{21}\\)\n= \\(113.14 cm^{3}\\)\n\\(mass = density \\times volume\\)\n\\(0.334 \\times 113.14 = 37.789g\\)\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(562, '2016', '\n\nQuestion\n\n(a) Two places X and Y on the equator are on longitudes 67°E and 123°E respectively. (i) What is the distance between them along the equator? (ii) How far from the North pole is X? [Take \\(\\pi = \\frac{22}{7}\\) and radius of earth = 6400km].\n(b)In the diagram, PQR is a circle centre O. N is the mid-point of chord PQ. |PQ| = 8cm, |ON| = 3cm and < ONR = 20°. Calculate the size of < ORN to the nearest degree.\n\n', '', '(a) Two places X and Y on the equator are on longitudes 67°E and 123°E respectively. (i) What is the distance between them along the equator? (ii) How far from the North pole is X? [Take \\(\\pi = \\frac{22}{7}\\) and radius of earth = 6400km].\n(b)In the diagram, PQR is a circle centre O. N is the mid-point of chord PQ. |PQ| = 8cm, |ON| = 3cm and < ONR = 20°. Calculate the size of < ORN to the nearest degree.', '(a) Two places X and Y on the equator are on longitudes 67°E and 123°E respectively. (i) What is the distance between them along the equator? (ii) How far from the North pole is X? [Take \\(\\pi = \\frac{22}{7}\\) and radius of earth = 6400km].~(b)In the diagram, PQR is a circle centre O. N is the mid-point of chord PQ. |PQ| = 8cm, |ON| = 3cm and < ONR = 20°. Calculate the size of < ORN to the nearest degree.', NULL, '\n\nExplanation\n(a) Longitude difference = 123° - 67° = 56°\n(i) Distance between X and Y along the equator = \\(\\frac{56}{360} \\times 2 \\times \\frac{22}{7} \\times 6400 \\cos 0\\)\n= \\(\\frac{56}{360} \\times 2 \\times \\frac{22}{7} \\times 6400\\)\n= \\(6257.78km \\approxeq 6258km\\)\n(ii) Latitude difference = 90°\nDistance from the North pole = \\(\\frac{90}{360} \\times 2 \\times \\frac{22}{7} \\times 6400\\)\n= \\(10057.14km \\approxeq 10057km\\)\n(b)\n\\(OP^{2} = PN^{2} + ON^{2}\\)\n\\(4^{2} + 3^{2} = 25\\)\n\\(OP = \\sqrt{25} = 5cm\\)\n\\(OP = OR = 5cm\\)\n\\(\\Delta NOR = scalene\\)\n\\(\\frac{OR}{\\sin 20} = \\frac{3}{\\sin < ORN}\\)\n\\(\\frac{5}{\\sin 20} = \\frac{3}{\\sin < ORN}\\)\n\\(\\sin < ORN = \\frac{3 \\times \\sin 20}{5}\\)\n\\(\\sin < ORN = 0.2052\\)\n\\(< ORN = \\sin^{-1} (0.2052)\\)\n\\(< ORN = 11.84° \\approxeq 12°\\)\n\n\n', '', 'Explanation\n(a) Longitude difference = 123° - 67° = 56°\n(i) Distance between X and Y along the equator = \\(\\frac{56}{360} \\times 2 \\times \\frac{22}{7} \\times 6400 \\cos 0\\)\n= \\(\\frac{56}{360} \\times 2 \\times \\frac{22}{7} \\times 6400\\)\n= \\(6257.78km \\approxeq 6258km\\)\n(ii) Latitude difference = 90°\nDistance from the North pole = \\(\\frac{90}{360} \\times 2 \\times \\frac{22}{7} \\times 6400\\)\n= \\(10057.14km \\approxeq 10057km\\)\n(b)\n\\(OP^{2} = PN^{2} + ON^{2}\\)\n\\(4^{2} + 3^{2} = 25\\)\n\\(OP = \\sqrt{25} = 5cm\\)\n\\(OP = OR = 5cm\\)\n\\(\\Delta NOR = scalene\\)\n\\(\\frac{OR}{\\sin 20} = \\frac{3}{\\sin < ORN}\\)\n\\(\\frac{5}{\\sin 20} = \\frac{3}{\\sin < ORN}\\)\n\\(\\sin < ORN = \\frac{3 \\times \\sin 20}{5}\\)\n\\(\\sin < ORN = 0.2052\\)\n\\(< ORN = \\sin^{-1} (0.2052)\\)\n\\(< ORN = 11.84° \\approxeq 12°\\)\n\n\n', '(a) Longitude difference = 123° - 67° = 56°\n(i) Distance between X and Y along the equator = \\(\\frac{56}{360} \\times 2 \\times \\frac{22}{7} \\times 6400 \\cos 0\\)\n= \\(\\frac{56}{360} \\times 2 \\times \\frac{22}{7} \\times 6400\\)\n= \\(6257.78km \\approxeq 6258km\\)\n(ii) Latitude difference = 90°\nDistance from the North pole = \\(\\frac{90}{360} \\times 2 \\times \\frac{22}{7} \\times 6400\\)\n= \\(10057.14km \\approxeq 10057km\\)~(b)\n\\(OP^{2} = PN^{2} + ON^{2}\\)\n\\(4^{2} + 3^{2} = 25\\)\n\\(OP = \\sqrt{25} = 5cm\\)\n\\(OP = OR = 5cm\\)\n\\(\\Delta NOR = scalene\\)\n\\(\\frac{OR}{\\sin 20} = \\frac{3}{\\sin < ORN}\\)\n\\(\\frac{5}{\\sin 20} = \\frac{3}{\\sin < ORN}\\)\n\\(\\sin < ORN = \\frac{3 \\times \\sin 20}{5}\\)\n\\(\\sin < ORN = 0.2052\\)\n\\(< ORN = \\sin^{-1} (0.2052)\\)\n\\(< ORN = 11.84° \\approxeq 12°\\)\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(563, '2016', '\n\nQuestion\n\n(a) Simplify \\((\\frac{4}{25})^{-\\frac{1}{2}} \\times 2^{4} \\div (\\frac{15}{2})^{-2}\\)\n(b) Evaluate \\(\\log_{5} (\\frac{3}{5}) + 3 \\log_{5} (\\frac{5}{2}) - \\log_{5} (\\frac{81}{8})\\).\n\n', '', '(a) Simplify \\((\\frac{4}{25})^{-\\frac{1}{2}} \\times 2^{4} \\div (\\frac{15}{2})^{-2}\\)\n(b) Evaluate \\(\\log_{5} (\\frac{3}{5}) + 3 \\log_{5} (\\frac{5}{2}) - \\log_{5} (\\frac{81}{8})\\).', '(a) Simplify \\((\\frac{4}{25})^{-\\frac{1}{2}} \\times 2^{4} \\div (\\frac{15}{2})^{-2}\\)~(b) Evaluate \\(\\log_{5} (\\frac{3}{5}) + 3 \\log_{5} (\\frac{5}{2}) - \\log_{5} (\\frac{81}{8})\\).', NULL, '\n\nExplanation\n(a) \\((\\frac{4}{25})^{-\\frac{1}{2}} \\times 2^{4} \\div (\\frac{15}{2})^{-2}\\)\n= \\((\\frac{25}{4})^{\\frac{1}{2}} \\times 2^{4} \\div (\\frac{2}{15})^{2}\\)\n= \\(\\frac{5}{2} \\times 16 \\times \\frac{225}{4} = 2250\\)\n(b) \\(\\log_{5} (\\frac{3}{5}) + 3\\log_{5} (\\frac{15}{2}) - \\log_{5} (\\frac{81}{8})\\)\n= \\(\\log_{5} 3 - \\log_{5} 5 + 3\\log_{5} 15 - 3\\log_{5} 2 - \\log_{5} 81 + \\log_{5} 8\\)\n= \\(\\log_{5} 3 - \\log_{5} 5 + 3\\log_{5} (3 \\times 5) - 3\\log_{5} 2 - \\log_{5} 3^{4} + \\log_{5} 2^{3}\\)\n= \\(\\log_{5} 3 - \\log_{5} 5 + 3\\log_{5} 3 + 3\\log_{5} 5 - 3\\log_{5} 2 - 4\\log_{5} 3 + 3\\log_{5} 2\\)\n= \\(2\\log_{5} 5 \\)\n= \\(2\\)\n\n', '', 'Explanation\n(a) \\((\\frac{4}{25})^{-\\frac{1}{2}} \\times 2^{4} \\div (\\frac{15}{2})^{-2}\\)\n= \\((\\frac{25}{4})^{\\frac{1}{2}} \\times 2^{4} \\div (\\frac{2}{15})^{2}\\)\n= \\(\\frac{5}{2} \\times 16 \\times \\frac{225}{4} = 2250\\)\n(b) \\(\\log_{5} (\\frac{3}{5}) + 3\\log_{5} (\\frac{15}{2}) - \\log_{5} (\\frac{81}{8})\\)\n= \\(\\log_{5} 3 - \\log_{5} 5 + 3\\log_{5} 15 - 3\\log_{5} 2 - \\log_{5} 81 + \\log_{5} 8\\)\n= \\(\\log_{5} 3 - \\log_{5} 5 + 3\\log_{5} (3 \\times 5) - 3\\log_{5} 2 - \\log_{5} 3^{4} + \\log_{5} 2^{3}\\)\n= \\(\\log_{5} 3 - \\log_{5} 5 + 3\\log_{5} 3 + 3\\log_{5} 5 - 3\\log_{5} 2 - 4\\log_{5} 3 + 3\\log_{5} 2\\)\n= \\(2\\log_{5} 5 \\)\n= \\(2\\)\n\n', '(a) \\((\\frac{4}{25})^{-\\frac{1}{2}} \\times 2^{4} \\div (\\frac{15}{2})^{-2}\\)\n= \\((\\frac{25}{4})^{\\frac{1}{2}} \\times 2^{4} \\div (\\frac{2}{15})^{2}\\)\n= \\(\\frac{5}{2} \\times 16 \\times \\frac{225}{4} = 2250\\)~(b) \\(\\log_{5} (\\frac{3}{5}) + 3\\log_{5} (\\frac{15}{2}) - \\log_{5} (\\frac{81}{8})\\)\n= \\(\\log_{5} 3 - \\log_{5} 5 + 3\\log_{5} 15 - 3\\log_{5} 2 - \\log_{5} 81 + \\log_{5} 8\\)\n= \\(\\log_{5} 3 - \\log_{5} 5 + 3\\log_{5} (3 \\times 5) - 3\\log_{5} 2 - \\log_{5} 3^{4} + \\log_{5} 2^{3}\\)\n= \\(\\log_{5} 3 - \\log_{5} 5 + 3\\log_{5} 3 + 3\\log_{5} 5 - 3\\log_{5} 2 - 4\\log_{5} 3 + 3\\log_{5} 2\\)\n= \\(2\\log_{5} 5 \\)\n= \\(2\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(564, '2016', '\n\nQuestion\n\n(a) If \\(\\varepsilon\\) is the set \\({1, 2, 3,..., 19, 20}\\) and A, B and C are subsets of \\(\\varepsilon\\) such that A = { multiples of five}, B = {multiples of four} and C = {multiples of three}, list the elements of (i) A ; (ii) B ; (iii) C ;\n(b) Find : (i) \\(A \\cap B\\) ; (ii) \\(A \\cap C\\) ; (iii) \\(B \\cup C\\).\n(c) Using your results in (b), show that \\((A \\cap B) \\cup (A \\cap C) = A \\cap (B \\cup C)\\).\n\n', '', '(a) If \\(\\varepsilon\\) is the set \\({1, 2, 3,..., 19, 20}\\) and A, B and C are subsets of \\(\\varepsilon\\) such that A = { multiples of five}, B = {multiples of four} and C = {multiples of three}, list the elements of (i) A ; (ii) B ; (iii) C ;\n(b) Find : (i) \\(A \\cap B\\) ; (ii) \\(A \\cap C\\) ; (iii) \\(B \\cup C\\).\n(c) Using your results in (b), show that \\((A \\cap B) \\cup (A \\cap C) = A \\cap (B \\cup C)\\).', '(a) If \\(\\varepsilon\\) is the set \\({1, 2, 3,..., 19, 20}\\) and A, B and C are subsets of \\(\\varepsilon\\) such that A = { multiples of five}, B = {multiples of four} and C = {multiples of three}, list the elements of (i) A ; (ii) B ; (iii) C ;~(b) Find : (i) \\(A \\cap B\\) ; (ii) \\(A \\cap C\\) ; (iii) \\(B \\cup C\\).\n(c) Using your results in (b), show that \\((A \\cap B) \\cup (A \\cap C) = A \\cap (B \\cup C)\\).~(c) Using your results in (b), show that \\((A \\cap B) \\cup (A \\cap C) = A \\cap (B \\cup C)\\).', NULL, '\n\nExplanation\n(a)(i) A = {5, 10, 15, 20}\n(ii) B = {4, 8, 12, 16, 20}\n(iii) C = {3, 6, 9, 12, 15, 18}\n(b) (i) \\(A \\cap B = {20}\\)\n(ii) \\(A \\cap C = {15}\\)\n(iii) \\(B \\cap C = {12}\\)\n(c) \\((A \\cap B) \\cup (A \\cap C) = A \\cap (B \\cup C)\\)\n\\((A \\cap B) \\cup (A \\cap C) = {15, 20}\\)\n\\(A \\cap (B \\cup C) = {5, 10, 15, 20} \\cap {3, 4, 6, 8, 9, 12, 15, 16, 18, 20}\\)\n= \\({15, 20}\\)\n\\(\\therefore (A \\cap B) \\cup (A \\cap C) = A \\cap (B \\cup C)\\)\n\n', '', 'Explanation\n(a)(i) A = {5, 10, 15, 20}\n(ii) B = {4, 8, 12, 16, 20}\n(iii) C = {3, 6, 9, 12, 15, 18}\n(b) (i) \\(A \\cap B = {20}\\)\n(ii) \\(A \\cap C = {15}\\)\n(iii) \\(B \\cap C = {12}\\)\n(c) \\((A \\cap B) \\cup (A \\cap C) = A \\cap (B \\cup C)\\)\n\\((A \\cap B) \\cup (A \\cap C) = {15, 20}\\)\n\\(A \\cap (B \\cup C) = {5, 10, 15, 20} \\cap {3, 4, 6, 8, 9, 12, 15, 16, 18, 20}\\)\n= \\({15, 20}\\)\n\\(\\therefore (A \\cap B) \\cup (A \\cap C) = A \\cap (B \\cup C)\\)\n\n', '(a)(i) A = {5, 10, 15, 20}\n(ii) B = {4, 8, 12, 16, 20}\n(iii) C = {3, 6, 9, 12, 15, 18}~(b) (i) \\(A \\cap B = {20}\\)\n(ii) \\(A \\cap C = {15}\\)\n(iii) \\(B \\cap C = {12}\\)\n(c) \\((A \\cap B) \\cup (A \\cap C) = A \\cap (B \\cup C)\\)\n\\((A \\cap B) \\cup (A \\cap C) = {15, 20}\\)\n\\(A \\cap (B \\cup C) = {5, 10, 15, 20} \\cap {3, 4, 6, 8, 9, 12, 15, 16, 18, 20}\\)\n= \\({15, 20}\\)\n\\(\\therefore (A \\cap B) \\cup (A \\cap C) = A \\cap (B \\cup C)\\)\n~(c) \\((A \\cap B) \\cup (A \\cap C) = A \\cap (B \\cup C)\\)\n\\((A \\cap B) \\cup (A \\cap C) = {15, 20}\\)\n\\(A \\cap (B \\cup C) = {5, 10, 15, 20} \\cap {3, 4, 6, 8, 9, 12, 15, 16, 18, 20}\\)\n= \\({15, 20}\\)\n\\(\\therefore (A \\cap B) \\cup (A \\cap C) = A \\cap (B \\cup C)\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(565, '2016', '\n\nQuestion\n\nABC is a triangle, right-angled at C. P is the mid-point of AC, < PBC = 37° and |BC| = 5 cm. Calculate :\n(a) |AC|, correct to 3 significant figures ;\n(b) < PBA.\n\n', '', 'ABC is a triangle, right-angled at C. P is the mid-point of AC, < PBC = 37° and |BC| = 5 cm. Calculate :\n(a) |AC|, correct to 3 significant figures ;\n(b) < PBA.', 'ABC is a triangle, right-angled at C. P is the mid-point of AC, < PBC = 37° and |BC| = 5 cm. Calculate :\n(a) |AC|, correct to 3 significant figures ;~(b) < PBA.', NULL, '\n\nExplanation\nLet |PC| = x cm; Hence, |AC| = 2x cm\n\\(\\tan 37° = \\frac{x}{5}\\)\n\\(x = 5 \\tan 37 \\)\n\\(x = 3.768 cm\\)\n\\(\\therefore |AC| = 2 \\times 3.768\\)\n= \\(7.536 cm\\)\n\\(\\approxeq 7.54 cm\\) (3 sig. figs)\n(b) From \\(\\Delta ABC\\),\n\\(\\tan < ABC = \\frac{7.536}{5} = 1.5072\\)\n\\(< ABC = \\tan^{-1} (1.5072) = 56.436°\\)\n\\(\\therefore < PBA = < ABC - < PBC\\)\n= \\(56.436° - 37°\\)\n= \\(19.436° \\approxeq 19.44°\\)\n\n', '', 'Explanation\nLet |PC| = x cm; Hence, |AC| = 2x cm\n\\(\\tan 37° = \\frac{x}{5}\\)\n\\(x = 5 \\tan 37 \\)\n\\(x = 3.768 cm\\)\n\\(\\therefore |AC| = 2 \\times 3.768\\)\n= \\(7.536 cm\\)\n\\(\\approxeq 7.54 cm\\) (3 sig. figs)\n(b) From \\(\\Delta ABC\\),\n\\(\\tan < ABC = \\frac{7.536}{5} = 1.5072\\)\n\\(< ABC = \\tan^{-1} (1.5072) = 56.436°\\)\n\\(\\therefore < PBA = < ABC - < PBC\\)\n= \\(56.436° - 37°\\)\n= \\(19.436° \\approxeq 19.44°\\)\n\n', '~(b) From \\(\\Delta ABC\\),\n\\(\\tan < ABC = \\frac{7.536}{5} = 1.5072\\)\n\\(< ABC = \\tan^{-1} (1.5072) = 56.436°\\)\n\\(\\therefore < PBA = < ABC - < PBC\\)\n= \\(56.436° - 37°\\)\n= \\(19.436° \\approxeq 19.44°\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(566, '2016', '\n\nQuestion\n\nIn the diagram, ABCD is a trapezium in which \\(AD \\parallel BC\\) and \\(< ABC\\) is a right angle. If |AD| = 15 cm, |BD| = 17 cm and |BC| = 9 cm, calculate :\n(a) |AB| ;\n(b) the area of the triangle BCD ;\n(c) |CD| ;\n(d) perimeter of the trapezium.\n\n\n', '', 'In the diagram, ABCD is a trapezium in which \\(AD \\parallel BC\\) and \\(< ABC\\) is a right angle. If |AD| = 15 cm, |BD| = 17 cm and |BC| = 9 cm, calculate :\n(a) |AB| ;\n(b) the area of the triangle BCD ;\n(c) |CD| ;\n(d) perimeter of the trapezium.\n', 'In the diagram, ABCD is a trapezium in which \\(AD \\parallel BC\\) and \\(< ABC\\) is a right angle. If |AD| = 15 cm, |BD| = 17 cm and |BC| = 9 cm, calculate :\n(a) |AB| ;~(b) the area of the triangle BCD ;\n(c) |CD| ;\n(d) perimeter of the trapezium.\n~(c) |CD| ;\n(d) perimeter of the trapezium.\n~(d) perimeter of the trapezium.\n', NULL, '\n\nExplanation\n(a) \\(|AB| = \\sqrt{|BD|^{2} - |AD|^{2}}\\)\n= \\(\\sqrt{17^{2} - 15^{2}} = \\sqrt{289 - 225}\\)\n= \\(\\sqrt{64} = 8 cm\\)\n(b) Area of triangle BCD = \\(\\frac{1}{2} \\times 9 \\times 8 \\)\n= \\(36 cm^{2}\\)\n(c) \\(|EC| = 6 cm\\)\n\\(|CD| = \\sqrt{8^{2} + 6^{2}} = \\sqrt{100}\\)\n= 10 cm.\n(d) Perimeter of the trapezium = 15 cm + 8 cm + 9 cm + 10 cm\n= 42 cm.\n\n', '', 'Explanation\n(a) \\(|AB| = \\sqrt{|BD|^{2} - |AD|^{2}}\\)\n= \\(\\sqrt{17^{2} - 15^{2}} = \\sqrt{289 - 225}\\)\n= \\(\\sqrt{64} = 8 cm\\)\n(b) Area of triangle BCD = \\(\\frac{1}{2} \\times 9 \\times 8 \\)\n= \\(36 cm^{2}\\)\n(c) \\(|EC| = 6 cm\\)\n\\(|CD| = \\sqrt{8^{2} + 6^{2}} = \\sqrt{100}\\)\n= 10 cm.\n(d) Perimeter of the trapezium = 15 cm + 8 cm + 9 cm + 10 cm\n= 42 cm.\n\n', '(a) \\(|AB| = \\sqrt{|BD|^{2} - |AD|^{2}}\\)\n= \\(\\sqrt{17^{2} - 15^{2}} = \\sqrt{289 - 225}\\)\n= \\(\\sqrt{64} = 8 cm\\)~(b) Area of triangle BCD = \\(\\frac{1}{2} \\times 9 \\times 8 \\)\n= \\(36 cm^{2}\\)\n(c) \\(|EC| = 6 cm\\)\n\\(|CD| = \\sqrt{8^{2} + 6^{2}} = \\sqrt{100}\\)\n= 10 cm.\n(d) Perimeter of the trapezium = 15 cm + 8 cm + 9 cm + 10 cm\n= 42 cm.\n~(c) \\(|EC| = 6 cm\\)\n\\(|CD| = \\sqrt{8^{2} + 6^{2}} = \\sqrt{100}\\)\n= 10 cm.\n(d) Perimeter of the trapezium = 15 cm + 8 cm + 9 cm + 10 cm\n= 42 cm.\n~(d) Perimeter of the trapezium = 15 cm + 8 cm + 9 cm + 10 cm\n= 42 cm.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(567, '2016', '\n\nQuestion\n\n(a) Solve the simultaneous equations 3y - 2x = 21 ; 4y + 5x = 5.\n(b) Six identical cards numbered 1 - 6 are placed face down. A card is to be picked at random. A person wins$60.00 if he picks the card numbered 6. If he picks any of the other cards, he loses$10.00 times the number on the card. Calculate the probability of (i) losing ; (ii) losing$20.00 after two picks.\n\n', '', '(a) Solve the simultaneous equations 3y - 2x = 21 ; 4y + 5x = 5.\n(b) Six identical cards numbered 1 - 6 are placed face down. A card is to be picked at random. A person wins$60.00 if he picks the card numbered 6. If he picks any of the other cards, he loses$10.00 times the number on the card. Calculate the probability of (i) losing ; (ii) losing$20.00 after two picks.', '(a) Solve the simultaneous equations 3y - 2x = 21 ; 4y + 5x = 5.~(b) Six identical cards numbered 1 - 6 are placed face down. A card is to be picked at random. A person wins$60.00 if he picks the card numbered 6. If he picks any of the other cards, he loses$10.00 times the number on the card. Calculate the probability of (i) losing ; (ii) losing$20.00 after two picks.', NULL, '\n\nExplanation\n(a) \\(3y - 2x = 21 ... (1)\\)\n\\(4y + 5x = 5 .... (2)\\)\nMultiply (1) by 4 and (2) by 3 so we have,\n\\(12y - 8x = 84 ... (3)\\)\n\\(12y + 15x= 15 ... (4)\\)\n(3) - (4) : \\(-8x- 15x = 69 \\implies -23x = 69\\)\n\\(x = \\frac{69}{-23} = -3\\)\nPut x = -3 in (1),\n\\(3y - 2(-3) = 21\\)\n\\(3y + 6 = 21 \\implies 3y = 21 - 6 = 15\\)\n\\(y = \\frac{15}{3} = 5\\)\n\\(x, y = -3, 5\\).\n(b)(i) Probability of losing\nProbability of winning = \\(\\frac{1}{6}\\)\n\\(\\therefore\\) Probability of losing = \\(1 - \\frac{1}{6} = \\frac{5}{6}\\).\n(ii) Losing$20.00 after two picks = picking the card numbered 1 twice.\n= \\(\\frac{1}{6} \\times \\frac{1}{6} = \\frac{1}{36}\\)\n\n', '', 'Explanation\n(a) \\(3y - 2x = 21 ... (1)\\)\n\\(4y + 5x = 5 .... (2)\\)\nMultiply (1) by 4 and (2) by 3 so we have,\n\\(12y - 8x = 84 ... (3)\\)\n\\(12y + 15x= 15 ... (4)\\)\n(3) - (4) : \\(-8x- 15x = 69 \\implies -23x = 69\\)\n\\(x = \\frac{69}{-23} = -3\\)\nPut x = -3 in (1),\n\\(3y - 2(-3) = 21\\)\n\\(3y + 6 = 21 \\implies 3y = 21 - 6 = 15\\)\n\\(y = \\frac{15}{3} = 5\\)\n\\(x, y = -3, 5\\).\n(b)(i) Probability of losing\nProbability of winning = \\(\\frac{1}{6}\\)\n\\(\\therefore\\) Probability of losing = \\(1 - \\frac{1}{6} = \\frac{5}{6}\\).\n(ii) Losing$20.00 after two picks = picking the card numbered 1 twice.\n= \\(\\frac{1}{6} \\times \\frac{1}{6} = \\frac{1}{36}\\)\n\n', '(a) \\(3y - 2x = 21 ... (1)\\)\n\\(4y + 5x = 5 .... (2)\\)\nMultiply (1) by 4 and (2) by 3 so we have,\n\\(12y - 8x = 84 ... (3)\\)\n\\(12y + 15x= 15 ... (4)\\)\n(3) - (4) : \\(-8x- 15x = 69 \\implies -23x = 69\\)\n\\(x = \\frac{69}{-23} = -3\\)\nPut x = -3 in (1),\n\\(3y - 2(-3) = 21\\)\n\\(3y + 6 = 21 \\implies 3y = 21 - 6 = 15\\)\n\\(y = \\frac{15}{3} = 5\\)\n\\(x, y = -3, 5\\).~(b)(i) Probability of losing\nProbability of winning = \\(\\frac{1}{6}\\)\n\\(\\therefore\\) Probability of losing = \\(1 - \\frac{1}{6} = \\frac{5}{6}\\).\n(ii) Losing$20.00 after two picks = picking the card numbered 1 twice.\n= \\(\\frac{1}{6} \\times \\frac{1}{6} = \\frac{1}{36}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(568, '2016', '\n\nQuestion\n\nThe table gives the frequency distribution of marks obtained by a group of students in a test.\n\n\n\nMarks\n3\n4\n5\n6\n7\n8\n\n\nFrequency\n5\nx - 1\nx\n9\n4\n1\n\n\n\nIf the mean is 5,\n(a) Calculate the value of x;\n(b) Find the : (i) mode ; (ii) median of the distribution.\n(c) If one of the students is selected at random, find the probability that he scored at least 7 marks.\n\n\n\nThe table gives the frequency distribution of marks obtained by a group of students in a test.\n\n\n\nMarks\n3\n4\n5\n6\n7\n8\n\n\nFrequency\n5\nx - 1\nx\n9\n4\n1\n\n\n', '', 'The table gives the frequency distribution of marks obtained by a group of students in a test.\n\n\n\nMarks\n3\n4\n5\n6\n7\n8\n\n\nFrequency\n5\nx - 1\nx\n9\n4\n1\n\n\n\nIf the mean is 5,\n(a) Calculate the value of x;\n(b) Find the : (i) mode ; (ii) median of the distribution.\n(c) If one of the students is selected at random, find the probability that he scored at least 7 marks.\n\n\n\nThe table gives the frequency distribution of marks obtained by a group of students in a test.\n\n\n\nMarks\n3\n4\n5\n6\n7\n8\n\n\nFrequency\n5\nx - 1\nx\n9\n4\n1\n', 'The table gives the frequency distribution of marks obtained by a group of students in a test.\n\n\n\nMarks\n3\n4\n5\n6\n7\n8\n\n\nFrequency\n5\nx - 1\nx\n9\n4\n1\n\n\n\nIf the mean is 5,\n(a) Calculate the value of x;~(b) Find the : (i) mode ; (ii) median of the distribution.\n(c) If one of the students is selected at random, find the probability that he scored at least 7 marks.\n\n\n\nThe table gives the frequency distribution of marks obtained by a group of students in a test.\n\n\n\nMarks\n3\n4\n5\n6\n7\n8\n\n\nFrequency\n5\nx - 1\nx\n9\n4\n1\n~(c) If one of the students is selected at random, find the probability that he scored at least 7 marks.\n\n\n\nThe table gives the frequency distribution of marks obtained by a group of students in a test.\n\n\n\nMarks\n3\n4\n5\n6\n7\n8\n\n\nFrequency\n5\nx - 1\nx\n9\n4\n1\n', NULL, '\n\nExplanation\nMarks\n3\n4\n5\n6\n7\n8\n\n\nFrequency\n5\nx - 1\nx\n9\n4\n1\n\n\nfx\n15\n4x - 4\n5x\n54\n28\n8\n\n\n\n\\(\\sum fx = 15 + 4x - 4 + 5x + 54 + 28 + 8 = 101 + 9x\\)\n\\(\\sum f = 5 + x - 1 + x + 9 + 4 + 1 = 18 + 2x\\)\n\\(\\bar{x} = \\frac{\\sum fx}{\\sum f}\\)\n\\(5 = \\frac{101 + 9x}{18 + 2x} \\implies 101 + 9x = 5(18 + 2x)\\)\n\\(101 + 9x = 90 + 10x \\implies 101 - 90 = 10x - 9x\\)\n\\(11 = x \\)\n(b)(i) Mode = 5.\n(ii) Median\nFrequency = 18 + 2(11) = 40.\nMedian position = \\(\\frac{40}{2} = 20\\)\n20th position = 5.\n(c) No of students that scored at least 7 marks = 4 + 1 = 5.\nProbability of scoring at least 7 marks = \\(\\frac{5}{40} = \\frac{1}{8}\\)\n\n', '', 'Explanation\nMarks\n3\n4\n5\n6\n7\n8\n\n\nFrequency\n5\nx - 1\nx\n9\n4\n1\n\n\nfx\n15\n4x - 4\n5x\n54\n28\n8\n\n\n\n\\(\\sum fx = 15 + 4x - 4 + 5x + 54 + 28 + 8 = 101 + 9x\\)\n\\(\\sum f = 5 + x - 1 + x + 9 + 4 + 1 = 18 + 2x\\)\n\\(\\bar{x} = \\frac{\\sum fx}{\\sum f}\\)\n\\(5 = \\frac{101 + 9x}{18 + 2x} \\implies 101 + 9x = 5(18 + 2x)\\)\n\\(101 + 9x = 90 + 10x \\implies 101 - 90 = 10x - 9x\\)\n\\(11 = x \\)\n(b)(i) Mode = 5.\n(ii) Median\nFrequency = 18 + 2(11) = 40.\nMedian position = \\(\\frac{40}{2} = 20\\)\n20th position = 5.\n(c) No of students that scored at least 7 marks = 4 + 1 = 5.\nProbability of scoring at least 7 marks = \\(\\frac{5}{40} = \\frac{1}{8}\\)\n\n', '~(b)(i) Mode = 5.\n(ii) Median\nFrequency = 18 + 2(11) = 40.\nMedian position = \\(\\frac{40}{2} = 20\\)\n20th position = 5.\n(c) No of students that scored at least 7 marks = 4 + 1 = 5.\nProbability of scoring at least 7 marks = \\(\\frac{5}{40} = \\frac{1}{8}\\)\n~(c) No of students that scored at least 7 marks = 4 + 1 = 5.\nProbability of scoring at least 7 marks = \\(\\frac{5}{40} = \\frac{1}{8}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(569, '2016', '\n\nQuestion\n\n(a) A cylindrical pipe is 28 metres long. Its internal radius is 3.5 cm and external radius 5 cm. Calaulate : (i) the volume, in cm\\(^{3}\\), of metal used in making the pipe ; (ii) the volume of water in litres that the pipe can hold when full, correct to 1 decimal place. [Take \\(\\pi = \\frac{22}{7}\\)]\n(b) In the diagram, MP is a tangent to the circle LMN at M. If the chord LN is parallel to MP, show that the triangle LMN is isosceles.\n\n', '', '(a) A cylindrical pipe is 28 metres long. Its internal radius is 3.5 cm and external radius 5 cm. Calaulate : (i) the volume, in cm\\(^{3}\\), of metal used in making the pipe ; (ii) the volume of water in litres that the pipe can hold when full, correct to 1 decimal place. [Take \\(\\pi = \\frac{22}{7}\\)]\n(b) In the diagram, MP is a tangent to the circle LMN at M. If the chord LN is parallel to MP, show that the triangle LMN is isosceles.', '(a) A cylindrical pipe is 28 metres long. Its internal radius is 3.5 cm and external radius 5 cm. Calaulate : (i) the volume, in cm\\(^{3}\\), of metal used in making the pipe ; (ii) the volume of water in litres that the pipe can hold when full, correct to 1 decimal place. [Take \\(\\pi = \\frac{22}{7}\\)]~(b) In the diagram, MP is a tangent to the circle LMN at M. If the chord LN is parallel to MP, show that the triangle LMN is isosceles.', NULL, '\n\nExplanation\n(a)(i) Volume of outer pipe = \\(\\pi r^{2} h\\)\n= \\(\\frac{22}{7} \\times (5.0)^{2} \\times 28 \\times 100\\)\nVolume of inner pipe = \\(\\frac{22}{7} \\times (3.5)^{2} \\times 28 \\times 100\\)\nVolume of metal used = Volume of outer pipe - Volume of inner pipe\n= \\(22 \\times 400 \\times (5.0^{2} - 3.5^{2})\\)\n= \\(22 \\times 400 \\times (5.0 - 3.5) \\times (5.0 + 3.5)\\)\n= \\(112,200 cm^{3}\\)\n(ii) Vol of water the pipe can hold = vol of inner pipe\n= \\(\\frac{22}{7} \\times (3.5)^{2} \\times 28 \\times 100\\)\n= \\(107,800 cm^{3}\\)\n= 107.8 litres.\n(b) < PMN = < LNM (alternate angles)\n< PMN = < LMN (angles in alternate segment)\n\\(\\therefore\\) < LNM = < NLM, then \\(\\Delta\\) LMN is isosceles.\n\n', '', 'Explanation\n(a)(i) Volume of outer pipe = \\(\\pi r^{2} h\\)\n= \\(\\frac{22}{7} \\times (5.0)^{2} \\times 28 \\times 100\\)\nVolume of inner pipe = \\(\\frac{22}{7} \\times (3.5)^{2} \\times 28 \\times 100\\)\nVolume of metal used = Volume of outer pipe - Volume of inner pipe\n= \\(22 \\times 400 \\times (5.0^{2} - 3.5^{2})\\)\n= \\(22 \\times 400 \\times (5.0 - 3.5) \\times (5.0 + 3.5)\\)\n= \\(112,200 cm^{3}\\)\n(ii) Vol of water the pipe can hold = vol of inner pipe\n= \\(\\frac{22}{7} \\times (3.5)^{2} \\times 28 \\times 100\\)\n= \\(107,800 cm^{3}\\)\n= 107.8 litres.\n(b) < PMN = < LNM (alternate angles)\n< PMN = < LMN (angles in alternate segment)\n\\(\\therefore\\) < LNM = < NLM, then \\(\\Delta\\) LMN is isosceles.\n\n', '(a)(i) Volume of outer pipe = \\(\\pi r^{2} h\\)\n= \\(\\frac{22}{7} \\times (5.0)^{2} \\times 28 \\times 100\\)\nVolume of inner pipe = \\(\\frac{22}{7} \\times (3.5)^{2} \\times 28 \\times 100\\)\nVolume of metal used = Volume of outer pipe - Volume of inner pipe\n= \\(22 \\times 400 \\times (5.0^{2} - 3.5^{2})\\)\n= \\(22 \\times 400 \\times (5.0 - 3.5) \\times (5.0 + 3.5)\\)\n= \\(112,200 cm^{3}\\)\n(ii) Vol of water the pipe can hold = vol of inner pipe\n= \\(\\frac{22}{7} \\times (3.5)^{2} \\times 28 \\times 100\\)\n= \\(107,800 cm^{3}\\)\n= 107.8 litres.~(b) < PMN = < LNM (alternate angles)\n< PMN = < LMN (angles in alternate segment)\n\\(\\therefore\\) < LNM = < NLM, then \\(\\Delta\\) LMN is isosceles.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(570, '2016', '\n\nQuestion\n\n(a) Given that \\(\\sin(A + B) = \\sin A \\cos B + \\cos A \\sin B\\). Without using mathematical tables or calculator, evaluate \\(\\sin 105°\\),leaving your answer in the surd form.\n(You may use 105° = 60° + 45°)\n(b) The houses on one side of a particular street are assigned odd numbers, starting from 11. If the sum of the numbers is 551, how many houses are there?\n(c) The 1st and 3rd terms of a Geometric Progression (G.P) are \\(2\\) and \\(\\frac{2}{9}\\) respectively. Find :\n(i) the common difference ; (ii) the 5th term.\n\n', '', '(a) Given that \\(\\sin(A + B) = \\sin A \\cos B + \\cos A \\sin B\\). Without using mathematical tables or calculator, evaluate \\(\\sin 105°\\),leaving your answer in the surd form.\n(You may use 105° = 60° + 45°)\n(b) The houses on one side of a particular street are assigned odd numbers, starting from 11. If the sum of the numbers is 551, how many houses are there?\n(c) The 1st and 3rd terms of a Geometric Progression (G.P) are \\(2\\) and \\(\\frac{2}{9}\\) respectively. Find :\n(i) the common difference ; (ii) the 5th term.', '(a) Given that \\(\\sin(A + B) = \\sin A \\cos B + \\cos A \\sin B\\). Without using mathematical tables or calculator, evaluate \\(\\sin 105°\\),leaving your answer in the surd form.\n(You may use 105° = 60° + 45°)~(b) The houses on one side of a particular street are assigned odd numbers, starting from 11. If the sum of the numbers is 551, how many houses are there?\n(c) The 1st and 3rd terms of a Geometric Progression (G.P) are \\(2\\) and \\(\\frac{2}{9}\\) respectively. Find :\n(i) the common difference ; (ii) the 5th term.~(c) The 1st and 3rd terms of a Geometric Progression (G.P) are \\(2\\) and \\(\\frac{2}{9}\\) respectively. Find :\n(i) the common difference ; (ii) the 5th term.', NULL, '\n\nExplanation\n(a) \\(\\sin 105° = \\sin(60 + 45)\\)\n= \\(\\sin 60° \\cos 45° + \\cos 60° \\sin 45°\\)\n= \\(\\frac{\\sqrt{3}}{2} \\times \\frac{\\sqrt{2}}{2} + \\frac{1}{2} \\times \\frac{\\sqrt{2}}{2}\\)\n= \\(\\frac{\\sqrt{6}}{4} + \\frac{\\sqrt{2}}{4}\\)\n(b) a = 11, d = 2.\n\\(S_{n} = \\frac{n}{2} (2a + (n - 1) d\\)\n\\(551= \\frac{n}{2} (2(11) + (n - 1) 2\\)\n\\(551= \\frac{n}{2} (22 + 2n - 2)\\)\n\\(551 = \\frac{n}{2} (2n + 20)\\)\n\\(1102 = 2n^{2} + 20n \\implies 2n^{2} + 20n - 1102 = 0\\)\nDivide through by 2.\n\\(n^{2} + 10n - 551 = 0\\)\n\\(n^{2} - 19n + 29n - 551 = 0\\)\n\\(n(n - 19) + 29(n - 19) = 0\\)\n\\(\\text{n = 19 or -29}\\)\nSince n cannot be negative, n = 19 houses.\n(c)(i) \\(T_{n} = ar^{n - 1}\\) (terms of a G.P)\n\\(a = 2\\)\n\\(T_{3} = 2 \\times r^{3 - 1} = \\frac{2}{9}\\)\n\\(r^{2} = \\frac{1}{9} \\implies r = \\frac{1}{3}\\)\n(ii) \\(T_{5} = ar^{5 - 1}\\)\n= \\(2 \\times (\\frac{1}{3})^{4}\\)\n= \\(2 \\times \\frac{1}{81}\\)\n= \\(\\frac{2}{81}\\)\n\n', '', 'Explanation\n(a) \\(\\sin 105° = \\sin(60 + 45)\\)\n= \\(\\sin 60° \\cos 45° + \\cos 60° \\sin 45°\\)\n= \\(\\frac{\\sqrt{3}}{2} \\times \\frac{\\sqrt{2}}{2} + \\frac{1}{2} \\times \\frac{\\sqrt{2}}{2}\\)\n= \\(\\frac{\\sqrt{6}}{4} + \\frac{\\sqrt{2}}{4}\\)\n(b) a = 11, d = 2.\n\\(S_{n} = \\frac{n}{2} (2a + (n - 1) d\\)\n\\(551= \\frac{n}{2} (2(11) + (n - 1) 2\\)\n\\(551= \\frac{n}{2} (22 + 2n - 2)\\)\n\\(551 = \\frac{n}{2} (2n + 20)\\)\n\\(1102 = 2n^{2} + 20n \\implies 2n^{2} + 20n - 1102 = 0\\)\nDivide through by 2.\n\\(n^{2} + 10n - 551 = 0\\)\n\\(n^{2} - 19n + 29n - 551 = 0\\)\n\\(n(n - 19) + 29(n - 19) = 0\\)\n\\(\\text{n = 19 or -29}\\)\nSince n cannot be negative, n = 19 houses.\n(c)(i) \\(T_{n} = ar^{n - 1}\\) (terms of a G.P)\n\\(a = 2\\)\n\\(T_{3} = 2 \\times r^{3 - 1} = \\frac{2}{9}\\)\n\\(r^{2} = \\frac{1}{9} \\implies r = \\frac{1}{3}\\)\n(ii) \\(T_{5} = ar^{5 - 1}\\)\n= \\(2 \\times (\\frac{1}{3})^{4}\\)\n= \\(2 \\times \\frac{1}{81}\\)\n= \\(\\frac{2}{81}\\)\n\n', '(a) \\(\\sin 105° = \\sin(60 + 45)\\)\n= \\(\\sin 60° \\cos 45° + \\cos 60° \\sin 45°\\)\n= \\(\\frac{\\sqrt{3}}{2} \\times \\frac{\\sqrt{2}}{2} + \\frac{1}{2} \\times \\frac{\\sqrt{2}}{2}\\)\n= \\(\\frac{\\sqrt{6}}{4} + \\frac{\\sqrt{2}}{4}\\)~(b) a = 11, d = 2.\n\\(S_{n} = \\frac{n}{2} (2a + (n - 1) d\\)\n\\(551= \\frac{n}{2} (2(11) + (n - 1) 2\\)\n\\(551= \\frac{n}{2} (22 + 2n - 2)\\)\n\\(551 = \\frac{n}{2} (2n + 20)\\)\n\\(1102 = 2n^{2} + 20n \\implies 2n^{2} + 20n - 1102 = 0\\)\nDivide through by 2.\n\\(n^{2} + 10n - 551 = 0\\)\n\\(n^{2} - 19n + 29n - 551 = 0\\)\n\\(n(n - 19) + 29(n - 19) = 0\\)\n\\(\\text{n = 19 or -29}\\)\nSince n cannot be negative, n = 19 houses.\n(c)(i) \\(T_{n} = ar^{n - 1}\\) (terms of a G.P)\n\\(a = 2\\)\n\\(T_{3} = 2 \\times r^{3 - 1} = \\frac{2}{9}\\)\n\\(r^{2} = \\frac{1}{9} \\implies r = \\frac{1}{3}\\)\n(ii) \\(T_{5} = ar^{5 - 1}\\)\n= \\(2 \\times (\\frac{1}{3})^{4}\\)\n= \\(2 \\times \\frac{1}{81}\\)\n= \\(\\frac{2}{81}\\)\n~(c)(i) \\(T_{n} = ar^{n - 1}\\) (terms of a G.P)\n\\(a = 2\\)\n\\(T_{3} = 2 \\times r^{3 - 1} = \\frac{2}{9}\\)\n\\(r^{2} = \\frac{1}{9} \\implies r = \\frac{1}{3}\\)\n(ii) \\(T_{5} = ar^{5 - 1}\\)\n= \\(2 \\times (\\frac{1}{3})^{4}\\)\n= \\(2 \\times \\frac{1}{81}\\)\n= \\(\\frac{2}{81}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(571, '2016', '\n\nQuestion\n\nIn the diagram, a ladder TF, 10 metres long is placed against a wall at an angle of 70° to the horizontal.\n(a) How high up the wall, correct to the nearest metre, does the ladder reach?\n(b) If the foot (F) of the ladder is pulled from the wall to F\\(^{1}\\) by 1 metre, (i) how far, correct to 2 significant figures, does the top T slide down the wall to T\\(^{1}\\).\n(ii) Calculate, correct to the nearest degree, \\(QF^{1}T^{1}\\).\n\n\n', '', 'In the diagram, a ladder TF, 10 metres long is placed against a wall at an angle of 70° to the horizontal.\n(a) How high up the wall, correct to the nearest metre, does the ladder reach?\n(b) If the foot (F) of the ladder is pulled from the wall to F\\(^{1}\\) by 1 metre, (i) how far, correct to 2 significant figures, does the top T slide down the wall to T\\(^{1}\\).\n(ii) Calculate, correct to the nearest degree, \\(QF^{1}T^{1}\\).\n', 'In the diagram, a ladder TF, 10 metres long is placed against a wall at an angle of 70° to the horizontal.\n(a) How high up the wall, correct to the nearest metre, does the ladder reach?~(b) If the foot (F) of the ladder is pulled from the wall to F\\(^{1}\\) by 1 metre, (i) how far, correct to 2 significant figures, does the top T slide down the wall to T\\(^{1}\\).\n(ii) Calculate, correct to the nearest degree, \\(QF^{1}T^{1}\\).\n', NULL, '\n\nExplanation\n(a) \\(TQ = 10 \\sin 70° = 10 \\times 0.9397 = 9.397m\\)\n\\(\\approxeq 9m\\)\n(b)(i) \\(QF = 10 \\cos 70° = 10 \\times 0.3420 = 3.42m\\)\n\\(QF^{1} = 3.42m + 1m = 4.42m\\)\n\\(QT^{1} = \\sqrt{10^{2} - 4.42^{2}} = \\sqrt{100 - 19.5364}\\)\n= \\(\\sqrt{80.4636} = 8.97m\\)\n\\(TT^{1} = 9.397 - 8.97 = 0.43m\\)\n(ii) \\(\\sin < QF^{1}T^{1} = \\frac{8,97}{10} = 0.897\\)\n\\(< QF^{1} T^{1} = 63.768° \\approxeq 64°\\)\n\n', '', 'Explanation\n(a) \\(TQ = 10 \\sin 70° = 10 \\times 0.9397 = 9.397m\\)\n\\(\\approxeq 9m\\)\n(b)(i) \\(QF = 10 \\cos 70° = 10 \\times 0.3420 = 3.42m\\)\n\\(QF^{1} = 3.42m + 1m = 4.42m\\)\n\\(QT^{1} = \\sqrt{10^{2} - 4.42^{2}} = \\sqrt{100 - 19.5364}\\)\n= \\(\\sqrt{80.4636} = 8.97m\\)\n\\(TT^{1} = 9.397 - 8.97 = 0.43m\\)\n(ii) \\(\\sin < QF^{1}T^{1} = \\frac{8,97}{10} = 0.897\\)\n\\(< QF^{1} T^{1} = 63.768° \\approxeq 64°\\)\n\n', '(a) \\(TQ = 10 \\sin 70° = 10 \\times 0.9397 = 9.397m\\)\n\\(\\approxeq 9m\\)~(b)(i) \\(QF = 10 \\cos 70° = 10 \\times 0.3420 = 3.42m\\)\n\\(QF^{1} = 3.42m + 1m = 4.42m\\)\n\\(QT^{1} = \\sqrt{10^{2} - 4.42^{2}} = \\sqrt{100 - 19.5364}\\)\n= \\(\\sqrt{80.4636} = 8.97m\\)\n\\(TT^{1} = 9.397 - 8.97 = 0.43m\\)\n(ii) \\(\\sin < QF^{1}T^{1} = \\frac{8,97}{10} = 0.897\\)\n\\(< QF^{1} T^{1} = 63.768° \\approxeq 64°\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1);
