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
(322, '2013', '\n\nQuestion\n\nIn a road worthiness test on 240 cars, 60% passed. The number that failed had faults in Clutch, Brakes and Steering as follows: Clutch only - 28, Clutch and Steering - 14; Clutch, Steering and Brakes - 8; Clutch and Brakes - 20; Brakes and Steering only - 6. The number of cars with faults in Steering only is twice the number of cars with faults in Brakes only.\n(a) Draw a Venn Diagram to illustrate this information.\n(b) How many cars had : (i) Faulty Brakes? (ii) Only one fault?\n\n', '', 'In a road worthiness test on 240 cars, 60% passed. The number that failed had faults in Clutch, Brakes and Steering as follows: Clutch only - 28, Clutch and Steering - 14; Clutch, Steering and Brakes - 8; Clutch and Brakes - 20; Brakes and Steering only - 6. The number of cars with faults in Steering only is twice the number of cars with faults in Brakes only.\n(a) Draw a Venn Diagram to illustrate this information.\n(b) How many cars had : (i) Faulty Brakes? (ii) Only one fault?', 'In a road worthiness test on 240 cars, 60% passed. The number that failed had faults in Clutch, Brakes and Steering as follows: Clutch only - 28, Clutch and Steering - 14; Clutch, Steering and Brakes - 8; Clutch and Brakes - 20; Brakes and Steering only - 6. The number of cars with faults in Steering only is twice the number of cars with faults in Brakes only.\n(a) Draw a Venn Diagram to illustrate this information.~(b) How many cars had : (i) Faulty Brakes? (ii) Only one fault?', NULL, '\n\nExplanation\n (a)\n(b) From the Venn Diagram above,\n28 + 2x + x + 12 + 6 + 6 + 8 = 96\n60 + 3x = 96\n3x = 36\nx = 12\n(i) Faulty brakes = 12 + 12 + 8 + 6 =38\n(ii) Cars with only one fault = 28 + 2x + x\n= 28 + 3x = 28 + 3(12)\n= 28 + 36 = 64\n\n', '', 'Explanation\n (a)\n(b) From the Venn Diagram above,\n28 + 2x + x + 12 + 6 + 6 + 8 = 96\n60 + 3x = 96\n3x = 36\nx = 12\n(i) Faulty brakes = 12 + 12 + 8 + 6 =38\n(ii) Cars with only one fault = 28 + 2x + x\n= 28 + 3x = 28 + 3(12)\n= 28 + 36 = 64\n\n', '~(b) From the Venn Diagram above,\n28 + 2x + x + 12 + 6 + 6 + 8 = 96\n60 + 3x = 96\n3x = 36\nx = 12\n(i) Faulty brakes = 12 + 12 + 8 + 6 =38\n(ii) Cars with only one fault = 28 + 2x + x\n= 28 + 3x = 28 + 3(12)\n= 28 + 36 = 64\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(323, '2013', '\n\nQuestion\n\n(a) Find the equation of the line passing through the points (2, 5) and (-4, -7).\n(b) Three ships P, Q and R are at sea. The bearing of Q from P is 030° and the bearing of P and R is 300°. If |PQ| = 5 km and |PR| = 8 km,\n(i) Illustrate the information in a diagram.\n(ii) Calculate, correct to three significant figures, the:\n(1) distance between Q and R\n(2) bearing of R from Q.\n\n', '', '(a) Find the equation of the line passing through the points (2, 5) and (-4, -7).\n(b) Three ships P, Q and R are at sea. The bearing of Q from P is 030° and the bearing of P and R is 300°. If |PQ| = 5 km and |PR| = 8 km,\n(i) Illustrate the information in a diagram.\n(ii) Calculate, correct to three significant figures, the:\n(1) distance between Q and R\n(2) bearing of R from Q.', '(a) Find the equation of the line passing through the points (2, 5) and (-4, -7).~(b) Three ships P, Q and R are at sea. The bearing of Q from P is 030° and the bearing of P and R is 300°. If |PQ| = 5 km and |PR| = 8 km,\n(i) Illustrate the information in a diagram.\n(ii) Calculate, correct to three significant figures, the:\n(1) distance between Q and R\n(2) bearing of R from Q.', NULL, '\n\nExplanation\n(a) Using the two- point form,\n\\(\\frac{y - y_{1}}{y_{2} - y_{1}} = \\frac{x - x_{1}}{x_{2} - x_{1}}\\)\n\\(\\frac{y - 5}{-7 - 5} = \\frac{x - 2}{-4 - 2}\\)\n\\(\\frac{y - 5}{- 12} = \\frac{x - 2}{- 6}\\)\n\\(y - 5 = 2(x - 2)\\)\n\\(y - 5 = 2x - 4 \\implies y - 2x = -4 + 5 = 1\\)\n\\(Equation : y = 2x + 1\\)\n(b) (i)\nIn \\(\\Delta PQR, <QPR = 60° + 30° = 90°\\). PQR is a right- angled triangle.\n(ii) (1) In \\(\\Delta PQR, |QR|^{2} = 5^{2} + 8^{2}\\)\n\\(25 + 64 = 89\\)\n\\(|QR| = \\sqrt{89} = 9.434 km \\approxeq 9.43 km\\)\n(2) In the diagram above, the bearing of R from Q is the obtuse angle NQR.\nBut \\(\\tan \\alpha = \\frac{8}{5} = 1.6\\)\n\\(\\alpha = \\tan^{-1} (1.6) = 58°\\)\nHence, angle NQR = 360° - (a + 60° + 90°)\n= 360° - (58° + 60° + 90°)\n= 360° - 208°\n= 152°\n\n', '', 'Explanation\n(a) Using the two- point form,\n\\(\\frac{y - y_{1}}{y_{2} - y_{1}} = \\frac{x - x_{1}}{x_{2} - x_{1}}\\)\n\\(\\frac{y - 5}{-7 - 5} = \\frac{x - 2}{-4 - 2}\\)\n\\(\\frac{y - 5}{- 12} = \\frac{x - 2}{- 6}\\)\n\\(y - 5 = 2(x - 2)\\)\n\\(y - 5 = 2x - 4 \\implies y - 2x = -4 + 5 = 1\\)\n\\(Equation : y = 2x + 1\\)\n(b) (i)\nIn \\(\\Delta PQR, <QPR = 60° + 30° = 90°\\). PQR is a right- angled triangle.\n(ii) (1) In \\(\\Delta PQR, |QR|^{2} = 5^{2} + 8^{2}\\)\n\\(25 + 64 = 89\\)\n\\(|QR| = \\sqrt{89} = 9.434 km \\approxeq 9.43 km\\)\n(2) In the diagram above, the bearing of R from Q is the obtuse angle NQR.\nBut \\(\\tan \\alpha = \\frac{8}{5} = 1.6\\)\n\\(\\alpha = \\tan^{-1} (1.6) = 58°\\)\nHence, angle NQR = 360° - (a + 60° + 90°)\n= 360° - (58° + 60° + 90°)\n= 360° - 208°\n= 152°\n\n', '(a) Using the two- point form,\n\\(\\frac{y - y_{1}}{y_{2} - y_{1}} = \\frac{x - x_{1}}{x_{2} - x_{1}}\\)\n\\(\\frac{y - 5}{-7 - 5} = \\frac{x - 2}{-4 - 2}\\)\n\\(\\frac{y - 5}{- 12} = \\frac{x - 2}{- 6}\\)\n\\(y - 5 = 2(x - 2)\\)\n\\(y - 5 = 2x - 4 \\implies y - 2x = -4 + 5 = 1\\)\n\\(Equation : y = 2x + 1\\)~(b) (i)\nIn \\(\\Delta PQR, <QPR = 60° + 30° = 90°\\). PQR is a right- angled triangle.\n(ii) (1) In \\(\\Delta PQR, |QR|^{2} = 5^{2} + 8^{2}\\)\n\\(25 + 64 = 89\\)\n\\(|QR| = \\sqrt{89} = 9.434 km \\approxeq 9.43 km\\)\n(2) In the diagram above, the bearing of R from Q is the obtuse angle NQR.\nBut \\(\\tan \\alpha = \\frac{8}{5} = 1.6\\)\n\\(\\alpha = \\tan^{-1} (1.6) = 58°\\)\nHence, angle NQR = 360° - (a + 60° + 90°)\n= 360° - (58° + 60° + 90°)\n= 360° - 208°\n= 152°\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(324, '2013', '\n\nQuestion\n\n(a) Lamin bought a book for N300.00 and sold it to Bola at a profit of x%. Bola then sold the same book at a profit of x%. If James paid \\(N(6x + \\frac{3}{4})\\) more for the book than Lamin paid, find the value of x.\n(b) Find the range of values of x which satisfies the inequality \\(3x - 2 < 10 + x < 2 + 5x\\).\n\n', '', '(a) Lamin bought a book for N300.00 and sold it to Bola at a profit of x%. Bola then sold the same book at a profit of x%. If James paid \\(N(6x + \\frac{3}{4})\\) more for the book than Lamin paid, find the value of x.\n(b) Find the range of values of x which satisfies the inequality \\(3x - 2 < 10 + x < 2 + 5x\\).', '(a) Lamin bought a book for N300.00 and sold it to Bola at a profit of x%. Bola then sold the same book at a profit of x%. If James paid \\(N(6x + \\frac{3}{4})\\) more for the book than Lamin paid, find the value of x.~(b) Find the range of values of x which satisfies the inequality \\(3x - 2 < 10 + x < 2 + 5x\\).', NULL, '\n\nExplanation\n(a) S.P = \\(C.P + \\frac{% profit}{100} \\times C.P\\)\n= \\(300 + (\\frac{x}{100} \\times 300)\\)\nS.P = N(300 + 3x)\nTherefore, Bola bought it at N(300 + 3x).\nJames paid \\(N(6x + \\frac{3}{4})\\) extra from what Lamin paid, therefore Bola\'s S.P = \\(N(300 +6x + \\frac{3}{4})\\)\n= N(300.75 + 6x).\nProfit for Bola = \\(N(300.75 + 6x - (300 + 3x)) = N(0.75 + 3x)\\)\n\\(\\frac{x}{100} \\times N(300 + 3x) = N(0.75 + 3x)\\)\n\\(300x + 3x^{2} = 75 + 300x\\)\n\\(\\implies 3x^{2} = 75\\)\n\\(x^{2} = 25 \\therefore x = 5\\)\n(b) \\(3x - 2 < 10 + x < 2 + 5x\\)\n\\(3x - 2 < 10 + x \\implies 3x - x < 10 + 2\\)\n\\(2x < 12 \\implies x < 6\\)\n\\(10 + x < 2 + 5x\\)\n\\(x - 5x < 2 - 10\\)\n\\(-4x < -8 \\implies x > 2\\)\nThe range = \\(2 < x < 6\\)\n\n\n', '', 'Explanation\n(a) S.P = \\(C.P + \\frac{% profit}{100} \\times C.P\\)\n= \\(300 + (\\frac{x}{100} \\times 300)\\)\nS.P = N(300 + 3x)\nTherefore, Bola bought it at N(300 + 3x).\nJames paid \\(N(6x + \\frac{3}{4})\\) extra from what Lamin paid, therefore Bola\'s S.P = \\(N(300 +6x + \\frac{3}{4})\\)\n= N(300.75 + 6x).\nProfit for Bola = \\(N(300.75 + 6x - (300 + 3x)) = N(0.75 + 3x)\\)\n\\(\\frac{x}{100} \\times N(300 + 3x) = N(0.75 + 3x)\\)\n\\(300x + 3x^{2} = 75 + 300x\\)\n\\(\\implies 3x^{2} = 75\\)\n\\(x^{2} = 25 \\therefore x = 5\\)\n(b) \\(3x - 2 < 10 + x < 2 + 5x\\)\n\\(3x - 2 < 10 + x \\implies 3x - x < 10 + 2\\)\n\\(2x < 12 \\implies x < 6\\)\n\\(10 + x < 2 + 5x\\)\n\\(x - 5x < 2 - 10\\)\n\\(-4x < -8 \\implies x > 2\\)\nThe range = \\(2 < x < 6\\)\n\n\n', '(a) S.P = \\(C.P + \\frac{% profit}{100} \\times C.P\\)\n= \\(300 + (\\frac{x}{100} \\times 300)\\)\nS.P = N(300 + 3x)\nTherefore, Bola bought it at N(300 + 3x).\nJames paid \\(N(6x + \\frac{3}{4})\\) extra from what Lamin paid, therefore Bola\'s S.P = \\(N(300 +6x + \\frac{3}{4})\\)\n= N(300.75 + 6x).\nProfit for Bola = \\(N(300.75 + 6x - (300 + 3x)) = N(0.75 + 3x)\\)\n\\(\\frac{x}{100} \\times N(300 + 3x) = N(0.75 + 3x)\\)\n\\(300x + 3x^{2} = 75 + 300x\\)\n\\(\\implies 3x^{2} = 75\\)\n\\(x^{2} = 25 \\therefore x = 5\\)~(b) \\(3x - 2 < 10 + x < 2 + 5x\\)\n\\(3x - 2 < 10 + x \\implies 3x - x < 10 + 2\\)\n\\(2x < 12 \\implies x < 6\\)\n\\(10 + x < 2 + 5x\\)\n\\(x - 5x < 2 - 10\\)\n\\(-4x < -8 \\implies x > 2\\)\nThe range = \\(2 < x < 6\\)\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(325, '2013', '\n\nQuestion\n\n\n\nIn the diagram, |PT| = 4 cm, |TS| = 6 cm, |PQ| = 6 cm and < SPR = 30°. Calculate, correct to the nearest whole number:\n(a) |SR| ;\n(b) area of TQRS.\n\n\n\n\n', '', 'In the diagram, |PT| = 4 cm, |TS| = 6 cm, |PQ| = 6 cm and < SPR = 30°. Calculate, correct to the nearest whole number:\n(a) |SR| ;\n(b) area of TQRS.\n\n\n', 'In the diagram, |PT| = 4 cm, |TS| = 6 cm, |PQ| = 6 cm and < SPR = 30°. Calculate, correct to the nearest whole number:\n(a) |SR| ;~(b) area of TQRS.\n\n\n', NULL, '\n\nExplanation\n|TQ|^{2} = 4^{2} + 6^{2} - 2 \\times 4 \\times 6 \\times \\cos 30°|TQ|^{2} = 4^{2} + 6^{2} - 2 \\times 4 \\times 6 \\times \\cos 30°\n= 16 + 36 - 48 \\times 0.866016 + 36 - 48 \\times 0.8660\n= 52 - 41.568 = 10.43252 - 41.568 = 10.432\n\\therefore |TQ| = \\sqrt{10.432} \\approxeq 3.23 cm\\therefore |TQ| = \\sqrt{10.432} \\approxeq 3.23 cm\n(a) By the rules of similar triangles,\n\\frac{|PT|}{|TQ|} = \\frac{|PS|}{|SR|}\\frac{|PT|}{|TQ|} = \\frac{|PS|}{|SR|}\n\\frac{4}{3.23} = \\frac{10}{|SR|}\\frac{4}{3.23} = \\frac{10}{|SR|}\n|SR| = \\frac{3.23 \\times 10}{4}|SR| = \\frac{3.23 \\times 10}{4}\n= 8.075 cm \\approxeq 8 cm8.075 cm \\approxeq 8 cm\n(b) Using sine rule,\n\\frac{|PQ|}{\\sin \\alpha} = \\frac{|TQ|}{\\sin 30}\\frac{|PQ|}{\\sin \\alpha} = \\frac{|TQ|}{\\sin 30}\n\\frac{6}{\\sin \\alpha} = \\frac{3.23}{\\sin 30}\\frac{6}{\\sin \\alpha} = \\frac{3.23}{\\sin 30}\n\\sin \\alpha = \\frac{6 \\sin 30}{3.23}\\sin \\alpha = \\frac{6 \\sin 30}{3.23}\nsin \\alpha\\alpha = 0.9288\nFrom the diagram, \\alpha\\alpha = \\beta\\beta\n\\sin \\alpha = \\sin \\beta = 0.9288\\sin \\alpha = \\sin \\beta = 0.9288\n\\sin \\beta = \\frac{h}{6}\\sin \\beta = \\frac{h}{6}\nh = 6 \\sin \\beta6 \\sin \\beta\nHence, area of quadrilateral\nTQRS = \\frac{1}{2} (TQ + SR) \\times h\\frac{1}{2} (TQ + SR) \\times h\n= \\frac{1}{2} (3.23 + 8.075) \\times 6\\sin \\beta\\frac{1}{2} (3.23 + 8.075) \\times 6\\sin \\beta\n= \\frac{1}{2} (11.305) (6 \\times 0.9288)\\frac{1}{2} (11.305) (6 \\times 0.9288)\n= 31.501 cm^2^2\n\\approxeq\\approxeq 32 cm^2^2 (to the nearest whole number)\n\n', '', 'Explanation\n|TQ|^{2} = 4^{2} + 6^{2} - 2 \\times 4 \\times 6 \\times \\cos 30°|TQ|^{2} = 4^{2} + 6^{2} - 2 \\times 4 \\times 6 \\times \\cos 30°\n= 16 + 36 - 48 \\times 0.866016 + 36 - 48 \\times 0.8660\n= 52 - 41.568 = 10.43252 - 41.568 = 10.432\n\\therefore |TQ| = \\sqrt{10.432} \\approxeq 3.23 cm\\therefore |TQ| = \\sqrt{10.432} \\approxeq 3.23 cm\n(a) By the rules of similar triangles,\n\\frac{|PT|}{|TQ|} = \\frac{|PS|}{|SR|}\\frac{|PT|}{|TQ|} = \\frac{|PS|}{|SR|}\n\\frac{4}{3.23} = \\frac{10}{|SR|}\\frac{4}{3.23} = \\frac{10}{|SR|}\n|SR| = \\frac{3.23 \\times 10}{4}|SR| = \\frac{3.23 \\times 10}{4}\n= 8.075 cm \\approxeq 8 cm8.075 cm \\approxeq 8 cm\n(b) Using sine rule,\n\\frac{|PQ|}{\\sin \\alpha} = \\frac{|TQ|}{\\sin 30}\\frac{|PQ|}{\\sin \\alpha} = \\frac{|TQ|}{\\sin 30}\n\\frac{6}{\\sin \\alpha} = \\frac{3.23}{\\sin 30}\\frac{6}{\\sin \\alpha} = \\frac{3.23}{\\sin 30}\n\\sin \\alpha = \\frac{6 \\sin 30}{3.23}\\sin \\alpha = \\frac{6 \\sin 30}{3.23}\nsin \\alpha\\alpha = 0.9288\nFrom the diagram, \\alpha\\alpha = \\beta\\beta\n\\sin \\alpha = \\sin \\beta = 0.9288\\sin \\alpha = \\sin \\beta = 0.9288\n\\sin \\beta = \\frac{h}{6}\\sin \\beta = \\frac{h}{6}\nh = 6 \\sin \\beta6 \\sin \\beta\nHence, area of quadrilateral\nTQRS = \\frac{1}{2} (TQ + SR) \\times h\\frac{1}{2} (TQ + SR) \\times h\n= \\frac{1}{2} (3.23 + 8.075) \\times 6\\sin \\beta\\frac{1}{2} (3.23 + 8.075) \\times 6\\sin \\beta\n= \\frac{1}{2} (11.305) (6 \\times 0.9288)\\frac{1}{2} (11.305) (6 \\times 0.9288)\n= 31.501 cm^2^2\n\\approxeq\\approxeq 32 cm^2^2 (to the nearest whole number)\n\n', '(a) By the rules of similar triangles,\n\\frac{|PT|}{|TQ|} = \\frac{|PS|}{|SR|}\\frac{|PT|}{|TQ|} = \\frac{|PS|}{|SR|}\n\\frac{4}{3.23} = \\frac{10}{|SR|}\\frac{4}{3.23} = \\frac{10}{|SR|}\n|SR| = \\frac{3.23 \\times 10}{4}|SR| = \\frac{3.23 \\times 10}{4}\n= 8.075 cm \\approxeq 8 cm8.075 cm \\approxeq 8 cm~(b) Using sine rule,\n\\frac{|PQ|}{\\sin \\alpha} = \\frac{|TQ|}{\\sin 30}\\frac{|PQ|}{\\sin \\alpha} = \\frac{|TQ|}{\\sin 30}\n\\frac{6}{\\sin \\alpha} = \\frac{3.23}{\\sin 30}\\frac{6}{\\sin \\alpha} = \\frac{3.23}{\\sin 30}\n\\sin \\alpha = \\frac{6 \\sin 30}{3.23}\\sin \\alpha = \\frac{6 \\sin 30}{3.23}\nsin \\alpha\\alpha = 0.9288\nFrom the diagram, \\alpha\\alpha = \\beta\\beta\n\\sin \\alpha = \\sin \\beta = 0.9288\\sin \\alpha = \\sin \\beta = 0.9288\n\\sin \\beta = \\frac{h}{6}\\sin \\beta = \\frac{h}{6}\nh = 6 \\sin \\beta6 \\sin \\beta\nHence, area of quadrilateral\nTQRS = \\frac{1}{2} (TQ + SR) \\times h\\frac{1}{2} (TQ + SR) \\times h\n= \\frac{1}{2} (3.23 + 8.075) \\times 6\\sin \\beta\\frac{1}{2} (3.23 + 8.075) \\times 6\\sin \\beta\n= \\frac{1}{2} (11.305) (6 \\times 0.9288)\\frac{1}{2} (11.305) (6 \\times 0.9288)\n= 31.501 cm^2^2\n\\approxeq\\approxeq 32 cm^2^2 (to the nearest whole number)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(326, '2013', '\n\nQuestion\n\n(a) In < PQS, |PQ| = 12 cm, |PS| = 5 cm, < SPQ = < PRQ = 90°, Find, correct to three significant figures, |PR|.\n(b) The length of two ladders, L and M are 10m and 12m respectively. They are placed against a wall such that each ladder makes angle with the horizontal ground. If the foot of L is 8m from the foot of the wall.\n(i) Draw a diagram to illustrate this information; (ii) Calculate the height at which M touches the wall.\n\n', '', '(a) In < PQS, |PQ| = 12 cm, |PS| = 5 cm, < SPQ = < PRQ = 90°, Find, correct to three significant figures, |PR|.\n(b) The length of two ladders, L and M are 10m and 12m respectively. They are placed against a wall such that each ladder makes angle with the horizontal ground. If the foot of L is 8m from the foot of the wall.\n(i) Draw a diagram to illustrate this information; (ii) Calculate the height at which M touches the wall.', '(a) In < PQS, |PQ| = 12 cm, |PS| = 5 cm, < SPQ = < PRQ = 90°, Find, correct to three significant figures, |PR|.~(b) The length of two ladders, L and M are 10m and 12m respectively. They are placed against a wall such that each ladder makes angle with the horizontal ground. If the foot of L is 8m from the foot of the wall.\n(i) Draw a diagram to illustrate this information; (ii) Calculate the height at which M touches the wall.', NULL, '\n\nExplanation\n(a)\nIn triangle SPQ, \\(|SQ|^{2} = 5^{2} + 12^{2}\\) (Pythagoras theorem)\n= \\(25 + 144 = 169\\)\n\\(|SQ| = \\sqrt{169} = 13 cm\\)\nAngle b is common to triangles SPQ and PRS are similar.\nUsing \\(\\sin b = \\frac{12}{|SQ|} = \\frac{|PR|}{5}\\)\n\\(\\sin b = \\frac{12}{13} = \\frac{|PR|}{5}\\)\n\\(|PR| = \\frac{12 \\times 5}{13} \\approxeq 4.62 cm\\) (to 3 s.f)\n(b)(i)\n\\(h^{2} = 10^{2} - 8^{2} = 36\\)\n\\(h = \\sqrt{36} = 6 cm\\)\n(ii) In the smaller triangle, \\(\\cos x = \\frac{8}{10} = 0.8\\)\n\\(\\cos^{-1} (0.8) = 36.87°\\)\nSince these are corresponding angles, x = x in the bigger triangle.\n\\(\\sin x = \\frac{y}{12}\\)\n\\(y = 12 \\sin x = 12 \\sin 36.87\\)\n= \\(12 \\times 0.6\\)\n= 7.20 m\nThe ladder M touches the wall at a height 7.2 m above the horizontal ground.\n\n', '', 'Explanation\n(a)\nIn triangle SPQ, \\(|SQ|^{2} = 5^{2} + 12^{2}\\) (Pythagoras theorem)\n= \\(25 + 144 = 169\\)\n\\(|SQ| = \\sqrt{169} = 13 cm\\)\nAngle b is common to triangles SPQ and PRS are similar.\nUsing \\(\\sin b = \\frac{12}{|SQ|} = \\frac{|PR|}{5}\\)\n\\(\\sin b = \\frac{12}{13} = \\frac{|PR|}{5}\\)\n\\(|PR| = \\frac{12 \\times 5}{13} \\approxeq 4.62 cm\\) (to 3 s.f)\n(b)(i)\n\\(h^{2} = 10^{2} - 8^{2} = 36\\)\n\\(h = \\sqrt{36} = 6 cm\\)\n(ii) In the smaller triangle, \\(\\cos x = \\frac{8}{10} = 0.8\\)\n\\(\\cos^{-1} (0.8) = 36.87°\\)\nSince these are corresponding angles, x = x in the bigger triangle.\n\\(\\sin x = \\frac{y}{12}\\)\n\\(y = 12 \\sin x = 12 \\sin 36.87\\)\n= \\(12 \\times 0.6\\)\n= 7.20 m\nThe ladder M touches the wall at a height 7.2 m above the horizontal ground.\n\n', '(a)\nIn triangle SPQ, \\(|SQ|^{2} = 5^{2} + 12^{2}\\) (Pythagoras theorem)\n= \\(25 + 144 = 169\\)\n\\(|SQ| = \\sqrt{169} = 13 cm\\)\nAngle b is common to triangles SPQ and PRS are similar.\nUsing \\(\\sin b = \\frac{12}{|SQ|} = \\frac{|PR|}{5}\\)\n\\(\\sin b = \\frac{12}{13} = \\frac{|PR|}{5}\\)\n\\(|PR| = \\frac{12 \\times 5}{13} \\approxeq 4.62 cm\\) (to 3 s.f)~(b)(i)\n\\(h^{2} = 10^{2} - 8^{2} = 36\\)\n\\(h = \\sqrt{36} = 6 cm\\)\n(ii) In the smaller triangle, \\(\\cos x = \\frac{8}{10} = 0.8\\)\n\\(\\cos^{-1} (0.8) = 36.87°\\)\nSince these are corresponding angles, x = x in the bigger triangle.\n\\(\\sin x = \\frac{y}{12}\\)\n\\(y = 12 \\sin x = 12 \\sin 36.87\\)\n= \\(12 \\times 0.6\\)\n= 7.20 m\nThe ladder M touches the wall at a height 7.2 m above the horizontal ground.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(327, '2013', '\n\nQuestion\n\n(a) Copy and complete the table of values for y = 2x\\(^{2}\\) + x - 10for -5 \\(\\leq\\) x \\(\\leq\\)4.\n\n\n\nx\n-5\n-4\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\ny\n\n\n5\n\n-9\n-10\n\n0\n\n\n\n\n\n(b) Using scales of 2cm to 1 unit on the x- axis and 2cm to 5 units on the y- axis, Draw the graph of y = 2x\\(^{2}\\)+ x - 10for -5 \\(\\leq\\)x \\(\\leq\\) 4.\n(c) Use the graph to find the solution of :\n(i) 2x\\(^{2}\\) + x = 10\n(ii) 2x\\(^{2}\\) + x - 10 = 2x\n\n\n(a) Copy and complete the table of values for y = 2x\\(^{2}\\) + x - 10for -5 \\(\\leq\\) x \\(\\leq\\)4.\n\n\n\nx\n-5\n-4\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\ny\n\n\n5\n\n-9\n-10\n\n0\n\n\n\n\n', '', '(a) Copy and complete the table of values for y = 2x\\(^{2}\\) + x - 10for -5 \\(\\leq\\) x \\(\\leq\\)4.\n\n\n\nx\n-5\n-4\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\ny\n\n\n5\n\n-9\n-10\n\n0\n\n\n\n\n\n(b) Using scales of 2cm to 1 unit on the x- axis and 2cm to 5 units on the y- axis, Draw the graph of y = 2x\\(^{2}\\)+ x - 10for -5 \\(\\leq\\)x \\(\\leq\\) 4.\n(c) Use the graph to find the solution of :\n(i) 2x\\(^{2}\\) + x = 10\n(ii) 2x\\(^{2}\\) + x - 10 = 2x\n\n\n(a) Copy and complete the table of values for y = 2x\\(^{2}\\) + x - 10for -5 \\(\\leq\\) x \\(\\leq\\)4.\n\n\n\nx\n-5\n-4\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\ny\n\n\n5\n\n-9\n-10\n\n0\n\n\n', '(a) Copy and complete the table of values for y = 2x\\(^{2}\\) + x - 10for -5 \\(\\leq\\) x \\(\\leq\\)4.\n\n\n\nx\n-5\n-4\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\ny\n\n\n5\n\n-9\n-10\n\n0\n\n\n\n\n~(b) Using scales of 2cm to 1 unit on the x- axis and 2cm to 5 units on the y- axis, Draw the graph of y = 2x\\(^{2}\\)+ x - 10for -5 \\(\\leq\\)x \\(\\leq\\) 4.\n(c) Use the graph to find the solution of :\n(i) 2x\\(^{2}\\) + x = 10\n(ii) 2x\\(^{2}\\) + x - 10 = 2x\n\n\n(a) Copy and complete the table of values for y = 2x\\(^{2}\\) + x - 10for -5 \\(\\leq\\) x \\(\\leq\\)4.\n\n\n\nx\n-5\n-4\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\ny\n\n\n5\n\n-9\n-10\n\n0\n\n\n~(c) Use the graph to find the solution of :\n(i) 2x\\(^{2}\\) + x = 10\n(ii) 2x\\(^{2}\\) + x - 10 = 2x\n\n\n(a) Copy and complete the table of values for y = 2x\\(^{2}\\) + x - 10for -5 \\(\\leq\\) x \\(\\leq\\)4.\n\n\n\nx\n-5\n-4\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\ny\n\n\n5\n\n-9\n-10\n\n0\n\n\n', NULL, '\n\nExplanation\n(a)\n\n\n\nx\n-5\n-4\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\ny\n35\n18\n5\n-4\n-9\n-10\n-7\n0\n11\n26\n\n\n\n\n(b)\n(c)(i) 2x\\(^{2}\\)+ x = 10 \\(\\equiv\\)2x\\(^{2}\\)+ x - 10 = 0\nThus, the solution of 2x\\(^{2}\\) + x - 10 = 0are the values of x at which the curve cuts the x- axis.\n  x = -2.5 or x = 2.\n(ii) 2x\\(^{2}\\)+ x - 10 = 2x\nTo solve the given equation, we first draw the graph of y = 2x: when x = -2, y = 2(-2) = -4; when x = 3, y = 2(3) = 6.\nThe values of x at which y = 2x\\(^{2}\\) + x - 10and y = 2xintersect give the solution of 2x\\(^{2}\\)+ x - 10 = 2x.\nThese are x = -2 or x = 2.5.\n\n', '', 'Explanation\n(a)\n\n\n\nx\n-5\n-4\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\ny\n35\n18\n5\n-4\n-9\n-10\n-7\n0\n11\n26\n\n\n\n\n(b)\n(c)(i) 2x\\(^{2}\\)+ x = 10 \\(\\equiv\\)2x\\(^{2}\\)+ x - 10 = 0\nThus, the solution of 2x\\(^{2}\\) + x - 10 = 0are the values of x at which the curve cuts the x- axis.\n  x = -2.5 or x = 2.\n(ii) 2x\\(^{2}\\)+ x - 10 = 2x\nTo solve the given equation, we first draw the graph of y = 2x: when x = -2, y = 2(-2) = -4; when x = 3, y = 2(3) = 6.\nThe values of x at which y = 2x\\(^{2}\\) + x - 10and y = 2xintersect give the solution of 2x\\(^{2}\\)+ x - 10 = 2x.\nThese are x = -2 or x = 2.5.\n\n', '(a)\n\n\n\nx\n-5\n-4\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\ny\n35\n18\n5\n-4\n-9\n-10\n-7\n0\n11\n26\n\n\n\n~(b)\n(c)(i) 2x\\(^{2}\\)+ x = 10 \\(\\equiv\\)2x\\(^{2}\\)+ x - 10 = 0\nThus, the solution of 2x\\(^{2}\\) + x - 10 = 0are the values of x at which the curve cuts the x- axis.\n  x = -2.5 or x = 2.\n(ii) 2x\\(^{2}\\)+ x - 10 = 2x\nTo solve the given equation, we first draw the graph of y = 2x: when x = -2, y = 2(-2) = -4; when x = 3, y = 2(3) = 6.\nThe values of x at which y = 2x\\(^{2}\\) + x - 10and y = 2xintersect give the solution of 2x\\(^{2}\\)+ x - 10 = 2x.\nThese are x = -2 or x = 2.5.\n~(c)(i) 2x\\(^{2}\\)+ x = 10 \\(\\equiv\\)2x\\(^{2}\\)+ x - 10 = 0\nThus, the solution of 2x\\(^{2}\\) + x - 10 = 0are the values of x at which the curve cuts the x- axis.\n  x = -2.5 or x = 2.\n(ii) 2x\\(^{2}\\)+ x - 10 = 2x\nTo solve the given equation, we first draw the graph of y = 2x: when x = -2, y = 2(-2) = -4; when x = 3, y = 2(3) = 6.\nThe values of x at which y = 2x\\(^{2}\\) + x - 10and y = 2xintersect give the solution of 2x\\(^{2}\\)+ x - 10 = 2x.\nThese are x = -2 or x = 2.5.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(328, '2013', '\n\nQuestion\n\n(a) If \\(x = \\begin{pmatrix} 2 \\\\ 3 \\end{pmatrix}, y = \\begin{pmatrix} 5 \\\\ -2 \\end{pmatrix}\\) and \\(z = \\begin{pmatrix} -4 \\\\ 13 \\end{pmatrix}\\), find the scalars p and q such that \\(px + qy = z\\).\n(b)(i) Using the scale of 2cm to 2 units on both axis, draw on a graph paper two perpendicular axis x and y for \\(-5 \\leq x \\leq 5, -5 \\leq y \\leq 5\\) respectively.\n(ii) Draw, on the graph paper, indicating clearly the vertices and their coordinates,\n(1) the quadrilateral WXYZ with W(2, 3), X(4, -1), Y(-3, -4) and Z(-3, 2).\n(2) the image \\(W_{1}X_{1}Y_{1}Z_{1}\\) of the quadrilateral WXYZ under an anti-clockwise rotation of 90° about the origin where \\(W \\to W_{1}, X \\to X_{1}, Y \\to Y_{1}\\) and \\(Z \\to Z_{1}\\).\n\n', '', '(a) If \\(x = \\begin{pmatrix} 2 \\\\ 3 \\end{pmatrix}, y = \\begin{pmatrix} 5 \\\\ -2 \\end{pmatrix}\\) and \\(z = \\begin{pmatrix} -4 \\\\ 13 \\end{pmatrix}\\), find the scalars p and q such that \\(px + qy = z\\).\n(b)(i) Using the scale of 2cm to 2 units on both axis, draw on a graph paper two perpendicular axis x and y for \\(-5 \\leq x \\leq 5, -5 \\leq y \\leq 5\\) respectively.\n(ii) Draw, on the graph paper, indicating clearly the vertices and their coordinates,\n(1) the quadrilateral WXYZ with W(2, 3), X(4, -1), Y(-3, -4) and Z(-3, 2).\n(2) the image \\(W_{1}X_{1}Y_{1}Z_{1}\\) of the quadrilateral WXYZ under an anti-clockwise rotation of 90° about the origin where \\(W \\to W_{1}, X \\to X_{1}, Y \\to Y_{1}\\) and \\(Z \\to Z_{1}\\).', '(a) If \\(x = \\begin{pmatrix} 2 \\\\ 3 \\end{pmatrix}, y = \\begin{pmatrix} 5 \\\\ -2 \\end{pmatrix}\\) and \\(z = \\begin{pmatrix} -4 \\\\ 13 \\end{pmatrix}\\), find the scalars p and q such that \\(px + qy = z\\).~(b)(i) Using the scale of 2cm to 2 units on both axis, draw on a graph paper two perpendicular axis x and y for \\(-5 \\leq x \\leq 5, -5 \\leq y \\leq 5\\) respectively.\n(ii) Draw, on the graph paper, indicating clearly the vertices and their coordinates,\n(1) the quadrilateral WXYZ with W(2, 3), X(4, -1), Y(-3, -4) and Z(-3, 2).\n(2) the image \\(W_{1}X_{1}Y_{1}Z_{1}\\) of the quadrilateral WXYZ under an anti-clockwise rotation of 90° about the origin where \\(W \\to W_{1}, X \\to X_{1}, Y \\to Y_{1}\\) and \\(Z \\to Z_{1}\\).', NULL, '\n\nExplanation\n(a) px + qy = z\n\\(p \\begin{pmatrix} 2 \\\\ 3 \\end{pmatrix} + q \\begin{pmatrix} 5 \\\\ -2 \\end{pmatrix} = \\begin{pmatrix} -4 \\\\ 13 \\end{pmatrix}\\)\n\\(\\begin{pmatrix} 2p + 5q \\\\ 3p - 2q \\end{pmatrix} = \\begin{pmatrix} -4 \\\\ 13 \\end{pmatrix}\\)\nThus,\n\\(2p + 5q = -4 .....(1)\\)\n\\(3p - 2q = 13 ......(2)\\)\nSolving the equations, we have p = 3 and q = -2.\n(b)(i)\n(ii) See attached graph for the quadrilateral W(2, 3), X(4, -1), Y(-3, -4) and Z(-3,2).\n(iii) See attached for the image for the quadrilateral WXYZ under an anticlockwise rotation of 90° is \\(W_{1}(-3, 2), X_{1}(1, 4), Y_{1}(4, -3)\\) and \\(Z_{1}(-2, -3)\\).\n\n', '', 'Explanation\n(a) px + qy = z\n\\(p \\begin{pmatrix} 2 \\\\ 3 \\end{pmatrix} + q \\begin{pmatrix} 5 \\\\ -2 \\end{pmatrix} = \\begin{pmatrix} -4 \\\\ 13 \\end{pmatrix}\\)\n\\(\\begin{pmatrix} 2p + 5q \\\\ 3p - 2q \\end{pmatrix} = \\begin{pmatrix} -4 \\\\ 13 \\end{pmatrix}\\)\nThus,\n\\(2p + 5q = -4 .....(1)\\)\n\\(3p - 2q = 13 ......(2)\\)\nSolving the equations, we have p = 3 and q = -2.\n(b)(i)\n(ii) See attached graph for the quadrilateral W(2, 3), X(4, -1), Y(-3, -4) and Z(-3,2).\n(iii) See attached for the image for the quadrilateral WXYZ under an anticlockwise rotation of 90° is \\(W_{1}(-3, 2), X_{1}(1, 4), Y_{1}(4, -3)\\) and \\(Z_{1}(-2, -3)\\).\n\n', '(a) px + qy = z\n\\(p \\begin{pmatrix} 2 \\\\ 3 \\end{pmatrix} + q \\begin{pmatrix} 5 \\\\ -2 \\end{pmatrix} = \\begin{pmatrix} -4 \\\\ 13 \\end{pmatrix}\\)\n\\(\\begin{pmatrix} 2p + 5q \\\\ 3p - 2q \\end{pmatrix} = \\begin{pmatrix} -4 \\\\ 13 \\end{pmatrix}\\)\nThus,\n\\(2p + 5q = -4 .....(1)\\)\n\\(3p - 2q = 13 ......(2)\\)\nSolving the equations, we have p = 3 and q = -2.~(b)(i)\n(ii) See attached graph for the quadrilateral W(2, 3), X(4, -1), Y(-3, -4) and Z(-3,2).\n(iii) See attached for the image for the quadrilateral WXYZ under an anticlockwise rotation of 90° is \\(W_{1}(-3, 2), X_{1}(1, 4), Y_{1}(4, -3)\\) and \\(Z_{1}(-2, -3)\\).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(329, '2013', '\n\nQuestion\n\n\n\nMarks\n10\n20\n30\n40\n50\n60\n70\n80\n90\n\n\nFrequency\n1\n1\nx\n5\ny\n1\n4\n3\n1\n\n\n\nThe frequency distribution shows the marks distribution of a class of 30 students in an examination.\nThe mean mark of the distribution is 52.\n(a) Find the values of x and y.\n(b) Construct a group frequency distribution table starting with a lower class limit of 1 and class interval of 10.\n(c) Draw a histogram for the distribution\n(d) Use the histogram to estimate the mode.\n\n\n\n\n\nMarks\n10\n20\n30\n40\n50\n60\n70\n80\n90\n\n\nFrequency\n1\n1\nx\n5\ny\n1\n4\n3\n1\n\n\n', '', 'Marks\n10\n20\n30\n40\n50\n60\n70\n80\n90\n\n\nFrequency\n1\n1\nx\n5\ny\n1\n4\n3\n1\n\n\n\nThe frequency distribution shows the marks distribution of a class of 30 students in an examination.\nThe mean mark of the distribution is 52.\n(a) Find the values of x and y.\n(b) Construct a group frequency distribution table starting with a lower class limit of 1 and class interval of 10.\n(c) Draw a histogram for the distribution\n(d) Use the histogram to estimate the mode.\n\n\n\n\n\nMarks\n10\n20\n30\n40\n50\n60\n70\n80\n90\n\n\nFrequency\n1\n1\nx\n5\ny\n1\n4\n3\n1\n', 'Marks\n10\n20\n30\n40\n50\n60\n70\n80\n90\n\n\nFrequency\n1\n1\nx\n5\ny\n1\n4\n3\n1\n\n\n\nThe frequency distribution shows the marks distribution of a class of 30 students in an examination.\nThe mean mark of the distribution is 52.\n(a) Find the values of x and y.~(b) Construct a group frequency distribution table starting with a lower class limit of 1 and class interval of 10.\n(c) Draw a histogram for the distribution\n(d) Use the histogram to estimate the mode.\n\n\n\n\n\nMarks\n10\n20\n30\n40\n50\n60\n70\n80\n90\n\n\nFrequency\n1\n1\nx\n5\ny\n1\n4\n3\n1\n~(c) Draw a histogram for the distribution\n(d) Use the histogram to estimate the mode.\n\n\n\n\n\nMarks\n10\n20\n30\n40\n50\n60\n70\n80\n90\n\n\nFrequency\n1\n1\nx\n5\ny\n1\n4\n3\n1\n~(d) Use the histogram to estimate the mode.\n\n\n\n\n\nMarks\n10\n20\n30\n40\n50\n60\n70\n80\n90\n\n\nFrequency\n1\n1\nx\n5\ny\n1\n4\n3\n1\n', NULL, '\n\nExplanation\n(a)\n\n\n\nMarks(x)\nFrequency (f)\nfx\n\n\n10\n1\n10\n\n\n20\n1\n20\n\n\n30\nx\n30x\n\n\n40\n5\n200\n\n\n50\ny\n50y\n\n\n60\n1\n60\n\n\n70\n4\n280\n\n\n80\n3\n240\n\n\n90\n1\n90\n\n\nTotal\n16 + x + y\n900 + 30x + 50y\n\n\n\n∑f=16+x+y=30\\sum f = 16 + x + y = 30\n⟹x+y=14...(1)\\implies x + y = 14 ... (1)\nˉx=∑fx∑f\\bar{x} = \\frac{\\sum fx}{\\sum f}\n52=900+30x+50y3052 = \\frac{900 + 30x + 50y}{30}\n1560=900+30x+50y⟹30x+50y=6601560 = 900 + 30x + 50y \\implies 30x + 50y = 660\n3x+5y=66....(2)3x + 5y = 66 .... (2)\nSolving equation (1) and (2),\nFrom (1), x = 14 - y\n3(14−y)+5y=42−3y+5y=663(14 - y) + 5y = 42 - 3y + 5y = 66\n2y=24⟹y=122y = 24 \\implies y = 12\nx=14−y=14−12=2x = 14 - y = 14 - 12 = 2\n(x, y) = (2, 12).\n(b)\n\n\n\nClass interval\nFrequency\nUpper class boundary\n\n\n1 - 10\n1\n10.5\n\n\n11 - 20\n1\n20.5\n\n\n21 - 30\n2\n30.5\n\n\n31 - 40\n5\n40.5\n\n\n41 - 50\n12\n50.5\n\n\n51 - 60\n1\n60.5\n\n\n61 - 70\n4\n70.5\n\n\n71 - 80\n3\n80.5\n\n\n81 - 90\n1\n90.5\n\n\n\n(c)\n(d) From the histogram, Mode = 44.\n\n', '', 'Explanation\n(a)\n\n\n\nMarks(x)\nFrequency (f)\nfx\n\n\n10\n1\n10\n\n\n20\n1\n20\n\n\n30\nx\n30x\n\n\n40\n5\n200\n\n\n50\ny\n50y\n\n\n60\n1\n60\n\n\n70\n4\n280\n\n\n80\n3\n240\n\n\n90\n1\n90\n\n\nTotal\n16 + x + y\n900 + 30x + 50y\n\n\n\n∑f=16+x+y=30\\sum f = 16 + x + y = 30\n⟹x+y=14...(1)\\implies x + y = 14 ... (1)\nˉx=∑fx∑f\\bar{x} = \\frac{\\sum fx}{\\sum f}\n52=900+30x+50y3052 = \\frac{900 + 30x + 50y}{30}\n1560=900+30x+50y⟹30x+50y=6601560 = 900 + 30x + 50y \\implies 30x + 50y = 660\n3x+5y=66....(2)3x + 5y = 66 .... (2)\nSolving equation (1) and (2),\nFrom (1), x = 14 - y\n3(14−y)+5y=42−3y+5y=663(14 - y) + 5y = 42 - 3y + 5y = 66\n2y=24⟹y=122y = 24 \\implies y = 12\nx=14−y=14−12=2x = 14 - y = 14 - 12 = 2\n(x, y) = (2, 12).\n(b)\n\n\n\nClass interval\nFrequency\nUpper class boundary\n\n\n1 - 10\n1\n10.5\n\n\n11 - 20\n1\n20.5\n\n\n21 - 30\n2\n30.5\n\n\n31 - 40\n5\n40.5\n\n\n41 - 50\n12\n50.5\n\n\n51 - 60\n1\n60.5\n\n\n61 - 70\n4\n70.5\n\n\n71 - 80\n3\n80.5\n\n\n81 - 90\n1\n90.5\n\n\n\n(c)\n(d) From the histogram, Mode = 44.\n\n', '(a)\n\n\n\nMarks(x)\nFrequency (f)\nfx\n\n\n10\n1\n10\n\n\n20\n1\n20\n\n\n30\nx\n30x\n\n\n40\n5\n200\n\n\n50\ny\n50y\n\n\n60\n1\n60\n\n\n70\n4\n280\n\n\n80\n3\n240\n\n\n90\n1\n90\n\n\nTotal\n16 + x + y\n900 + 30x + 50y\n\n\n\n∑f=16+x+y=30\\sum f = 16 + x + y = 30\n⟹x+y=14...(1)\\implies x + y = 14 ... (1)\nˉx=∑fx∑f\\bar{x} = \\frac{\\sum fx}{\\sum f}\n52=900+30x+50y3052 = \\frac{900 + 30x + 50y}{30}\n1560=900+30x+50y⟹30x+50y=6601560 = 900 + 30x + 50y \\implies 30x + 50y = 660\n3x+5y=66....(2)3x + 5y = 66 .... (2)\nSolving equation (1) and (2),\nFrom (1), x = 14 - y\n3(14−y)+5y=42−3y+5y=663(14 - y) + 5y = 42 - 3y + 5y = 66\n2y=24⟹y=122y = 24 \\implies y = 12\nx=14−y=14−12=2x = 14 - y = 14 - 12 = 2\n(x, y) = (2, 12).~(b)\n\n\n\nClass interval\nFrequency\nUpper class boundary\n\n\n1 - 10\n1\n10.5\n\n\n11 - 20\n1\n20.5\n\n\n21 - 30\n2\n30.5\n\n\n31 - 40\n5\n40.5\n\n\n41 - 50\n12\n50.5\n\n\n51 - 60\n1\n60.5\n\n\n61 - 70\n4\n70.5\n\n\n71 - 80\n3\n80.5\n\n\n81 - 90\n1\n90.5\n\n\n\n(c)\n(d) From the histogram, Mode = 44.\n~(c)\n(d) From the histogram, Mode = 44.\n~(d) From the histogram, Mode = 44.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(330, '2013', '\n\nQuestion\n\n(a) If \\((y - 1)\\log_{10}4 = y\\log_{10}16\\), without using Mathematics tables or calculator, find the value of y.\n(b) When I walk from my house at 4km/h, I will get to my office 30mins later than when I walk at 5km/h. Calculate the distance between my house and office.\n\n', '', '(a) If \\((y - 1)\\log_{10}4 = y\\log_{10}16\\), without using Mathematics tables or calculator, find the value of y.\n(b) When I walk from my house at 4km/h, I will get to my office 30mins later than when I walk at 5km/h. Calculate the distance between my house and office.', '(a) If \\((y - 1)\\log_{10}4 = y\\log_{10}16\\), without using Mathematics tables or calculator, find the value of y.~(b) When I walk from my house at 4km/h, I will get to my office 30mins later than when I walk at 5km/h. Calculate the distance between my house and office.', NULL, '\n\nExplanation\n(a) \\((y - 1)\\log_{10} 4 = y\\log_{10} 16\\)\n\\((y - 1)\\log_{10} 4 = y \\log_{10} 4^{2}\\)\n\\((y - 1)\\log_{10} 4 = 2y\\log_{10} 4\\)\nEquating both sides, we have\n\\(y - 1 = 2y \\implies -1 = 2y - y\\)\n\\(\\therefore y = -1\\)\n(b) Let the distance from my house to the office = c.\nAt 4km/h, the time taken to get to the office from the house = \\(\\frac{c}{4} hr\\)\nAt 5km/h, the time taken to get to the office from the house = \\(\\frac{c}{5} hr\\)\n\\(\\frac{c}{4} = \\frac{c}{5} + \\frac{30}{60}\\)\n\\(\\frac{c}{4} - \\frac{c}{5} = \\frac{1}{2}\\)\n\\(\\frac{c}{20} = \\frac{1}{2} \\implies c = 10km\\)\n\n', '', 'Explanation\n(a) \\((y - 1)\\log_{10} 4 = y\\log_{10} 16\\)\n\\((y - 1)\\log_{10} 4 = y \\log_{10} 4^{2}\\)\n\\((y - 1)\\log_{10} 4 = 2y\\log_{10} 4\\)\nEquating both sides, we have\n\\(y - 1 = 2y \\implies -1 = 2y - y\\)\n\\(\\therefore y = -1\\)\n(b) Let the distance from my house to the office = c.\nAt 4km/h, the time taken to get to the office from the house = \\(\\frac{c}{4} hr\\)\nAt 5km/h, the time taken to get to the office from the house = \\(\\frac{c}{5} hr\\)\n\\(\\frac{c}{4} = \\frac{c}{5} + \\frac{30}{60}\\)\n\\(\\frac{c}{4} - \\frac{c}{5} = \\frac{1}{2}\\)\n\\(\\frac{c}{20} = \\frac{1}{2} \\implies c = 10km\\)\n\n', '(a) \\((y - 1)\\log_{10} 4 = y\\log_{10} 16\\)\n\\((y - 1)\\log_{10} 4 = y \\log_{10} 4^{2}\\)\n\\((y - 1)\\log_{10} 4 = 2y\\log_{10} 4\\)\nEquating both sides, we have\n\\(y - 1 = 2y \\implies -1 = 2y - y\\)\n\\(\\therefore y = -1\\)~(b) Let the distance from my house to the office = c.\nAt 4km/h, the time taken to get to the office from the house = \\(\\frac{c}{4} hr\\)\nAt 5km/h, the time taken to get to the office from the house = \\(\\frac{c}{5} hr\\)\n\\(\\frac{c}{4} = \\frac{c}{5} + \\frac{30}{60}\\)\n\\(\\frac{c}{4} - \\frac{c}{5} = \\frac{1}{2}\\)\n\\(\\frac{c}{20} = \\frac{1}{2} \\implies c = 10km\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(331, '2013', '\n\nQuestion\n\n(a) Solve the equation : \\(\\frac{2}{3}(3x - 5) - \\frac{3}{5}(2x - 3) = 3\\)\n(b)\nIn the diagram, < STQ = m, < TUQ = 80°, < UPQ = r, < PQU = n and < RQT = 88°. Find the value of (m + n).\n\n', '', '(a) Solve the equation : \\(\\frac{2}{3}(3x - 5) - \\frac{3}{5}(2x - 3) = 3\\)\n(b)\nIn the diagram, < STQ = m, < TUQ = 80°, < UPQ = r, < PQU = n and < RQT = 88°. Find the value of (m + n).', '(a) Solve the equation : \\(\\frac{2}{3}(3x - 5) - \\frac{3}{5}(2x - 3) = 3\\)~(b)\nIn the diagram, < STQ = m, < TUQ = 80°, < UPQ = r, < PQU = n and < RQT = 88°. Find the value of (m + n).', NULL, '\n\nExplanation\n(a) \\(\\frac{2}{3}(3x - 5) - \\frac{3}{5}(2x - 3) = 3\\)\n\\(\\frac{10(3x - 5) - 9(2x - 3)}{15} = 3\\)\n\\(30x - 50 - 18x + 27 = 3 \\times 15 = 45\\)\n\\(12x = 45 + 23 = 68\\)\n\\(x = \\frac{68}{12} = 5\\frac{2}{3}\\)\n(b) \\(n + r = 80° .... (1)\\)\n\\(180° - m + r = 88° .... (2)\\)\nFrom (1), r = 80° - n.\nPutting into (2), we have\n\\(180 - m + 80 - n = 88 \\implies - m - n = 88 - 260 = - 172°\\)\n\\(-(m + n) = - 172 \\implies m + n = 172°\\).\n\n', '', 'Explanation\n(a) \\(\\frac{2}{3}(3x - 5) - \\frac{3}{5}(2x - 3) = 3\\)\n\\(\\frac{10(3x - 5) - 9(2x - 3)}{15} = 3\\)\n\\(30x - 50 - 18x + 27 = 3 \\times 15 = 45\\)\n\\(12x = 45 + 23 = 68\\)\n\\(x = \\frac{68}{12} = 5\\frac{2}{3}\\)\n(b) \\(n + r = 80° .... (1)\\)\n\\(180° - m + r = 88° .... (2)\\)\nFrom (1), r = 80° - n.\nPutting into (2), we have\n\\(180 - m + 80 - n = 88 \\implies - m - n = 88 - 260 = - 172°\\)\n\\(-(m + n) = - 172 \\implies m + n = 172°\\).\n\n', '(a) \\(\\frac{2}{3}(3x - 5) - \\frac{3}{5}(2x - 3) = 3\\)\n\\(\\frac{10(3x - 5) - 9(2x - 3)}{15} = 3\\)\n\\(30x - 50 - 18x + 27 = 3 \\times 15 = 45\\)\n\\(12x = 45 + 23 = 68\\)\n\\(x = \\frac{68}{12} = 5\\frac{2}{3}\\)~(b) \\(n + r = 80° .... (1)\\)\n\\(180° - m + r = 88° .... (2)\\)\nFrom (1), r = 80° - n.\nPutting into (2), we have\n\\(180 - m + 80 - n = 88 \\implies - m - n = 88 - 260 = - 172°\\)\n\\(-(m + n) = - 172 \\implies m + n = 172°\\).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(332, '2013', '\n\nQuestion\n\n(a) The angle of depression of a point P on the ground from the top T of a building is 23.6°. If the distance from P to the foot of the building is 50m, calculate, correct to the nearest metre, the height if the building.\n(b)\nIn the diagram, \\(PT // SU, QS // TR, /SR/ = 6cm\\)and \\(/RU/ = 10 cm\\). If the area of \\(\\Delta TRU = 45 cm^{2}\\), calculate the area of the trapezium QTUS.\n\n', '', '(a) The angle of depression of a point P on the ground from the top T of a building is 23.6°. If the distance from P to the foot of the building is 50m, calculate, correct to the nearest metre, the height if the building.\n(b)\nIn the diagram, \\(PT // SU, QS // TR, /SR/ = 6cm\\)and \\(/RU/ = 10 cm\\). If the area of \\(\\Delta TRU = 45 cm^{2}\\), calculate the area of the trapezium QTUS.', '(a) The angle of depression of a point P on the ground from the top T of a building is 23.6°. If the distance from P to the foot of the building is 50m, calculate, correct to the nearest metre, the height if the building.~(b)\nIn the diagram, \\(PT // SU, QS // TR, /SR/ = 6cm\\)and \\(/RU/ = 10 cm\\). If the area of \\(\\Delta TRU = 45 cm^{2}\\), calculate the area of the trapezium QTUS.', NULL, '\n\nExplanation\n(a)\n\\(< P = 23.6°\\) (Alternate angle)\n\\(\\tan 23.6 = \\frac{x}{50}\\)\n\\(x = 50\\tan 23.6\\)\n= \\(50 \\times 0.4369\\)\n= \\(21.845 m \\approxeq 22 m\\) (to the nearest metre)\n(b) Given that \\(\\Delta TRU = 45cm^{2}\\)\n\\(\\implies \\frac{1}{2}(bh) = 45\\)\n\\(10h= 45 \\times 2 = 90\\)\n\\(h= 9cm\\)\nArea of trapezium = \\(\\frac{1}{2}(a + b)h\\), where \\(a = 6cm; b = 10 + 6 = 16cm; h = 9cm\\)\n= \\(\\frac{1}{2}(6 + 16) \\times 9\\)\n= \\(11 \\times 9\\)\n= \\(99 cm^{2}\\)\n\n', '', 'Explanation\n(a)\n\\(< P = 23.6°\\) (Alternate angle)\n\\(\\tan 23.6 = \\frac{x}{50}\\)\n\\(x = 50\\tan 23.6\\)\n= \\(50 \\times 0.4369\\)\n= \\(21.845 m \\approxeq 22 m\\) (to the nearest metre)\n(b) Given that \\(\\Delta TRU = 45cm^{2}\\)\n\\(\\implies \\frac{1}{2}(bh) = 45\\)\n\\(10h= 45 \\times 2 = 90\\)\n\\(h= 9cm\\)\nArea of trapezium = \\(\\frac{1}{2}(a + b)h\\), where \\(a = 6cm; b = 10 + 6 = 16cm; h = 9cm\\)\n= \\(\\frac{1}{2}(6 + 16) \\times 9\\)\n= \\(11 \\times 9\\)\n= \\(99 cm^{2}\\)\n\n', '(a)\n\\(< P = 23.6°\\) (Alternate angle)\n\\(\\tan 23.6 = \\frac{x}{50}\\)\n\\(x = 50\\tan 23.6\\)\n= \\(50 \\times 0.4369\\)\n= \\(21.845 m \\approxeq 22 m\\) (to the nearest metre)~(b) Given that \\(\\Delta TRU = 45cm^{2}\\)\n\\(\\implies \\frac{1}{2}(bh) = 45\\)\n\\(10h= 45 \\times 2 = 90\\)\n\\(h= 9cm\\)\nArea of trapezium = \\(\\frac{1}{2}(a + b)h\\), where \\(a = 6cm; b = 10 + 6 = 16cm; h = 9cm\\)\n= \\(\\frac{1}{2}(6 + 16) \\times 9\\)\n= \\(11 \\times 9\\)\n= \\(99 cm^{2}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(333, '2013', '\n\nQuestion\n\nIf the sixth term of an Arithmetic Progression (A.P) is 37 and the sum of the first six terms is 147, find the\n(a) first term;\n(b) sum of the first fifteen terms.\n\n', '', 'If the sixth term of an Arithmetic Progression (A.P) is 37 and the sum of the first six terms is 147, find the\n(a) first term;\n(b) sum of the first fifteen terms.', 'If the sixth term of an Arithmetic Progression (A.P) is 37 and the sum of the first six terms is 147, find the\n(a) first term;~(b) sum of the first fifteen terms.', NULL, '\n\nExplanation\n(a) \\(T_{n} = a + (n - 1)d\\) (For an AP series)\n\\(T_{6} = 37 = a + (6 - 1) d = a + 5d\\)\n\\(a + 5d = 37 .... (1)\\)\n\\(S_{n} = \\frac{n}{2}(2a + (n - 1) d)\\)\nor\n\\(S_{n} = \\frac{n}{2}(a + l)\\) where a = first term; and l = last term.\nSince, we are given the sum of the first 6 terms and also given the sixth term which is the last term in the sum, we can use the second formula where the sixth term is the last term in this case.\n\\(S_{6} = 147 = \\frac{6}{2} (a + 37)\\)\n\\(147 = 3(a + 37) \\implies 147 = 3a + 111 \\)\n\\(3a = 147 - 111 = 36 \\implies a = 12\\)\n(b) From equation (1) above,\n\\(a + 5d = 37\\)\n\\(12 + 5d = 37\\)\n\\(5d = 37 - 12 = 25\\)\n\\(d = 5\\)\n\\(S_{15} = \\frac{15}{2} (2(12) + (15 - 1)(5))\\)\n\\(\\frac{15}{2} (24 + 70) = \\frac{15}{2} (94)\\)\n= \\(15 \\times 47\\)\n= \\(705\\).\n\n', '', 'Explanation\n(a) \\(T_{n} = a + (n - 1)d\\) (For an AP series)\n\\(T_{6} = 37 = a + (6 - 1) d = a + 5d\\)\n\\(a + 5d = 37 .... (1)\\)\n\\(S_{n} = \\frac{n}{2}(2a + (n - 1) d)\\)\nor\n\\(S_{n} = \\frac{n}{2}(a + l)\\) where a = first term; and l = last term.\nSince, we are given the sum of the first 6 terms and also given the sixth term which is the last term in the sum, we can use the second formula where the sixth term is the last term in this case.\n\\(S_{6} = 147 = \\frac{6}{2} (a + 37)\\)\n\\(147 = 3(a + 37) \\implies 147 = 3a + 111 \\)\n\\(3a = 147 - 111 = 36 \\implies a = 12\\)\n(b) From equation (1) above,\n\\(a + 5d = 37\\)\n\\(12 + 5d = 37\\)\n\\(5d = 37 - 12 = 25\\)\n\\(d = 5\\)\n\\(S_{15} = \\frac{15}{2} (2(12) + (15 - 1)(5))\\)\n\\(\\frac{15}{2} (24 + 70) = \\frac{15}{2} (94)\\)\n= \\(15 \\times 47\\)\n= \\(705\\).\n\n', '(a) \\(T_{n} = a + (n - 1)d\\) (For an AP series)\n\\(T_{6} = 37 = a + (6 - 1) d = a + 5d\\)\n\\(a + 5d = 37 .... (1)\\)\n\\(S_{n} = \\frac{n}{2}(2a + (n - 1) d)\\)\nor\n\\(S_{n} = \\frac{n}{2}(a + l)\\) where a = first term; and l = last term.\nSince, we are given the sum of the first 6 terms and also given the sixth term which is the last term in the sum, we can use the second formula where the sixth term is the last term in this case.\n\\(S_{6} = 147 = \\frac{6}{2} (a + 37)\\)\n\\(147 = 3(a + 37) \\implies 147 = 3a + 111 \\)\n\\(3a = 147 - 111 = 36 \\implies a = 12\\)~(b) From equation (1) above,\n\\(a + 5d = 37\\)\n\\(12 + 5d = 37\\)\n\\(5d = 37 - 12 = 25\\)\n\\(d = 5\\)\n\\(S_{15} = \\frac{15}{2} (2(12) + (15 - 1)(5))\\)\n\\(\\frac{15}{2} (24 + 70) = \\frac{15}{2} (94)\\)\n= \\(15 \\times 47\\)\n= \\(705\\).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(334, '2013', '\n\nQuestion\n\nOut of 120 customers in a shop, 45 bought both bags and shoes. If all the customers bought either bags or shoes and 11 more customers bought shoes than bags:\n(a) Illustrate the this information in a diagram;\n(b) find the number of customers who bought shoes;\n(c) calculate the probability that a customer selected at random bought bags.\n\n', '', 'Out of 120 customers in a shop, 45 bought both bags and shoes. If all the customers bought either bags or shoes and 11 more customers bought shoes than bags:\n(a) Illustrate the this information in a diagram;\n(b) find the number of customers who bought shoes;\n(c) calculate the probability that a customer selected at random bought bags.', 'Out of 120 customers in a shop, 45 bought both bags and shoes. If all the customers bought either bags or shoes and 11 more customers bought shoes than bags:\n(a) Illustrate the this information in a diagram;~(b) find the number of customers who bought shoes;\n(c) calculate the probability that a customer selected at random bought bags.~(c) calculate the probability that a customer selected at random bought bags.', NULL, '\n\nExplanation\n(a)\n(b) \\(x + 45 + x + 11 = 120 \\implies 2x + 56 = 120\\)\n\\(2x = 120 - 56 = 64\\)\n\\(x = 32\\)\nTherefore, the number of customers that bought bags = 32.\n(c) \\(P(\\text{a random customer bought bags}) = \\frac{\\text{no of customers that bought bags}}{\\text{total no of customers}}\\)\nNo of customers that bought bags = 32 + 45 = 77\n\\(P(\\text{customer bought bag}) = \\frac{77}{120}\\).\n\n', '', 'Explanation\n(a)\n(b) \\(x + 45 + x + 11 = 120 \\implies 2x + 56 = 120\\)\n\\(2x = 120 - 56 = 64\\)\n\\(x = 32\\)\nTherefore, the number of customers that bought bags = 32.\n(c) \\(P(\\text{a random customer bought bags}) = \\frac{\\text{no of customers that bought bags}}{\\text{total no of customers}}\\)\nNo of customers that bought bags = 32 + 45 = 77\n\\(P(\\text{customer bought bag}) = \\frac{77}{120}\\).\n\n', '~(b) \\(x + 45 + x + 11 = 120 \\implies 2x + 56 = 120\\)\n\\(2x = 120 - 56 = 64\\)\n\\(x = 32\\)\nTherefore, the number of customers that bought bags = 32.\n(c) \\(P(\\text{a random customer bought bags}) = \\frac{\\text{no of customers that bought bags}}{\\text{total no of customers}}\\)\nNo of customers that bought bags = 32 + 45 = 77\n\\(P(\\text{customer bought bag}) = \\frac{77}{120}\\).\n~(c) \\(P(\\text{a random customer bought bags}) = \\frac{\\text{no of customers that bought bags}}{\\text{total no of customers}}\\)\nNo of customers that bought bags = 32 + 45 = 77\n\\(P(\\text{customer bought bag}) = \\frac{77}{120}\\).\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
(335, '2013', '\n\nQuestion\n\n(a) A manufacturing company requires 3 hours of direct labour to process N87.00 worth of raw materials. If the company uses N30,450.00 worth of raw materials, what amount should it budget at N18.25 per hour?\n(b) An investor invested Nx in bank M at the rate of 6% simple interest per annum and Ny in bank N at the rate of 8% simple interest per annum. If a total of N8,000,000.00 was invested in the two banks and the investor received a total of N2,320,000.00 as interest from the two banks after 4 years, calculate the:\n(i) values of x and y\n(ii) interest paid by the second bank.\n\n', '', '(a) A manufacturing company requires 3 hours of direct labour to process N87.00 worth of raw materials. If the company uses N30,450.00 worth of raw materials, what amount should it budget at N18.25 per hour?\n(b) An investor invested Nx in bank M at the rate of 6% simple interest per annum and Ny in bank N at the rate of 8% simple interest per annum. If a total of N8,000,000.00 was invested in the two banks and the investor received a total of N2,320,000.00 as interest from the two banks after 4 years, calculate the:\n(i) values of x and y\n(ii) interest paid by the second bank.', '(a) A manufacturing company requires 3 hours of direct labour to process N87.00 worth of raw materials. If the company uses N30,450.00 worth of raw materials, what amount should it budget at N18.25 per hour?~(b) An investor invested Nx in bank M at the rate of 6% simple interest per annum and Ny in bank N at the rate of 8% simple interest per annum. If a total of N8,000,000.00 was invested in the two banks and the investor received a total of N2,320,000.00 as interest from the two banks after 4 years, calculate the:\n(i) values of x and y\n(ii) interest paid by the second bank.', NULL, '\n\nExplanation\n(a) Worth of raw materials = N30,450.00\nWorth of raw material to get processed materials = N87.00\nAmount of raw material to get processed material = \\(\\frac{30,450}{87}\\)\nTime required for direct labour = 3 hours\nAmount that will be budgeted for direct labour at N18.25 per hour =\n\\(3 \\times 18.25 \\times \\frac{30,450}{87}\\)\n= \\(\\frac{1667137.5}{87}\\)\n= N19,162.50\n(b) Total amount invested \\(N(x + y) = N8,000,000 ....(1)\\)\nInterest from bank M = \\(\\frac{x \\times 4 \\times 6}{100} = 0.24x\\)\nInterest from bank N = \\(\\frac{y \\times 4\\times 8}{100} = 0.32y\\)\n\\(0.24x + 0.32y = 2,320,000 .... (2)\\)\n(i) Multiplying (1) by 0.24 in order to eliminate x, we have\n\\(0.24(x + y) = 0.24(8,000,000.00)\\)\n\\(0.24x + 0.24y = 1,920,000.00 .... (3)\\)\n(2) - (3) : \\(0.32y - 0.24y = 2,320,000 - 1,920,000 = 400,000\\)\n\\(0.08y = 400,000 \\implies y = \\frac{400,000}{0.08} = N5,000,000\\)\n\\(x = 8,000,000 - y\\)\n= \\(8,000,000 - 5,000,000\\)\n= \\(N3,000,000\\)\nTherefore, x = N3,000,000 and y = N5,000,000.\n(ii) Interest paid by bank N = 0.32y\n= \\(0.32 \\times N5,000,000\\)\n= \\(N1,600,000\\)\n\n', '', 'Explanation\n(a) Worth of raw materials = N30,450.00\nWorth of raw material to get processed materials = N87.00\nAmount of raw material to get processed material = \\(\\frac{30,450}{87}\\)\nTime required for direct labour = 3 hours\nAmount that will be budgeted for direct labour at N18.25 per hour =\n\\(3 \\times 18.25 \\times \\frac{30,450}{87}\\)\n= \\(\\frac{1667137.5}{87}\\)\n= N19,162.50\n(b) Total amount invested \\(N(x + y) = N8,000,000 ....(1)\\)\nInterest from bank M = \\(\\frac{x \\times 4 \\times 6}{100} = 0.24x\\)\nInterest from bank N = \\(\\frac{y \\times 4\\times 8}{100} = 0.32y\\)\n\\(0.24x + 0.32y = 2,320,000 .... (2)\\)\n(i) Multiplying (1) by 0.24 in order to eliminate x, we have\n\\(0.24(x + y) = 0.24(8,000,000.00)\\)\n\\(0.24x + 0.24y = 1,920,000.00 .... (3)\\)\n(2) - (3) : \\(0.32y - 0.24y = 2,320,000 - 1,920,000 = 400,000\\)\n\\(0.08y = 400,000 \\implies y = \\frac{400,000}{0.08} = N5,000,000\\)\n\\(x = 8,000,000 - y\\)\n= \\(8,000,000 - 5,000,000\\)\n= \\(N3,000,000\\)\nTherefore, x = N3,000,000 and y = N5,000,000.\n(ii) Interest paid by bank N = 0.32y\n= \\(0.32 \\times N5,000,000\\)\n= \\(N1,600,000\\)\n\n', '(a) Worth of raw materials = N30,450.00\nWorth of raw material to get processed materials = N87.00\nAmount of raw material to get processed material = \\(\\frac{30,450}{87}\\)\nTime required for direct labour = 3 hours\nAmount that will be budgeted for direct labour at N18.25 per hour =\n\\(3 \\times 18.25 \\times \\frac{30,450}{87}\\)\n= \\(\\frac{1667137.5}{87}\\)\n= N19,162.50~(b) Total amount invested \\(N(x + y) = N8,000,000 ....(1)\\)\nInterest from bank M = \\(\\frac{x \\times 4 \\times 6}{100} = 0.24x\\)\nInterest from bank N = \\(\\frac{y \\times 4\\times 8}{100} = 0.32y\\)\n\\(0.24x + 0.32y = 2,320,000 .... (2)\\)\n(i) Multiplying (1) by 0.24 in order to eliminate x, we have\n\\(0.24(x + y) = 0.24(8,000,000.00)\\)\n\\(0.24x + 0.24y = 1,920,000.00 .... (3)\\)\n(2) - (3) : \\(0.32y - 0.24y = 2,320,000 - 1,920,000 = 400,000\\)\n\\(0.08y = 400,000 \\implies y = \\frac{400,000}{0.08} = N5,000,000\\)\n\\(x = 8,000,000 - y\\)\n= \\(8,000,000 - 5,000,000\\)\n= \\(N3,000,000\\)\nTherefore, x = N3,000,000 and y = N5,000,000.\n(ii) Interest paid by bank N = 0.32y\n= \\(0.32 \\times N5,000,000\\)\n= \\(N1,600,000\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(336, '2013', '\n\nQuestion\n\n(a) Copy and complete the table of values for the equation \\(y = 2x^{2} - 7x - 9\\) for \\(-3 \\leq x \\leq 6\\).\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n6\n\n\ny\n\n13\n\n-9\n-14\n\n-12\n\n6\n\n\n\n\n(b) Using scales of 2cm to 1 unit on the x- axis and 2cm to 4 units on the y- axis, draw the graphs of \\(y = 2x^{2} - 7x - 9\\) for \\(-3 \\leq x \\leq 6\\).\n(c) Use the graph to estimate the :\n(i) roots of the equation \\(2x^{2} - 7x = 26\\);\n(ii) coordinates of the minimum point of y;\n(iii) range of values for which \\(2x^{2} - 7x < 9\\).\n\n\n(a) Copy and complete the table of values for the equation \\(y = 2x^{2} - 7x - 9\\) for \\(-3 \\leq x \\leq 6\\).\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n6\n\n\ny\n\n13\n\n-9\n-14\n\n-12\n\n6\n\n\n\n', '', '(a) Copy and complete the table of values for the equation \\(y = 2x^{2} - 7x - 9\\) for \\(-3 \\leq x \\leq 6\\).\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n6\n\n\ny\n\n13\n\n-9\n-14\n\n-12\n\n6\n\n\n\n\n(b) Using scales of 2cm to 1 unit on the x- axis and 2cm to 4 units on the y- axis, draw the graphs of \\(y = 2x^{2} - 7x - 9\\) for \\(-3 \\leq x \\leq 6\\).\n(c) Use the graph to estimate the :\n(i) roots of the equation \\(2x^{2} - 7x = 26\\);\n(ii) coordinates of the minimum point of y;\n(iii) range of values for which \\(2x^{2} - 7x < 9\\).\n\n\n(a) Copy and complete the table of values for the equation \\(y = 2x^{2} - 7x - 9\\) for \\(-3 \\leq x \\leq 6\\).\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n6\n\n\ny\n\n13\n\n-9\n-14\n\n-12\n\n6\n\n', '(a) Copy and complete the table of values for the equation \\(y = 2x^{2} - 7x - 9\\) for \\(-3 \\leq x \\leq 6\\).\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n6\n\n\ny\n\n13\n\n-9\n-14\n\n-12\n\n6\n\n\n\n~(b) Using scales of 2cm to 1 unit on the x- axis and 2cm to 4 units on the y- axis, draw the graphs of \\(y = 2x^{2} - 7x - 9\\) for \\(-3 \\leq x \\leq 6\\).\n(c) Use the graph to estimate the :\n(i) roots of the equation \\(2x^{2} - 7x = 26\\);\n(ii) coordinates of the minimum point of y;\n(iii) range of values for which \\(2x^{2} - 7x < 9\\).\n\n\n(a) Copy and complete the table of values for the equation \\(y = 2x^{2} - 7x - 9\\) for \\(-3 \\leq x \\leq 6\\).\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n6\n\n\ny\n\n13\n\n-9\n-14\n\n-12\n\n6\n\n~(c) Use the graph to estimate the :\n(i) roots of the equation \\(2x^{2} - 7x = 26\\);\n(ii) coordinates of the minimum point of y;\n(iii) range of values for which \\(2x^{2} - 7x < 9\\).\n\n\n(a) Copy and complete the table of values for the equation \\(y = 2x^{2} - 7x - 9\\) for \\(-3 \\leq x \\leq 6\\).\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n6\n\n\ny\n\n13\n\n-9\n-14\n\n-12\n\n6\n\n', NULL, '\n\nExplanation\n(a)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n6\n\n\ny\n30\n13\n0\n-9\n-14\n-15\n-12\n-5\n6\n21\n\n\n\n(b)\n\n\n(c) (i) From the graph, the roots of the equation 2x\\(^2\\) - 7x = 26 can be obtained by drawing a line through y = 17 to meet the curve 2x\\(^2\\) - 7x - 9.\nThe roots of the equation 2x\\(^2\\) - 7x = 26 : -2.2 and 5.8.\n(ii) The coordinates of the minimum point of y : 1.8, -15.\n(iii) The range of values for which 2x\\(^{2}\\) - 7x < 9 : - 0.9 < x < 4.5.\n\n', '', 'Explanation\n(a)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n6\n\n\ny\n30\n13\n0\n-9\n-14\n-15\n-12\n-5\n6\n21\n\n\n\n(b)\n\n\n(c) (i) From the graph, the roots of the equation 2x\\(^2\\) - 7x = 26 can be obtained by drawing a line through y = 17 to meet the curve 2x\\(^2\\) - 7x - 9.\nThe roots of the equation 2x\\(^2\\) - 7x = 26 : -2.2 and 5.8.\n(ii) The coordinates of the minimum point of y : 1.8, -15.\n(iii) The range of values for which 2x\\(^{2}\\) - 7x < 9 : - 0.9 < x < 4.5.\n\n', '(a)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n6\n\n\ny\n30\n13\n0\n-9\n-14\n-15\n-12\n-5\n6\n21\n\n\n~(b)\n\n\n(c) (i) From the graph, the roots of the equation 2x\\(^2\\) - 7x = 26 can be obtained by drawing a line through y = 17 to meet the curve 2x\\(^2\\) - 7x - 9.\nThe roots of the equation 2x\\(^2\\) - 7x = 26 : -2.2 and 5.8.\n(ii) The coordinates of the minimum point of y : 1.8, -15.\n(iii) The range of values for which 2x\\(^{2}\\) - 7x < 9 : - 0.9 < x < 4.5.\n~(c) (i) From the graph, the roots of the equation 2x\\(^2\\) - 7x = 26 can be obtained by drawing a line through y = 17 to meet the curve 2x\\(^2\\) - 7x - 9.\nThe roots of the equation 2x\\(^2\\) - 7x = 26 : -2.2 and 5.8.\n(ii) The coordinates of the minimum point of y : 1.8, -15.\n(iii) The range of values for which 2x\\(^{2}\\) - 7x < 9 : - 0.9 < x < 4.5.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(337, '2013', '\n\nQuestion\n\n\n\nMarks\n1\n2\n3\n4\n5\n\n\nNumber of students\nm + 2\nm - 1\n2m - 3\nm + 5\n3m - 4\n\n\n\nThe table shows the distribution of marks scored by some students in a test.\n(a) If the mean mark is \\(3\\frac{6}{23}\\), find the value of m.\n(b) Find the : (i) interquartile range\n(ii) probability of selecting a student who scored at least 4 marks in the test.\n\n\n\n\n\nMarks\n1\n2\n3\n4\n5\n\n\nNumber of students\nm + 2\nm - 1\n2m - 3\nm + 5\n3m - 4\n\n\n', '', 'Marks\n1\n2\n3\n4\n5\n\n\nNumber of students\nm + 2\nm - 1\n2m - 3\nm + 5\n3m - 4\n\n\n\nThe table shows the distribution of marks scored by some students in a test.\n(a) If the mean mark is \\(3\\frac{6}{23}\\), find the value of m.\n(b) Find the : (i) interquartile range\n(ii) probability of selecting a student who scored at least 4 marks in the test.\n\n\n\n\n\nMarks\n1\n2\n3\n4\n5\n\n\nNumber of students\nm + 2\nm - 1\n2m - 3\nm + 5\n3m - 4\n', 'Marks\n1\n2\n3\n4\n5\n\n\nNumber of students\nm + 2\nm - 1\n2m - 3\nm + 5\n3m - 4\n\n\n\nThe table shows the distribution of marks scored by some students in a test.\n(a) If the mean mark is \\(3\\frac{6}{23}\\), find the value of m.~(b) Find the : (i) interquartile range\n(ii) probability of selecting a student who scored at least 4 marks in the test.\n\n\n\n\n\nMarks\n1\n2\n3\n4\n5\n\n\nNumber of students\nm + 2\nm - 1\n2m - 3\nm + 5\n3m - 4\n', NULL, '\n\nExplanation\nMark (x)\n1\n2\n3\n4\n5\nTotal\n\n\nFrequency (f)\nm + 2\nm - 1\n2m - 3\nm + 5\n3m - 4\n8m - 1\n\n\nfx\nm + 2\n2m - 2\n6m - 9\n4m + 20\n15m - 20\n28m - 9\n\n\n\n(a) Mean \\(\\bar{x} = \\frac{\\sum fx}{\\sum f}\\)\n\\(\\frac{75}{23} = \\frac{28m - 9}{8m - 1}\\)\n\\(75(8m - 1) = 23(28m - 9) \\implies 600m - 75 = 644m - 207\\)\n\\(-75 + 207 = 644m - 600m\\)\n\\(132 = 44m \\implies m = 3\\)\n(b)(i) Interquartile range = Third quartile - First quartile\nFrequency = 8(3) - 1 = 24 - 1 = 23\n\\(Q_{3} = \\frac{3}{4} \\times 23 = 17.25th\\) position = 4\n\\(Q_{1} = \\frac{1}{4} \\times 23 = 5.75th\\) position = 2\nInterquartile range : 4 - 2 = 2\n(ii) P(at least 4 marks) = \\(\\frac{(m + 5 + 3m - 4)}{23} = \\frac{4m + 1}{23}\\)\n= \\(\\frac{4(3) + 1}{23} \\)\n= \\(\\frac{13}{23}\\)\n\n', '', 'Explanation\nMark (x)\n1\n2\n3\n4\n5\nTotal\n\n\nFrequency (f)\nm + 2\nm - 1\n2m - 3\nm + 5\n3m - 4\n8m - 1\n\n\nfx\nm + 2\n2m - 2\n6m - 9\n4m + 20\n15m - 20\n28m - 9\n\n\n\n(a) Mean \\(\\bar{x} = \\frac{\\sum fx}{\\sum f}\\)\n\\(\\frac{75}{23} = \\frac{28m - 9}{8m - 1}\\)\n\\(75(8m - 1) = 23(28m - 9) \\implies 600m - 75 = 644m - 207\\)\n\\(-75 + 207 = 644m - 600m\\)\n\\(132 = 44m \\implies m = 3\\)\n(b)(i) Interquartile range = Third quartile - First quartile\nFrequency = 8(3) - 1 = 24 - 1 = 23\n\\(Q_{3} = \\frac{3}{4} \\times 23 = 17.25th\\) position = 4\n\\(Q_{1} = \\frac{1}{4} \\times 23 = 5.75th\\) position = 2\nInterquartile range : 4 - 2 = 2\n(ii) P(at least 4 marks) = \\(\\frac{(m + 5 + 3m - 4)}{23} = \\frac{4m + 1}{23}\\)\n= \\(\\frac{4(3) + 1}{23} \\)\n= \\(\\frac{13}{23}\\)\n\n', '(a) Mean \\(\\bar{x} = \\frac{\\sum fx}{\\sum f}\\)\n\\(\\frac{75}{23} = \\frac{28m - 9}{8m - 1}\\)\n\\(75(8m - 1) = 23(28m - 9) \\implies 600m - 75 = 644m - 207\\)\n\\(-75 + 207 = 644m - 600m\\)\n\\(132 = 44m \\implies m = 3\\)~(b)(i) Interquartile range = Third quartile - First quartile\nFrequency = 8(3) - 1 = 24 - 1 = 23\n\\(Q_{3} = \\frac{3}{4} \\times 23 = 17.25th\\) position = 4\n\\(Q_{1} = \\frac{1}{4} \\times 23 = 5.75th\\) position = 2\nInterquartile range : 4 - 2 = 2\n(ii) P(at least 4 marks) = \\(\\frac{(m + 5 + 3m - 4)}{23} = \\frac{4m + 1}{23}\\)\n= \\(\\frac{4(3) + 1}{23} \\)\n= \\(\\frac{13}{23}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(338, '2013', '\n\nQuestion\n\n(a) PQ is a tangent to a circle RST at the point S. PRT is a straight line, < TPS = 34° and < TSQ = 65°.\n(i) Illustrate the information in a diagram; (ii) find the value of : (a) < RTS ; (b) < SRP.\n(b)\nIn the diagram, /VZ/ = /YZ/, < YXZ = 20° and < ZVY = 52°. Calculate the size of < WYZ.\n\n', '', '(a) PQ is a tangent to a circle RST at the point S. PRT is a straight line, < TPS = 34° and < TSQ = 65°.\n(i) Illustrate the information in a diagram; (ii) find the value of : (a) < RTS ; (b) < SRP.\n(b)\nIn the diagram, /VZ/ = /YZ/, < YXZ = 20° and < ZVY = 52°. Calculate the size of < WYZ.', '(a) PQ is a tangent to a circle RST at the point S. PRT is a straight line, < TPS = 34° and < TSQ = 65°.\n(i) Illustrate the information in a diagram; (ii) find the value of : (a) < RTS ; (b) < SRP.~(b) < SRP.\n(b)\nIn the diagram, /VZ/ = /YZ/, < YXZ = 20° and < ZVY = 52°. Calculate the size of < WYZ.', NULL, '\n\nExplanation\n(ii) (a) < RTS = 65° - 34° (exterior angles of triangle PST)\n= 31°\n(ii) (b) < SRP = 65° (angles in alternate segment are equal)\n\\(\\therefore < SRP = 180° - 65° = 115°\\).\n(b) <XYZ = 180 - 52 = 128° (angles on a straight line)\nIf < XYZ = 128°\nthen < XZY = 128° + 20° = 148°\n< XZV = 180° - 148° = 32°\n< YVZ = < VYZ = 52° (Isoceles triangle)\n< WYZ = 52° - 32° = 20°\n\n', '', 'Explanation\n(ii) (a) < RTS = 65° - 34° (exterior angles of triangle PST)\n= 31°\n(ii) (b) < SRP = 65° (angles in alternate segment are equal)\n\\(\\therefore < SRP = 180° - 65° = 115°\\).\n(b) <XYZ = 180 - 52 = 128° (angles on a straight line)\nIf < XYZ = 128°\nthen < XZY = 128° + 20° = 148°\n< XZV = 180° - 148° = 32°\n< YVZ = < VYZ = 52° (Isoceles triangle)\n< WYZ = 52° - 32° = 20°\n\n', 'i) (a) < RTS = 65° - 34° (exterior angles of triangle PST)\n= 31°\n(ii) (b) < SRP = 65° (angles in alternate segment are equal)\n\\(\\therefore < SRP = 180° - 65° = 115°\\).~(b) < SRP = 65° (angles in alternate segment are equal)\n\\(\\therefore < SRP = 180° - 65° = 115°\\).\n(b) <XYZ = 180 - 52 = 128° (angles on a straight line)\nIf < XYZ = 128°\nthen < XZY = 128° + 20° = 148°\n< XZV = 180° - 148° = 32°\n< YVZ = < VYZ = 52° (Isoceles triangle)\n< WYZ = 52° - 32° = 20°\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(339, '2013', '\n\nQuestion\n\n(a) Given that \\(\\sin x = \\frac{5}{13}, 0° < x < 90°\\), find \\(\\frac{\\cos x - 2\\sin x}{2\\tan x}\\).\n(b) A ladder, LA, leans against a vertical pole at a point L which is 9.6metres above the groung. Another ladder, LB, 12 metres long, leans on the opposite side of the pole and at the same point L. If A and B are 10 metres apart and on the same straight line as the foot of the pole, calculate, correct to 2 significant figures, the :\n(i) length of ladder LA (ii) angle which LA makes with the ground.\n\n', '', '(a) Given that \\(\\sin x = \\frac{5}{13}, 0° < x < 90°\\), find \\(\\frac{\\cos x - 2\\sin x}{2\\tan x}\\).\n(b) A ladder, LA, leans against a vertical pole at a point L which is 9.6metres above the groung. Another ladder, LB, 12 metres long, leans on the opposite side of the pole and at the same point L. If A and B are 10 metres apart and on the same straight line as the foot of the pole, calculate, correct to 2 significant figures, the :\n(i) length of ladder LA (ii) angle which LA makes with the ground.', '(a) Given that \\(\\sin x = \\frac{5}{13}, 0° < x < 90°\\), find \\(\\frac{\\cos x - 2\\sin x}{2\\tan x}\\).~(b) A ladder, LA, leans against a vertical pole at a point L which is 9.6metres above the groung. Another ladder, LB, 12 metres long, leans on the opposite side of the pole and at the same point L. If A and B are 10 metres apart and on the same straight line as the foot of the pole, calculate, correct to 2 significant figures, the :\n(i) length of ladder LA (ii) angle which LA makes with the ground.', NULL, '\n\nExplanation\n(a) \\(\\sin x = \\frac{5}{13}\\)\nUsing SOHCAHTOA, then Opp = 5 and Hyp = 13\n\\(13^{2} = 5^{2} + Adj^{2}\\)\n\\(169 = 25 + Adj^{2}\\)\n\\(Adj = \\sqrt{169 - 25} = \\sqrt{144} = 12\\)\n\\(\\cos x = \\frac{12}{13}\\)\n\\(\\tan x = \\frac{5}{12}\\)\n\\(\\frac{\\cos x - 2\\sin x}{2\\tan x} = \\frac{\\frac{12}{13} - 2(\\frac{5}{13})}{2(\\frac{5}{12})}\\)\n= \\(\\frac{\\frac{2}{13}}{\\frac{5}{6}}\\)\n= \\(\\frac{12}{65}\\)\n(b)\nTaking \\(\\Delta LPB\\),\n\nUsing Pythagoras theorem, \\(/BP/^{2} = /BL/^{2} - /LP/^{2}\\)\n\\(i.e. /BP/ = \\sqrt{12^{2} - 9.6^{2}}\\)\n= \\(\\sqrt{144 - 92.16} = \\sqrt{51.84} = 7.2m\\)\n\\(/PA/ = 10m - 7.2m = 2.8m\\)\n\\(/LA/^{2} = x^{2}\\)\n\\(x^{2} = /LP/^{2} + /PA/^{2}\\)\n\\(x^{2} = 9.6^{2} + 2.8^{2}\\)\n= \\(92.16 + 7.84\\)\n\\(x^{2} = 100 \\implies x = 10m\\)\n(ii) \\(\\tan \\theta_{y} = \\frac{9.6}{2.8} = 3.4286\\)\n\\(\\theta_{y} = \\tan^{-1} (3.4286) \\approxeq 74°\\)\n\n\n', '', 'Explanation\n(a) \\(\\sin x = \\frac{5}{13}\\)\nUsing SOHCAHTOA, then Opp = 5 and Hyp = 13\n\\(13^{2} = 5^{2} + Adj^{2}\\)\n\\(169 = 25 + Adj^{2}\\)\n\\(Adj = \\sqrt{169 - 25} = \\sqrt{144} = 12\\)\n\\(\\cos x = \\frac{12}{13}\\)\n\\(\\tan x = \\frac{5}{12}\\)\n\\(\\frac{\\cos x - 2\\sin x}{2\\tan x} = \\frac{\\frac{12}{13} - 2(\\frac{5}{13})}{2(\\frac{5}{12})}\\)\n= \\(\\frac{\\frac{2}{13}}{\\frac{5}{6}}\\)\n= \\(\\frac{12}{65}\\)\n(b)\nTaking \\(\\Delta LPB\\),\n\nUsing Pythagoras theorem, \\(/BP/^{2} = /BL/^{2} - /LP/^{2}\\)\n\\(i.e. /BP/ = \\sqrt{12^{2} - 9.6^{2}}\\)\n= \\(\\sqrt{144 - 92.16} = \\sqrt{51.84} = 7.2m\\)\n\\(/PA/ = 10m - 7.2m = 2.8m\\)\n\\(/LA/^{2} = x^{2}\\)\n\\(x^{2} = /LP/^{2} + /PA/^{2}\\)\n\\(x^{2} = 9.6^{2} + 2.8^{2}\\)\n= \\(92.16 + 7.84\\)\n\\(x^{2} = 100 \\implies x = 10m\\)\n(ii) \\(\\tan \\theta_{y} = \\frac{9.6}{2.8} = 3.4286\\)\n\\(\\theta_{y} = \\tan^{-1} (3.4286) \\approxeq 74°\\)\n\n\n', '(a) \\(\\sin x = \\frac{5}{13}\\)\nUsing SOHCAHTOA, then Opp = 5 and Hyp = 13\n\\(13^{2} = 5^{2} + Adj^{2}\\)\n\\(169 = 25 + Adj^{2}\\)\n\\(Adj = \\sqrt{169 - 25} = \\sqrt{144} = 12\\)\n\\(\\cos x = \\frac{12}{13}\\)\n\\(\\tan x = \\frac{5}{12}\\)\n\\(\\frac{\\cos x - 2\\sin x}{2\\tan x} = \\frac{\\frac{12}{13} - 2(\\frac{5}{13})}{2(\\frac{5}{12})}\\)\n= \\(\\frac{\\frac{2}{13}}{\\frac{5}{6}}\\)\n= \\(\\frac{12}{65}\\)~(b)\nTaking \\(\\Delta LPB\\),\n\nUsing Pythagoras theorem, \\(/BP/^{2} = /BL/^{2} - /LP/^{2}\\)\n\\(i.e. /BP/ = \\sqrt{12^{2} - 9.6^{2}}\\)\n= \\(\\sqrt{144 - 92.16} = \\sqrt{51.84} = 7.2m\\)\n\\(/PA/ = 10m - 7.2m = 2.8m\\)\n\\(/LA/^{2} = x^{2}\\)\n\\(x^{2} = /LP/^{2} + /PA/^{2}\\)\n\\(x^{2} = 9.6^{2} + 2.8^{2}\\)\n= \\(92.16 + 7.84\\)\n\\(x^{2} = 100 \\implies x = 10m\\)\n(ii) \\(\\tan \\theta_{y} = \\frac{9.6}{2.8} = 3.4286\\)\n\\(\\theta_{y} = \\tan^{-1} (3.4286) \\approxeq 74°\\)\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(340, '2013', '\n\nQuestion\n\n(a) It takes 8 students two- thirds of an hour to fill 12 tanks with water. How many tanks of water will 4 students fill in one- third of an hour at the same rate?\n(b) A chord, 20 cm long, is 12 cm from the centre of the circle. Calculate, correct to one decimal place, the :\n(i) angle subtended by the chord at the centre of the circle;\n(ii) perimeter of the minor segment cut off by the chord. [Take \\(\\pi = 3.142\\)].\n\n', '', '(a) It takes 8 students two- thirds of an hour to fill 12 tanks with water. How many tanks of water will 4 students fill in one- third of an hour at the same rate?\n(b) A chord, 20 cm long, is 12 cm from the centre of the circle. Calculate, correct to one decimal place, the :\n(i) angle subtended by the chord at the centre of the circle;\n(ii) perimeter of the minor segment cut off by the chord. [Take \\(\\pi = 3.142\\)].', '(a) It takes 8 students two- thirds of an hour to fill 12 tanks with water. How many tanks of water will 4 students fill in one- third of an hour at the same rate?~(b) A chord, 20 cm long, is 12 cm from the centre of the circle. Calculate, correct to one decimal place, the :\n(i) angle subtended by the chord at the centre of the circle;\n(ii) perimeter of the minor segment cut off by the chord. [Take \\(\\pi = 3.142\\)].', NULL, '\n\nExplanation\n(a) \\(Time = \\frac{2}{3} hour = \\frac{2}{3} \\times 60 = 40 mins\\)\n4 students will take \\(\\frac{8 \\times 40}{4} = 80 minutes\\)\n\\(Time = \\frac{1}{3} hour = \\frac{1}{3} \\times 60 = 20 minutes\\)\nIt will take 20 minutes to fill \\(\\frac{12 \\times 20}{80} = 3 tanks\\)\n(b)\n\\(\\tan (\\frac{\\theta}{2}) = \\frac{10}{12}\\)\n\\(\\frac{\\theta}{2} = \\tan^{-1} (0.8337) = 39.8°\\)\n\\(\\theta = 39.8° \\times 2 = 79.6°\\)\n(ii) Perimeter of minor segment = Length of chord + length of arc\nBut \\(r^{2} = 12^{2} + 10^{2}\\)\n= \\(144 + 100 = 244\\)\n\\(r = \\sqrt{244} = 15.62 cm\\)\nLength of arc = \\(\\frac{79.6}{360} \\times 2 \\times 3.142 \\times 15.62 = 21.71cm\\)\nPerimeter of minor segment = 20 + 21.71 = 41.71cm\n\n', '', 'Explanation\n(a) \\(Time = \\frac{2}{3} hour = \\frac{2}{3} \\times 60 = 40 mins\\)\n4 students will take \\(\\frac{8 \\times 40}{4} = 80 minutes\\)\n\\(Time = \\frac{1}{3} hour = \\frac{1}{3} \\times 60 = 20 minutes\\)\nIt will take 20 minutes to fill \\(\\frac{12 \\times 20}{80} = 3 tanks\\)\n(b)\n\\(\\tan (\\frac{\\theta}{2}) = \\frac{10}{12}\\)\n\\(\\frac{\\theta}{2} = \\tan^{-1} (0.8337) = 39.8°\\)\n\\(\\theta = 39.8° \\times 2 = 79.6°\\)\n(ii) Perimeter of minor segment = Length of chord + length of arc\nBut \\(r^{2} = 12^{2} + 10^{2}\\)\n= \\(144 + 100 = 244\\)\n\\(r = \\sqrt{244} = 15.62 cm\\)\nLength of arc = \\(\\frac{79.6}{360} \\times 2 \\times 3.142 \\times 15.62 = 21.71cm\\)\nPerimeter of minor segment = 20 + 21.71 = 41.71cm\n\n', '(a) \\(Time = \\frac{2}{3} hour = \\frac{2}{3} \\times 60 = 40 mins\\)\n4 students will take \\(\\frac{8 \\times 40}{4} = 80 minutes\\)\n\\(Time = \\frac{1}{3} hour = \\frac{1}{3} \\times 60 = 20 minutes\\)\nIt will take 20 minutes to fill \\(\\frac{12 \\times 20}{80} = 3 tanks\\)~(b)\n\\(\\tan (\\frac{\\theta}{2}) = \\frac{10}{12}\\)\n\\(\\frac{\\theta}{2} = \\tan^{-1} (0.8337) = 39.8°\\)\n\\(\\theta = 39.8° \\times 2 = 79.6°\\)\n(ii) Perimeter of minor segment = Length of chord + length of arc\nBut \\(r^{2} = 12^{2} + 10^{2}\\)\n= \\(144 + 100 = 244\\)\n\\(r = \\sqrt{244} = 15.62 cm\\)\nLength of arc = \\(\\frac{79.6}{360} \\times 2 \\times 3.142 \\times 15.62 = 21.71cm\\)\nPerimeter of minor segment = 20 + 21.71 = 41.71cm\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(341, '2013', '\n\nQuestion\n\n(a) Using completing the square method, solve, correct to 2 decimal places, the equation \\(3y^{2} - 5y + 2 = 0\\).\n(b) Given that \\(M = \\begin{pmatrix} 1 & 2 \\\\ 4 & 3 \\end{pmatrix}, N = \\begin{pmatrix} m & x \\\\ n & y \\end{pmatrix}\\) and \\(MN = \\begin{pmatrix} 2 & 1 \\\\ 3 & 4 \\end{pmatrix}\\), find the matrix N.\n\n', '', '(a) Using completing the square method, solve, correct to 2 decimal places, the equation \\(3y^{2} - 5y + 2 = 0\\).\n(b) Given that \\(M = \\begin{pmatrix} 1 & 2 \\\\ 4 & 3 \\end{pmatrix}, N = \\begin{pmatrix} m & x \\\\ n & y \\end{pmatrix}\\) and \\(MN = \\begin{pmatrix} 2 & 1 \\\\ 3 & 4 \\end{pmatrix}\\), find the matrix N.', '(a) Using completing the square method, solve, correct to 2 decimal places, the equation \\(3y^{2} - 5y + 2 = 0\\).~(b) Given that \\(M = \\begin{pmatrix} 1 & 2 \\\\ 4 & 3 \\end{pmatrix}, N = \\begin{pmatrix} m & x \\\\ n & y \\end{pmatrix}\\) and \\(MN = \\begin{pmatrix} 2 & 1 \\\\ 3 & 4 \\end{pmatrix}\\), find the matrix N.', NULL, '\n\nExplanation\n(a) \\(3y^{2} - 5y + 2 = 0\\)\n\\(\\frac{3y^{2}}{3} - \\frac{5y}{3} = \\frac{-2}{3}\\)\n\\(y^{2} - \\frac{5}{3}y = \\frac{-2}{3}\\)\n\\(\\frac{1}{2} of \\frac{-5}{3} = \\frac{-5}{6}\\)\n\\(y^{2} - \\frac{5}{3}y + (-\\frac{5}{6})^{2} = \\frac{-2}{3} + (\\frac{-5}{6})^{2}\\)\n\\((y- (\\frac{5}{6}))^{2} = \\frac{1}{36}\\)\nTaking square root of both sides,\n\\(\\sqrt{(y - (\\frac{5}{6}))^{2}} = \\sqrt{\\frac{1}{36}}\\)\n\\(y - \\frac{5}{6} =\\pm {\\frac{1}{6}}\\)\n\\(y = \\frac{5}{6} \\pm \\frac{1}{6}\\)\n\\(y = \\frac{5}{6} + \\frac{1}{6} = 1.00\\) or \\(y = \\frac{5}{6} - \\frac{1}{6} = \\frac{2}{3} = 0.67\\).\n(b) \\(M = \\begin{pmatrix} 1 & 2 \\\\ 4 & 3 \\end{pmatrix}, N = \\begin{pmatrix} m & x \\\\ n & y \\end{pmatrix}\\)\n\\(MN = \\begin{pmatrix} m + 2n & x + 2y \\\\ 4m + 2n & 4x + 2y \\end{pmatrix} = \\begin{pmatrix} 2 & 1 \\\\ 3 & 4 \\end{pmatrix}\\)\n\\(m + 2n = 2 ... (1)\\)\n\\(4m + 2n = 3 ... (2)\\)\nMultiplying (1) by 4, we have\n\\(4m + 8n = 8 ... (3)\\)\n(3) - (1) : \\(8n - 2n = 8 - 2 = 6\\)\n\\(6n = 6 \\implies n = 1\\)\n\\(m + 2n = m + 2(1) = m + 2 = 2\\)\n\\(m = 2 - 2 = 0\\)\n\\(x + 2y = 1 ... (1)\\)\n\\(4x + y = 4 ... (2)\\)\nMultiply (1) by 4,\n\\(4x + 8y = 4 ... (3)\\)\n(3) - (2) : \\(8y - y = 4 - 4 = 0\\)\n\\(\\implies y = 0\\)\n\\(x + 2y = x + 2(0) = x + 0 = 1\\)\n\\(x = 1 - 0 = 1\\)\n\\(\\therefore N = \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\end{pmatrix}\\).\n\n', '', 'Explanation\n(a) \\(3y^{2} - 5y + 2 = 0\\)\n\\(\\frac{3y^{2}}{3} - \\frac{5y}{3} = \\frac{-2}{3}\\)\n\\(y^{2} - \\frac{5}{3}y = \\frac{-2}{3}\\)\n\\(\\frac{1}{2} of \\frac{-5}{3} = \\frac{-5}{6}\\)\n\\(y^{2} - \\frac{5}{3}y + (-\\frac{5}{6})^{2} = \\frac{-2}{3} + (\\frac{-5}{6})^{2}\\)\n\\((y- (\\frac{5}{6}))^{2} = \\frac{1}{36}\\)\nTaking square root of both sides,\n\\(\\sqrt{(y - (\\frac{5}{6}))^{2}} = \\sqrt{\\frac{1}{36}}\\)\n\\(y - \\frac{5}{6} =\\pm {\\frac{1}{6}}\\)\n\\(y = \\frac{5}{6} \\pm \\frac{1}{6}\\)\n\\(y = \\frac{5}{6} + \\frac{1}{6} = 1.00\\) or \\(y = \\frac{5}{6} - \\frac{1}{6} = \\frac{2}{3} = 0.67\\).\n(b) \\(M = \\begin{pmatrix} 1 & 2 \\\\ 4 & 3 \\end{pmatrix}, N = \\begin{pmatrix} m & x \\\\ n & y \\end{pmatrix}\\)\n\\(MN = \\begin{pmatrix} m + 2n & x + 2y \\\\ 4m + 2n & 4x + 2y \\end{pmatrix} = \\begin{pmatrix} 2 & 1 \\\\ 3 & 4 \\end{pmatrix}\\)\n\\(m + 2n = 2 ... (1)\\)\n\\(4m + 2n = 3 ... (2)\\)\nMultiplying (1) by 4, we have\n\\(4m + 8n = 8 ... (3)\\)\n(3) - (1) : \\(8n - 2n = 8 - 2 = 6\\)\n\\(6n = 6 \\implies n = 1\\)\n\\(m + 2n = m + 2(1) = m + 2 = 2\\)\n\\(m = 2 - 2 = 0\\)\n\\(x + 2y = 1 ... (1)\\)\n\\(4x + y = 4 ... (2)\\)\nMultiply (1) by 4,\n\\(4x + 8y = 4 ... (3)\\)\n(3) - (2) : \\(8y - y = 4 - 4 = 0\\)\n\\(\\implies y = 0\\)\n\\(x + 2y = x + 2(0) = x + 0 = 1\\)\n\\(x = 1 - 0 = 1\\)\n\\(\\therefore N = \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\end{pmatrix}\\).\n\n', '(a) \\(3y^{2} - 5y + 2 = 0\\)\n\\(\\frac{3y^{2}}{3} - \\frac{5y}{3} = \\frac{-2}{3}\\)\n\\(y^{2} - \\frac{5}{3}y = \\frac{-2}{3}\\)\n\\(\\frac{1}{2} of \\frac{-5}{3} = \\frac{-5}{6}\\)\n\\(y^{2} - \\frac{5}{3}y + (-\\frac{5}{6})^{2} = \\frac{-2}{3} + (\\frac{-5}{6})^{2}\\)\n\\((y- (\\frac{5}{6}))^{2} = \\frac{1}{36}\\)\nTaking square root of both sides,\n\\(\\sqrt{(y - (\\frac{5}{6}))^{2}} = \\sqrt{\\frac{1}{36}}\\)\n\\(y - \\frac{5}{6} =\\pm {\\frac{1}{6}}\\)\n\\(y = \\frac{5}{6} \\pm \\frac{1}{6}\\)\n\\(y = \\frac{5}{6} + \\frac{1}{6} = 1.00\\) or \\(y = \\frac{5}{6} - \\frac{1}{6} = \\frac{2}{3} = 0.67\\).~(b) \\(M = \\begin{pmatrix} 1 & 2 \\\\ 4 & 3 \\end{pmatrix}, N = \\begin{pmatrix} m & x \\\\ n & y \\end{pmatrix}\\)\n\\(MN = \\begin{pmatrix} m + 2n & x + 2y \\\\ 4m + 2n & 4x + 2y \\end{pmatrix} = \\begin{pmatrix} 2 & 1 \\\\ 3 & 4 \\end{pmatrix}\\)\n\\(m + 2n = 2 ... (1)\\)\n\\(4m + 2n = 3 ... (2)\\)\nMultiplying (1) by 4, we have\n\\(4m + 8n = 8 ... (3)\\)\n(3) - (1) : \\(8n - 2n = 8 - 2 = 6\\)\n\\(6n = 6 \\implies n = 1\\)\n\\(m + 2n = m + 2(1) = m + 2 = 2\\)\n\\(m = 2 - 2 = 0\\)\n\\(x + 2y = 1 ... (1)\\)\n\\(4x + y = 4 ... (2)\\)\nMultiply (1) by 4,\n\\(4x + 8y = 4 ... (3)\\)\n(3) - (2) : \\(8y - y = 4 - 4 = 0\\)\n\\(\\implies y = 0\\)\n\\(x + 2y = x + 2(0) = x + 0 = 1\\)\n\\(x = 1 - 0 = 1\\)\n\\(\\therefore N = \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\end{pmatrix}\\).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(342, '2013', '\n\nQuestion\n\n(a) The operation (*) is defined on the set of real numbers, R, by \\(x * y = \\frac{x + y}{2}, x, y \\in R\\).\n(i) Evaluate \\(3 * \\frac{2}{5}\\).\n(ii) If \\(8 * y = 8\\frac{1}{4}\\), find the value of y.\n(b) In \\(\\Delta ABC, \\overline{AB} = \\begin{pmatrix} -4 \\\\ 6 \\end{pmatrix}\\) and \\(\\overline{AC} = \\begin{pmatrix} 3 \\\\ -8 \\end{pmatrix}\\). If P is the midpoint of \\(\\overline{AB}\\), express \\(\\overline{CP}\\) as a column vector.\n\n', '', '(a) The operation (*) is defined on the set of real numbers, R, by \\(x * y = \\frac{x + y}{2}, x, y \\in R\\).\n(i) Evaluate \\(3 * \\frac{2}{5}\\).\n(ii) If \\(8 * y = 8\\frac{1}{4}\\), find the value of y.\n(b) In \\(\\Delta ABC, \\overline{AB} = \\begin{pmatrix} -4 \\\\ 6 \\end{pmatrix}\\) and \\(\\overline{AC} = \\begin{pmatrix} 3 \\\\ -8 \\end{pmatrix}\\). If P is the midpoint of \\(\\overline{AB}\\), express \\(\\overline{CP}\\) as a column vector.', '(a) The operation (*) is defined on the set of real numbers, R, by \\(x * y = \\frac{x + y}{2}, x, y \\in R\\).\n(i) Evaluate \\(3 * \\frac{2}{5}\\).\n(ii) If \\(8 * y = 8\\frac{1}{4}\\), find the value of y.~(b) In \\(\\Delta ABC, \\overline{AB} = \\begin{pmatrix} -4 \\\\ 6 \\end{pmatrix}\\) and \\(\\overline{AC} = \\begin{pmatrix} 3 \\\\ -8 \\end{pmatrix}\\). If P is the midpoint of \\(\\overline{AB}\\), express \\(\\overline{CP}\\) as a column vector.', NULL, '\n\nExplanation\n(a) (i) \\(3 * \\frac{2}{5} = \\frac{3 + \\frac{2}{5}}{2}\\)\n= \\(\\frac{\\frac{17}{5}}{2}\\)\n= \\(\\frac{17}{10}\\)\n(ii) \\(8 * y = 8\\frac{1}{4}\\)\n\\(\\frac{8 + y}{2} = \\frac{33}{4}\\)\n\\(66 = 32 + 4y \\implies 4y = 66 - 32 = 34\\)\n\\(y = \\frac{34}{4} = 8.5\\)\n(b) \\(\\overrightarrow{AP} = \\frac{1}{2}(\\overrightarrow{AB})\\)\n= \\(\\frac{1}{2} \\begin{pmatrix} -4 \\\\ 6 \\end{pmatrix}\\)\n= \\(\\begin{pmatrix} -2 \\\\ 3 \\end{pmatrix}\\)\nMidpoint = \\(\\overrightarrow{AP} + \\overrightarrow{PC} = \\overrightarrow{AC}\\)\n\\(\\overrightarrow{PC} = \\overrightarrow{AC} - \\overrightarrow{AP}\\)\n= \\(\\begin{pmatrix} 3 \\\\ -8 \\end{pmatrix} - \\begin{pmatrix} -2 \\\\ 3 \\end{pmatrix}\\)\n= \\(\\begin{pmatrix} 5 \\\\ -11 \\end{pmatrix}\\)\n\\(\\overrightarrow{CP} = -1 \\times \\begin{pmatrix} 5 \\\\ -11 \\end{pmatrix}\\)\n= \\(\\begin{pmatrix} -5 \\\\ 11 \\end{pmatrix}\\)\n\n', '', 'Explanation\n(a) (i) \\(3 * \\frac{2}{5} = \\frac{3 + \\frac{2}{5}}{2}\\)\n= \\(\\frac{\\frac{17}{5}}{2}\\)\n= \\(\\frac{17}{10}\\)\n(ii) \\(8 * y = 8\\frac{1}{4}\\)\n\\(\\frac{8 + y}{2} = \\frac{33}{4}\\)\n\\(66 = 32 + 4y \\implies 4y = 66 - 32 = 34\\)\n\\(y = \\frac{34}{4} = 8.5\\)\n(b) \\(\\overrightarrow{AP} = \\frac{1}{2}(\\overrightarrow{AB})\\)\n= \\(\\frac{1}{2} \\begin{pmatrix} -4 \\\\ 6 \\end{pmatrix}\\)\n= \\(\\begin{pmatrix} -2 \\\\ 3 \\end{pmatrix}\\)\nMidpoint = \\(\\overrightarrow{AP} + \\overrightarrow{PC} = \\overrightarrow{AC}\\)\n\\(\\overrightarrow{PC} = \\overrightarrow{AC} - \\overrightarrow{AP}\\)\n= \\(\\begin{pmatrix} 3 \\\\ -8 \\end{pmatrix} - \\begin{pmatrix} -2 \\\\ 3 \\end{pmatrix}\\)\n= \\(\\begin{pmatrix} 5 \\\\ -11 \\end{pmatrix}\\)\n\\(\\overrightarrow{CP} = -1 \\times \\begin{pmatrix} 5 \\\\ -11 \\end{pmatrix}\\)\n= \\(\\begin{pmatrix} -5 \\\\ 11 \\end{pmatrix}\\)\n\n', '(a) (i) \\(3 * \\frac{2}{5} = \\frac{3 + \\frac{2}{5}}{2}\\)\n= \\(\\frac{\\frac{17}{5}}{2}\\)\n= \\(\\frac{17}{10}\\)\n(ii) \\(8 * y = 8\\frac{1}{4}\\)\n\\(\\frac{8 + y}{2} = \\frac{33}{4}\\)\n\\(66 = 32 + 4y \\implies 4y = 66 - 32 = 34\\)\n\\(y = \\frac{34}{4} = 8.5\\)~(b) \\(\\overrightarrow{AP} = \\frac{1}{2}(\\overrightarrow{AB})\\)\n= \\(\\frac{1}{2} \\begin{pmatrix} -4 \\\\ 6 \\end{pmatrix}\\)\n= \\(\\begin{pmatrix} -2 \\\\ 3 \\end{pmatrix}\\)\nMidpoint = \\(\\overrightarrow{AP} + \\overrightarrow{PC} = \\overrightarrow{AC}\\)\n\\(\\overrightarrow{PC} = \\overrightarrow{AC} - \\overrightarrow{AP}\\)\n= \\(\\begin{pmatrix} 3 \\\\ -8 \\end{pmatrix} - \\begin{pmatrix} -2 \\\\ 3 \\end{pmatrix}\\)\n= \\(\\begin{pmatrix} 5 \\\\ -11 \\end{pmatrix}\\)\n\\(\\overrightarrow{CP} = -1 \\times \\begin{pmatrix} 5 \\\\ -11 \\end{pmatrix}\\)\n= \\(\\begin{pmatrix} -5 \\\\ 11 \\end{pmatrix}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(343, '2013', '\n\nQuestion\n\n(a) Without using Mathematical tables or calculators, evaluate \\(\\frac{0.09 \\times 1.21}{3.3 \\times 0.00025}\\), leaving the answer in standard form (Scientific Notation).\n(b) A principal of GH¢5,600 was deposited for 3 years at compound interest. If the interest earned wasGH¢1,200, find, correct to 3 significant figures, the interest rate per annum.\n\n', '', '(a) Without using Mathematical tables or calculators, evaluate \\(\\frac{0.09 \\times 1.21}{3.3 \\times 0.00025}\\), leaving the answer in standard form (Scientific Notation).\n(b) A principal of GH¢5,600 was deposited for 3 years at compound interest. If the interest earned wasGH¢1,200, find, correct to 3 significant figures, the interest rate per annum.', '(a) Without using Mathematical tables or calculators, evaluate \\(\\frac{0.09 \\times 1.21}{3.3 \\times 0.00025}\\), leaving the answer in standard form (Scientific Notation).~(b) A principal of GH¢5,600 was deposited for 3 years at compound interest. If the interest earned wasGH¢1,200, find, correct to 3 significant figures, the interest rate per annum.', NULL, '\n\nExplanation\n(a) \\(\\frac{0.09 \\times 1.21}{3.3 \\times 0.00025} = \\frac{9 \\times 10^{-2} \\times 121 \\times 10^{-2}}{33 \\times 10^{-1} \\times 25 \\times 10^{-5}}\\)\n= \\(\\frac{33 \\times 10^{-4}}{25 \\times 10^{-6}}\\)\n= \\(1.32 \\times 10^{2}\\)\n(b) Amount, A = Principal + Interest\n=GH¢(5,600 + 1,200)\n=GH¢6,800\nCompound Interest formula : \\(A = P(1 + \\frac{r}{100})^{3}\\)\n\\(6,800 = 5,600(1 + \\frac{r}{100})^{3}\\)\n\\((1 + \\frac{r}{100})^{3} = \\frac{6800}{5600} = 1.214\\)\n\\(1 + \\frac{r}{100} = \\sqrt[3]{1.214} = 1.06686\\)\n\\(\\frac{r}{100} = 1.06686 - 1 = 0.06686\\)\n\\(r = 0.06686 \\times 100 = 6.686 \\approxeq 6.69\\) (3 significant figures)\n\n', '', 'Explanation\n(a) \\(\\frac{0.09 \\times 1.21}{3.3 \\times 0.00025} = \\frac{9 \\times 10^{-2} \\times 121 \\times 10^{-2}}{33 \\times 10^{-1} \\times 25 \\times 10^{-5}}\\)\n= \\(\\frac{33 \\times 10^{-4}}{25 \\times 10^{-6}}\\)\n= \\(1.32 \\times 10^{2}\\)\n(b) Amount, A = Principal + Interest\n=GH¢(5,600 + 1,200)\n=GH¢6,800\nCompound Interest formula : \\(A = P(1 + \\frac{r}{100})^{3}\\)\n\\(6,800 = 5,600(1 + \\frac{r}{100})^{3}\\)\n\\((1 + \\frac{r}{100})^{3} = \\frac{6800}{5600} = 1.214\\)\n\\(1 + \\frac{r}{100} = \\sqrt[3]{1.214} = 1.06686\\)\n\\(\\frac{r}{100} = 1.06686 - 1 = 0.06686\\)\n\\(r = 0.06686 \\times 100 = 6.686 \\approxeq 6.69\\) (3 significant figures)\n\n', '(a) \\(\\frac{0.09 \\times 1.21}{3.3 \\times 0.00025} = \\frac{9 \\times 10^{-2} \\times 121 \\times 10^{-2}}{33 \\times 10^{-1} \\times 25 \\times 10^{-5}}\\)\n= \\(\\frac{33 \\times 10^{-4}}{25 \\times 10^{-6}}\\)\n= \\(1.32 \\times 10^{2}\\)~(b) Amount, A = Principal + Interest\n=GH¢(5,600 + 1,200)\n=GH¢6,800\nCompound Interest formula : \\(A = P(1 + \\frac{r}{100})^{3}\\)\n\\(6,800 = 5,600(1 + \\frac{r}{100})^{3}\\)\n\\((1 + \\frac{r}{100})^{3} = \\frac{6800}{5600} = 1.214\\)\n\\(1 + \\frac{r}{100} = \\sqrt[3]{1.214} = 1.06686\\)\n\\(\\frac{r}{100} = 1.06686 - 1 = 0.06686\\)\n\\(r = 0.06686 \\times 100 = 6.686 \\approxeq 6.69\\) (3 significant figures)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(344, '2013', '\n\nQuestion\n\n(a) Solve : \\(7(x + 4) - \\frac{2}{3}(x - 6) \\leq 2[x - 3(x + 5)]\\)\n(b) A transport company has a total of 20 vehicles made up of tricycle and taxicabs. Each tricycle carries 2 passengers while each taxicab carries four passengers. If the 20 vehicles carry a total of 66 passengers at a time, how many tricycles does the company have?\n\n', '', '(a) Solve : \\(7(x + 4) - \\frac{2}{3}(x - 6) \\leq 2[x - 3(x + 5)]\\)\n(b) A transport company has a total of 20 vehicles made up of tricycle and taxicabs. Each tricycle carries 2 passengers while each taxicab carries four passengers. If the 20 vehicles carry a total of 66 passengers at a time, how many tricycles does the company have?', '(a) Solve : \\(7(x + 4) - \\frac{2}{3}(x - 6) \\leq 2[x - 3(x + 5)]\\)~(b) A transport company has a total of 20 vehicles made up of tricycle and taxicabs. Each tricycle carries 2 passengers while each taxicab carries four passengers. If the 20 vehicles carry a total of 66 passengers at a time, how many tricycles does the company have?', NULL, '\n\nExplanation\n(a) \\(7(x + 4) - \\frac{2}{3}(x - 6) \\leq 2[x - 3(x + 5)]\\)\nMultiply through by 3 to clear the fraction :\n\\(21(x + 4) - 2(x - 6) \\leq 6[x - 3(x + 5)]\\)\n\\(21x + 84 - 2x + 12 \\leq 6x - 18x - 90\\)\n\\(19x + 96 \\leq -12x - 90\\)\n\n\\(19x + 12x \\leq -90 - 96\\)\n\\(31x \\leq - 186\\)\nHence, \\(x \\leq -6\\).\n(b) Let : t represent number of tricycles; and b represent number of taxicabs\nThen \\(t + b = 20 ... (1)\\)\n\\(2t + 4b = 66 .... (2)\\)\nFrom (1), b= 20 - t. Put that into eqtn (2), we have:\n\\(2t + 4(20 - t) = 2t + 80 - 4t = 66\\)\n\\(80 - 66 = 14 = 2t \\)\n\\(t = 7\\)\nThe transport company has 7 tricycles.\n\n', '', 'Explanation\n(a) \\(7(x + 4) - \\frac{2}{3}(x - 6) \\leq 2[x - 3(x + 5)]\\)\nMultiply through by 3 to clear the fraction :\n\\(21(x + 4) - 2(x - 6) \\leq 6[x - 3(x + 5)]\\)\n\\(21x + 84 - 2x + 12 \\leq 6x - 18x - 90\\)\n\\(19x + 96 \\leq -12x - 90\\)\n\n\\(19x + 12x \\leq -90 - 96\\)\n\\(31x \\leq - 186\\)\nHence, \\(x \\leq -6\\).\n(b) Let : t represent number of tricycles; and b represent number of taxicabs\nThen \\(t + b = 20 ... (1)\\)\n\\(2t + 4b = 66 .... (2)\\)\nFrom (1), b= 20 - t. Put that into eqtn (2), we have:\n\\(2t + 4(20 - t) = 2t + 80 - 4t = 66\\)\n\\(80 - 66 = 14 = 2t \\)\n\\(t = 7\\)\nThe transport company has 7 tricycles.\n\n', '(a) \\(7(x + 4) - \\frac{2}{3}(x - 6) \\leq 2[x - 3(x + 5)]\\)\nMultiply through by 3 to clear the fraction :\n\\(21(x + 4) - 2(x - 6) \\leq 6[x - 3(x + 5)]\\)\n\\(21x + 84 - 2x + 12 \\leq 6x - 18x - 90\\)\n\\(19x + 96 \\leq -12x - 90\\)\n\n\\(19x + 12x \\leq -90 - 96\\)\n\\(31x \\leq - 186\\)\nHence, \\(x \\leq -6\\).~(b) Let : t represent number of tricycles; and b represent number of taxicabs\nThen \\(t + b = 20 ... (1)\\)\n\\(2t + 4b = 66 .... (2)\\)\nFrom (1), b= 20 - t. Put that into eqtn (2), we have:\n\\(2t + 4(20 - t) = 2t + 80 - 4t = 66\\)\n\\(80 - 66 = 14 = 2t \\)\n\\(t = 7\\)\nThe transport company has 7 tricycles.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(345, '2013', '\n\nQuestion\n\n(a)\nIn the diagram, < RTS = 28°, < VRM = 46°, MQ is a tangent to the circle VRSTU at the point R. Find < VUS.\n(b) A cylinder tin, 7cm high, is closed at one end. If its total surface area is 462\\(cm^{2}\\), calculate its radius. [Take \\(\\pi = \\frac{22}{7}\\)].\n\n', '', '(a)\nIn the diagram, < RTS = 28°, < VRM = 46°, MQ is a tangent to the circle VRSTU at the point R. Find < VUS.\n(b) A cylinder tin, 7cm high, is closed at one end. If its total surface area is 462\\(cm^{2}\\), calculate its radius. [Take \\(\\pi = \\frac{22}{7}\\)].', '(a)\nIn the diagram, < RTS = 28°, < VRM = 46°, MQ is a tangent to the circle VRSTU at the point R. Find < VUS.~(b) A cylinder tin, 7cm high, is closed at one end. If its total surface area is 462\\(cm^{2}\\), calculate its radius. [Take \\(\\pi = \\frac{22}{7}\\)].', NULL, '\n\nExplanation\n(a)\nIn the diagram above, \\(\\alpha = 28°\\) (angles in alternate segment)\n\\(46° + \\beta + \\alpha = 180°\\) (angles on a straight line)\n\\(46° + \\beta + 28° = 74° + \\beta = 180°\\)\n\\(\\beta = 180° - 74° = 106°\\)\n\\(x + \\beta = 180°\\) (opposite angles of a cyclic quadrilateral)\n\\(x + 106° = 180°\\)\n\\(x = 180° - 106° = 74°\\)\nHence, < VUS = 74°.\n(b) Total surface area = curved surface area + area of closed end\n= \\(2\\pi rh + \\pi r^{2}\\)\n= \\(\\pi r(2h + r)\\)\n\\(462 = \\frac{22r}{7}((2 \\times 7) + r)\\)\n\\(\\frac{462 \\times 7}{22} = r(14 + r)\\)\n\\(147 = 14r + r^{2}\\)\n\\(r^{2} + 14r - 147 = 0\\)\n\\(r^{2} - 7r + 21r - 147 = 0\\)\n\\(r(r - 7) + 21(r - 7) = 0\\)\n\\((r + 21)(r - 7) = 0\\)\n\\(r = \\text{-21 or 7}\\)\nThe radius cannot be negative, therefore, r = 7 cm.\n\n', '', 'Explanation\n(a)\nIn the diagram above, \\(\\alpha = 28°\\) (angles in alternate segment)\n\\(46° + \\beta + \\alpha = 180°\\) (angles on a straight line)\n\\(46° + \\beta + 28° = 74° + \\beta = 180°\\)\n\\(\\beta = 180° - 74° = 106°\\)\n\\(x + \\beta = 180°\\) (opposite angles of a cyclic quadrilateral)\n\\(x + 106° = 180°\\)\n\\(x = 180° - 106° = 74°\\)\nHence, < VUS = 74°.\n(b) Total surface area = curved surface area + area of closed end\n= \\(2\\pi rh + \\pi r^{2}\\)\n= \\(\\pi r(2h + r)\\)\n\\(462 = \\frac{22r}{7}((2 \\times 7) + r)\\)\n\\(\\frac{462 \\times 7}{22} = r(14 + r)\\)\n\\(147 = 14r + r^{2}\\)\n\\(r^{2} + 14r - 147 = 0\\)\n\\(r^{2} - 7r + 21r - 147 = 0\\)\n\\(r(r - 7) + 21(r - 7) = 0\\)\n\\((r + 21)(r - 7) = 0\\)\n\\(r = \\text{-21 or 7}\\)\nThe radius cannot be negative, therefore, r = 7 cm.\n\n', '(a)\nIn the diagram above, \\(\\alpha = 28°\\) (angles in alternate segment)\n\\(46° + \\beta + \\alpha = 180°\\) (angles on a straight line)\n\\(46° + \\beta + 28° = 74° + \\beta = 180°\\)\n\\(\\beta = 180° - 74° = 106°\\)\n\\(x + \\beta = 180°\\) (opposite angles of a cyclic quadrilateral)\n\\(x + 106° = 180°\\)\n\\(x = 180° - 106° = 74°\\)\nHence, < VUS = 74°.~(b) Total surface area = curved surface area + area of closed end\n= \\(2\\pi rh + \\pi r^{2}\\)\n= \\(\\pi r(2h + r)\\)\n\\(462 = \\frac{22r}{7}((2 \\times 7) + r)\\)\n\\(\\frac{462 \\times 7}{22} = r(14 + r)\\)\n\\(147 = 14r + r^{2}\\)\n\\(r^{2} + 14r - 147 = 0\\)\n\\(r^{2} - 7r + 21r - 147 = 0\\)\n\\(r(r - 7) + 21(r - 7) = 0\\)\n\\((r + 21)(r - 7) = 0\\)\n\\(r = \\text{-21 or 7}\\)\nThe radius cannot be negative, therefore, r = 7 cm.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(346, '2013', '\n\nQuestion\n\n\n\nScores\n1\n2\n3\n4\n5\n6\n\n\nFrequency\n25\n30\nx\n28\n40\n32\n\n\n\nThe table shows the outcome when a die is thrown a number of times. If the probability of obtaining a 3 is 0.225;\n(a) How many times was the die thrown?\n(b) Calculate the probability that a trial chosen at random gives a score of an even number or a prime number.\n\n\n\n\n\nScores\n1\n2\n3\n4\n5\n6\n\n\nFrequency\n25\n30\nx\n28\n40\n32\n\n\n', '', 'Scores\n1\n2\n3\n4\n5\n6\n\n\nFrequency\n25\n30\nx\n28\n40\n32\n\n\n\nThe table shows the outcome when a die is thrown a number of times. If the probability of obtaining a 3 is 0.225;\n(a) How many times was the die thrown?\n(b) Calculate the probability that a trial chosen at random gives a score of an even number or a prime number.\n\n\n\n\n\nScores\n1\n2\n3\n4\n5\n6\n\n\nFrequency\n25\n30\nx\n28\n40\n32\n', 'Scores\n1\n2\n3\n4\n5\n6\n\n\nFrequency\n25\n30\nx\n28\n40\n32\n\n\n\nThe table shows the outcome when a die is thrown a number of times. If the probability of obtaining a 3 is 0.225;\n(a) How many times was the die thrown?~(b) Calculate the probability that a trial chosen at random gives a score of an even number or a prime number.\n\n\n\n\n\nScores\n1\n2\n3\n4\n5\n6\n\n\nFrequency\n25\n30\nx\n28\n40\n32\n', NULL, '\n\nExplanation\n(a) From the table, \\(\\sum f = x + 155\\)\nP(obtaining a 3) = \\(\\frac{\\text{no of times a 3 was obtained}}{\\sum f}\\)\n\\(\\frac{x}{x + 155} = 0.225\\)\n\\(x = 0.255x + (155 \\times 0.225)\\)\n\\(x - 0.225x = 34.875 \\)\n\\(0.775x = 34.875\\)\n\\(x = \\frac{39.875}{0.775}\\)\n\\(x = 45\\)\n\\(\\sum f = 155 + x = 155 + 45 = 200\\)\nHence, the die was thrown 200 times.\n(b) Set of even numbers = {2, 4, 6}\nSet of prime numbers = {2, 4, 5}\nTherefore, \\(N \\cup M = (2, 3, 4,5, 6)\\)\n\\(n(N \\cup M) = 30 + 45 + 28 + 40 + 32 = 175\\)\n\\(P(N \\cup M) = \\frac{175}{200} = 0.875\\)\n\n', '', 'Explanation\n(a) From the table, \\(\\sum f = x + 155\\)\nP(obtaining a 3) = \\(\\frac{\\text{no of times a 3 was obtained}}{\\sum f}\\)\n\\(\\frac{x}{x + 155} = 0.225\\)\n\\(x = 0.255x + (155 \\times 0.225)\\)\n\\(x - 0.225x = 34.875 \\)\n\\(0.775x = 34.875\\)\n\\(x = \\frac{39.875}{0.775}\\)\n\\(x = 45\\)\n\\(\\sum f = 155 + x = 155 + 45 = 200\\)\nHence, the die was thrown 200 times.\n(b) Set of even numbers = {2, 4, 6}\nSet of prime numbers = {2, 4, 5}\nTherefore, \\(N \\cup M = (2, 3, 4,5, 6)\\)\n\\(n(N \\cup M) = 30 + 45 + 28 + 40 + 32 = 175\\)\n\\(P(N \\cup M) = \\frac{175}{200} = 0.875\\)\n\n', '(a) From the table, \\(\\sum f = x + 155\\)\nP(obtaining a 3) = \\(\\frac{\\text{no of times a 3 was obtained}}{\\sum f}\\)\n\\(\\frac{x}{x + 155} = 0.225\\)\n\\(x = 0.255x + (155 \\times 0.225)\\)\n\\(x - 0.225x = 34.875 \\)\n\\(0.775x = 34.875\\)\n\\(x = \\frac{39.875}{0.775}\\)\n\\(x = 45\\)\n\\(\\sum f = 155 + x = 155 + 45 = 200\\)\nHence, the die was thrown 200 times.~(b) Set of even numbers = {2, 4, 6}\nSet of prime numbers = {2, 4, 5}\nTherefore, \\(N \\cup M = (2, 3, 4,5, 6)\\)\n\\(n(N \\cup M) = 30 + 45 + 28 + 40 + 32 = 175\\)\n\\(P(N \\cup M) = \\frac{175}{200} = 0.875\\)\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
(347, '2013', '\n\nQuestion\n\n(a)\nIn the diagram, PQST is a parallelogram, PR is a straight line, |TS| = 8cm, |SM| = 6cm and area of triangle PSR = \\(36 cm^{2}\\). Find the value of |QR|.\n(b) A tree and a flagpole are on the same horizontal ground. A bird on top of the tree observes the top and bottom of the flagpole below it at angle of 45° and 60° respectively. If the tree is 10.65m high, calculate, correct to 3 significant figures, the height of the flagpole.\n\n', '', '(a)\nIn the diagram, PQST is a parallelogram, PR is a straight line, |TS| = 8cm, |SM| = 6cm and area of triangle PSR = \\(36 cm^{2}\\). Find the value of |QR|.\n(b) A tree and a flagpole are on the same horizontal ground. A bird on top of the tree observes the top and bottom of the flagpole below it at angle of 45° and 60° respectively. If the tree is 10.65m high, calculate, correct to 3 significant figures, the height of the flagpole.', '(a)\nIn the diagram, PQST is a parallelogram, PR is a straight line, |TS| = 8cm, |SM| = 6cm and area of triangle PSR = \\(36 cm^{2}\\). Find the value of |QR|.~(b) A tree and a flagpole are on the same horizontal ground. A bird on top of the tree observes the top and bottom of the flagpole below it at angle of 45° and 60° respectively. If the tree is 10.65m high, calculate, correct to 3 significant figures, the height of the flagpole.', NULL, '\n\nExplanation\n(a)\nIn the diagram above, area of \\(\\Delta PSR = \\frac{1}{2} \\times |PR| \\times 6\\)\n\\(36 = \\frac{1}{2} \\times |PR| \\times 6\\)\n\\(36 = 3 \\times |PR|\\)\n\\(|PR| = 12 cm\\)\nBut |PQ| = |TS| = 8cm (opp. sides of a parallelogram)\nHence, |QR| = |PR| - |PQ| = 12 - 8 = 4cm.\n(b)\nIn the diagram above, \\(\\beta = 60°\\) (alternate angles)\n\\(\\tan \\beta = \\frac{10.65}{|QB|}\\)\n\\(|QB| = \\frac{10.65}{\\tan 60}\\)\n= \\(\\frac{10.65}{1.732} = 6.149m\\)\n|PO| = |QB| = 6.149m (opp. sides of a rectangle)\nIn \\(\\Delta POT, \\tan 45 = \\frac{|TO|}{6.149}\\)\n\\(|TO| = \\tan 45 \\times 6.149 = 1 \\times 6.149 = 6.149m\\)\n|OB| = |PQ| = 10.65m (opp sides of a rectangle)\n\\(h = |OB| - |OT| \\)\n= \\(10.65 - 6.149\\)\n= \\(4.501 m \\approxeq 4.50m\\) (3 sig. figs)\nHence, the height of the flagpole = 4.50m\n\n', '', 'Explanation\n(a)\nIn the diagram above, area of \\(\\Delta PSR = \\frac{1}{2} \\times |PR| \\times 6\\)\n\\(36 = \\frac{1}{2} \\times |PR| \\times 6\\)\n\\(36 = 3 \\times |PR|\\)\n\\(|PR| = 12 cm\\)\nBut |PQ| = |TS| = 8cm (opp. sides of a parallelogram)\nHence, |QR| = |PR| - |PQ| = 12 - 8 = 4cm.\n(b)\nIn the diagram above, \\(\\beta = 60°\\) (alternate angles)\n\\(\\tan \\beta = \\frac{10.65}{|QB|}\\)\n\\(|QB| = \\frac{10.65}{\\tan 60}\\)\n= \\(\\frac{10.65}{1.732} = 6.149m\\)\n|PO| = |QB| = 6.149m (opp. sides of a rectangle)\nIn \\(\\Delta POT, \\tan 45 = \\frac{|TO|}{6.149}\\)\n\\(|TO| = \\tan 45 \\times 6.149 = 1 \\times 6.149 = 6.149m\\)\n|OB| = |PQ| = 10.65m (opp sides of a rectangle)\n\\(h = |OB| - |OT| \\)\n= \\(10.65 - 6.149\\)\n= \\(4.501 m \\approxeq 4.50m\\) (3 sig. figs)\nHence, the height of the flagpole = 4.50m\n\n', '(a)\nIn the diagram above, area of \\(\\Delta PSR = \\frac{1}{2} \\times |PR| \\times 6\\)\n\\(36 = \\frac{1}{2} \\times |PR| \\times 6\\)\n\\(36 = 3 \\times |PR|\\)\n\\(|PR| = 12 cm\\)\nBut |PQ| = |TS| = 8cm (opp. sides of a parallelogram)\nHence, |QR| = |PR| - |PQ| = 12 - 8 = 4cm.~(b)\nIn the diagram above, \\(\\beta = 60°\\) (alternate angles)\n\\(\\tan \\beta = \\frac{10.65}{|QB|}\\)\n\\(|QB| = \\frac{10.65}{\\tan 60}\\)\n= \\(\\frac{10.65}{1.732} = 6.149m\\)\n|PO| = |QB| = 6.149m (opp. sides of a rectangle)\nIn \\(\\Delta POT, \\tan 45 = \\frac{|TO|}{6.149}\\)\n\\(|TO| = \\tan 45 \\times 6.149 = 1 \\times 6.149 = 6.149m\\)\n|OB| = |PQ| = 10.65m (opp sides of a rectangle)\n\\(h = |OB| - |OT| \\)\n= \\(10.65 - 6.149\\)\n= \\(4.501 m \\approxeq 4.50m\\) (3 sig. figs)\nHence, the height of the flagpole = 4.50m\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(348, '2013', '\n\nQuestion\n\n(a) Find the sum of the Arithmetic Progression (AP) 1, 3, 5,..., 101.\n(b) Out of the 95 travellers interviewed, 7 travelled by bus and train only, 3 by train and car only and 8 travelled by all 3 means of transport. The number, x, of travellerswho travelled by bus only was equal to the number who travelled by bus and car only. If 47 people travelled by bus and 30 by train :\n(i) represent this information in a Venn diagram;\n(ii) calculate the\n(1) value of x ; (2) number who travelled by at least two means.\n\n', '', '(a) Find the sum of the Arithmetic Progression (AP) 1, 3, 5,..., 101.\n(b) Out of the 95 travellers interviewed, 7 travelled by bus and train only, 3 by train and car only and 8 travelled by all 3 means of transport. The number, x, of travellerswho travelled by bus only was equal to the number who travelled by bus and car only. If 47 people travelled by bus and 30 by train :\n(i) represent this information in a Venn diagram;\n(ii) calculate the\n(1) value of x ; (2) number who travelled by at least two means.', '(a) Find the sum of the Arithmetic Progression (AP) 1, 3, 5,..., 101.~(b) Out of the 95 travellers interviewed, 7 travelled by bus and train only, 3 by train and car only and 8 travelled by all 3 means of transport. The number, x, of travellerswho travelled by bus only was equal to the number who travelled by bus and car only. If 47 people travelled by bus and 30 by train :\n(i) represent this information in a Venn diagram;\n(ii) calculate the\n(1) value of x ; (2) number who travelled by at least two means.', NULL, '\n\nExplanation\n(a)\n(b)(i) From the Venn diagram,\nx + x + 7 + 8 = n(B)\n2x + 15 = 47\n2x = 47 - 15 = 32\nx = 16\n(ii) Number of people who travelled by at least 2 means\n= 16 + 7 + 8 + 3 = 34\n\n', '', 'Explanation\n(a)\n(b)(i) From the Venn diagram,\nx + x + 7 + 8 = n(B)\n2x + 15 = 47\n2x = 47 - 15 = 32\nx = 16\n(ii) Number of people who travelled by at least 2 means\n= 16 + 7 + 8 + 3 = 34\n\n', '~(b)(i) From the Venn diagram,\nx + x + 7 + 8 = n(B)\n2x + 15 = 47\n2x = 47 - 15 = 32\nx = 16\n(ii) Number of people who travelled by at least 2 means\n= 16 + 7 + 8 + 3 = 34\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(349, '2013', '\n\nQuestion\n\n(a) Using the method of completing the square, solve, correct to 2 decimal places, \\(\\frac{x - 2}{4} = \\frac{x + 2}{2x}\\).\n(b)\nIn the diagram, PQRST is a circle with centre O. If PS is a diameter, RS//QT, and < QTS = 52°, find :\n(i) < SQT ; (ii) < PQT.\n\n', '', '(a) Using the method of completing the square, solve, correct to 2 decimal places, \\(\\frac{x - 2}{4} = \\frac{x + 2}{2x}\\).\n(b)\nIn the diagram, PQRST is a circle with centre O. If PS is a diameter, RS//QT, and < QTS = 52°, find :\n(i) < SQT ; (ii) < PQT.', '(a) Using the method of completing the square, solve, correct to 2 decimal places, \\(\\frac{x - 2}{4} = \\frac{x + 2}{2x}\\).~(b)\nIn the diagram, PQRST is a circle with centre O. If PS is a diameter, RS//QT, and < QTS = 52°, find :\n(i) < SQT ; (ii) < PQT.', NULL, '\n\nExplanation\n(a) \\(\\frac{x - 2}{4} = \\frac{x + 2}{2x}\\)\n\\(2x(x - 2) = 4(x + 2)\\)\n\\(2x^{2} - 4x = 4x + 8\\)\n\\(2x^{2} - 4x - 4x - 8 = 2x^{2} - 8x - 8 = 0\\)\nDivide through by 2, we have\n\\(x^{2} - 4x - 4 = 0\\)\n\\(x^{2} - 4x = 4\\)\nTaking the square of \\(\\frac{b}{2}\\) and add to both sides,\n\\(x^{2} - 4x + (-2)^{2} = 4 + (-2)^{2}\\)\n\\((x - 2)^{2} = 8\\)\n\\(x - 2 = \\pm {\\sqrt{8}}\\)\n\\(x = 2 \\pm \\sqrt{8}\\)\n\\(x = 2 \\pm 2.828\\)\n\\(x = 4.828\\) or \\(x = -0.828\\).\nHence, x = 4.83 or -0.83 (2 decimal place).\n(b)(i)\nIn the diagram above, < QRS = 180° - 52° = 128° (opp. angles of a cyclic quadrilateral).\n\\(x_{1} = x_{2}\\) (base angles of isosceles triangle)\n\\(x_{1} + x_{2} + 128° = 180°\\) (sum of angles of a triangle)\n\\(2x_{2} = 180° - 128° = 52°\\)\n\\(x_{2} = 26°\\)\nHence, < SQT = 26°.\n(ii) < PQS = 90° (angles in a semi-circle)\nHence, < PQT = 90° - 26° = 64°.\n\n\n', '', 'Explanation\n(a) \\(\\frac{x - 2}{4} = \\frac{x + 2}{2x}\\)\n\\(2x(x - 2) = 4(x + 2)\\)\n\\(2x^{2} - 4x = 4x + 8\\)\n\\(2x^{2} - 4x - 4x - 8 = 2x^{2} - 8x - 8 = 0\\)\nDivide through by 2, we have\n\\(x^{2} - 4x - 4 = 0\\)\n\\(x^{2} - 4x = 4\\)\nTaking the square of \\(\\frac{b}{2}\\) and add to both sides,\n\\(x^{2} - 4x + (-2)^{2} = 4 + (-2)^{2}\\)\n\\((x - 2)^{2} = 8\\)\n\\(x - 2 = \\pm {\\sqrt{8}}\\)\n\\(x = 2 \\pm \\sqrt{8}\\)\n\\(x = 2 \\pm 2.828\\)\n\\(x = 4.828\\) or \\(x = -0.828\\).\nHence, x = 4.83 or -0.83 (2 decimal place).\n(b)(i)\nIn the diagram above, < QRS = 180° - 52° = 128° (opp. angles of a cyclic quadrilateral).\n\\(x_{1} = x_{2}\\) (base angles of isosceles triangle)\n\\(x_{1} + x_{2} + 128° = 180°\\) (sum of angles of a triangle)\n\\(2x_{2} = 180° - 128° = 52°\\)\n\\(x_{2} = 26°\\)\nHence, < SQT = 26°.\n(ii) < PQS = 90° (angles in a semi-circle)\nHence, < PQT = 90° - 26° = 64°.\n\n\n', '(a) \\(\\frac{x - 2}{4} = \\frac{x + 2}{2x}\\)\n\\(2x(x - 2) = 4(x + 2)\\)\n\\(2x^{2} - 4x = 4x + 8\\)\n\\(2x^{2} - 4x - 4x - 8 = 2x^{2} - 8x - 8 = 0\\)\nDivide through by 2, we have\n\\(x^{2} - 4x - 4 = 0\\)\n\\(x^{2} - 4x = 4\\)\nTaking the square of \\(\\frac{b}{2}\\) and add to both sides,\n\\(x^{2} - 4x + (-2)^{2} = 4 + (-2)^{2}\\)\n\\((x - 2)^{2} = 8\\)\n\\(x - 2 = \\pm {\\sqrt{8}}\\)\n\\(x = 2 \\pm \\sqrt{8}\\)\n\\(x = 2 \\pm 2.828\\)\n\\(x = 4.828\\) or \\(x = -0.828\\).\nHence, x = 4.83 or -0.83 (2 decimal place).~(b)(i)\nIn the diagram above, < QRS = 180° - 52° = 128° (opp. angles of a cyclic quadrilateral).\n\\(x_{1} = x_{2}\\) (base angles of isosceles triangle)\n\\(x_{1} + x_{2} + 128° = 180°\\) (sum of angles of a triangle)\n\\(2x_{2} = 180° - 128° = 52°\\)\n\\(x_{2} = 26°\\)\nHence, < SQT = 26°.\n(ii) < PQS = 90° (angles in a semi-circle)\nHence, < PQT = 90° - 26° = 64°.\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(350, '2013', '\n\nQuestion\n\n(a)\nIn the diagram, < KLM = x, < LMK = y, < KJH = r and < KGF = 110°. If 2x = r = y, find the value of x.\n(b) Ten boys and twelve girls collected donations for a project. The total amount collected by the boys was N600.00 gretaer than that collected by the girls. If the average collection of the boys was N100.00 greater than the average collection of the girls, how much was collected by the two groups?\n\n', '', '(a)\nIn the diagram, < KLM = x, < LMK = y, < KJH = r and < KGF = 110°. If 2x = r = y, find the value of x.\n(b) Ten boys and twelve girls collected donations for a project. The total amount collected by the boys was N600.00 gretaer than that collected by the girls. If the average collection of the boys was N100.00 greater than the average collection of the girls, how much was collected by the two groups?', '(a)\nIn the diagram, < KLM = x, < LMK = y, < KJH = r and < KGF = 110°. If 2x = r = y, find the value of x.~(b) Ten boys and twelve girls collected donations for a project. The total amount collected by the boys was N600.00 gretaer than that collected by the girls. If the average collection of the boys was N100.00 greater than the average collection of the girls, how much was collected by the two groups?', NULL, '\n\nExplanation\n(a)\nIn the diagram above, \\(\\theta = y\\) (vertically opposite angles)\n\\(\\alpha = 180° - 110° = 70°\\) (angles on a straight line)\nThus, \\(\\beta = \\theta + \\alpha\\) (sum of interior opp. angles)\n\\(\\beta = y + 70° ..... (1)\\)\n\\(x + r + \\beta = 180° ..... (2)\\)\nBut 2x = r = y\nHence, \\(x + 2x + (2x + 70) = 5x + 70 = 180\\)\n\\(5x = 180 - 70 = 110\\)\n\\(x = 22°\\)\n(b) Let Nx represent the donation collected by the girls. Then, the donation collected by the boys = N(x + 600).\nAlso, let y represent the average collection, \\(x_{g}\\), of the girls. Then the average collection, \\(x_{b}\\), of the boys = N(y + 100).\nThus, \\(x_{g} = \\frac{x}{12} = y\\)\n\\(x = 12y ... (1)\\)\n\\(x_{b} = \\frac{x + 600}{10} = y + 100\\)\n\\(x + 600 = 10(y + 100) .... (2)\\)\nSubstitute 12y for x in (2) :\n\\(12y + 600 = 10y + 1000\\)\n\\(12y - 10y = 1000 - 600 \\)\n\\(2y = 400 \\implies y = 200\\)\nFrom (1), \\(x = 12y\\)\n\\(\\therefore x = 12 \\times 200 = N2,400\\).\nThus, the girls collected N2,400\nThe boys collected N2,400 + N600 = N3000\nTotal = N(2,400 + 3,000) = N5,400\n\n', '', 'Explanation\n(a)\nIn the diagram above, \\(\\theta = y\\) (vertically opposite angles)\n\\(\\alpha = 180° - 110° = 70°\\) (angles on a straight line)\nThus, \\(\\beta = \\theta + \\alpha\\) (sum of interior opp. angles)\n\\(\\beta = y + 70° ..... (1)\\)\n\\(x + r + \\beta = 180° ..... (2)\\)\nBut 2x = r = y\nHence, \\(x + 2x + (2x + 70) = 5x + 70 = 180\\)\n\\(5x = 180 - 70 = 110\\)\n\\(x = 22°\\)\n(b) Let Nx represent the donation collected by the girls. Then, the donation collected by the boys = N(x + 600).\nAlso, let y represent the average collection, \\(x_{g}\\), of the girls. Then the average collection, \\(x_{b}\\), of the boys = N(y + 100).\nThus, \\(x_{g} = \\frac{x}{12} = y\\)\n\\(x = 12y ... (1)\\)\n\\(x_{b} = \\frac{x + 600}{10} = y + 100\\)\n\\(x + 600 = 10(y + 100) .... (2)\\)\nSubstitute 12y for x in (2) :\n\\(12y + 600 = 10y + 1000\\)\n\\(12y - 10y = 1000 - 600 \\)\n\\(2y = 400 \\implies y = 200\\)\nFrom (1), \\(x = 12y\\)\n\\(\\therefore x = 12 \\times 200 = N2,400\\).\nThus, the girls collected N2,400\nThe boys collected N2,400 + N600 = N3000\nTotal = N(2,400 + 3,000) = N5,400\n\n', '(a)\nIn the diagram above, \\(\\theta = y\\) (vertically opposite angles)\n\\(\\alpha = 180° - 110° = 70°\\) (angles on a straight line)\nThus, \\(\\beta = \\theta + \\alpha\\) (sum of interior opp. angles)\n\\(\\beta = y + 70° ..... (1)\\)\n\\(x + r + \\beta = 180° ..... (2)\\)\nBut 2x = r = y\nHence, \\(x + 2x + (2x + 70) = 5x + 70 = 180\\)\n\\(5x = 180 - 70 = 110\\)\n\\(x = 22°\\)~(b) Let Nx represent the donation collected by the girls. Then, the donation collected by the boys = N(x + 600).\nAlso, let y represent the average collection, \\(x_{g}\\), of the girls. Then the average collection, \\(x_{b}\\), of the boys = N(y + 100).\nThus, \\(x_{g} = \\frac{x}{12} = y\\)\n\\(x = 12y ... (1)\\)\n\\(x_{b} = \\frac{x + 600}{10} = y + 100\\)\n\\(x + 600 = 10(y + 100) .... (2)\\)\nSubstitute 12y for x in (2) :\n\\(12y + 600 = 10y + 1000\\)\n\\(12y - 10y = 1000 - 600 \\)\n\\(2y = 400 \\implies y = 200\\)\nFrom (1), \\(x = 12y\\)\n\\(\\therefore x = 12 \\times 200 = N2,400\\).\nThus, the girls collected N2,400\nThe boys collected N2,400 + N600 = N3000\nTotal = N(2,400 + 3,000) = N5,400\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(351, '2013', '\n\nQuestion\n\nThe weight (in kg) of 50 contestants at a competition is as follows:\n65 66 67 66 64 66 65 63 65 68 64 62 66 64 67 65 64 66 65 67 65 67 66 64 65 64 66 65 64 65 66 65 64 65 63 63 67 65 63 64 66 64 68 65 63 65 64 67 66 64.\n(a) Construct a frequenct table for the discrete data.\n(b) Calculate, correct to 2 decimal places, the;\n(i) mean ; (ii) standard deviation of the data.\n\n', '', 'The weight (in kg) of 50 contestants at a competition is as follows:\n65 66 67 66 64 66 65 63 65 68 64 62 66 64 67 65 64 66 65 67 65 67 66 64 65 64 66 65 64 65 66 65 64 65 63 63 67 65 63 64 66 64 68 65 63 65 64 67 66 64.\n(a) Construct a frequenct table for the discrete data.\n(b) Calculate, correct to 2 decimal places, the;\n(i) mean ; (ii) standard deviation of the data.', 'The weight (in kg) of 50 contestants at a competition is as follows:\n65 66 67 66 64 66 65 63 65 68 64 62 66 64 67 65 64 66 65 67 65 67 66 64 65 64 66 65 64 65 66 65 64 65 63 63 67 65 63 64 66 64 68 65 63 65 64 67 66 64.\n(a) Construct a frequenct table for the discrete data.~(b) Calculate, correct to 2 decimal places, the;\n(i) mean ; (ii) standard deviation of the data.', NULL, '\n\nExplanation\n(a)\n\n\n\nWeight (kg)\nFrequency\n\n\n62\n1\n\n\n63\n5\n\n\n64\n12\n\n\n65\n14\n\n\n66\n10\n\n\n67\n6\n\n\n68\n2\n\n\n\n(b) Using an assumed mean of 65kg\n\n\n\n\\(x\\)\n\\(f\\)\n\\(fx\\)\n\\(d = x - A\\)\n\\(fd\\)\n\\(fd^{2}\\)\n\n\n62\n1\n62\n-3\n-3\n9\n\n\n63\n5\n315\n-2\n-10\n20\n\n\n64\n12\n768\n-1\n-12\n12\n\n\n65\n14\n910\n0\n0\n0\n\n\n66\n10\n660\n1\n10\n10\n\n\n67\n6\n402\n2\n12\n24\n\n\n68\n2\n136\n3\n6\n18\n\n\n\\(\\sum\\) =\n50\n3253\n\n3\n93\n\n\n\n(i) Mean \\(\\bar{x} = \\frac{\\sum fx}{\\sum f}\\)\n= \\(\\frac{3253}{50}\\)\n= 65.06 kg\n(ii) \\(SD = \\sqrt{\\frac{\\sum fd^{2}}{\\sum f} - (\\frac{\\sum fd^{2}}{\\sum f})^{2}}\\)\n\\(SD = \\sqrt{\\frac{93}{50} - (\\frac{3}{50})^{2}}\\)\n= \\(\\sqrt{\\frac{93}{50} - \\frac{9}{2500}}\\)\n= \\(\\sqrt{\\frac{4650 - 9}{2500}}\\)\n= \\(\\sqrt{\\frac{4641}{2500}}\\)\n= \\(1.3625 \\approxeq 1.36\\) (2 d.p.)\n\n', '', 'Explanation\n(a)\n\n\n\nWeight (kg)\nFrequency\n\n\n62\n1\n\n\n63\n5\n\n\n64\n12\n\n\n65\n14\n\n\n66\n10\n\n\n67\n6\n\n\n68\n2\n\n\n\n(b) Using an assumed mean of 65kg\n\n\n\n\\(x\\)\n\\(f\\)\n\\(fx\\)\n\\(d = x - A\\)\n\\(fd\\)\n\\(fd^{2}\\)\n\n\n62\n1\n62\n-3\n-3\n9\n\n\n63\n5\n315\n-2\n-10\n20\n\n\n64\n12\n768\n-1\n-12\n12\n\n\n65\n14\n910\n0\n0\n0\n\n\n66\n10\n660\n1\n10\n10\n\n\n67\n6\n402\n2\n12\n24\n\n\n68\n2\n136\n3\n6\n18\n\n\n\\(\\sum\\) =\n50\n3253\n\n3\n93\n\n\n\n(i) Mean \\(\\bar{x} = \\frac{\\sum fx}{\\sum f}\\)\n= \\(\\frac{3253}{50}\\)\n= 65.06 kg\n(ii) \\(SD = \\sqrt{\\frac{\\sum fd^{2}}{\\sum f} - (\\frac{\\sum fd^{2}}{\\sum f})^{2}}\\)\n\\(SD = \\sqrt{\\frac{93}{50} - (\\frac{3}{50})^{2}}\\)\n= \\(\\sqrt{\\frac{93}{50} - \\frac{9}{2500}}\\)\n= \\(\\sqrt{\\frac{4650 - 9}{2500}}\\)\n= \\(\\sqrt{\\frac{4641}{2500}}\\)\n= \\(1.3625 \\approxeq 1.36\\) (2 d.p.)\n\n', '(a)\n\n\n\nWeight (kg)\nFrequency\n\n\n62\n1\n\n\n63\n5\n\n\n64\n12\n\n\n65\n14\n\n\n66\n10\n\n\n67\n6\n\n\n68\n2\n\n\n~(b) Using an assumed mean of 65kg\n\n\n\n\\(x\\)\n\\(f\\)\n\\(fx\\)\n\\(d = x - A\\)\n\\(fd\\)\n\\(fd^{2}\\)\n\n\n62\n1\n62\n-3\n-3\n9\n\n\n63\n5\n315\n-2\n-10\n20\n\n\n64\n12\n768\n-1\n-12\n12\n\n\n65\n14\n910\n0\n0\n0\n\n\n66\n10\n660\n1\n10\n10\n\n\n67\n6\n402\n2\n12\n24\n\n\n68\n2\n136\n3\n6\n18\n\n\n\\(\\sum\\) =\n50\n3253\n\n3\n93\n\n\n\n(i) Mean \\(\\bar{x} = \\frac{\\sum fx}{\\sum f}\\)\n= \\(\\frac{3253}{50}\\)\n= 65.06 kg\n(ii) \\(SD = \\sqrt{\\frac{\\sum fd^{2}}{\\sum f} - (\\frac{\\sum fd^{2}}{\\sum f})^{2}}\\)\n\\(SD = \\sqrt{\\frac{93}{50} - (\\frac{3}{50})^{2}}\\)\n= \\(\\sqrt{\\frac{93}{50} - \\frac{9}{2500}}\\)\n= \\(\\sqrt{\\frac{4650 - 9}{2500}}\\)\n= \\(\\sqrt{\\frac{4641}{2500}}\\)\n= \\(1.3625 \\approxeq 1.36\\) (2 d.p.)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(352, '2013', '\n\nQuestion\n\nUsing ruler and a pair of compasses only,\n(a) construct :\n(i) \\(\\Delta\\)XYZ such that |XY| = 10cm, < XYZ = 30° and < YXZ = 45°.\n(ii) locus, \\(l_{1}\\), of points equidistant from Y and Z.\n(iii) locus, \\(l_{2}\\), of points parallel to XY through Z.\n(b) Locate M, the point of intersection of \\(l_{1}\\) and \\(l_{2}\\).\n(c) Measure < ZMY.\n\n', '', 'Using ruler and a pair of compasses only,\n(a) construct :\n(i) \\(\\Delta\\)XYZ such that |XY| = 10cm, < XYZ = 30° and < YXZ = 45°.\n(ii) locus, \\(l_{1}\\), of points equidistant from Y and Z.\n(iii) locus, \\(l_{2}\\), of points parallel to XY through Z.\n(b) Locate M, the point of intersection of \\(l_{1}\\) and \\(l_{2}\\).\n(c) Measure < ZMY.', 'Using ruler and a pair of compasses only,\n(a) construct :\n(i) \\(\\Delta\\)XYZ such that |XY| = 10cm, < XYZ = 30° and < YXZ = 45°.\n(ii) locus, \\(l_{1}\\), of points equidistant from Y and Z.\n(iii) locus, \\(l_{2}\\), of points parallel to XY through Z.~(b) Locate M, the point of intersection of \\(l_{1}\\) and \\(l_{2}\\).\n(c) Measure < ZMY.~(c) Measure < ZMY.', NULL, '\n\nExplanation\n(a), (b)\n(c) < ZMY = 120°.\n\n', '', 'Explanation\n(a), (b)\n(c) < ZMY = 120°.\n\n', '~(b)\n(c) < ZMY = 120°.\n~(c) < ZMY = 120°.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(353, '2013', '\n\nQuestion\n\n(a) If \\(\\frac{3p + 4q}{3p - 4q} = 2\\), find \\(p : q\\).\n(b)\nThe diagram shows the cross section of a bridge with a semi-circular hollow in the middle. If the perimeter of the cross section is 34 cm, calculate the :\n(i) length PQ; (ii) area of the cross section.\n[Take \\(\\pi = \\frac{22}{7}\\)].\n\n', '', '(a) If \\(\\frac{3p + 4q}{3p - 4q} = 2\\), find \\(p : q\\).\n(b)\nThe diagram shows the cross section of a bridge with a semi-circular hollow in the middle. If the perimeter of the cross section is 34 cm, calculate the :\n(i) length PQ; (ii) area of the cross section.\n[Take \\(\\pi = \\frac{22}{7}\\)].', '(a) If \\(\\frac{3p + 4q}{3p - 4q} = 2\\), find \\(p : q\\).~(b)\nThe diagram shows the cross section of a bridge with a semi-circular hollow in the middle. If the perimeter of the cross section is 34 cm, calculate the :\n(i) length PQ; (ii) area of the cross section.\n[Take \\(\\pi = \\frac{22}{7}\\)].', NULL, '\n\nExplanation\n(a) \\(\\frac{3p + 4q}{3p - 4q} = 2\\)\n\\(3p + 4q = 2(3p - 4q)\\)\n\\(3p + 4q = 6p - 8q\\)\n\\(3p - 6p = - 8q - 4q\\)\n\\(-3p = - 12q\\)\n\\(p = 4q\\)\n\\(\\frac{p}{q} = \\frac{4}{1}\\)\n\\(\\therefore p : q = 4 : 1\\).\n(b)(i)\nPerimeter of cross- section = 4 + x + 4 + 2 + y + 2\ni.e 34 = 12 + x + y\nx + y = 22 ..... (1)\nFrom the diagram, |PQ| = |UR| (opp. sides of a rectangle)\ni.e. x = 2 + d + 2\nx - 4 = d ..... (2)\nFrom (1), y = 22 - x is the circumference of the semi-circle.\n\\(22 - x = \\frac{2\\pi r}{2} = \\pi r\\)\n\\(r = \\frac{d}{2} \\)\n\\(22 - x = \\frac{22}{7} \\times \\frac{(x - 4)}{2}\\)\n\\(154 - 7x = 11x - 44\\)\n\\(154 + 44 = 11x + 7x \\)\n\\(198 = 18x\\)\n\\(x = \\frac{198}{18} = 11 m\\)\nHence, |PQ| = 11 m.\n(ii) Area of cross section = Area of rectangle PQRU - area of semi-circle\nArea of rectangle = \\(11 \\times 4 = 44 m^{2}\\)\nArea of semi-circle = \\(\\frac{\\pi r^{2}}{2} \\)\n= \\(\\frac{22}{7} \\times (\\frac{(11 - 4)}{2})^{2} \\times \\frac{1}{2}\\)\n= \\(19.25 m^{2}\\)\nArea of cross section = \\(44 - 19.25 = 24.75 m^{2}\\).\n\n', '', 'Explanation\n(a) \\(\\frac{3p + 4q}{3p - 4q} = 2\\)\n\\(3p + 4q = 2(3p - 4q)\\)\n\\(3p + 4q = 6p - 8q\\)\n\\(3p - 6p = - 8q - 4q\\)\n\\(-3p = - 12q\\)\n\\(p = 4q\\)\n\\(\\frac{p}{q} = \\frac{4}{1}\\)\n\\(\\therefore p : q = 4 : 1\\).\n(b)(i)\nPerimeter of cross- section = 4 + x + 4 + 2 + y + 2\ni.e 34 = 12 + x + y\nx + y = 22 ..... (1)\nFrom the diagram, |PQ| = |UR| (opp. sides of a rectangle)\ni.e. x = 2 + d + 2\nx - 4 = d ..... (2)\nFrom (1), y = 22 - x is the circumference of the semi-circle.\n\\(22 - x = \\frac{2\\pi r}{2} = \\pi r\\)\n\\(r = \\frac{d}{2} \\)\n\\(22 - x = \\frac{22}{7} \\times \\frac{(x - 4)}{2}\\)\n\\(154 - 7x = 11x - 44\\)\n\\(154 + 44 = 11x + 7x \\)\n\\(198 = 18x\\)\n\\(x = \\frac{198}{18} = 11 m\\)\nHence, |PQ| = 11 m.\n(ii) Area of cross section = Area of rectangle PQRU - area of semi-circle\nArea of rectangle = \\(11 \\times 4 = 44 m^{2}\\)\nArea of semi-circle = \\(\\frac{\\pi r^{2}}{2} \\)\n= \\(\\frac{22}{7} \\times (\\frac{(11 - 4)}{2})^{2} \\times \\frac{1}{2}\\)\n= \\(19.25 m^{2}\\)\nArea of cross section = \\(44 - 19.25 = 24.75 m^{2}\\).\n\n', '(a) \\(\\frac{3p + 4q}{3p - 4q} = 2\\)\n\\(3p + 4q = 2(3p - 4q)\\)\n\\(3p + 4q = 6p - 8q\\)\n\\(3p - 6p = - 8q - 4q\\)\n\\(-3p = - 12q\\)\n\\(p = 4q\\)\n\\(\\frac{p}{q} = \\frac{4}{1}\\)\n\\(\\therefore p : q = 4 : 1\\).~(b)(i)\nPerimeter of cross- section = 4 + x + 4 + 2 + y + 2\ni.e 34 = 12 + x + y\nx + y = 22 ..... (1)\nFrom the diagram, |PQ| = |UR| (opp. sides of a rectangle)\ni.e. x = 2 + d + 2\nx - 4 = d ..... (2)\nFrom (1), y = 22 - x is the circumference of the semi-circle.\n\\(22 - x = \\frac{2\\pi r}{2} = \\pi r\\)\n\\(r = \\frac{d}{2} \\)\n\\(22 - x = \\frac{22}{7} \\times \\frac{(x - 4)}{2}\\)\n\\(154 - 7x = 11x - 44\\)\n\\(154 + 44 = 11x + 7x \\)\n\\(198 = 18x\\)\n\\(x = \\frac{198}{18} = 11 m\\)\nHence, |PQ| = 11 m.\n(ii) Area of cross section = Area of rectangle PQRU - area of semi-circle\nArea of rectangle = \\(11 \\times 4 = 44 m^{2}\\)\nArea of semi-circle = \\(\\frac{\\pi r^{2}}{2} \\)\n= \\(\\frac{22}{7} \\times (\\frac{(11 - 4)}{2})^{2} \\times \\frac{1}{2}\\)\n= \\(19.25 m^{2}\\)\nArea of cross section = \\(44 - 19.25 = 24.75 m^{2}\\).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(354, '2013', '\n\nQuestion\n\n(a) Copy and complete the table of values, correct to one decimal place, for the relation \\(y = 3\\sin x + 2\\cos x\\) for \\(0° \\leq x \\leq 360°\\).\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n240°\n270°\n300°\n330°\n360°\n\n\ny\n\n\n\n3.0\n1.6\n\n-2.0\n\n-3.6\n-3.0\n\n\n2.0\n\n\n\n(b) Using scales of 2cm to 30°mon the x- axis and 2cm to 1 unit on the y- axis, draw the graph of the relation \\(y = 3\\sin x + 2\\cos x\\) for \\(0°\\leq x \\leq 360°\\).\n(c) Use the graph to solve :\n(i) \\(3\\sin x + 2\\cos x = 0\\)\n(ii) \\(2 + 2\\cos x + 3\\sin x = 0\\).\n\n\n(a) Copy and complete the table of values, correct to one decimal place, for the relation \\(y = 3\\sin x + 2\\cos x\\) for \\(0° \\leq x \\leq 360°\\).\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n240°\n270°\n300°\n330°\n360°\n\n\ny\n\n\n\n3.0\n1.6\n\n-2.0\n\n-3.6\n-3.0\n\n\n2.0\n\n\n', '', '(a) Copy and complete the table of values, correct to one decimal place, for the relation \\(y = 3\\sin x + 2\\cos x\\) for \\(0° \\leq x \\leq 360°\\).\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n240°\n270°\n300°\n330°\n360°\n\n\ny\n\n\n\n3.0\n1.6\n\n-2.0\n\n-3.6\n-3.0\n\n\n2.0\n\n\n\n(b) Using scales of 2cm to 30°mon the x- axis and 2cm to 1 unit on the y- axis, draw the graph of the relation \\(y = 3\\sin x + 2\\cos x\\) for \\(0°\\leq x \\leq 360°\\).\n(c) Use the graph to solve :\n(i) \\(3\\sin x + 2\\cos x = 0\\)\n(ii) \\(2 + 2\\cos x + 3\\sin x = 0\\).\n\n\n(a) Copy and complete the table of values, correct to one decimal place, for the relation \\(y = 3\\sin x + 2\\cos x\\) for \\(0° \\leq x \\leq 360°\\).\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n240°\n270°\n300°\n330°\n360°\n\n\ny\n\n\n\n3.0\n1.6\n\n-2.0\n\n-3.6\n-3.0\n\n\n2.0\n', '(a) Copy and complete the table of values, correct to one decimal place, for the relation \\(y = 3\\sin x + 2\\cos x\\) for \\(0° \\leq x \\leq 360°\\).\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n240°\n270°\n300°\n330°\n360°\n\n\ny\n\n\n\n3.0\n1.6\n\n-2.0\n\n-3.6\n-3.0\n\n\n2.0\n\n\n~(b) Using scales of 2cm to 30°mon the x- axis and 2cm to 1 unit on the y- axis, draw the graph of the relation \\(y = 3\\sin x + 2\\cos x\\) for \\(0°\\leq x \\leq 360°\\).\n(c) Use the graph to solve :\n(i) \\(3\\sin x + 2\\cos x = 0\\)\n(ii) \\(2 + 2\\cos x + 3\\sin x = 0\\).\n\n\n(a) Copy and complete the table of values, correct to one decimal place, for the relation \\(y = 3\\sin x + 2\\cos x\\) for \\(0° \\leq x \\leq 360°\\).\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n240°\n270°\n300°\n330°\n360°\n\n\ny\n\n\n\n3.0\n1.6\n\n-2.0\n\n-3.6\n-3.0\n\n\n2.0\n~(c) Use the graph to solve :\n(i) \\(3\\sin x + 2\\cos x = 0\\)\n(ii) \\(2 + 2\\cos x + 3\\sin x = 0\\).\n\n\n(a) Copy and complete the table of values, correct to one decimal place, for the relation \\(y = 3\\sin x + 2\\cos x\\) for \\(0° \\leq x \\leq 360°\\).\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n240°\n270°\n300°\n330°\n360°\n\n\ny\n\n\n\n3.0\n1.6\n\n-2.0\n\n-3.6\n-3.0\n\n\n2.0\n', NULL, '\n\nExplanation\n(a)\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n240°\n270°\n300°\n330°\n360°\n\n\ny\n2.0\n3.2\n3.6\n3.0\n1.6\n-0.2\n-2.0\n-3.2\n-3.6\n-3.0\n-1.6\n0.2\n2.0\n\n\n\n(b)\n(c)(i) The equation, \\(3\\sin x + 2\\cos x = 0\\) has solution where the curve cuts the x- axis, i.e. at A(x = 147°) and B(x = 325.5°).\n(ii)First, rearrange \\(2 + 2\\cos x + 3\\sin x = 0\\) to have the main graph content \\(3\\sin x + 2\\cos x\\) on one side of the equation; i.e. \\(3\\sin x + 2\\cos x = -2\\).\nAdd the line \\(y = -2\\) to the graph. The line intersects the curve at the points (x = 180°) and (x = 292.5°). Hence, these are the solutions.\n\n', '', 'Explanation\n(a)\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n240°\n270°\n300°\n330°\n360°\n\n\ny\n2.0\n3.2\n3.6\n3.0\n1.6\n-0.2\n-2.0\n-3.2\n-3.6\n-3.0\n-1.6\n0.2\n2.0\n\n\n\n(b)\n(c)(i) The equation, \\(3\\sin x + 2\\cos x = 0\\) has solution where the curve cuts the x- axis, i.e. at A(x = 147°) and B(x = 325.5°).\n(ii)First, rearrange \\(2 + 2\\cos x + 3\\sin x = 0\\) to have the main graph content \\(3\\sin x + 2\\cos x\\) on one side of the equation; i.e. \\(3\\sin x + 2\\cos x = -2\\).\nAdd the line \\(y = -2\\) to the graph. The line intersects the curve at the points (x = 180°) and (x = 292.5°). Hence, these are the solutions.\n\n', '(a)\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n240°\n270°\n300°\n330°\n360°\n\n\ny\n2.0\n3.2\n3.6\n3.0\n1.6\n-0.2\n-2.0\n-3.2\n-3.6\n-3.0\n-1.6\n0.2\n2.0\n\n\n~(b)\n(c)(i) The equation, \\(3\\sin x + 2\\cos x = 0\\) has solution where the curve cuts the x- axis, i.e. at A(x = 147°) and B(x = 325.5°).\n(ii)First, rearrange \\(2 + 2\\cos x + 3\\sin x = 0\\) to have the main graph content \\(3\\sin x + 2\\cos x\\) on one side of the equation; i.e. \\(3\\sin x + 2\\cos x = -2\\).\nAdd the line \\(y = -2\\) to the graph. The line intersects the curve at the points (x = 180°) and (x = 292.5°). Hence, these are the solutions.\n~(c)(i) The equation, \\(3\\sin x + 2\\cos x = 0\\) has solution where the curve cuts the x- axis, i.e. at A(x = 147°) and B(x = 325.5°).\n(ii)First, rearrange \\(2 + 2\\cos x + 3\\sin x = 0\\) to have the main graph content \\(3\\sin x + 2\\cos x\\) on one side of the equation; i.e. \\(3\\sin x + 2\\cos x = -2\\).\nAdd the line \\(y = -2\\) to the graph. The line intersects the curve at the points (x = 180°) and (x = 292.5°). Hence, these are the solutions.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(355, '2013', '\n\nQuestion\n\n(a) Find the equation of a straight line which passes through the point (2, -3) and is parallel to the line \\(2x + y = 6\\).\n(b) The operation \\(\\Delta\\) is defined on the set T = {2, 3, 5, 7} by \\(x \\Delta y = (x + y + xy) mod 8\\).\n(i) Construct modulo 8 table for the operation \\(\\Delta\\) on the set T.\n(ii) Use the the table to find: (a) \\(2 \\Delta (5 \\Delta 7)\\) ; (b) \\(2 \\Delta n = 5 \\Delta 7\\).\n\n', '', '(a) Find the equation of a straight line which passes through the point (2, -3) and is parallel to the line \\(2x + y = 6\\).\n(b) The operation \\(\\Delta\\) is defined on the set T = {2, 3, 5, 7} by \\(x \\Delta y = (x + y + xy) mod 8\\).\n(i) Construct modulo 8 table for the operation \\(\\Delta\\) on the set T.\n(ii) Use the the table to find: (a) \\(2 \\Delta (5 \\Delta 7)\\) ; (b) \\(2 \\Delta n = 5 \\Delta 7\\).', '(a) Find the equation of a straight line which passes through the point (2, -3) and is parallel to the line \\(2x + y = 6\\).~(b) The operation \\(\\Delta\\) is defined on the set T = {2, 3, 5, 7} by \\(x \\Delta y = (x + y + xy) mod 8\\).\n(i) Construct modulo 8 table for the operation \\(\\Delta\\) on the set T.\n(ii) Use the the table to find: (a) \\(2 \\Delta (5 \\Delta 7)\\) ; (b) \\(2 \\Delta n = 5 \\Delta 7\\).', NULL, '\n\nExplanation\n(a) First, change the given linear equation to the gradient- intercept form:\n\\(2x + y = 6 \\implies y = 6 - 2x\\)\nThus, \\(m_{1} = -2, c = 6\\). Thus, for a line to be parallel, the gradient \\(m_{2} = -2\\) (condition for parallelism).\nUsing \\(y - y_{1} = m(x - x_{1})\\), we have\n\\(y - (-3) = -2(x - 2)\\)\n\\(y + 3 = - 2x + 4\\)\n\\(2x + y = 4 - 3 = 1\\)\nThus, the line passing through (2, -3) parallel to 2x + y = 6 is 2x + y = 1.\n(b) (i)\n\n\n\n\\(\\Delta\\)\n2\n3\n5\n7\n\n\n2\n0\n3\n1\n7\n\n\n3\n3\n7\n7\n7\n\n\n5\n1\n7\n3\n7\n\n\n7\n7\n7\n7\n7\n\n\n\n(ii) (a) \\(2 \\Delta (5 \\Delta 7) = 2 \\Delta 7 = 7\\)\n(b) \\(2 \\Delta n = 5 \\Delta 7\\)\n\\(2 \\Delta n = 7\\)\nBy comparison, n = 7.\n\n', '', 'Explanation\n(a) First, change the given linear equation to the gradient- intercept form:\n\\(2x + y = 6 \\implies y = 6 - 2x\\)\nThus, \\(m_{1} = -2, c = 6\\). Thus, for a line to be parallel, the gradient \\(m_{2} = -2\\) (condition for parallelism).\nUsing \\(y - y_{1} = m(x - x_{1})\\), we have\n\\(y - (-3) = -2(x - 2)\\)\n\\(y + 3 = - 2x + 4\\)\n\\(2x + y = 4 - 3 = 1\\)\nThus, the line passing through (2, -3) parallel to 2x + y = 6 is 2x + y = 1.\n(b) (i)\n\n\n\n\\(\\Delta\\)\n2\n3\n5\n7\n\n\n2\n0\n3\n1\n7\n\n\n3\n3\n7\n7\n7\n\n\n5\n1\n7\n3\n7\n\n\n7\n7\n7\n7\n7\n\n\n\n(ii) (a) \\(2 \\Delta (5 \\Delta 7) = 2 \\Delta 7 = 7\\)\n(b) \\(2 \\Delta n = 5 \\Delta 7\\)\n\\(2 \\Delta n = 7\\)\nBy comparison, n = 7.\n\n', '(a) First, change the given linear equation to the gradient- intercept form:\n\\(2x + y = 6 \\implies y = 6 - 2x\\)\nThus, \\(m_{1} = -2, c = 6\\). Thus, for a line to be parallel, the gradient \\(m_{2} = -2\\) (condition for parallelism).\nUsing \\(y - y_{1} = m(x - x_{1})\\), we have\n\\(y - (-3) = -2(x - 2)\\)\n\\(y + 3 = - 2x + 4\\)\n\\(2x + y = 4 - 3 = 1\\)\nThus, the line passing through (2, -3) parallel to 2x + y = 6 is 2x + y = 1.\n(b) (i)\n\n\n\n\\(\\Delta\\)\n2\n3\n5\n7\n\n\n2\n0\n3\n1\n7\n\n\n3\n3\n7\n7\n7\n\n\n5\n1\n7\n3\n7\n\n\n7\n7\n7\n7\n7\n\n\n\n(ii) (a) \\(2 \\Delta (5 \\Delta 7) = 2 \\Delta 7 = 7\\)~(b) (i)\n\n\n\n\\(\\Delta\\)\n2\n3\n5\n7\n\n\n2\n0\n3\n1\n7\n\n\n3\n3\n7\n7\n7\n\n\n5\n1\n7\n3\n7\n\n\n7\n7\n7\n7\n7\n\n\n\n(ii) (a) \\(2 \\Delta (5 \\Delta 7) = 2 \\Delta 7 = 7\\)\n(b) \\(2 \\Delta n = 5 \\Delta 7\\)\n\\(2 \\Delta n = 7\\)\nBy comparison, n = 7.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(356, '2013', '\n\nQuestion\n\n(a) Without using Mathematical tables or calculators, simplify:\n\\(3\\frac{4}{9} \\div (5\\frac{1}{3} - 2\\frac{3}{4}) + 5\\frac{9}{10}\\)\n(b) A number is selected at random from each of the sets {2, 3, 4} and {1, 3, 5}. Find the probability that the sum of the two numbers is greater than 3 and less than 7.\n\n', '', '(a) Without using Mathematical tables or calculators, simplify:\n\\(3\\frac{4}{9} \\div (5\\frac{1}{3} - 2\\frac{3}{4}) + 5\\frac{9}{10}\\)\n(b) A number is selected at random from each of the sets {2, 3, 4} and {1, 3, 5}. Find the probability that the sum of the two numbers is greater than 3 and less than 7.', '(a) Without using Mathematical tables or calculators, simplify:\n\\(3\\frac{4}{9} \\div (5\\frac{1}{3} - 2\\frac{3}{4}) + 5\\frac{9}{10}\\)~(b) A number is selected at random from each of the sets {2, 3, 4} and {1, 3, 5}. Find the probability that the sum of the two numbers is greater than 3 and less than 7.', NULL, '\n\nExplanation\n(a) \\(3\\frac{4}{9} \\div (5\\frac{1}{3} - 2\\frac{3}{4}) + 5\\frac{9}{10}\\)\n\\(\\frac{31}{9} \\div (\\frac{16}{3} - \\frac{11}{4}) + \\frac{59}{10}\\)\n\\(\\frac{31}{9} \\div (\\frac{64 - 33}{12}) + \\frac{59}{10}\\)\n\\((\\frac{31}{9} \\div \\frac{31}{12}) + \\frac{59}{10}\\)\n\\((\\frac{31}{9} \\times \\frac{12}{31}) + \\frac{59}{10}\\)\n\\(\\frac{12}{9} + \\frac{59}{10} = \\frac{120 + 531}{90}\\)\n\\(\\frac{651}{90} = \\frac{217}{30}\\).\n(b)\n\n\n\n+\n1\n3\n5\n\n\n2\n3\n5\n7\n\n\n3\n4\n6\n8\n\n\n4\n5\n7\n9\n\n\n\nLet E be the event of the sum being greater than 3 and less than 7 and S be the total sample space.\nn(E) = 4; and n(S) = 9.\nP(E) = \\(\\frac{n(E)}{n(S)} = \\frac{4}{9}\\)\n\n', '', 'Explanation\n(a) \\(3\\frac{4}{9} \\div (5\\frac{1}{3} - 2\\frac{3}{4}) + 5\\frac{9}{10}\\)\n\\(\\frac{31}{9} \\div (\\frac{16}{3} - \\frac{11}{4}) + \\frac{59}{10}\\)\n\\(\\frac{31}{9} \\div (\\frac{64 - 33}{12}) + \\frac{59}{10}\\)\n\\((\\frac{31}{9} \\div \\frac{31}{12}) + \\frac{59}{10}\\)\n\\((\\frac{31}{9} \\times \\frac{12}{31}) + \\frac{59}{10}\\)\n\\(\\frac{12}{9} + \\frac{59}{10} = \\frac{120 + 531}{90}\\)\n\\(\\frac{651}{90} = \\frac{217}{30}\\).\n(b)\n\n\n\n+\n1\n3\n5\n\n\n2\n3\n5\n7\n\n\n3\n4\n6\n8\n\n\n4\n5\n7\n9\n\n\n\nLet E be the event of the sum being greater than 3 and less than 7 and S be the total sample space.\nn(E) = 4; and n(S) = 9.\nP(E) = \\(\\frac{n(E)}{n(S)} = \\frac{4}{9}\\)\n\n', '(a) \\(3\\frac{4}{9} \\div (5\\frac{1}{3} - 2\\frac{3}{4}) + 5\\frac{9}{10}\\)\n\\(\\frac{31}{9} \\div (\\frac{16}{3} - \\frac{11}{4}) + \\frac{59}{10}\\)\n\\(\\frac{31}{9} \\div (\\frac{64 - 33}{12}) + \\frac{59}{10}\\)\n\\((\\frac{31}{9} \\div \\frac{31}{12}) + \\frac{59}{10}\\)\n\\((\\frac{31}{9} \\times \\frac{12}{31}) + \\frac{59}{10}\\)\n\\(\\frac{12}{9} + \\frac{59}{10} = \\frac{120 + 531}{90}\\)\n\\(\\frac{651}{90} = \\frac{217}{30}\\).~(b)\n\n\n\n+\n1\n3\n5\n\n\n2\n3\n5\n7\n\n\n3\n4\n6\n8\n\n\n4\n5\n7\n9\n\n\n\nLet E be the event of the sum being greater than 3 and less than 7 and S be the total sample space.\nn(E) = 4; and n(S) = 9.\nP(E) = \\(\\frac{n(E)}{n(S)} = \\frac{4}{9}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(357, '2013', '\n\nQuestion\n\n(a) Solve the inequality : \\(4 + \\frac{3}{4}(x + 2) \\leq \\frac{3}{8}x + 1\\)\n(b)\nThe diagram shows a rectangle PQRS from which a square of side x cm has been cut. If the area of the shaded portion is 484\\(cm^{2}\\), find the values of x.\n\n', '', '(a) Solve the inequality : \\(4 + \\frac{3}{4}(x + 2) \\leq \\frac{3}{8}x + 1\\)\n(b)\nThe diagram shows a rectangle PQRS from which a square of side x cm has been cut. If the area of the shaded portion is 484\\(cm^{2}\\), find the values of x.', '(a) Solve the inequality : \\(4 + \\frac{3}{4}(x + 2) \\leq \\frac{3}{8}x + 1\\)~(b)\nThe diagram shows a rectangle PQRS from which a square of side x cm has been cut. If the area of the shaded portion is 484\\(cm^{2}\\), find the values of x.', NULL, '\n\nExplanation\n(a) \\(4 + \\frac{3}{4}(x + 2) \\leq \\frac{3}{8}x + 1\\)\nMultiply through by the LCM of 4 and 8, to clear fractions :\n\\(8 \\times 4 + 8 \\times \\frac{3}{4}(x + 2) \\leq 8 \\times \\frac{3}{8}x + 8\\)\n\\(32 + 6(x + 2) \\leq 3x + 8 \\)\n\\(32 + 6x + 12 \\leq 3x + 8 \\)\n\\(44 + 6x \\leq 3x + 8\\)\n\\(6x - 3x \\leq 8 - 44 \\implies 3x \\leq -36\\)\n\\(x \\leq - 12\\).\n(b)\nArea \\((A_{1})\\), of rectangle PQRS = \\(20(20 + x) cm^{2}\\)\nArea \\((A_{2})\\), of square = \\(x \\times x = x^{2} cm^{2}\\)\n\\(\\implies 20(20 + x) - x^{2} = 484\\)\n\\(400 + 20x - x^{2} = 484\\)\n\\(x^{2} - 20x + 484 - 400 = x^{2} - 20x + 84 = 0\\)\n\\(x^{2} - 14x - 6x + 84 = 0\\)\n\\(x( x - 14) - 6(x - 14) = 0 \\)\n\\((x - 14)(x - 6) = 0\\)\n\\(\\therefore \\text{x = 6 cm or 14 cm}\\)\n\n', '', 'Explanation\n(a) \\(4 + \\frac{3}{4}(x + 2) \\leq \\frac{3}{8}x + 1\\)\nMultiply through by the LCM of 4 and 8, to clear fractions :\n\\(8 \\times 4 + 8 \\times \\frac{3}{4}(x + 2) \\leq 8 \\times \\frac{3}{8}x + 8\\)\n\\(32 + 6(x + 2) \\leq 3x + 8 \\)\n\\(32 + 6x + 12 \\leq 3x + 8 \\)\n\\(44 + 6x \\leq 3x + 8\\)\n\\(6x - 3x \\leq 8 - 44 \\implies 3x \\leq -36\\)\n\\(x \\leq - 12\\).\n(b)\nArea \\((A_{1})\\), of rectangle PQRS = \\(20(20 + x) cm^{2}\\)\nArea \\((A_{2})\\), of square = \\(x \\times x = x^{2} cm^{2}\\)\n\\(\\implies 20(20 + x) - x^{2} = 484\\)\n\\(400 + 20x - x^{2} = 484\\)\n\\(x^{2} - 20x + 484 - 400 = x^{2} - 20x + 84 = 0\\)\n\\(x^{2} - 14x - 6x + 84 = 0\\)\n\\(x( x - 14) - 6(x - 14) = 0 \\)\n\\((x - 14)(x - 6) = 0\\)\n\\(\\therefore \\text{x = 6 cm or 14 cm}\\)\n\n', '(a) \\(4 + \\frac{3}{4}(x + 2) \\leq \\frac{3}{8}x + 1\\)\nMultiply through by the LCM of 4 and 8, to clear fractions :\n\\(8 \\times 4 + 8 \\times \\frac{3}{4}(x + 2) \\leq 8 \\times \\frac{3}{8}x + 8\\)\n\\(32 + 6(x + 2) \\leq 3x + 8 \\)\n\\(32 + 6x + 12 \\leq 3x + 8 \\)\n\\(44 + 6x \\leq 3x + 8\\)\n\\(6x - 3x \\leq 8 - 44 \\implies 3x \\leq -36\\)\n\\(x \\leq - 12\\).~(b)\nArea \\((A_{1})\\), of rectangle PQRS = \\(20(20 + x) cm^{2}\\)\nArea \\((A_{2})\\), of square = \\(x \\times x = x^{2} cm^{2}\\)\n\\(\\implies 20(20 + x) - x^{2} = 484\\)\n\\(400 + 20x - x^{2} = 484\\)\n\\(x^{2} - 20x + 484 - 400 = x^{2} - 20x + 84 = 0\\)\n\\(x^{2} - 14x - 6x + 84 = 0\\)\n\\(x( x - 14) - 6(x - 14) = 0 \\)\n\\((x - 14)(x - 6) = 0\\)\n\\(\\therefore \\text{x = 6 cm or 14 cm}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(358, '2013', '\n\nQuestion\n\n(a) The ratio of the interior angle to the exterior angle of a regular polygon is 5 : 2, Find the number of sides of the polygon.\n(b)\nThe diagram shows a circle PQRS with centre O, < UQR = 68°, < TPS = 74° and < QSR = 40°. Calculate the value of < PRS.\n\n', '', '(a) The ratio of the interior angle to the exterior angle of a regular polygon is 5 : 2, Find the number of sides of the polygon.\n(b)\nThe diagram shows a circle PQRS with centre O, < UQR = 68°, < TPS = 74° and < QSR = 40°. Calculate the value of < PRS.', '(a) The ratio of the interior angle to the exterior angle of a regular polygon is 5 : 2, Find the number of sides of the polygon.~(b)\nThe diagram shows a circle PQRS with centre O, < UQR = 68°, < TPS = 74° and < QSR = 40°. Calculate the value of < PRS.', NULL, '\n\nExplanation\n(a) Let the diagram below represent a section of the polygon\n\nAlso, let :\n\\(i\\) represent the size of an interior angle; \\(e\\)represent the size of an exterior angle.\nThen \\(\\frac{i}{e} = \\frac{5}{2}\\)\n\\(\\implies i = \\frac{5}{2}e\\)\nBut \\(i + e = 180°\\) (sum of angles on a straight line)\nSubstitute \\(\\frac{5}{2}e\\) for \\(i\\) in the equation\n\\(\\frac{5}{2}e + e = 180°\\)\n\\(\\frac{7}{2}e = 180°\\)\n\\(e = \\frac{180° \\times 2}{7} = \\frac{360°}{7}\\)\nNumber of sides of the polygon = \\(\\frac{360°}{\\text{size of one exterior angle}}\\)\n= \\(360° \\div \\frac{360°}{7}\\)\n\\(360° \\times \\frac{7}{360°} = 7 sides\\).\n(b)\nIn the diagram above, < PSR = 68° (interior angle of a cyclic quad = opp exterior angle)\n< PSQ = 68° - 40° = 28°\n< PRQ = 28° (angles in same segment)\n< SRQ = 74° (interior angle of a cyclic quad = opp exterior angle)\nhence, < PRS = 74° - 28° = 46°\n\n', '', 'Explanation\n(a) Let the diagram below represent a section of the polygon\n\nAlso, let :\n\\(i\\) represent the size of an interior angle; \\(e\\)represent the size of an exterior angle.\nThen \\(\\frac{i}{e} = \\frac{5}{2}\\)\n\\(\\implies i = \\frac{5}{2}e\\)\nBut \\(i + e = 180°\\) (sum of angles on a straight line)\nSubstitute \\(\\frac{5}{2}e\\) for \\(i\\) in the equation\n\\(\\frac{5}{2}e + e = 180°\\)\n\\(\\frac{7}{2}e = 180°\\)\n\\(e = \\frac{180° \\times 2}{7} = \\frac{360°}{7}\\)\nNumber of sides of the polygon = \\(\\frac{360°}{\\text{size of one exterior angle}}\\)\n= \\(360° \\div \\frac{360°}{7}\\)\n\\(360° \\times \\frac{7}{360°} = 7 sides\\).\n(b)\nIn the diagram above, < PSR = 68° (interior angle of a cyclic quad = opp exterior angle)\n< PSQ = 68° - 40° = 28°\n< PRQ = 28° (angles in same segment)\n< SRQ = 74° (interior angle of a cyclic quad = opp exterior angle)\nhence, < PRS = 74° - 28° = 46°\n\n', '(a) Let the diagram below represent a section of the polygon\n\nAlso, let :\n\\(i\\) represent the size of an interior angle; \\(e\\)represent the size of an exterior angle.\nThen \\(\\frac{i}{e} = \\frac{5}{2}\\)\n\\(\\implies i = \\frac{5}{2}e\\)\nBut \\(i + e = 180°\\) (sum of angles on a straight line)\nSubstitute \\(\\frac{5}{2}e\\) for \\(i\\) in the equation\n\\(\\frac{5}{2}e + e = 180°\\)\n\\(\\frac{7}{2}e = 180°\\)\n\\(e = \\frac{180° \\times 2}{7} = \\frac{360°}{7}\\)\nNumber of sides of the polygon = \\(\\frac{360°}{\\text{size of one exterior angle}}\\)\n= \\(360° \\div \\frac{360°}{7}\\)\n\\(360° \\times \\frac{7}{360°} = 7 sides\\).~(b)\nIn the diagram above, < PSR = 68° (interior angle of a cyclic quad = opp exterior angle)\n< PSQ = 68° - 40° = 28°\n< PRQ = 28° (angles in same segment)\n< SRQ = 74° (interior angle of a cyclic quad = opp exterior angle)\nhence, < PRS = 74° - 28° = 46°\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(359, '2013', '\n\nQuestion\n\n(a) By how much is the sum of \\(3\\frac{2}{3}\\) and \\(2\\frac{1}{5}\\) less than 7?\n(b) The height, h m, of a dock above sea level is given by \\(h = 6 + 4\\cos (15p)°, 0 < p < 6\\). Find :\n(i) the value of h when p = 4 ; (ii) correct to two significant figures, the value of p when h = 9 m.\n\n', '', '(a) By how much is the sum of \\(3\\frac{2}{3}\\) and \\(2\\frac{1}{5}\\) less than 7?\n(b) The height, h m, of a dock above sea level is given by \\(h = 6 + 4\\cos (15p)°, 0 < p < 6\\). Find :\n(i) the value of h when p = 4 ; (ii) correct to two significant figures, the value of p when h = 9 m.', '(a) By how much is the sum of \\(3\\frac{2}{3}\\) and \\(2\\frac{1}{5}\\) less than 7?~(b) The height, h m, of a dock above sea level is given by \\(h = 6 + 4\\cos (15p)°, 0 < p < 6\\). Find :\n(i) the value of h when p = 4 ; (ii) correct to two significant figures, the value of p when h = 9 m.', NULL, '\n\nExplanation\n(a) Let the sum of \\(3\\frac{2}{3}\\) and \\(2\\frac{1}{5}\\) be less than 7 by z.\nThen z = \\(7 - (3\\frac{2}{3} + 2\\frac{1}{5})\\)\n= \\(7 - (\\frac{11}{3} + \\frac{11}{5})\\)\n= \\(7 - \\frac{88}{15}\\)\n= \\(\\frac{17}{15} = 1\\frac{2}{15}\\)\nHence, the sum of \\(3\\frac{2}{3}\\) and \\(2\\frac{1}{5}\\) is less than 7 by \\(1\\frac{2}{15}\\).\n(b) \\(h = 6 + 4\\cos (15p)°\\) (Given)\n(i) When p = 4,\n\\(h = 6 + 4\\cos (15 \\times 4)°\\)\n\\(h = 6 + 4 \\cos 60°\\)\n= \\(6 + 4 \\times 0.5\\)\n= \\(6 + 2 = 8 m\\)\n(ii) When h = 9 m, the given equation becomes\n\\(9 = 6 + 4\\cos (15p)°\\)\n\\(9 - 6 = 4 \\cos (15p)°\\)\n\\(\\cos (15p)° = \\frac{3}{4}\\)\n\\((15p)° = \\cos^{-1} (0.75)\\)\n\\((15p)° = 41.4°\\)\n\\(p° = \\frac{41.4}{15}\\)\n\\(p° = 2.76°\\)\n\\(p \\approxeq 2.8°\\) (2 significant figures).\n\n', '', 'Explanation\n(a) Let the sum of \\(3\\frac{2}{3}\\) and \\(2\\frac{1}{5}\\) be less than 7 by z.\nThen z = \\(7 - (3\\frac{2}{3} + 2\\frac{1}{5})\\)\n= \\(7 - (\\frac{11}{3} + \\frac{11}{5})\\)\n= \\(7 - \\frac{88}{15}\\)\n= \\(\\frac{17}{15} = 1\\frac{2}{15}\\)\nHence, the sum of \\(3\\frac{2}{3}\\) and \\(2\\frac{1}{5}\\) is less than 7 by \\(1\\frac{2}{15}\\).\n(b) \\(h = 6 + 4\\cos (15p)°\\) (Given)\n(i) When p = 4,\n\\(h = 6 + 4\\cos (15 \\times 4)°\\)\n\\(h = 6 + 4 \\cos 60°\\)\n= \\(6 + 4 \\times 0.5\\)\n= \\(6 + 2 = 8 m\\)\n(ii) When h = 9 m, the given equation becomes\n\\(9 = 6 + 4\\cos (15p)°\\)\n\\(9 - 6 = 4 \\cos (15p)°\\)\n\\(\\cos (15p)° = \\frac{3}{4}\\)\n\\((15p)° = \\cos^{-1} (0.75)\\)\n\\((15p)° = 41.4°\\)\n\\(p° = \\frac{41.4}{15}\\)\n\\(p° = 2.76°\\)\n\\(p \\approxeq 2.8°\\) (2 significant figures).\n\n', '(a) Let the sum of \\(3\\frac{2}{3}\\) and \\(2\\frac{1}{5}\\) be less than 7 by z.\nThen z = \\(7 - (3\\frac{2}{3} + 2\\frac{1}{5})\\)\n= \\(7 - (\\frac{11}{3} + \\frac{11}{5})\\)\n= \\(7 - \\frac{88}{15}\\)\n= \\(\\frac{17}{15} = 1\\frac{2}{15}\\)\nHence, the sum of \\(3\\frac{2}{3}\\) and \\(2\\frac{1}{5}\\) is less than 7 by \\(1\\frac{2}{15}\\).~(b) \\(h = 6 + 4\\cos (15p)°\\) (Given)\n(i) When p = 4,\n\\(h = 6 + 4\\cos (15 \\times 4)°\\)\n\\(h = 6 + 4 \\cos 60°\\)\n= \\(6 + 4 \\times 0.5\\)\n= \\(6 + 2 = 8 m\\)\n(ii) When h = 9 m, the given equation becomes\n\\(9 = 6 + 4\\cos (15p)°\\)\n\\(9 - 6 = 4 \\cos (15p)°\\)\n\\(\\cos (15p)° = \\frac{3}{4}\\)\n\\((15p)° = \\cos^{-1} (0.75)\\)\n\\((15p)° = 41.4°\\)\n\\(p° = \\frac{41.4}{15}\\)\n\\(p° = 2.76°\\)\n\\(p \\approxeq 2.8°\\) (2 significant figures).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(360, '2013', '\n\nQuestion\n\nA trapezium PQRS is such that PQ // RS and the perpendicular P to RS is 40 cm. If |PQ| = 20 cm, |SP| = 50 cm and |SR| = 60 cm. Calculate, correct to 2 significant figures, the\n(a) Area of the trapezium ; (b) < QRS.\n\n', '', 'A trapezium PQRS is such that PQ // RS and the perpendicular P to RS is 40 cm. If |PQ| = 20 cm, |SP| = 50 cm and |SR| = 60 cm. Calculate, correct to 2 significant figures, the\n(a) Area of the trapezium ; (b) < QRS.', '~(b) < QRS.', NULL, '\n\nExplanation\n(a) Area of trapezium PQRS = \\(\\frac{1}{2}(PQ + RS) \\times 40\\)\n= \\(\\frac{1}{2}(20 + 60) \\times 40\\)\n= \\(40 \\times 40\\)\n= \\(1600 cm^{2}\\)\n(b) In \\(\\Delta\\) SPT,\n\\(|SP|^{2} = |ST|^{2} + |TP|^{2}\\) (Pythagoras theorem)\n\n\\(|ST|^{2} = |SP|^{2} - |TP|^{2}\\)\n= \\(50^{2} - 40^{2}\\)\n= \\(2500 - 1600\\)\n\\(|ST|^{2} = 900\\)\n\\(\\therefore |ST| = 30 cm\\)\n\\(|UR| = 60 cm - (30 + 20) cm\\)\n= 10 cm\nLet < QRS = \\(\\alpha\\),\n\\(\\tan \\alpha = \\frac{40}{10} = 4\\)\n\\(\\alpha = \\tan^{-1} (4)\\)\n= \\(75.96° \\approxeq 76°\\)\n\n', '', 'Explanation\n(a) Area of trapezium PQRS = \\(\\frac{1}{2}(PQ + RS) \\times 40\\)\n= \\(\\frac{1}{2}(20 + 60) \\times 40\\)\n= \\(40 \\times 40\\)\n= \\(1600 cm^{2}\\)\n(b) In \\(\\Delta\\) SPT,\n\\(|SP|^{2} = |ST|^{2} + |TP|^{2}\\) (Pythagoras theorem)\n\n\\(|ST|^{2} = |SP|^{2} - |TP|^{2}\\)\n= \\(50^{2} - 40^{2}\\)\n= \\(2500 - 1600\\)\n\\(|ST|^{2} = 900\\)\n\\(\\therefore |ST| = 30 cm\\)\n\\(|UR| = 60 cm - (30 + 20) cm\\)\n= 10 cm\nLet < QRS = \\(\\alpha\\),\n\\(\\tan \\alpha = \\frac{40}{10} = 4\\)\n\\(\\alpha = \\tan^{-1} (4)\\)\n= \\(75.96° \\approxeq 76°\\)\n\n', '(a) Area of trapezium PQRS = \\(\\frac{1}{2}(PQ + RS) \\times 40\\)\n= \\(\\frac{1}{2}(20 + 60) \\times 40\\)\n= \\(40 \\times 40\\)\n= \\(1600 cm^{2}\\)~(b) In \\(\\Delta\\) SPT,\n\\(|SP|^{2} = |ST|^{2} + |TP|^{2}\\) (Pythagoras theorem)\n\n\\(|ST|^{2} = |SP|^{2} - |TP|^{2}\\)\n= \\(50^{2} - 40^{2}\\)\n= \\(2500 - 1600\\)\n\\(|ST|^{2} = 900\\)\n\\(\\therefore |ST| = 30 cm\\)\n\\(|UR| = 60 cm - (30 + 20) cm\\)\n= 10 cm\nLet < QRS = \\(\\alpha\\),\n\\(\\tan \\alpha = \\frac{40}{10} = 4\\)\n\\(\\alpha = \\tan^{-1} (4)\\)\n= \\(75.96° \\approxeq 76°\\)\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
(361, '2013', '\n\nQuestion\n\n(a) (i) Illustrate the following statements in a Venn diagram : All good Literature students in a school are in the General Arts class.\n(ii) Use ths diagram to determine whether or not the following are valid conclusions from the given statement.\n(1) Vivian is in the General Arts class therefore she is a good Literature student.\n(2) Audu is not a good Literature student therefore he is not in the General Arts class;\n(3) Kweku is not in the General Arts class therefore he is not a good Literature student.\n(b) The cost (c) of producing n bricks is the sum of a fixed amount, h, and a variable amount, y, where y varies directly as n. If it costs GH¢950.00 to produce 600 bricks andGH¢ 1,030.00 to produce 1000 bricks,\n(i) Find the relationship between c, h and n ; (ii) Calculate the cost of producing 500 bricks.\n\n', '', '(a) (i) Illustrate the following statements in a Venn diagram : All good Literature students in a school are in the General Arts class.\n(ii) Use ths diagram to determine whether or not the following are valid conclusions from the given statement.\n(1) Vivian is in the General Arts class therefore she is a good Literature student.\n(2) Audu is not a good Literature student therefore he is not in the General Arts class;\n(3) Kweku is not in the General Arts class therefore he is not a good Literature student.\n(b) The cost (c) of producing n bricks is the sum of a fixed amount, h, and a variable amount, y, where y varies directly as n. If it costs GH¢950.00 to produce 600 bricks andGH¢ 1,030.00 to produce 1000 bricks,\n(i) Find the relationship between c, h and n ; (ii) Calculate the cost of producing 500 bricks.', '(a) (i) Illustrate the following statements in a Venn diagram : All good Literature students in a school are in the General Arts class.\n(ii) Use ths diagram to determine whether or not the following are valid conclusions from the given statement.\n(1) Vivian is in the General Arts class therefore she is a good Literature student.\n(2) Audu is not a good Literature student therefore he is not in the General Arts class;\n(3) Kweku is not in the General Arts class therefore he is not a good Literature student.~(b) The cost (c) of producing n bricks is the sum of a fixed amount, h, and a variable amount, y, where y varies directly as n. If it costs GH¢950.00 to produce 600 bricks andGH¢ 1,030.00 to produce 1000 bricks,\n(i) Find the relationship between c, h and n ; (ii) Calculate the cost of producing 500 bricks.~(c) of producing n bricks is the sum of a fixed amount, h, and a variable amount, y, where y varies directly as n. If it costs GH¢950.00 to produce 600 bricks andGH¢ 1,030.00 to produce 1000 bricks,\n(i) Find the relationship between c, h and n ; (ii) Calculate the cost of producing 500 bricks.', NULL, '\n\nExplanation\n(a) (i) Let :\nL = { all good literature students in the school}\nA = {all students in the General Arts class}\nThen the Venn diagram below represents the statement : \'All good Literature students in a school are in the General Arts class\'.\n \nwhere \\(\\in\\) = {all students in the school}.\n(ii) (1) Invalid conclusion\n(2) Invalid conclusion\n(3) Valid conclusion\n(b) (i) c = h + y\nwhere \\(y \\propto n \\implies y = kn\\)\n\\(\\implies c = h + kn\\)\nWhen n = 600 bricks, c = GH¢ 950.00\n\\(950.00 = h + 600k ....... (1)\\)\nWhen n = 1000 bricks, c =GH¢ 1,030.00\n\\(1,030.00 = h + 1000k ....... (2)\\)\nSubtracting (1) from (2), we have\n\\(1,030 - 950 = 1000k - 600k\\)\n\\(80 = 400k\\)\n\\(k = \\frac{80}{400} = \\frac{1}{5}\\)\n\\(\\therefore c = h + \\frac{1}{5}n\\)\nFrom (1), \\(950 = h + 600k\\)\n\\(950 = h + (600 \\times \\frac{1}{5})\\)\n\\(950 = h + 120\\)\n\\(h = 950 - 120 = 830\\)\n\\(c = 830 + \\frac{1}{5}n\\)\n(ii) Using\n\\(c = 830 + \\frac{1}{5}n\\), we have\n\\(c = 830 + \\frac{1}{5} \\times 500\\)\n= \\(830 + 100\\)\n= \\(GH¢ 930.00\\)\nHence, the cost of producing 500 bricks =GH¢ 930.00.\n\n', '', 'Explanation\n(a) (i) Let :\nL = { all good literature students in the school}\nA = {all students in the General Arts class}\nThen the Venn diagram below represents the statement : \'All good Literature students in a school are in the General Arts class\'.\n \nwhere \\(\\in\\) = {all students in the school}.\n(ii) (1) Invalid conclusion\n(2) Invalid conclusion\n(3) Valid conclusion\n(b) (i) c = h + y\nwhere \\(y \\propto n \\implies y = kn\\)\n\\(\\implies c = h + kn\\)\nWhen n = 600 bricks, c = GH¢ 950.00\n\\(950.00 = h + 600k ....... (1)\\)\nWhen n = 1000 bricks, c =GH¢ 1,030.00\n\\(1,030.00 = h + 1000k ....... (2)\\)\nSubtracting (1) from (2), we have\n\\(1,030 - 950 = 1000k - 600k\\)\n\\(80 = 400k\\)\n\\(k = \\frac{80}{400} = \\frac{1}{5}\\)\n\\(\\therefore c = h + \\frac{1}{5}n\\)\nFrom (1), \\(950 = h + 600k\\)\n\\(950 = h + (600 \\times \\frac{1}{5})\\)\n\\(950 = h + 120\\)\n\\(h = 950 - 120 = 830\\)\n\\(c = 830 + \\frac{1}{5}n\\)\n(ii) Using\n\\(c = 830 + \\frac{1}{5}n\\), we have\n\\(c = 830 + \\frac{1}{5} \\times 500\\)\n= \\(830 + 100\\)\n= \\(GH¢ 930.00\\)\nHence, the cost of producing 500 bricks =GH¢ 930.00.\n\n', '(a) (i) Let :\nL = { all good literature students in the school}\nA = {all students in the General Arts class}\nThen the Venn diagram below represents the statement : \'All good Literature students in a school are in the General Arts class\'.\n \nwhere \\(\\in\\) = {all students in the school}.\n(ii) (1) Invalid conclusion\n(2) Invalid conclusion\n(3) Valid conclusion~(b) (i) c = h + y\nwhere \\(y \\propto n \\implies y = kn\\)\n\\(\\implies c = h + kn\\)\nWhen n = 600 bricks, c = GH¢ 950.00\n\\(950.00 = h + 600k ....... (1)\\)\nWhen n = 1000 bricks, c =GH¢ 1,030.00\n\\(1,030.00 = h + 1000k ....... (2)\\)\nSubtracting (1) from (2), we have\n\\(1,030 - 950 = 1000k - 600k\\)\n\\(80 = 400k\\)\n\\(k = \\frac{80}{400} = \\frac{1}{5}\\)\n\\(\\therefore c = h + \\frac{1}{5}n\\)\nFrom (1), \\(950 = h + 600k\\)\n\\(950 = h + (600 \\times \\frac{1}{5})\\)\n\\(950 = h + 120\\)\n\\(h = 950 - 120 = 830\\)\n\\(c = 830 + \\frac{1}{5}n\\)\n(ii) Using\n\\(c = 830 + \\frac{1}{5}n\\), we have\n\\(c = 830 + \\frac{1}{5} \\times 500\\)\n= \\(830 + 100\\)\n= \\(GH¢ 930.00\\)\nHence, the cost of producing 500 bricks =GH¢ 930.00.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(362, '2013', '\n\nQuestion\n\nThe table is for the relation \\(y = px^{2} - 5x + q\\).\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n21\n6\n\n-12\n\n\n\n0\n13\n\n\n\n(a)(i) Use the table to find the values of p and q.\n(ii) Copy and complete the table.\n(b) Using scales of 2cm to 1 unit on the x- axis and 2 cm to 5 units on the y- axis, draw the graph of the relation for \\(-3 \\leq x \\leq 5\\).\n(c) Use the graph to find :\n(i) y when x = 1.8 ; (ii) x when y = -8.\n\n\nThe table is for the relation \\(y = px^{2} - 5x + q\\).\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n21\n6\n\n-12\n\n\n\n0\n13\n\n\n', '', 'The table is for the relation \\(y = px^{2} - 5x + q\\).\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n21\n6\n\n-12\n\n\n\n0\n13\n\n\n\n(a)(i) Use the table to find the values of p and q.\n(ii) Copy and complete the table.\n(b) Using scales of 2cm to 1 unit on the x- axis and 2 cm to 5 units on the y- axis, draw the graph of the relation for \\(-3 \\leq x \\leq 5\\).\n(c) Use the graph to find :\n(i) y when x = 1.8 ; (ii) x when y = -8.\n\n\nThe table is for the relation \\(y = px^{2} - 5x + q\\).\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n21\n6\n\n-12\n\n\n\n0\n13\n', 'The table is for the relation \\(y = px^{2} - 5x + q\\).\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n21\n6\n\n-12\n\n\n\n0\n13\n\n\n\n(a)(i) Use the table to find the values of p and q.\n(ii) Copy and complete the table.~(b) Using scales of 2cm to 1 unit on the x- axis and 2 cm to 5 units on the y- axis, draw the graph of the relation for \\(-3 \\leq x \\leq 5\\).\n(c) Use the graph to find :\n(i) y when x = 1.8 ; (ii) x when y = -8.\n\n\nThe table is for the relation \\(y = px^{2} - 5x + q\\).\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n21\n6\n\n-12\n\n\n\n0\n13\n~(c) Use the graph to find :\n(i) y when x = 1.8 ; (ii) x when y = -8.\n\n\nThe table is for the relation \\(y = px^{2} - 5x + q\\).\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n21\n6\n\n-12\n\n\n\n0\n13\n', NULL, '\n\nExplanation\n(a) (i) \\(y = px^{2} - 5x + q\\)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n21\n6\n\n-12\n\n\n\n0\n13\n\n\n\nFrom the table,\ny = -12 when x = 0, i.e. \\(-12 = p(0^{2}) - 5(0) + q\\)\n\\(q = -12\\).\ny = 0 when x = 4, i.e. \\(0 = p(4^{2}) - 5(4) + q\\)\n\\(0 = 16p - 20 + q \\implies 20 = 16p + q .... (1) \\)\nSubstitute q = -12 in (1),\n\\(20 = 16p - 12\\)\n\\(20 + 12 = 16p \\implies p = \\frac{32}{16} = 2\\)\nEquation : \\(y = 2x^{2} - 5x - 12 \\).\n(ii) Given the equation \\(y = 2x^{2} - 5x - 12\\). The table becomes\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n21\n6\n-5\n-12\n-15\n-14\n-9\n0\n13\n\n\n\n(b)\n(c) From the graph,\n(i) When x = 1.8, y = -14.5 ; (ii) When y = -8, x = -0.65 or 3.15 or 3.2.\n\n', '', 'Explanation\n(a) (i) \\(y = px^{2} - 5x + q\\)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n21\n6\n\n-12\n\n\n\n0\n13\n\n\n\nFrom the table,\ny = -12 when x = 0, i.e. \\(-12 = p(0^{2}) - 5(0) + q\\)\n\\(q = -12\\).\ny = 0 when x = 4, i.e. \\(0 = p(4^{2}) - 5(4) + q\\)\n\\(0 = 16p - 20 + q \\implies 20 = 16p + q .... (1) \\)\nSubstitute q = -12 in (1),\n\\(20 = 16p - 12\\)\n\\(20 + 12 = 16p \\implies p = \\frac{32}{16} = 2\\)\nEquation : \\(y = 2x^{2} - 5x - 12 \\).\n(ii) Given the equation \\(y = 2x^{2} - 5x - 12\\). The table becomes\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n21\n6\n-5\n-12\n-15\n-14\n-9\n0\n13\n\n\n\n(b)\n(c) From the graph,\n(i) When x = 1.8, y = -14.5 ; (ii) When y = -8, x = -0.65 or 3.15 or 3.2.\n\n', '(a) (i) \\(y = px^{2} - 5x + q\\)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n21\n6\n\n-12\n\n\n\n0\n13\n\n\n\nFrom the table,\ny = -12 when x = 0, i.e. \\(-12 = p(0^{2}) - 5(0) + q\\)\n\\(q = -12\\).\ny = 0 when x = 4, i.e. \\(0 = p(4^{2}) - 5(4) + q\\)\n\\(0 = 16p - 20 + q \\implies 20 = 16p + q .... (1) \\)\nSubstitute q = -12 in (1),\n\\(20 = 16p - 12\\)\n\\(20 + 12 = 16p \\implies p = \\frac{32}{16} = 2\\)\nEquation : \\(y = 2x^{2} - 5x - 12 \\).\n(ii) Given the equation \\(y = 2x^{2} - 5x - 12\\). The table becomes\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n5\n\n\ny\n21\n6\n-5\n-12\n-15\n-14\n-9\n0\n13\n\n\n~(b)\n(c) From the graph,\n(i) When x = 1.8, y = -14.5 ; (ii) When y = -8, x = -0.65 or 3.15 or 3.2.\n~(c) From the graph,\n(i) When x = 1.8, y = -14.5 ; (ii) When y = -8, x = -0.65 or 3.15 or 3.2.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(363, '2013', '\n\nQuestion\n\n(a) Using ruler and a pair of compasses only, construct a :\n(i) Trapezium WXYZ such that |WX| = 8 cm, |XY| = 5.5 cm, |YZ| = 8.3 cm, < WXY = 60° and WX // ZY;\n(ii) rectangle PQYZ where P and Q are on WX\n(b) Measure : (i) |QX| ; (ii) < XWZ.\n\n', '', '(a) Using ruler and a pair of compasses only, construct a :\n(i) Trapezium WXYZ such that |WX| = 8 cm, |XY| = 5.5 cm, |YZ| = 8.3 cm, < WXY = 60° and WX // ZY;\n(ii) rectangle PQYZ where P and Q are on WX\n(b) Measure : (i) |QX| ; (ii) < XWZ.', '(a) Using ruler and a pair of compasses only, construct a :\n(i) Trapezium WXYZ such that |WX| = 8 cm, |XY| = 5.5 cm, |YZ| = 8.3 cm, < WXY = 60° and WX // ZY;\n(ii) rectangle PQYZ where P and Q are on WX~(b) Measure : (i) |QX| ; (ii) < XWZ.', NULL, '\n\nExplanation\n(a) (i) , (ii) \n(b) (i) |QX| = 2.7 cm ; (ii) < XWZ = 75°\n\n', '', 'Explanation\n(a) (i) , (ii) \n(b) (i) |QX| = 2.7 cm ; (ii) < XWZ = 75°\n\n', '(a) (i) , (ii) ~(b) (i) |QX| = 2.7 cm ; (ii) < XWZ = 75°\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(364, '2013', '\n\nQuestion\n\n(a) The first term of an Arithmetic Progression (AP) is 8, the ratio of the 7th term to the 9th term is 5 : 8, find the common difference of the AP.\n(b) A trader bought 30 baskets of pawpaw and 100 baskets of mangoes for N2,450.00. She sold the pawpaw at a profit of 40% and the mangoes at a profit of 30%. If her profit on the entire transaction was N855.00, find the (i) cost price of a basket of pawpaw ; (ii) selling price of the 100 baskets of mangoes.\n\n', '', '(a) The first term of an Arithmetic Progression (AP) is 8, the ratio of the 7th term to the 9th term is 5 : 8, find the common difference of the AP.\n(b) A trader bought 30 baskets of pawpaw and 100 baskets of mangoes for N2,450.00. She sold the pawpaw at a profit of 40% and the mangoes at a profit of 30%. If her profit on the entire transaction was N855.00, find the (i) cost price of a basket of pawpaw ; (ii) selling price of the 100 baskets of mangoes.', '(a) The first term of an Arithmetic Progression (AP) is 8, the ratio of the 7th term to the 9th term is 5 : 8, find the common difference of the AP.~(b) A trader bought 30 baskets of pawpaw and 100 baskets of mangoes for N2,450.00. She sold the pawpaw at a profit of 40% and the mangoes at a profit of 30%. If her profit on the entire transaction was N855.00, find the (i) cost price of a basket of pawpaw ; (ii) selling price of the 100 baskets of mangoes.', NULL, '\n\nExplanation\n(a) \\(T_{1} = a = -8\\)\n\\(\\frac{T_{7}}{T_{9}} = \\frac{5}{8}\\)\n\\(\\frac{a + 6d}{a + 8d} = \\frac{5}{8} \\implies \\frac{-8 + 6d}{-8 + 8d} = \\frac{5}{8}\\)\n\\(-8 + 6d = \\frac{5}{8}(-8 + 8d) \\implies -8 + 6d = -5 + 5d\\)\n\\(6d - 5d = -5 + 8\\)\n\\(d = 3\\)\n(b) Let :\nthe cost price \\(CP_{1}\\) of a basket of pawpaw be w;\nthe cost price \\(CP_{2}\\) of a basket of mangoes be m.\nThen \\(30w + 100m = N2,450 ..... (1)\\)\nAlso, let :\nThe selling price of the baskets of pawpaw be \\(SP_{1}\\);\nthe selling price of the baskets of mangoes be \\(SP_{2}\\).\nThen \\(SP_{1} + SP_{2} = N(2,450 +855) = N3,305 ...... (2)\\)\n%age profit on the sale of pawpaw = \\(\\frac{SP_{1} - CP_{1}}{CP_{1}} \\times 100%\\)\ni.e. \\(40% = \\frac{SP_{1} - 30w}{30w} \\times 100%\\)\n\\(\\implies 40 \\times 3w = (SP_{1} - 30w) \\times 10\\)\n\\(12w = SP_{1} - 30w\\)\n\\(SP_{1} = 12w + 30w = 42w\\)\n%age profit on the sale of mangoes = \\(\\frac{SP_{2} - CP_{2}}{CP_{2}} \\times 100%\\)\ni.e. \\(30% = \\frac{SP_{2} - 100m}{100m} \\times 100%\\)\n\\(\\implies 30m = SP_{2} - 100m\\)\n\\(SP_{2} = 30m + 100m = 130m\\)\nBut \\(SP_{1} + SP_{2} = N3,305\\)\n\\(\\implies 42w + 130m = 3305 .... (2)\\)\nFrom (1), \\(m = \\frac{2450 - 30w}{100}\\)\n\\(m = \\frac{245 - 3w}{10}\\)\nSubstitute \\(m = \\frac{245 - 3w}{10}\\) in (3)\n\\(42w + 130(\\frac{245 - 3w}{10}) = 3305\\)\n\\(42w + 3,185 - 39w = 3305\\)\n\\(3w = 3305 - 3185 = 120\\)\n\\(w = N40.00\\)\n(i) Hence, the cost price of a basket of pawpaw = N40.00.\n(ii) From (2),\n\\(SP_{1} + SP_{2} = 3305\\)\n\\(42w + 130m = 3305 \\implies 42\\times 40 + 130m = 3305\\)\n\\(1680 + 130m = 3305\\)\n\\(130m = SP_{2} = 3305 - 1680\\)\n= \\(N1,625.00\\)\nHence, the selling point of 100 baskets of mangoes = N1,625.00.\n\n', '', 'Explanation\n(a) \\(T_{1} = a = -8\\)\n\\(\\frac{T_{7}}{T_{9}} = \\frac{5}{8}\\)\n\\(\\frac{a + 6d}{a + 8d} = \\frac{5}{8} \\implies \\frac{-8 + 6d}{-8 + 8d} = \\frac{5}{8}\\)\n\\(-8 + 6d = \\frac{5}{8}(-8 + 8d) \\implies -8 + 6d = -5 + 5d\\)\n\\(6d - 5d = -5 + 8\\)\n\\(d = 3\\)\n(b) Let :\nthe cost price \\(CP_{1}\\) of a basket of pawpaw be w;\nthe cost price \\(CP_{2}\\) of a basket of mangoes be m.\nThen \\(30w + 100m = N2,450 ..... (1)\\)\nAlso, let :\nThe selling price of the baskets of pawpaw be \\(SP_{1}\\);\nthe selling price of the baskets of mangoes be \\(SP_{2}\\).\nThen \\(SP_{1} + SP_{2} = N(2,450 +855) = N3,305 ...... (2)\\)\n%age profit on the sale of pawpaw = \\(\\frac{SP_{1} - CP_{1}}{CP_{1}} \\times 100%\\)\ni.e. \\(40% = \\frac{SP_{1} - 30w}{30w} \\times 100%\\)\n\\(\\implies 40 \\times 3w = (SP_{1} - 30w) \\times 10\\)\n\\(12w = SP_{1} - 30w\\)\n\\(SP_{1} = 12w + 30w = 42w\\)\n%age profit on the sale of mangoes = \\(\\frac{SP_{2} - CP_{2}}{CP_{2}} \\times 100%\\)\ni.e. \\(30% = \\frac{SP_{2} - 100m}{100m} \\times 100%\\)\n\\(\\implies 30m = SP_{2} - 100m\\)\n\\(SP_{2} = 30m + 100m = 130m\\)\nBut \\(SP_{1} + SP_{2} = N3,305\\)\n\\(\\implies 42w + 130m = 3305 .... (2)\\)\nFrom (1), \\(m = \\frac{2450 - 30w}{100}\\)\n\\(m = \\frac{245 - 3w}{10}\\)\nSubstitute \\(m = \\frac{245 - 3w}{10}\\) in (3)\n\\(42w + 130(\\frac{245 - 3w}{10}) = 3305\\)\n\\(42w + 3,185 - 39w = 3305\\)\n\\(3w = 3305 - 3185 = 120\\)\n\\(w = N40.00\\)\n(i) Hence, the cost price of a basket of pawpaw = N40.00.\n(ii) From (2),\n\\(SP_{1} + SP_{2} = 3305\\)\n\\(42w + 130m = 3305 \\implies 42\\times 40 + 130m = 3305\\)\n\\(1680 + 130m = 3305\\)\n\\(130m = SP_{2} = 3305 - 1680\\)\n= \\(N1,625.00\\)\nHence, the selling point of 100 baskets of mangoes = N1,625.00.\n\n', '(a) \\(T_{1} = a = -8\\)\n\\(\\frac{T_{7}}{T_{9}} = \\frac{5}{8}\\)\n\\(\\frac{a + 6d}{a + 8d} = \\frac{5}{8} \\implies \\frac{-8 + 6d}{-8 + 8d} = \\frac{5}{8}\\)\n\\(-8 + 6d = \\frac{5}{8}(-8 + 8d) \\implies -8 + 6d = -5 + 5d\\)\n\\(6d - 5d = -5 + 8\\)\n\\(d = 3\\)~(b) Let :\nthe cost price \\(CP_{1}\\) of a basket of pawpaw be w;\nthe cost price \\(CP_{2}\\) of a basket of mangoes be m.\nThen \\(30w + 100m = N2,450 ..... (1)\\)\nAlso, let :\nThe selling price of the baskets of pawpaw be \\(SP_{1}\\);\nthe selling price of the baskets of mangoes be \\(SP_{2}\\).\nThen \\(SP_{1} + SP_{2} = N(2,450 +855) = N3,305 ...... (2)\\)\n%age profit on the sale of pawpaw = \\(\\frac{SP_{1} - CP_{1}}{CP_{1}} \\times 100%\\)\ni.e. \\(40% = \\frac{SP_{1} - 30w}{30w} \\times 100%\\)\n\\(\\implies 40 \\times 3w = (SP_{1} - 30w) \\times 10\\)\n\\(12w = SP_{1} - 30w\\)\n\\(SP_{1} = 12w + 30w = 42w\\)\n%age profit on the sale of mangoes = \\(\\frac{SP_{2} - CP_{2}}{CP_{2}} \\times 100%\\)\ni.e. \\(30% = \\frac{SP_{2} - 100m}{100m} \\times 100%\\)\n\\(\\implies 30m = SP_{2} - 100m\\)\n\\(SP_{2} = 30m + 100m = 130m\\)\nBut \\(SP_{1} + SP_{2} = N3,305\\)\n\\(\\implies 42w + 130m = 3305 .... (2)\\)\nFrom (1), \\(m = \\frac{2450 - 30w}{100}\\)\n\\(m = \\frac{245 - 3w}{10}\\)\nSubstitute \\(m = \\frac{245 - 3w}{10}\\) in (3)\n\\(42w + 130(\\frac{245 - 3w}{10}) = 3305\\)\n\\(42w + 3,185 - 39w = 3305\\)\n\\(3w = 3305 - 3185 = 120\\)\n\\(w = N40.00\\)\n(i) Hence, the cost price of a basket of pawpaw = N40.00.\n(ii) From (2),\n\\(SP_{1} + SP_{2} = 3305\\)\n\\(42w + 130m = 3305 \\implies 42\\times 40 + 130m = 3305\\)\n\\(1680 + 130m = 3305\\)\n\\(130m = SP_{2} = 3305 - 1680\\)\n= \\(N1,625.00\\)\nHence, the selling point of 100 baskets of mangoes = N1,625.00.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(365, '2013', '\n\nQuestion\n\n(a) Without using Mathematical tables or calculators, simplify : \\(\\frac{2\\tan 60° + \\cos 30°}{\\sin 60°}\\)\n(b) From an aeroplane in the air and at a horizontal distance of 1050m, the angles of depression of the top and base of a control tower at an instance are 36° and 41° respectively. Calculate, correct to the nearest meter, the :\n(i) height of the control tower ; (ii) shortest distance between the aeroplane and the base of the control tower.\n\n', '', '(a) Without using Mathematical tables or calculators, simplify : \\(\\frac{2\\tan 60° + \\cos 30°}{\\sin 60°}\\)\n(b) From an aeroplane in the air and at a horizontal distance of 1050m, the angles of depression of the top and base of a control tower at an instance are 36° and 41° respectively. Calculate, correct to the nearest meter, the :\n(i) height of the control tower ; (ii) shortest distance between the aeroplane and the base of the control tower.', '(a) Without using Mathematical tables or calculators, simplify : \\(\\frac{2\\tan 60° + \\cos 30°}{\\sin 60°}\\)~(b) From an aeroplane in the air and at a horizontal distance of 1050m, the angles of depression of the top and base of a control tower at an instance are 36° and 41° respectively. Calculate, correct to the nearest meter, the :\n(i) height of the control tower ; (ii) shortest distance between the aeroplane and the base of the control tower.', NULL, '\n\nExplanation\n(a)\nFrom \\(\\Delta\\) AMC, \\(\\sin 60° = \\frac{\\sqrt{3}}{2}\\)\n\\(\\cos 30° = \\frac{\\sqrt{3}}{2}\\)\n\\(\\tan 60° = \\frac{\\sqrt{3}}{1} = \\sqrt{3}\\)\nHence, \\(\\frac{2 \\tan 60° + \\cos 30°}{\\sin 60} = \\frac{2 \\times \\sqrt{3} + \\frac{\\sqrt{3}}{2}}{\\frac{\\sqrt{3}}{2}}\\)\n= \\(\\frac{4\\sqrt{3} + \\sqrt{3}}{2} \\div \\frac{\\sqrt{3}}{2}\\)\n= \\(\\frac{5\\sqrt{3}}{2} \\times \\frac{2}{\\sqrt{3}} = 5\\)\n(b)\nIn \\(\\Delta\\) POT, \\(\\tan 36° = \\frac{|TO|}{1050}\\)\n\\(|OT| = 1050 \\tan 36° = 1050 \\times 0.7265\\)\n= 762.87m\nIn \\(\\Delta\\) POB, \\(\\tan 41° = \\frac{|OB|}{1050}\\)\n\\(|OB| = 1050 \\tan 41° = 1050 \\times 0.8693\\)\n= 912.751m\n\\(|TB| = 912.751m - 762.87m = 149.88m\\)\nHence, the height of the control tower = 150 m (to the nearest meter).\n(ii) In \\(\\Delta\\) POB, \\(\\cos 41° = \\frac{1050}{|PB|}\\)\n\\(|PB| = \\frac{1050}{\\cos 41°} = \\frac{1050}{0.7547}\\)\n= 1,391.28m\nThe shortest distance between the aeroplane and the base of the control tower = 1,391m (to the nearest metre).\n\n', '', 'Explanation\n(a)\nFrom \\(\\Delta\\) AMC, \\(\\sin 60° = \\frac{\\sqrt{3}}{2}\\)\n\\(\\cos 30° = \\frac{\\sqrt{3}}{2}\\)\n\\(\\tan 60° = \\frac{\\sqrt{3}}{1} = \\sqrt{3}\\)\nHence, \\(\\frac{2 \\tan 60° + \\cos 30°}{\\sin 60} = \\frac{2 \\times \\sqrt{3} + \\frac{\\sqrt{3}}{2}}{\\frac{\\sqrt{3}}{2}}\\)\n= \\(\\frac{4\\sqrt{3} + \\sqrt{3}}{2} \\div \\frac{\\sqrt{3}}{2}\\)\n= \\(\\frac{5\\sqrt{3}}{2} \\times \\frac{2}{\\sqrt{3}} = 5\\)\n(b)\nIn \\(\\Delta\\) POT, \\(\\tan 36° = \\frac{|TO|}{1050}\\)\n\\(|OT| = 1050 \\tan 36° = 1050 \\times 0.7265\\)\n= 762.87m\nIn \\(\\Delta\\) POB, \\(\\tan 41° = \\frac{|OB|}{1050}\\)\n\\(|OB| = 1050 \\tan 41° = 1050 \\times 0.8693\\)\n= 912.751m\n\\(|TB| = 912.751m - 762.87m = 149.88m\\)\nHence, the height of the control tower = 150 m (to the nearest meter).\n(ii) In \\(\\Delta\\) POB, \\(\\cos 41° = \\frac{1050}{|PB|}\\)\n\\(|PB| = \\frac{1050}{\\cos 41°} = \\frac{1050}{0.7547}\\)\n= 1,391.28m\nThe shortest distance between the aeroplane and the base of the control tower = 1,391m (to the nearest metre).\n\n', '(a)\nFrom \\(\\Delta\\) AMC, \\(\\sin 60° = \\frac{\\sqrt{3}}{2}\\)\n\\(\\cos 30° = \\frac{\\sqrt{3}}{2}\\)\n\\(\\tan 60° = \\frac{\\sqrt{3}}{1} = \\sqrt{3}\\)\nHence, \\(\\frac{2 \\tan 60° + \\cos 30°}{\\sin 60} = \\frac{2 \\times \\sqrt{3} + \\frac{\\sqrt{3}}{2}}{\\frac{\\sqrt{3}}{2}}\\)\n= \\(\\frac{4\\sqrt{3} + \\sqrt{3}}{2} \\div \\frac{\\sqrt{3}}{2}\\)\n= \\(\\frac{5\\sqrt{3}}{2} \\times \\frac{2}{\\sqrt{3}} = 5\\)~(b)\nIn \\(\\Delta\\) POT, \\(\\tan 36° = \\frac{|TO|}{1050}\\)\n\\(|OT| = 1050 \\tan 36° = 1050 \\times 0.7265\\)\n= 762.87m\nIn \\(\\Delta\\) POB, \\(\\tan 41° = \\frac{|OB|}{1050}\\)\n\\(|OB| = 1050 \\tan 41° = 1050 \\times 0.8693\\)\n= 912.751m\n\\(|TB| = 912.751m - 762.87m = 149.88m\\)\nHence, the height of the control tower = 150 m (to the nearest meter).\n(ii) In \\(\\Delta\\) POB, \\(\\cos 41° = \\frac{1050}{|PB|}\\)\n\\(|PB| = \\frac{1050}{\\cos 41°} = \\frac{1050}{0.7547}\\)\n= 1,391.28m\nThe shortest distance between the aeroplane and the base of the control tower = 1,391m (to the nearest metre).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(366, '2013', '\n\nQuestion\n\n(a) Make m the subject of the relations \\(h = \\frac{mt}{d(m + p)}\\).\n(b)\nIn the diagram, WY and WZ are straight lines, O is the centre of circle WXM and < XWM = 48°. Calculate the value of < WYZ.\n(c) An operation \\(\\star\\) is defind on the set X = {1, 3, 5, 6} by \\(m \\star n = m + n + 2 (mod 7)\\) where \\(m, n \\in X\\).\n(i) Draw a table for the operation.\n(ii) Using the table, find the truth set of : (I) \\(3 \\star n = 3\\) ; (II) \\(n \\star n = 3\\).\n\n', '', '(a) Make m the subject of the relations \\(h = \\frac{mt}{d(m + p)}\\).\n(b)\nIn the diagram, WY and WZ are straight lines, O is the centre of circle WXM and < XWM = 48°. Calculate the value of < WYZ.\n(c) An operation \\(\\star\\) is defind on the set X = {1, 3, 5, 6} by \\(m \\star n = m + n + 2 (mod 7)\\) where \\(m, n \\in X\\).\n(i) Draw a table for the operation.\n(ii) Using the table, find the truth set of : (I) \\(3 \\star n = 3\\) ; (II) \\(n \\star n = 3\\).', '(a) Make m the subject of the relations \\(h = \\frac{mt}{d(m + p)}\\).~(b)\nIn the diagram, WY and WZ are straight lines, O is the centre of circle WXM and < XWM = 48°. Calculate the value of < WYZ.\n(c) An operation \\(\\star\\) is defind on the set X = {1, 3, 5, 6} by \\(m \\star n = m + n + 2 (mod 7)\\) where \\(m, n \\in X\\).\n(i) Draw a table for the operation.\n(ii) Using the table, find the truth set of : (I) \\(3 \\star n = 3\\) ; (II) \\(n \\star n = 3\\).~(c) An operation \\(\\star\\) is defind on the set X = {1, 3, 5, 6} by \\(m \\star n = m + n + 2 (mod 7)\\) where \\(m, n \\in X\\).\n(i) Draw a table for the operation.\n(ii) Using the table, find the truth set of : (I) \\(3 \\star n = 3\\) ; (II) \\(n \\star n = 3\\).', NULL, '\n\nExplanation\n(a) \\(h = \\frac{mt}{d(m + p)}\\)\n\\(dh(m + p) = mt\\)\n\\(dhm + dhp = mt \\implies dhp = mt - dhm\\)\n\\(dhp = m(t - dh) \\implies m = \\frac{dhp}{t - dh}\\)\n(b)\nIn the diagram above, < WXM = 90° (angle in a semicircle)\n< WMX = 180° - (90° + 48°)\n= 42°\n< XMZ = 180° - 42° (angles on a straight line)\n= 138°\n< WYZ = 180° - 138° (opp. angles of a cyclic quadrilateral)\n= 42°\n(c)\n\n\n\n\\(\\star\\)\n1\n3\n5\n6\n\n\n1\n4\n6\n1\n2\n\n\n3\n6\n1\n3\n4\n\n\n5\n1\n3\n5\n6\n\n\n6\n2\n4\n6\n0\n\n\n\n(ii) From the table, the truth set of :\n(I) \\(3 \\star n = 3; n = {5}\\)\n(II) \\(n \\star n = 3; n = { }\\)\n\n', '', 'Explanation\n(a) \\(h = \\frac{mt}{d(m + p)}\\)\n\\(dh(m + p) = mt\\)\n\\(dhm + dhp = mt \\implies dhp = mt - dhm\\)\n\\(dhp = m(t - dh) \\implies m = \\frac{dhp}{t - dh}\\)\n(b)\nIn the diagram above, < WXM = 90° (angle in a semicircle)\n< WMX = 180° - (90° + 48°)\n= 42°\n< XMZ = 180° - 42° (angles on a straight line)\n= 138°\n< WYZ = 180° - 138° (opp. angles of a cyclic quadrilateral)\n= 42°\n(c)\n\n\n\n\\(\\star\\)\n1\n3\n5\n6\n\n\n1\n4\n6\n1\n2\n\n\n3\n6\n1\n3\n4\n\n\n5\n1\n3\n5\n6\n\n\n6\n2\n4\n6\n0\n\n\n\n(ii) From the table, the truth set of :\n(I) \\(3 \\star n = 3; n = {5}\\)\n(II) \\(n \\star n = 3; n = { }\\)\n\n', '(a) \\(h = \\frac{mt}{d(m + p)}\\)\n\\(dh(m + p) = mt\\)\n\\(dhm + dhp = mt \\implies dhp = mt - dhm\\)\n\\(dhp = m(t - dh) \\implies m = \\frac{dhp}{t - dh}\\)~(b)\nIn the diagram above, < WXM = 90° (angle in a semicircle)\n< WMX = 180° - (90° + 48°)\n= 42°\n< XMZ = 180° - 42° (angles on a straight line)\n= 138°\n< WYZ = 180° - 138° (opp. angles of a cyclic quadrilateral)\n= 42°\n(c)\n\n\n\n\\(\\star\\)\n1\n3\n5\n6\n\n\n1\n4\n6\n1\n2\n\n\n3\n6\n1\n3\n4\n\n\n5\n1\n3\n5\n6\n\n\n6\n2\n4\n6\n0\n\n\n\n(ii) From the table, the truth set of :\n(I) \\(3 \\star n = 3; n = {5}\\)\n(II) \\(n \\star n = 3; n = { }\\)\n~(c)\n\n\n\n\\(\\star\\)\n1\n3\n5\n6\n\n\n1\n4\n6\n1\n2\n\n\n3\n6\n1\n3\n4\n\n\n5\n1\n3\n5\n6\n\n\n6\n2\n4\n6\n0\n\n\n\n(ii) From the table, the truth set of :\n(I) \\(3 \\star n = 3; n = {5}\\)\n(II) \\(n \\star n = 3; n = { }\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(367, '2013', '\n\nQuestion\n\nA water reservoir in the form of a cone mounted on a hemisphere is built such that the plane face of the hemisphere fits exactly to the base of the cone and the height of the cone is 6 times thr radius of its base.\n(a) Illustrate this information in a diagram.\n(b) If the volume of the reservoir is \\(333\\frac{1}{3}\\pi m^{3}\\), calculate, correct to the nearest whole number, the :\n(I) volume of the hemisphere ; (II) Total surface area of the reservoir. [Take \\(\\pi = \\frac{22}{7}\\)].\n\n', '', 'A water reservoir in the form of a cone mounted on a hemisphere is built such that the plane face of the hemisphere fits exactly to the base of the cone and the height of the cone is 6 times thr radius of its base.\n(a) Illustrate this information in a diagram.\n(b) If the volume of the reservoir is \\(333\\frac{1}{3}\\pi m^{3}\\), calculate, correct to the nearest whole number, the :\n(I) volume of the hemisphere ; (II) Total surface area of the reservoir. [Take \\(\\pi = \\frac{22}{7}\\)].', 'A water reservoir in the form of a cone mounted on a hemisphere is built such that the plane face of the hemisphere fits exactly to the base of the cone and the height of the cone is 6 times thr radius of its base.\n(a) Illustrate this information in a diagram.~(b) If the volume of the reservoir is \\(333\\frac{1}{3}\\pi m^{3}\\), calculate, correct to the nearest whole number, the :\n(I) volume of the hemisphere ; (II) Total surface area of the reservoir. [Take \\(\\pi = \\frac{22}{7}\\)].', NULL, '\n\nExplanation\n(a)\n(i) Using volume of reservoir = volume of hemisphere + volume of cone.\n\\(\\implies 333\\frac{1}{3}\\pi = \\frac{2}{3}\\pi r^{3} + \\frac{1}{3}\\pi r^{2} h\\)\n\\(\\implies \\frac{1000}{3}\\pi = \\frac{1}{3}\\pi (2r^{3} + r^{2}h)\\)\nLet the radius of the hemisphere be x m;\nThen \\(\\frac{1000}{3} = \\frac{1}{3}(2x^{3} + x^{2}(6x))\\)\n\\(1000 = 2x^{3} + 6x^{3}\\)\n\\(1000 = 8x^{3} \\implies x^{3} = \\frac{1000}{8} = 125\\)\n\\(x = \\sqrt[3]{125} = 5 m\\)\nHence, the volume of the hemisphere = \\(\\frac{2}{3}\\pi r^{3}\\)\n(where r = x = 5 m)\n= \\(\\frac{2}{3} \\times \\frac{22}{7} \\times 5^{3} m^{3}\\)\n= \\(\\frac{5500}{21} m^{3} \\approxeq 261.905 m^{3}\\)\n= \\(262 m^{3} \\) (nearest whole number)\n(ii) Total surface area of reservoir\n= surface area of hemisphere + surface area of cone\n= \\(2\\pi r^{2} + \\pi rl\\)\n\nFrom the diagram above, \\(l^{2} = 30^{2} + 5^{2}\\)\n\\(l^{2} = 900 + 25 = 925\\)\n\\(l = \\sqrt{925} = 30.41 m\\)\nHence, the total surface area of the reservoir = \\(2 \\times \\frac{22}{7} \\times 5 \\times 5 + \\frac{22}{7} \\times 5 \\times 30.41\\)\n= \\(157.143 + 477.871\\)\n= \\(635.104 m^{2}\\)\n\\(\\approxeq 635m^{2}\\) (to the nearest whole number).\n\n', '', 'Explanation\n(a)\n(i) Using volume of reservoir = volume of hemisphere + volume of cone.\n\\(\\implies 333\\frac{1}{3}\\pi = \\frac{2}{3}\\pi r^{3} + \\frac{1}{3}\\pi r^{2} h\\)\n\\(\\implies \\frac{1000}{3}\\pi = \\frac{1}{3}\\pi (2r^{3} + r^{2}h)\\)\nLet the radius of the hemisphere be x m;\nThen \\(\\frac{1000}{3} = \\frac{1}{3}(2x^{3} + x^{2}(6x))\\)\n\\(1000 = 2x^{3} + 6x^{3}\\)\n\\(1000 = 8x^{3} \\implies x^{3} = \\frac{1000}{8} = 125\\)\n\\(x = \\sqrt[3]{125} = 5 m\\)\nHence, the volume of the hemisphere = \\(\\frac{2}{3}\\pi r^{3}\\)\n(where r = x = 5 m)\n= \\(\\frac{2}{3} \\times \\frac{22}{7} \\times 5^{3} m^{3}\\)\n= \\(\\frac{5500}{21} m^{3} \\approxeq 261.905 m^{3}\\)\n= \\(262 m^{3} \\) (nearest whole number)\n(ii) Total surface area of reservoir\n= surface area of hemisphere + surface area of cone\n= \\(2\\pi r^{2} + \\pi rl\\)\n\nFrom the diagram above, \\(l^{2} = 30^{2} + 5^{2}\\)\n\\(l^{2} = 900 + 25 = 925\\)\n\\(l = \\sqrt{925} = 30.41 m\\)\nHence, the total surface area of the reservoir = \\(2 \\times \\frac{22}{7} \\times 5 \\times 5 + \\frac{22}{7} \\times 5 \\times 30.41\\)\n= \\(157.143 + 477.871\\)\n= \\(635.104 m^{2}\\)\n\\(\\approxeq 635m^{2}\\) (to the nearest whole number).\n\n', '', 'WAEC-PK-1', 'WAEC-MATH', 1),
(368, '2013', '\n\nQuestion\n\nThe table shows the marks scored by some candidates in an examination.\n\n\n\nMarks (%)\n0-9\n10-19\n20-29\n30-39\n40-49\n50-59\n60-69\n70-79\n80-89\n90-99\n\n\nFrequency\n7\n11\n17\n20\n29\n34\n30\n25\n21\n6\n\n\n\n(a) Construct a cumulative frequency table for the distribution and draw a cumulative frequency curve.\n(b) Use the curve to estimate, correct to one decimal place, the :\n(i) Lowest mark for distinction if 5% of the candidates passed with distinction ; (ii) probability of selecting a candidate who scored at most 45%.\n\n\nThe table shows the marks scored by some candidates in an examination.\n\n\n\nMarks (%)\n0-9\n10-19\n20-29\n30-39\n40-49\n50-59\n60-69\n70-79\n80-89\n90-99\n\n\nFrequency\n7\n11\n17\n20\n29\n34\n30\n25\n21\n6\n\n\n', '', 'The table shows the marks scored by some candidates in an examination.\n\n\n\nMarks (%)\n0-9\n10-19\n20-29\n30-39\n40-49\n50-59\n60-69\n70-79\n80-89\n90-99\n\n\nFrequency\n7\n11\n17\n20\n29\n34\n30\n25\n21\n6\n\n\n\n(a) Construct a cumulative frequency table for the distribution and draw a cumulative frequency curve.\n(b) Use the curve to estimate, correct to one decimal place, the :\n(i) Lowest mark for distinction if 5% of the candidates passed with distinction ; (ii) probability of selecting a candidate who scored at most 45%.\n\n\nThe table shows the marks scored by some candidates in an examination.\n\n\n\nMarks (%)\n0-9\n10-19\n20-29\n30-39\n40-49\n50-59\n60-69\n70-79\n80-89\n90-99\n\n\nFrequency\n7\n11\n17\n20\n29\n34\n30\n25\n21\n6\n', 'The table shows the marks scored by some candidates in an examination.\n\n\n\nMarks (%)\n0-9\n10-19\n20-29\n30-39\n40-49\n50-59\n60-69\n70-79\n80-89\n90-99\n\n\nFrequency\n7\n11\n17\n20\n29\n34\n30\n25\n21\n6\n\n\n\n(a) Construct a cumulative frequency table for the distribution and draw a cumulative frequency curve.~(b) Use the curve to estimate, correct to one decimal place, the :\n(i) Lowest mark for distinction if 5% of the candidates passed with distinction ; (ii) probability of selecting a candidate who scored at most 45%.\n\n\nThe table shows the marks scored by some candidates in an examination.\n\n\n\nMarks (%)\n0-9\n10-19\n20-29\n30-39\n40-49\n50-59\n60-69\n70-79\n80-89\n90-99\n\n\nFrequency\n7\n11\n17\n20\n29\n34\n30\n25\n21\n6\n', NULL, '\n\nExplanation\nMarks\n(%)\n\n\nfrequency\n(f)\n\n\nCumulative\nfrequency\n\n\nUpper class\nboundaries\n\n\n\n0 - 9\n7\n7\n9.5\n\n\n10 - 19\n11\n18\n19.5\n\n\n20 - 29\n17\n35\n29.5\n\n\n30 - 39\n20\n55\n39.5\n\n\n40 - 49\n29\n84\n49.5\n\n\n50 - 59\n34\n118\n59.5\n\n\n60 - 69\n30\n148\n69.5\n\n\n70 - 79\n25\n173\n79.5\n\n\n80 - 89\n21\n194\n89.5\n\n\n90 - 99\n6\n200\n99.5\n\n\n\n\n(b) (i) If 5% of the candidates passed the examination, then (100 - 5)% = 95% passed with a mark \\(\\leq\\) the lowest mark for distinction.\n\\(\\text{95% of 200} = \\frac{95}{100} \\times 200 \\)\n= 190 candidates. From the ogive, 190 corresponds to 79.5 + 8 = 87.5 marks (to one decimal place)\n(ii) From the ogive, the number of candidates who scored at most 45% is 69. Hence, the probability of selecting a candidate who scored at most 45%\n= \\(\\frac{69}{200} = 0.345 \\approxeq 0.3\\) (1 d.p)\n\n', '', 'Explanation\nMarks\n(%)\n\n\nfrequency\n(f)\n\n\nCumulative\nfrequency\n\n\nUpper class\nboundaries\n\n\n\n0 - 9\n7\n7\n9.5\n\n\n10 - 19\n11\n18\n19.5\n\n\n20 - 29\n17\n35\n29.5\n\n\n30 - 39\n20\n55\n39.5\n\n\n40 - 49\n29\n84\n49.5\n\n\n50 - 59\n34\n118\n59.5\n\n\n60 - 69\n30\n148\n69.5\n\n\n70 - 79\n25\n173\n79.5\n\n\n80 - 89\n21\n194\n89.5\n\n\n90 - 99\n6\n200\n99.5\n\n\n\n\n(b) (i) If 5% of the candidates passed the examination, then (100 - 5)% = 95% passed with a mark \\(\\leq\\) the lowest mark for distinction.\n\\(\\text{95% of 200} = \\frac{95}{100} \\times 200 \\)\n= 190 candidates. From the ogive, 190 corresponds to 79.5 + 8 = 87.5 marks (to one decimal place)\n(ii) From the ogive, the number of candidates who scored at most 45% is 69. Hence, the probability of selecting a candidate who scored at most 45%\n= \\(\\frac{69}{200} = 0.345 \\approxeq 0.3\\) (1 d.p)\n\n', '~(b) (i) If 5% of the candidates passed the examination, then (100 - 5)% = 95% passed with a mark \\(\\leq\\) the lowest mark for distinction.\n\\(\\text{95% of 200} = \\frac{95}{100} \\times 200 \\)\n= 190 candidates. From the ogive, 190 corresponds to 79.5 + 8 = 87.5 marks (to one decimal place)\n(ii) From the ogive, the number of candidates who scored at most 45% is 69. Hence, the probability of selecting a candidate who scored at most 45%\n= \\(\\frac{69}{200} = 0.345 \\approxeq 0.3\\) (1 d.p)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(369, '2013', '\n\nQuestion\n\n(a) Without using tables or calculator, simplify : \\(\\frac{0.6 \\times 32 \\times 0.004}{1.2 \\times 0.008 \\times 0.16}\\), leaving the answer in standard form (scientific notation).\n(b)\nIn the diagram, \\(\\overline{EF}\\) is parallel to \\(\\overline{GH}\\). If \\(< AEF = 3x°, < ABC = 120°\\) and \\(< CHG = 7x°\\), find the value of \\(< GHB\\).\n\n', '', '(a) Without using tables or calculator, simplify : \\(\\frac{0.6 \\times 32 \\times 0.004}{1.2 \\times 0.008 \\times 0.16}\\), leaving the answer in standard form (scientific notation).\n(b)\nIn the diagram, \\(\\overline{EF}\\) is parallel to \\(\\overline{GH}\\). If \\(< AEF = 3x°, < ABC = 120°\\) and \\(< CHG = 7x°\\), find the value of \\(< GHB\\).', '(a) Without using tables or calculator, simplify : \\(\\frac{0.6 \\times 32 \\times 0.004}{1.2 \\times 0.008 \\times 0.16}\\), leaving the answer in standard form (scientific notation).~(b)\nIn the diagram, \\(\\overline{EF}\\) is parallel to \\(\\overline{GH}\\). If \\(< AEF = 3x°, < ABC = 120°\\) and \\(< CHG = 7x°\\), find the value of \\(< GHB\\).', NULL, '\n\nExplanation\n(a) \\(\\frac{0.6 \\times 32 \\times 0.004}{1.2 \\times 0.008 \\times 0.16}\\)\n= \\(\\frac{6 \\times 10^{-1} \\times 32 \\times 4 \\times 10^{-3}}{12 \\times 10^{-1} \\times 8 \\times 10^{-3} \\times 16 \\times 10^{-2}}\\)\n= \\(\\frac{4}{2 \\times 4 \\times 10^{-2}}\\)\n= \\(\\frac{1}{2} \\times \\frac{1}{10^{-2}\\)\n= \\(0.5 \\times 10^{2} \\)\n= \\(5 \\times 10^{1}\\)\n(b)\nIn the diagram above, \\(< GHB = 180 - 7x°\\) (angles on a straight line)\n\\(< HBN = 180 - 7x°\\) (alternate angles)\n\\(< EBN = 3x°\\) (corresponding angles)\nThen \\(3x° + (180 - 7x)° = 120°\\)\n\\(\\implies 3x° - 7x° + 180° = 120°\\)\n\\(\\implies 180° - 120° = 7x° - 3x°\\)\n\\(60° = 4x°\\)\n\\(x = 15°\\)\nHence, \\(< GHB = 180° - (7 \\times 15°) = 180° - 105°\\)\n= \\(75°\\).\n\n', '', 'Explanation\n(a) \\(\\frac{0.6 \\times 32 \\times 0.004}{1.2 \\times 0.008 \\times 0.16}\\)\n= \\(\\frac{6 \\times 10^{-1} \\times 32 \\times 4 \\times 10^{-3}}{12 \\times 10^{-1} \\times 8 \\times 10^{-3} \\times 16 \\times 10^{-2}}\\)\n= \\(\\frac{4}{2 \\times 4 \\times 10^{-2}}\\)\n= \\(\\frac{1}{2} \\times \\frac{1}{10^{-2}\\)\n= \\(0.5 \\times 10^{2} \\)\n= \\(5 \\times 10^{1}\\)\n(b)\nIn the diagram above, \\(< GHB = 180 - 7x°\\) (angles on a straight line)\n\\(< HBN = 180 - 7x°\\) (alternate angles)\n\\(< EBN = 3x°\\) (corresponding angles)\nThen \\(3x° + (180 - 7x)° = 120°\\)\n\\(\\implies 3x° - 7x° + 180° = 120°\\)\n\\(\\implies 180° - 120° = 7x° - 3x°\\)\n\\(60° = 4x°\\)\n\\(x = 15°\\)\nHence, \\(< GHB = 180° - (7 \\times 15°) = 180° - 105°\\)\n= \\(75°\\).\n\n', '(a) \\(\\frac{0.6 \\times 32 \\times 0.004}{1.2 \\times 0.008 \\times 0.16}\\)\n= \\(\\frac{6 \\times 10^{-1} \\times 32 \\times 4 \\times 10^{-3}}{12 \\times 10^{-1} \\times 8 \\times 10^{-3} \\times 16 \\times 10^{-2}}\\)\n= \\(\\frac{4}{2 \\times 4 \\times 10^{-2}}\\)\n= \\(\\frac{1}{2} \\times \\frac{1}{10^{-2}\\)\n= \\(0.5 \\times 10^{2} \\)\n= \\(5 \\times 10^{1}\\)~(b)\nIn the diagram above, \\(< GHB = 180 - 7x°\\) (angles on a straight line)\n\\(< HBN = 180 - 7x°\\) (alternate angles)\n\\(< EBN = 3x°\\) (corresponding angles)\nThen \\(3x° + (180 - 7x)° = 120°\\)\n\\(\\implies 3x° - 7x° + 180° = 120°\\)\n\\(\\implies 180° - 120° = 7x° - 3x°\\)\n\\(60° = 4x°\\)\n\\(x = 15°\\)\nHence, \\(< GHB = 180° - (7 \\times 15°) = 180° - 105°\\)\n= \\(75°\\).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(370, '2013', '\n\nQuestion\n\n(a) Simplify : \\(3\\sqrt{75} - \\sqrt{12} + \\sqrt{108}\\), leaving the answer in surd form (radicals).\n(b) If \\(124_{n} = 232_{five}\\), find n.\n\n', '', '(a) Simplify : \\(3\\sqrt{75} - \\sqrt{12} + \\sqrt{108}\\), leaving the answer in surd form (radicals).\n(b) If \\(124_{n} = 232_{five}\\), find n.', '(a) Simplify : \\(3\\sqrt{75} - \\sqrt{12} + \\sqrt{108}\\), leaving the answer in surd form (radicals).~(b) If \\(124_{n} = 232_{five}\\), find n.', NULL, '\n\nExplanation\n(a) \\(3\\sqrt{75} = 3 \\times \\sqrt{25 \\times 3} = 3(5\\sqrt{3}) = 15\\sqrt{3}\\)\n\\(\\sqrt{12} = \\sqrt{4 \\times 3} = 2\\sqrt{3}\\)\n\\(\\sqrt{108} = \\sqrt{36 \\times 3} = 6\\sqrt{3}\\)\n\\(\\therefore 3\\sqrt{75} - \\sqrt{12} + \\sqrt{108} \\equiv 15\\sqrt{3} - 2\\sqrt{3} + 6\\sqrt{3}\\)\n= \\((15 - 2 + 6)\\sqrt{3}\\)\n= \\(19\\sqrt{3}\\).\n(b) \\(124_{n} = 232_{five}\\)\n\\(124_{n} = (1 \\times n^{2}) + (2 \\times n^{1}) + (4 \\times n^{0}) \\)\n= \\(n^{2} + 2n + 4\\)\n\\(232_{five} = (2 \\times 5^{2}) + (3 \\times 5^{1}) + (2 \\times 5^{0})\\)\n= \\(50 + 15 + 2\\)\n= \\(67\\)\n\\(\\implies n^{2} + 2n + 4 = 67\\)\n\\(n^{2} + 2n + 4 - 67 = 0\\)\n\\(n^{2} + 2n - 63 = 0\\)\n\\(n^{2} - 7n + 9n - 63 = 0\\)\n\\(n(n - 7) + 9(n - 7) = 0\\)\n\\((n +9)(n - 7) = 0\\)\n\\(\\text{n = -9 or 7}\\)\nHence, \\(n = 7\\)\n\\(124_{seven} = 232_{five}\\)\n\n', '', 'Explanation\n(a) \\(3\\sqrt{75} = 3 \\times \\sqrt{25 \\times 3} = 3(5\\sqrt{3}) = 15\\sqrt{3}\\)\n\\(\\sqrt{12} = \\sqrt{4 \\times 3} = 2\\sqrt{3}\\)\n\\(\\sqrt{108} = \\sqrt{36 \\times 3} = 6\\sqrt{3}\\)\n\\(\\therefore 3\\sqrt{75} - \\sqrt{12} + \\sqrt{108} \\equiv 15\\sqrt{3} - 2\\sqrt{3} + 6\\sqrt{3}\\)\n= \\((15 - 2 + 6)\\sqrt{3}\\)\n= \\(19\\sqrt{3}\\).\n(b) \\(124_{n} = 232_{five}\\)\n\\(124_{n} = (1 \\times n^{2}) + (2 \\times n^{1}) + (4 \\times n^{0}) \\)\n= \\(n^{2} + 2n + 4\\)\n\\(232_{five} = (2 \\times 5^{2}) + (3 \\times 5^{1}) + (2 \\times 5^{0})\\)\n= \\(50 + 15 + 2\\)\n= \\(67\\)\n\\(\\implies n^{2} + 2n + 4 = 67\\)\n\\(n^{2} + 2n + 4 - 67 = 0\\)\n\\(n^{2} + 2n - 63 = 0\\)\n\\(n^{2} - 7n + 9n - 63 = 0\\)\n\\(n(n - 7) + 9(n - 7) = 0\\)\n\\((n +9)(n - 7) = 0\\)\n\\(\\text{n = -9 or 7}\\)\nHence, \\(n = 7\\)\n\\(124_{seven} = 232_{five}\\)\n\n', '(a) \\(3\\sqrt{75} = 3 \\times \\sqrt{25 \\times 3} = 3(5\\sqrt{3}) = 15\\sqrt{3}\\)\n\\(\\sqrt{12} = \\sqrt{4 \\times 3} = 2\\sqrt{3}\\)\n\\(\\sqrt{108} = \\sqrt{36 \\times 3} = 6\\sqrt{3}\\)\n\\(\\therefore 3\\sqrt{75} - \\sqrt{12} + \\sqrt{108} \\equiv 15\\sqrt{3} - 2\\sqrt{3} + 6\\sqrt{3}\\)\n= \\((15 - 2 + 6)\\sqrt{3}\\)\n= \\(19\\sqrt{3}\\).~(b) \\(124_{n} = 232_{five}\\)\n\\(124_{n} = (1 \\times n^{2}) + (2 \\times n^{1}) + (4 \\times n^{0}) \\)\n= \\(n^{2} + 2n + 4\\)\n\\(232_{five} = (2 \\times 5^{2}) + (3 \\times 5^{1}) + (2 \\times 5^{0})\\)\n= \\(50 + 15 + 2\\)\n= \\(67\\)\n\\(\\implies n^{2} + 2n + 4 = 67\\)\n\\(n^{2} + 2n + 4 - 67 = 0\\)\n\\(n^{2} + 2n - 63 = 0\\)\n\\(n^{2} - 7n + 9n - 63 = 0\\)\n\\(n(n - 7) + 9(n - 7) = 0\\)\n\\((n +9)(n - 7) = 0\\)\n\\(\\text{n = -9 or 7}\\)\nHence, \\(n = 7\\)\n\\(124_{seven} = 232_{five}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(371, '2013', '\n\nQuestion\n\n(a) Solve the simultaneous equation : \\(\\frac{1}{x} + \\frac{1}{y} = 5 ; \\frac{1}{y} - \\frac{1}{x} = 1\\).\n(b) A man drives from Ibadan to Oyo, a distance of 48km in 45 minutes. If he drives at 72 km/h where the surface is good and 48 km/h where it is bad, find the number of kilometers of good surface.\n\n', '', '(a) Solve the simultaneous equation : \\(\\frac{1}{x} + \\frac{1}{y} = 5 ; \\frac{1}{y} - \\frac{1}{x} = 1\\).\n(b) A man drives from Ibadan to Oyo, a distance of 48km in 45 minutes. If he drives at 72 km/h where the surface is good and 48 km/h where it is bad, find the number of kilometers of good surface.', '(a) Solve the simultaneous equation : \\(\\frac{1}{x} + \\frac{1}{y} = 5 ; \\frac{1}{y} - \\frac{1}{x} = 1\\).~(b) A man drives from Ibadan to Oyo, a distance of 48km in 45 minutes. If he drives at 72 km/h where the surface is good and 48 km/h where it is bad, find the number of kilometers of good surface.', NULL, '\n\nExplanation\n(a) \\(\\frac{1}{x} + \\frac{1}{y} = 5 ..... (1)\\)\n\\(\\frac{1}{y} - \\frac{1}{x} = 1 ........ (2)\\)\nSubstitute p and q for \\(\\frac{1}{x}\\) and \\(\\frac{1}{y}\\) respectively.\n\\(p + q = 5 ......... (1a)\\)\n\\(q - p = 1 ....... (2a)\\)\nFrom (2a), q = 1 + p.\nPutting into (1a), we have:\n\\(p + (1 + p) = 5\\)\n\\(2p + 1 = 5 \\implies 2p = 5 - 1 = 4\\)\n\\(p = 2\\)\n\\(q = 1 + p = 1 + 2 = 3\\)\n\\(p = \\frac{1}{x} = 2\\)\n\\(\\implies x = \\frac{1}{2}\\)\n\\(q = \\frac{1}{y} = 3\\)\n\\(\\implies y = \\frac{1}{3}\\)\nHence, \\(x = \\frac{1}{2} ; y = \\frac{1}{3}\\).\n(b) Total distance travelled = 48 km\nTotal time taken = 45 minutes = \\(\\frac{45}{60} = 0.75 hr\\)\nLet the man travel a distance of x km on a good surface. Then he travel a distance of (48 - x) km on bad surface.\nLet the time taken to travel on good surface be t hours, then the time taken to travel on bad surface = (0.75 - t) hrs.\nUsing \\(speed = \\frac{distance}{time}\\) in each case\nOn good surface : \\(72 = \\frac{x}{t}\\)\n\\(\\implies x = 72t ... (1)\\)\nOn bad surface : \\(48 = \\frac{48 - x}{0.75 - t}\\)\n\\(48(0.75 - t) = 48 - x\\)\n\\(\\implies 36 - 48t = 48 - x ...... (2)\\)\nSubstitute 72t = x in (2) :\n\\(36 - 48t = 48 - 72t \\implies 72t - 48t = 48 - 36\\)\n\\(24t = 12 \\implies t = \\frac{12}{24} = 0.5 hours\\)\nFrom (1), x = 72t\n\\(x = 72 \\times 0.5 = 36 km\\)\nHence, the man travels a distance of 36 km on good road surface.\n\n', '', 'Explanation\n(a) \\(\\frac{1}{x} + \\frac{1}{y} = 5 ..... (1)\\)\n\\(\\frac{1}{y} - \\frac{1}{x} = 1 ........ (2)\\)\nSubstitute p and q for \\(\\frac{1}{x}\\) and \\(\\frac{1}{y}\\) respectively.\n\\(p + q = 5 ......... (1a)\\)\n\\(q - p = 1 ....... (2a)\\)\nFrom (2a), q = 1 + p.\nPutting into (1a), we have:\n\\(p + (1 + p) = 5\\)\n\\(2p + 1 = 5 \\implies 2p = 5 - 1 = 4\\)\n\\(p = 2\\)\n\\(q = 1 + p = 1 + 2 = 3\\)\n\\(p = \\frac{1}{x} = 2\\)\n\\(\\implies x = \\frac{1}{2}\\)\n\\(q = \\frac{1}{y} = 3\\)\n\\(\\implies y = \\frac{1}{3}\\)\nHence, \\(x = \\frac{1}{2} ; y = \\frac{1}{3}\\).\n(b) Total distance travelled = 48 km\nTotal time taken = 45 minutes = \\(\\frac{45}{60} = 0.75 hr\\)\nLet the man travel a distance of x km on a good surface. Then he travel a distance of (48 - x) km on bad surface.\nLet the time taken to travel on good surface be t hours, then the time taken to travel on bad surface = (0.75 - t) hrs.\nUsing \\(speed = \\frac{distance}{time}\\) in each case\nOn good surface : \\(72 = \\frac{x}{t}\\)\n\\(\\implies x = 72t ... (1)\\)\nOn bad surface : \\(48 = \\frac{48 - x}{0.75 - t}\\)\n\\(48(0.75 - t) = 48 - x\\)\n\\(\\implies 36 - 48t = 48 - x ...... (2)\\)\nSubstitute 72t = x in (2) :\n\\(36 - 48t = 48 - 72t \\implies 72t - 48t = 48 - 36\\)\n\\(24t = 12 \\implies t = \\frac{12}{24} = 0.5 hours\\)\nFrom (1), x = 72t\n\\(x = 72 \\times 0.5 = 36 km\\)\nHence, the man travels a distance of 36 km on good road surface.\n\n', '(a) \\(\\frac{1}{x} + \\frac{1}{y} = 5 ..... (1)\\)\n\\(\\frac{1}{y} - \\frac{1}{x} = 1 ........ (2)\\)\nSubstitute p and q for \\(\\frac{1}{x}\\) and \\(\\frac{1}{y}\\) respectively.\n\\(p + q = 5 ......... (1a)\\)\n\\(q - p = 1 ....... (2a)\\)\nFrom (2a), q = 1 + p.\nPutting into (1a), we have:\n\\(p + (1 + p) = 5\\)\n\\(2p + 1 = 5 \\implies 2p = 5 - 1 = 4\\)\n\\(p = 2\\)\n\\(q = 1 + p = 1 + 2 = 3\\)\n\\(p = \\frac{1}{x} = 2\\)\n\\(\\implies x = \\frac{1}{2}\\)\n\\(q = \\frac{1}{y} = 3\\)\n\\(\\implies y = \\frac{1}{3}\\)\nHence, \\(x = \\frac{1}{2} ; y = \\frac{1}{3}\\).~(b) Total distance travelled = 48 km\nTotal time taken = 45 minutes = \\(\\frac{45}{60} = 0.75 hr\\)\nLet the man travel a distance of x km on a good surface. Then he travel a distance of (48 - x) km on bad surface.\nLet the time taken to travel on good surface be t hours, then the time taken to travel on bad surface = (0.75 - t) hrs.\nUsing \\(speed = \\frac{distance}{time}\\) in each case\nOn good surface : \\(72 = \\frac{x}{t}\\)\n\\(\\implies x = 72t ... (1)\\)\nOn bad surface : \\(48 = \\frac{48 - x}{0.75 - t}\\)\n\\(48(0.75 - t) = 48 - x\\)\n\\(\\implies 36 - 48t = 48 - x ...... (2)\\)\nSubstitute 72t = x in (2) :\n\\(36 - 48t = 48 - 72t \\implies 72t - 48t = 48 - 36\\)\n\\(24t = 12 \\implies t = \\frac{12}{24} = 0.5 hours\\)\nFrom (1), x = 72t\n\\(x = 72 \\times 0.5 = 36 km\\)\nHence, the man travels a distance of 36 km on good road surface.\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
(372, '2013', '\n\nQuestion\n\n(a)\nIn the diagram, O is the centre of the circle radius r cm and < XOY = 90°.If the area of the shaded part is 504\\(cm^{2}\\), calculate the value of r. [Take \\(\\pi = \\frac{22}{7}\\)].\n(b) Two isosceles triangles PQR and PQS are drawn on opposite sides of a common base PQ. If \\(< PQR = 66°\\) and \\(< PSQ = 109°\\), calculate the value of \\(< RQS\\).\n\n', '', '(a)\nIn the diagram, O is the centre of the circle radius r cm and < XOY = 90°.If the area of the shaded part is 504\\(cm^{2}\\), calculate the value of r. [Take \\(\\pi = \\frac{22}{7}\\)].\n(b) Two isosceles triangles PQR and PQS are drawn on opposite sides of a common base PQ. If \\(< PQR = 66°\\) and \\(< PSQ = 109°\\), calculate the value of \\(< RQS\\).', '(a)\nIn the diagram, O is the centre of the circle radius r cm and < XOY = 90°.If the area of the shaded part is 504\\(cm^{2}\\), calculate the value of r. [Take \\(\\pi = \\frac{22}{7}\\)].~(b) Two isosceles triangles PQR and PQS are drawn on opposite sides of a common base PQ. If \\(< PQR = 66°\\) and \\(< PSQ = 109°\\), calculate the value of \\(< RQS\\).', NULL, '\n\nExplanation\n(a) Area of the shaded segment = area of sector XOY - area of triangle OXY.\n\\(\\implies 504 = \\frac{\\theta}{360°} \\times \\pi r^{2} - \\frac{1}{2} r^{2} \\sin \\theta\\)\n\\(\\implies 504 = \\frac{90}{360} \\times \\frac{22}{7} \\times r^{2} - \\frac{1}{2} \\times r^{2} \\sin 90\\)\n\\(504 = r^{2}(\\frac{22}{28} - \\frac{1}{2})\\)\n\\(504 = \\frac{2r^{2}}{7} \\implies r^{2} = \\frac{504 \\times 7}{2} = 1764\\)\n\\(r = \\sqrt{1764} = 42 cm\\).\n(b)In the diagram, \\(a_{1} = a_{2}\\) (base angles of an isosceles triangle)\n\\(\\therefore a_{1} + a_{2} = 2a_{1}\\)\n\\(\\therefore 2a_{1} + 109° = 180° \\implies 2a_{1} = 180° - 109° = 71°\\)\n\\(a_{1} = a_{2} = \\frac{71°}{2} = 35.5°\\)\nHence, \\(< RQS = 66° + 35.5°\\)\n= \\(101.5°\\)\n\n\n', '', 'Explanation\n(a) Area of the shaded segment = area of sector XOY - area of triangle OXY.\n\\(\\implies 504 = \\frac{\\theta}{360°} \\times \\pi r^{2} - \\frac{1}{2} r^{2} \\sin \\theta\\)\n\\(\\implies 504 = \\frac{90}{360} \\times \\frac{22}{7} \\times r^{2} - \\frac{1}{2} \\times r^{2} \\sin 90\\)\n\\(504 = r^{2}(\\frac{22}{28} - \\frac{1}{2})\\)\n\\(504 = \\frac{2r^{2}}{7} \\implies r^{2} = \\frac{504 \\times 7}{2} = 1764\\)\n\\(r = \\sqrt{1764} = 42 cm\\).\n(b)In the diagram, \\(a_{1} = a_{2}\\) (base angles of an isosceles triangle)\n\\(\\therefore a_{1} + a_{2} = 2a_{1}\\)\n\\(\\therefore 2a_{1} + 109° = 180° \\implies 2a_{1} = 180° - 109° = 71°\\)\n\\(a_{1} = a_{2} = \\frac{71°}{2} = 35.5°\\)\nHence, \\(< RQS = 66° + 35.5°\\)\n= \\(101.5°\\)\n\n\n', '(a) Area of the shaded segment = area of sector XOY - area of triangle OXY.\n\\(\\implies 504 = \\frac{\\theta}{360°} \\times \\pi r^{2} - \\frac{1}{2} r^{2} \\sin \\theta\\)\n\\(\\implies 504 = \\frac{90}{360} \\times \\frac{22}{7} \\times r^{2} - \\frac{1}{2} \\times r^{2} \\sin 90\\)\n\\(504 = r^{2}(\\frac{22}{28} - \\frac{1}{2})\\)\n\\(504 = \\frac{2r^{2}}{7} \\implies r^{2} = \\frac{504 \\times 7}{2} = 1764\\)\n\\(r = \\sqrt{1764} = 42 cm\\).~(b)In the diagram, \\(a_{1} = a_{2}\\) (base angles of an isosceles triangle)\n\\(\\therefore a_{1} + a_{2} = 2a_{1}\\)\n\\(\\therefore 2a_{1} + 109° = 180° \\implies 2a_{1} = 180° - 109° = 71°\\)\n\\(a_{1} = a_{2} = \\frac{71°}{2} = 35.5°\\)\nHence, \\(< RQS = 66° + 35.5°\\)\n= \\(101.5°\\)\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(373, '2013', '\n\nQuestion\n\nA building contractor tendered for two independent contracts, X and Y. The probabilities that he will win contract X is 0.5 and not win contract Y is 0.3, What is the probability that he will win :\n(a) both contracts ;\n(b) exactly one of the contracts ;\n(c) neither of the contracts?\n\n', '', 'A building contractor tendered for two independent contracts, X and Y. The probabilities that he will win contract X is 0.5 and not win contract Y is 0.3, What is the probability that he will win :\n(a) both contracts ;\n(b) exactly one of the contracts ;\n(c) neither of the contracts?', 'A building contractor tendered for two independent contracts, X and Y. The probabilities that he will win contract X is 0.5 and not win contract Y is 0.3, What is the probability that he will win :\n(a) both contracts ;~(b) exactly one of the contracts ;\n(c) neither of the contracts?~(c) neither of the contracts?', NULL, '\n\nExplanation\nLet A and B denote the events that the man wins contracts X and Y respectively.\nThen P(A) = 0.5\nP(A\') = 1 - 0.5 = 0.5\nP(B\') = 0.3\nP(B) = 1 - 0.3 = 0.7\n(a) The probability that the man wins both contracts = \\(0.5 \\times 0.7 = 0.35\\).\n(b) The probability that the man wins exactly one of the contracts is \\(P(A) \\times P(B\') + P(B) \\times P(A\')\\)\n= \\(0.5 \\times 0.3 + 0.7 \\times 0.5\\)\n= \\(0.15 + 0.35\\)\n= \\(0.50\\)\n(c) Neither of the contracts (i.e not X, not Y) = \\(P(A\') \\times P(B\')\\)\n= \\(0.5 \\times 0.3\\)\n= \\(0.15\\)\n\n', '', 'Explanation\nLet A and B denote the events that the man wins contracts X and Y respectively.\nThen P(A) = 0.5\nP(A\') = 1 - 0.5 = 0.5\nP(B\') = 0.3\nP(B) = 1 - 0.3 = 0.7\n(a) The probability that the man wins both contracts = \\(0.5 \\times 0.7 = 0.35\\).\n(b) The probability that the man wins exactly one of the contracts is \\(P(A) \\times P(B\') + P(B) \\times P(A\')\\)\n= \\(0.5 \\times 0.3 + 0.7 \\times 0.5\\)\n= \\(0.15 + 0.35\\)\n= \\(0.50\\)\n(c) Neither of the contracts (i.e not X, not Y) = \\(P(A\') \\times P(B\')\\)\n= \\(0.5 \\times 0.3\\)\n= \\(0.15\\)\n\n', '(a) The probability that the man wins both contracts = \\(0.5 \\times 0.7 = 0.35\\).~(b) The probability that the man wins exactly one of the contracts is \\(P(A) \\times P(B\') + P(B) \\times P(A\')\\)\n= \\(0.5 \\times 0.3 + 0.7 \\times 0.5\\)\n= \\(0.15 + 0.35\\)\n= \\(0.50\\)\n(c) Neither of the contracts (i.e not X, not Y) = \\(P(A\') \\times P(B\')\\)\n= \\(0.5 \\times 0.3\\)\n= \\(0.15\\)\n~(c) Neither of the contracts (i.e not X, not Y) = \\(P(A\') \\times P(B\')\\)\n= \\(0.5 \\times 0.3\\)\n= \\(0.15\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(374, '2013', '\n\nQuestion\n\n(a) If \\(\\frac{3}{2p - \\frac{1}{2}} = \\frac{\\frac{1}{3}}{\\frac{1}{4}p + 1}\\), find p.\n(b) A television set was marked for sale at GH¢ 760.00 in order to make a profit of 20%. The television set was actually sold at a discount of 5%. Calculate, correct to 2 significant figures, the actual percentage profit.\n\n', '', '(a) If \\(\\frac{3}{2p - \\frac{1}{2}} = \\frac{\\frac{1}{3}}{\\frac{1}{4}p + 1}\\), find p.\n(b) A television set was marked for sale at GH¢ 760.00 in order to make a profit of 20%. The television set was actually sold at a discount of 5%. Calculate, correct to 2 significant figures, the actual percentage profit.', '(a) If \\(\\frac{3}{2p - \\frac{1}{2}} = \\frac{\\frac{1}{3}}{\\frac{1}{4}p + 1}\\), find p.~(b) A television set was marked for sale at GH¢ 760.00 in order to make a profit of 20%. The television set was actually sold at a discount of 5%. Calculate, correct to 2 significant figures, the actual percentage profit.', NULL, '\n\nExplanation\n(a) \\(\\frac{3}{2p - \\frac{1}{2}} = \\frac{\\frac{1}{3}}{\\frac{1}{4}p + 1}\\)\n\\(\\implies 3(\\frac{1}{4}p + 1) = \\frac{1}{3}(2p - \\frac{1}{2})\\)\n\\(\\frac{3}{4}p + 3 = \\frac{2}{3}p - \\frac{1}{6}\\)\n\\(\\frac{3}{4}p - \\frac{2}{3}p = - 3 - \\frac{1}{6}\\)\n\\(\\frac{1}{12}p = - 3\\frac{1}{6}\\)\n\\(p = \\frac{-19}{6} \\div \\frac{1}{12}\\)\n\\(p = \\frac{-19}{6} \\times 12 = -38\\)\n(b) \\(5% \\times¢760.00 = \\frac{5}{100} \\times¢760 =¢38.00\\)\nSo the TV set was actually sold for¢(760 - 38) =¢722.00\nUsing,\n\\(% profit = \\frac{\\text{SP - CP}}{CP} \\times 100%\\)\n\\(20% = \\frac{760 - x}{x} \\times 100%\\)\n\\(20% x = (760 - x) \\times 100%\\)\n\\(x = 5(760 - x) \\implies x = 3800 - 5x\\)\n\\(6x = 3800 \\implies x =¢633.33\\)\nThus, the actual profit = actual SP - CP\n=¢(722 - 633.33)\n=¢88.67\nHence, actual %age profit = \\(\\frac{88.67}{633.33} \\times 100%\\)\n= 14.0006%\n= 14% (2 sig. figs)\n\n', '', 'Explanation\n(a) \\(\\frac{3}{2p - \\frac{1}{2}} = \\frac{\\frac{1}{3}}{\\frac{1}{4}p + 1}\\)\n\\(\\implies 3(\\frac{1}{4}p + 1) = \\frac{1}{3}(2p - \\frac{1}{2})\\)\n\\(\\frac{3}{4}p + 3 = \\frac{2}{3}p - \\frac{1}{6}\\)\n\\(\\frac{3}{4}p - \\frac{2}{3}p = - 3 - \\frac{1}{6}\\)\n\\(\\frac{1}{12}p = - 3\\frac{1}{6}\\)\n\\(p = \\frac{-19}{6} \\div \\frac{1}{12}\\)\n\\(p = \\frac{-19}{6} \\times 12 = -38\\)\n(b) \\(5% \\times¢760.00 = \\frac{5}{100} \\times¢760 =¢38.00\\)\nSo the TV set was actually sold for¢(760 - 38) =¢722.00\nUsing,\n\\(% profit = \\frac{\\text{SP - CP}}{CP} \\times 100%\\)\n\\(20% = \\frac{760 - x}{x} \\times 100%\\)\n\\(20% x = (760 - x) \\times 100%\\)\n\\(x = 5(760 - x) \\implies x = 3800 - 5x\\)\n\\(6x = 3800 \\implies x =¢633.33\\)\nThus, the actual profit = actual SP - CP\n=¢(722 - 633.33)\n=¢88.67\nHence, actual %age profit = \\(\\frac{88.67}{633.33} \\times 100%\\)\n= 14.0006%\n= 14% (2 sig. figs)\n\n', '(a) \\(\\frac{3}{2p - \\frac{1}{2}} = \\frac{\\frac{1}{3}}{\\frac{1}{4}p + 1}\\)\n\\(\\implies 3(\\frac{1}{4}p + 1) = \\frac{1}{3}(2p - \\frac{1}{2})\\)\n\\(\\frac{3}{4}p + 3 = \\frac{2}{3}p - \\frac{1}{6}\\)\n\\(\\frac{3}{4}p - \\frac{2}{3}p = - 3 - \\frac{1}{6}\\)\n\\(\\frac{1}{12}p = - 3\\frac{1}{6}\\)\n\\(p = \\frac{-19}{6} \\div \\frac{1}{12}\\)\n\\(p = \\frac{-19}{6} \\times 12 = -38\\)~(b) \\(5% \\times¢760.00 = \\frac{5}{100} \\times¢760 =¢38.00\\)\nSo the TV set was actually sold for¢(760 - 38) =¢722.00\nUsing,\n\\(% profit = \\frac{\\text{SP - CP}}{CP} \\times 100%\\)\n\\(20% = \\frac{760 - x}{x} \\times 100%\\)\n\\(20% x = (760 - x) \\times 100%\\)\n\\(x = 5(760 - x) \\implies x = 3800 - 5x\\)\n\\(6x = 3800 \\implies x =¢633.33\\)\nThus, the actual profit = actual SP - CP\n=¢(722 - 633.33)\n=¢88.67\nHence, actual %age profit = \\(\\frac{88.67}{633.33} \\times 100%\\)\n= 14.0006%\n= 14% (2 sig. figs)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(375, '2013', '\n\nQuestion\n\n(a) Copy and complete the table of values for the relation \\(y = 2 \\sin x + 1\\)\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n240°\n\n270°\n\n\n\ny\n1.0\n\n\n\n2.7\n\n\n0.0\n-0.7\n\n\n\n\n(b) Using scales of 2 cm to 30° on the x- axis and 2 cm to 1 unit on the y- axis, draw the graph of \\(y = 2 \\sin x + 1, 0° \\leq x \\leq 270°\\).\n(c) Use the graph to find the values of x for which \\(\\sin x = \\frac{1}{4}\\).\n\n\n(a) Copy and complete the table of values for the relation \\(y = 2 \\sin x + 1\\)\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n240°\n\n270°\n\n\n\ny\n1.0\n\n\n\n2.7\n\n\n0.0\n-0.7\n\n\n\n', '', '(a) Copy and complete the table of values for the relation \\(y = 2 \\sin x + 1\\)\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n240°\n\n270°\n\n\n\ny\n1.0\n\n\n\n2.7\n\n\n0.0\n-0.7\n\n\n\n\n(b) Using scales of 2 cm to 30° on the x- axis and 2 cm to 1 unit on the y- axis, draw the graph of \\(y = 2 \\sin x + 1, 0° \\leq x \\leq 270°\\).\n(c) Use the graph to find the values of x for which \\(\\sin x = \\frac{1}{4}\\).\n\n\n(a) Copy and complete the table of values for the relation \\(y = 2 \\sin x + 1\\)\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n240°\n\n270°\n\n\n\ny\n1.0\n\n\n\n2.7\n\n\n0.0\n-0.7\n\n', '(a) Copy and complete the table of values for the relation \\(y = 2 \\sin x + 1\\)\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n240°\n\n270°\n\n\n\ny\n1.0\n\n\n\n2.7\n\n\n0.0\n-0.7\n\n\n\n~(b) Using scales of 2 cm to 30° on the x- axis and 2 cm to 1 unit on the y- axis, draw the graph of \\(y = 2 \\sin x + 1, 0° \\leq x \\leq 270°\\).\n(c) Use the graph to find the values of x for which \\(\\sin x = \\frac{1}{4}\\).\n\n\n(a) Copy and complete the table of values for the relation \\(y = 2 \\sin x + 1\\)\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n240°\n\n270°\n\n\n\ny\n1.0\n\n\n\n2.7\n\n\n0.0\n-0.7\n\n~(c) Use the graph to find the values of x for which \\(\\sin x = \\frac{1}{4}\\).\n\n\n(a) Copy and complete the table of values for the relation \\(y = 2 \\sin x + 1\\)\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n240°\n\n270°\n\n\n\ny\n1.0\n\n\n\n2.7\n\n\n0.0\n-0.7\n\n', NULL, '\n\nExplanation\n(a)\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n240°\n\n270°\n\n\n\ny\n1.0\n2.0\n2.7\n3.0\n2.7\n2.0\n1.0\n0.0\n-0.7\n-1.0\n\n\n\n(b)\n(c) \\(\\sin x = \\frac{1}{4}\\) (Given)\nMultiply through by 2 ; \\(2 \\sin x = 2 \\times \\frac{1}{4} = \\frac{1}{2}\\)\n\\(2 \\sin x = \\frac{1}{2} \\implies 2 \\sin x - \\frac{1}{2} = 0\\)\nAdd \\(1\\frac{1}{2}\\) to both sides ;\n\\(2 \\sin x - \\frac{1}{2} + 1\\frac{1}{2} = 0 + 1\\frac{1}{2}\\)\n\\(\\implies 2 \\sin x + 1 = 1\\frac{1}{2}\\)\nDraw the line \\(y = 1\\frac{1}{2}\\) on the same axis as \\(y = 2 \\sin x + 1\\). The line \\(y = 1\\frac{1}{2}\\) cuts the curve at points P and Q where x = 15° and x = 168°. Hence, the values of x for which \\(\\sin x = \\frac{1}{4}\\)are 15° and 168°.\n\n', '', 'Explanation\n(a)\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n240°\n\n270°\n\n\n\ny\n1.0\n2.0\n2.7\n3.0\n2.7\n2.0\n1.0\n0.0\n-0.7\n-1.0\n\n\n\n(b)\n(c) \\(\\sin x = \\frac{1}{4}\\) (Given)\nMultiply through by 2 ; \\(2 \\sin x = 2 \\times \\frac{1}{4} = \\frac{1}{2}\\)\n\\(2 \\sin x = \\frac{1}{2} \\implies 2 \\sin x - \\frac{1}{2} = 0\\)\nAdd \\(1\\frac{1}{2}\\) to both sides ;\n\\(2 \\sin x - \\frac{1}{2} + 1\\frac{1}{2} = 0 + 1\\frac{1}{2}\\)\n\\(\\implies 2 \\sin x + 1 = 1\\frac{1}{2}\\)\nDraw the line \\(y = 1\\frac{1}{2}\\) on the same axis as \\(y = 2 \\sin x + 1\\). The line \\(y = 1\\frac{1}{2}\\) cuts the curve at points P and Q where x = 15° and x = 168°. Hence, the values of x for which \\(\\sin x = \\frac{1}{4}\\)are 15° and 168°.\n\n', '(a)\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n240°\n\n270°\n\n\n\ny\n1.0\n2.0\n2.7\n3.0\n2.7\n2.0\n1.0\n0.0\n-0.7\n-1.0\n\n\n~(b)\n(c) \\(\\sin x = \\frac{1}{4}\\) (Given)\nMultiply through by 2 ; \\(2 \\sin x = 2 \\times \\frac{1}{4} = \\frac{1}{2}\\)\n\\(2 \\sin x = \\frac{1}{2} \\implies 2 \\sin x - \\frac{1}{2} = 0\\)\nAdd \\(1\\frac{1}{2}\\) to both sides ;\n\\(2 \\sin x - \\frac{1}{2} + 1\\frac{1}{2} = 0 + 1\\frac{1}{2}\\)\n\\(\\implies 2 \\sin x + 1 = 1\\frac{1}{2}\\)\nDraw the line \\(y = 1\\frac{1}{2}\\) on the same axis as \\(y = 2 \\sin x + 1\\). The line \\(y = 1\\frac{1}{2}\\) cuts the curve at points P and Q where x = 15° and x = 168°. Hence, the values of x for which \\(\\sin x = \\frac{1}{4}\\)are 15° and 168°.\n~(c) \\(\\sin x = \\frac{1}{4}\\) (Given)\nMultiply through by 2 ; \\(2 \\sin x = 2 \\times \\frac{1}{4} = \\frac{1}{2}\\)\n\\(2 \\sin x = \\frac{1}{2} \\implies 2 \\sin x - \\frac{1}{2} = 0\\)\nAdd \\(1\\frac{1}{2}\\) to both sides ;\n\\(2 \\sin x - \\frac{1}{2} + 1\\frac{1}{2} = 0 + 1\\frac{1}{2}\\)\n\\(\\implies 2 \\sin x + 1 = 1\\frac{1}{2}\\)\nDraw the line \\(y = 1\\frac{1}{2}\\) on the same axis as \\(y = 2 \\sin x + 1\\). The line \\(y = 1\\frac{1}{2}\\) cuts the curve at points P and Q where x = 15° and x = 168°. Hence, the values of x for which \\(\\sin x = \\frac{1}{4}\\)are 15° and 168°.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(376, '2013', '\n\nQuestion\n\n(a) Copy and complete the following table for multiplication modulo 11.\n\n\n\n\\(\\otimes\\)\n1\n5\n9\n10\n\n\n1\n1\n5\n9\n10\n\n\n5\n5\n\n\n\n\n\n9\n9\n\n\n\n\n\n10\n10\n\n\n\n\n\n\nUse the table to : (i) evaluate \\((9 \\otimes 5) \\otimes (10 \\otimes 10)\\);\n(ii) find the truth set of :(1) \\(10 \\otimes m = 2\\); (2) \\(n \\otimes n = 4\\)\n(b) When a fraction is reduced to its lowest term, it is equal to \\(\\frac{3}{4}\\). The numerator of the fraction when doubled would be 34 greater than the denominator. Find the fraction.\n\n\n(a) Copy and complete the following table for multiplication modulo 11.\n\n\n\n\\(\\otimes\\)\n1\n5\n9\n10\n\n\n1\n1\n5\n9\n10\n\n\n5\n5\n\n\n\n\n\n9\n9\n\n\n\n\n\n10\n10\n\n\n\n\n\n', '', '(a) Copy and complete the following table for multiplication modulo 11.\n\n\n\n\\(\\otimes\\)\n1\n5\n9\n10\n\n\n1\n1\n5\n9\n10\n\n\n5\n5\n\n\n\n\n\n9\n9\n\n\n\n\n\n10\n10\n\n\n\n\n\n\nUse the table to : (i) evaluate \\((9 \\otimes 5) \\otimes (10 \\otimes 10)\\);\n(ii) find the truth set of :(1) \\(10 \\otimes m = 2\\); (2) \\(n \\otimes n = 4\\)\n(b) When a fraction is reduced to its lowest term, it is equal to \\(\\frac{3}{4}\\). The numerator of the fraction when doubled would be 34 greater than the denominator. Find the fraction.\n\n\n(a) Copy and complete the following table for multiplication modulo 11.\n\n\n\n\\(\\otimes\\)\n1\n5\n9\n10\n\n\n1\n1\n5\n9\n10\n\n\n5\n5\n\n\n\n\n\n9\n9\n\n\n\n\n\n10\n10\n\n\n\n', '(a) Copy and complete the following table for multiplication modulo 11.\n\n\n\n\\(\\otimes\\)\n1\n5\n9\n10\n\n\n1\n1\n5\n9\n10\n\n\n5\n5\n\n\n\n\n\n9\n9\n\n\n\n\n\n10\n10\n\n\n\n\n\n\nUse the table to : (i) evaluate \\((9 \\otimes 5) \\otimes (10 \\otimes 10)\\);\n(ii) find the truth set of :(1) \\(10 \\otimes m = 2\\); (2) \\(n \\otimes n = 4\\)~(b) When a fraction is reduced to its lowest term, it is equal to \\(\\frac{3}{4}\\). The numerator of the fraction when doubled would be 34 greater than the denominator. Find the fraction.\n\n\n(a) Copy and complete the following table for multiplication modulo 11.\n\n\n\n\\(\\otimes\\)\n1\n5\n9\n10\n\n\n1\n1\n5\n9\n10\n\n\n5\n5\n\n\n\n\n\n9\n9\n\n\n\n\n\n10\n10\n\n\n\n', NULL, '\n\nExplanation\n(a)\n\n\n\n\\(\\otimes\\)\n1\n5\n9\n10\n\n\n1\n1\n5\n9\n10\n\n\n5\n5\n3\n1\n6\n\n\n9\n9\n1\n4\n2\n\n\n10\n10\n6\n2\n1\n\n\n\n(i) \\((9 \\otimes 5) \\otimes (10 \\otimes 10) = 1 \\otimes 1 = 1\\)\n(ii) (1) \\(10 \\otimes m = 2\\)\nBy comparison, \\(10 \\otimes 9 = 2\\).\n(2) \\(n \\otimes n = 4\\)\nFrom the table, \\(9 \\otimes 9 = 4\\)\nHence, n = 9.\n(b) Let the fraction = \\(\\frac{m}{n}\\)\n\\(\\frac{m}{n} = \\frac{3}{4}..... (1)\\)\n\\(\\implies n = \\frac{4m}{3} ....... (2)\\)\n\\(2m = n + 34 ...... (3)\\)\nPut (2) in (3),\n\\(2m = \\frac{4m}{3} + 34\\)\n\\(2m - \\frac{4m}{3} = 34 \\implies \\frac{2m}{3} = 34\\)\n\\(m = \\frac{34 \\times 3}{2}= 51\\)\n\\(n = \\frac{4m}{3} = \\frac{4 \\times 51}{3} = 68\\)\nHence, the fraction = \\(\\frac{51}{68}\\).\n\n\n', '', 'Explanation\n(a)\n\n\n\n\\(\\otimes\\)\n1\n5\n9\n10\n\n\n1\n1\n5\n9\n10\n\n\n5\n5\n3\n1\n6\n\n\n9\n9\n1\n4\n2\n\n\n10\n10\n6\n2\n1\n\n\n\n(i) \\((9 \\otimes 5) \\otimes (10 \\otimes 10) = 1 \\otimes 1 = 1\\)\n(ii) (1) \\(10 \\otimes m = 2\\)\nBy comparison, \\(10 \\otimes 9 = 2\\).\n(2) \\(n \\otimes n = 4\\)\nFrom the table, \\(9 \\otimes 9 = 4\\)\nHence, n = 9.\n(b) Let the fraction = \\(\\frac{m}{n}\\)\n\\(\\frac{m}{n} = \\frac{3}{4}..... (1)\\)\n\\(\\implies n = \\frac{4m}{3} ....... (2)\\)\n\\(2m = n + 34 ...... (3)\\)\nPut (2) in (3),\n\\(2m = \\frac{4m}{3} + 34\\)\n\\(2m - \\frac{4m}{3} = 34 \\implies \\frac{2m}{3} = 34\\)\n\\(m = \\frac{34 \\times 3}{2}= 51\\)\n\\(n = \\frac{4m}{3} = \\frac{4 \\times 51}{3} = 68\\)\nHence, the fraction = \\(\\frac{51}{68}\\).\n\n\n', '(a)\n\n\n\n\\(\\otimes\\)\n1\n5\n9\n10\n\n\n1\n1\n5\n9\n10\n\n\n5\n5\n3\n1\n6\n\n\n9\n9\n1\n4\n2\n\n\n10\n10\n6\n2\n1\n\n\n\n(i) \\((9 \\otimes 5) \\otimes (10 \\otimes 10) = 1 \\otimes 1 = 1\\)\n(ii) (1) \\(10 \\otimes m = 2\\)\nBy comparison, \\(10 \\otimes 9 = 2\\).\n(2) \\(n \\otimes n = 4\\)\nFrom the table, \\(9 \\otimes 9 = 4\\)\nHence, n = 9.~(b) Let the fraction = \\(\\frac{m}{n}\\)\n\\(\\frac{m}{n} = \\frac{3}{4}..... (1)\\)\n\\(\\implies n = \\frac{4m}{3} ....... (2)\\)\n\\(2m = n + 34 ...... (3)\\)\nPut (2) in (3),\n\\(2m = \\frac{4m}{3} + 34\\)\n\\(2m - \\frac{4m}{3} = 34 \\implies \\frac{2m}{3} = 34\\)\n\\(m = \\frac{34 \\times 3}{2}= 51\\)\n\\(n = \\frac{4m}{3} = \\frac{4 \\times 51}{3} = 68\\)\nHence, the fraction = \\(\\frac{51}{68}\\).\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(377, '2013', '\n\nQuestion\n\n(a)\nIn the Venn diagram, P, Qand R are subsets of the universal set U. If n(U) = 125, find : (i) the value of x ; (ii) n(\\(P \\cup Q \\cap R\'\\)).\n(b)In the diagram, O is the centre of the circle. If WX is parallel to YZ and < WXY = 50°, find the value of (i) , WYZ\n(ii) < YEZ.\n\n', '', '(a)\nIn the Venn diagram, P, Qand R are subsets of the universal set U. If n(U) = 125, find : (i) the value of x ; (ii) n(\\(P \\cup Q \\cap R\'\\)).\n(b)In the diagram, O is the centre of the circle. If WX is parallel to YZ and < WXY = 50°, find the value of (i) , WYZ\n(ii) < YEZ.', '(a)\nIn the Venn diagram, P, Qand R are subsets of the universal set U. If n(U) = 125, find : (i) the value of x ; (ii) n(\\(P \\cup Q \\cap R\'\\)).~(b)In the diagram, O is the centre of the circle. If WX is parallel to YZ and < WXY = 50°, find the value of (i) , WYZ\n(ii) < YEZ.', NULL, '\n\nExplanation\n(a)\n(i) \\((16 - 2x) + (6 + x) + (19 - 3x) + 5x + 7x + 8x + 4x + 4 = 125\\)\n\\(45 + 20x = 125 \\implies 20x = 125 - 45 = 80\\)\n\\(x = \\frac{80}{20} = 4\\)\n(ii) \\(n(P \\cup Q \\cap R\') = (16 - 2x) + 5x + (6 + x)\\)\n= \\(22 + 4x\\)\n= \\(22 + 4(4) = 22 + 16 = 38\\)\n(b) In the diagram above, < XYW = 90° (angle in a semi-circle; XW is a diameter)\n< XWY + 90° + 50° = 180° (sum of angles in a triangle)\n< XWY = 180° - 140° = 40°\n(i) < WYZ = 40° (alternate angles; \\(\\overline{XW} || \\overline{YZ}\\))\n(ii) < WOZ = \\(2 \\times 40° \\) (angle at centre = twice angle at circumference)\n= 80°\n\\(\\therefore < WEO = 180° - (40° + 80°)\\)\n= \\(180° - 120° \\)\n= \\(60°\\)\n\\(\\therefore < YEZ = < WEO = 60°\\) (vertically opposite angles).\n\n', '', 'Explanation\n(a)\n(i) \\((16 - 2x) + (6 + x) + (19 - 3x) + 5x + 7x + 8x + 4x + 4 = 125\\)\n\\(45 + 20x = 125 \\implies 20x = 125 - 45 = 80\\)\n\\(x = \\frac{80}{20} = 4\\)\n(ii) \\(n(P \\cup Q \\cap R\') = (16 - 2x) + 5x + (6 + x)\\)\n= \\(22 + 4x\\)\n= \\(22 + 4(4) = 22 + 16 = 38\\)\n(b) In the diagram above, < XYW = 90° (angle in a semi-circle; XW is a diameter)\n< XWY + 90° + 50° = 180° (sum of angles in a triangle)\n< XWY = 180° - 140° = 40°\n(i) < WYZ = 40° (alternate angles; \\(\\overline{XW} || \\overline{YZ}\\))\n(ii) < WOZ = \\(2 \\times 40° \\) (angle at centre = twice angle at circumference)\n= 80°\n\\(\\therefore < WEO = 180° - (40° + 80°)\\)\n= \\(180° - 120° \\)\n= \\(60°\\)\n\\(\\therefore < YEZ = < WEO = 60°\\) (vertically opposite angles).\n\n', '(a)\n(i) \\((16 - 2x) + (6 + x) + (19 - 3x) + 5x + 7x + 8x + 4x + 4 = 125\\)\n\\(45 + 20x = 125 \\implies 20x = 125 - 45 = 80\\)\n\\(x = \\frac{80}{20} = 4\\)\n(ii) \\(n(P \\cup Q \\cap R\') = (16 - 2x) + 5x + (6 + x)\\)\n= \\(22 + 4x\\)\n= \\(22 + 4(4) = 22 + 16 = 38\\)~(b) In the diagram above, < XYW = 90° (angle in a semi-circle; XW is a diameter)\n< XWY + 90° + 50° = 180° (sum of angles in a triangle)\n< XWY = 180° - 140° = 40°\n(i) < WYZ = 40° (alternate angles; \\(\\overline{XW} || \\overline{YZ}\\))\n(ii) < WOZ = \\(2 \\times 40° \\) (angle at centre = twice angle at circumference)\n= 80°\n\\(\\therefore < WEO = 180° - (40° + 80°)\\)\n= \\(180° - 120° \\)\n= \\(60°\\)\n\\(\\therefore < YEZ = < WEO = 60°\\) (vertically opposite angles).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(378, '2020', '\n\nQuestion\n\n(a) Solve : \\((x - 2)(x - 3) = 12\\).\n(b)In the diagram, M and N are the centres of two circles of equal radii 7cm. The circle intercept at P and Q. If < PMQ = < PNQ = 60°, calculate, correct to the nearest whole number, the area of the shaded portion. [Take \\(\\pi = \\frac{22}{7}\\)].\n\n', '', '(a) Solve : \\((x - 2)(x - 3) = 12\\).\n(b)In the diagram, M and N are the centres of two circles of equal radii 7cm. The circle intercept at P and Q. If < PMQ = < PNQ = 60°, calculate, correct to the nearest whole number, the area of the shaded portion. [Take \\(\\pi = \\frac{22}{7}\\)].', '(a) Solve : \\((x - 2)(x - 3) = 12\\).~(b)In the diagram, M and N are the centres of two circles of equal radii 7cm. The circle intercept at P and Q. If < PMQ = < PNQ = 60°, calculate, correct to the nearest whole number, the area of the shaded portion. [Take \\(\\pi = \\frac{22}{7}\\)].', NULL, '\n\nExplanation\n(a) \\((x - 2)(x - 3) = 12\\)\n\\(x^{2} - 3x - 2x + 6 = 12 \\implies x^{2} - 5x + 6 - 12 = 0\\)\n\\(x^{2} - 5x - 6 = 0\\)\n\\(x^{2} - 6x + x - 6 = 0 \\implies x(x - 6) + 1(x - 6) = 0\\)\n\\( (x - 6)(x +1) = 0 \\implies \\text{x = 6 or -1}\\)\n(b) The shaded portion comprises two shaded segments labelled A and B.\nArea of segment A = \\(\\frac{60°}{360°} \\times \\pi r^{2} - \\frac{1}{2} r^{2} \\sin 60°\\)\n\\(\\frac{1}{6} \\times \\frac{22}{7} \\times 7 \\times 7 - \\frac{1}{2} \\times 7 \\times 7 \\times 0.8660\\)\n= \\(25.667 - 21.217\\)\n= \\(4.45 cm^{2}\\)\nAlso, the area of the segment B = \\(4.45 cm^{2}\\).\nHence, the area of the shaded portion = \\(4.45 cm^{2} + 4.45 cm^{2}\\)\n= \\(8.9 cm^{2}\\)\n\\(\\approxeq 9 cm^{2}\\) (to the nearest whole number).\n\n', '', 'Explanation\n(a) \\((x - 2)(x - 3) = 12\\)\n\\(x^{2} - 3x - 2x + 6 = 12 \\implies x^{2} - 5x + 6 - 12 = 0\\)\n\\(x^{2} - 5x - 6 = 0\\)\n\\(x^{2} - 6x + x - 6 = 0 \\implies x(x - 6) + 1(x - 6) = 0\\)\n\\( (x - 6)(x +1) = 0 \\implies \\text{x = 6 or -1}\\)\n(b) The shaded portion comprises two shaded segments labelled A and B.\nArea of segment A = \\(\\frac{60°}{360°} \\times \\pi r^{2} - \\frac{1}{2} r^{2} \\sin 60°\\)\n\\(\\frac{1}{6} \\times \\frac{22}{7} \\times 7 \\times 7 - \\frac{1}{2} \\times 7 \\times 7 \\times 0.8660\\)\n= \\(25.667 - 21.217\\)\n= \\(4.45 cm^{2}\\)\nAlso, the area of the segment B = \\(4.45 cm^{2}\\).\nHence, the area of the shaded portion = \\(4.45 cm^{2} + 4.45 cm^{2}\\)\n= \\(8.9 cm^{2}\\)\n\\(\\approxeq 9 cm^{2}\\) (to the nearest whole number).\n\n', '(a) \\((x - 2)(x - 3) = 12\\)\n\\(x^{2} - 3x - 2x + 6 = 12 \\implies x^{2} - 5x + 6 - 12 = 0\\)\n\\(x^{2} - 5x - 6 = 0\\)\n\\(x^{2} - 6x + x - 6 = 0 \\implies x(x - 6) + 1(x - 6) = 0\\)\n\\( (x - 6)(x +1) = 0 \\implies \\text{x = 6 or -1}\\)~(b) The shaded portion comprises two shaded segments labelled A and B.\nArea of segment A = \\(\\frac{60°}{360°} \\times \\pi r^{2} - \\frac{1}{2} r^{2} \\sin 60°\\)\n\\(\\frac{1}{6} \\times \\frac{22}{7} \\times 7 \\times 7 - \\frac{1}{2} \\times 7 \\times 7 \\times 0.8660\\)\n= \\(25.667 - 21.217\\)\n= \\(4.45 cm^{2}\\)\nAlso, the area of the segment B = \\(4.45 cm^{2}\\).\nHence, the area of the shaded portion = \\(4.45 cm^{2} + 4.45 cm^{2}\\)\n= \\(8.9 cm^{2}\\)\n\\(\\approxeq 9 cm^{2}\\) (to the nearest whole number).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(379, '2020', '\n\nQuestion\n\n\n\nScores\n1\n2\n3\n4\n5\n6\n\n\nFrequency\n2\n5\n13\n11\n9\n10\n\n\n\nThe table shows the distribution of outcomes when a die is thrown 50 times. Calculate the :\n(a) Mean deviation of the distribution ; (b) probability that a score selected at random is at least a 4.\n\n\n\n\n\nScores\n1\n2\n3\n4\n5\n6\n\n\nFrequency\n2\n5\n13\n11\n9\n10\n\n\n', '', 'Scores\n1\n2\n3\n4\n5\n6\n\n\nFrequency\n2\n5\n13\n11\n9\n10\n\n\n\nThe table shows the distribution of outcomes when a die is thrown 50 times. Calculate the :\n(a) Mean deviation of the distribution ; (b) probability that a score selected at random is at least a 4.\n\n\n\n\n\nScores\n1\n2\n3\n4\n5\n6\n\n\nFrequency\n2\n5\n13\n11\n9\n10\n', '~(b) probability that a score selected at random is at least a 4.\n\n\n\n\n\nScores\n1\n2\n3\n4\n5\n6\n\n\nFrequency\n2\n5\n13\n11\n9\n10\n', NULL, '\n\nExplanation\n(a)\n\n\n\nScores (x)\nFrequency (f)\n\\(fx\\)\n\n\n1\n2\n2\n\n\n2\n5\n10\n\n\n3\n13\n39\n\n\n4\n11\n44\n\n\n5\n9\n45\n\n\n6\n10\n60\n\n\n\n\\(\\sum f = 50\\)\n\\(\\sum fx = 200\\)\n\n\n\n\\(Mean (\\bar{x}) = \\frac{\\sum fx}{\\sum f} = \\frac{200}{50} = 4\\)\n\n\n\n\\(d = x - 4\\)\n\\(|d|\\)\nf\n\\(f|d|\\)\n\n\n-3\n3\n2\n6\n\n\n-2\n2\n5\n10\n\n\n-1\n1\n13\n13\n\n\n0\n0\n11\n0\n\n\n1\n1\n9\n9\n\n\n2\n2\n10\n20\n\n\n\n\n\n\\(\\sum f|d| = 69\\)\n\n\n\nHence, Mean Deviation = \\(\\frac{\\sum f|d|}{\\sum f} = \\frac{69}{50} \\)\n= \\(1.38\\)\n(b) Let E denote the event of getting a score of at least 4.\n\\(n(E) = 11 + 9 + 10 = 30\\)\n\\(p(E) = \\frac{n(E)}{n(S)} = \\frac{30}{50}\\)\n= \\(0.6\\)\n\n', '', 'Explanation\n(a)\n\n\n\nScores (x)\nFrequency (f)\n\\(fx\\)\n\n\n1\n2\n2\n\n\n2\n5\n10\n\n\n3\n13\n39\n\n\n4\n11\n44\n\n\n5\n9\n45\n\n\n6\n10\n60\n\n\n\n\\(\\sum f = 50\\)\n\\(\\sum fx = 200\\)\n\n\n\n\\(Mean (\\bar{x}) = \\frac{\\sum fx}{\\sum f} = \\frac{200}{50} = 4\\)\n\n\n\n\\(d = x - 4\\)\n\\(|d|\\)\nf\n\\(f|d|\\)\n\n\n-3\n3\n2\n6\n\n\n-2\n2\n5\n10\n\n\n-1\n1\n13\n13\n\n\n0\n0\n11\n0\n\n\n1\n1\n9\n9\n\n\n2\n2\n10\n20\n\n\n\n\n\n\\(\\sum f|d| = 69\\)\n\n\n\nHence, Mean Deviation = \\(\\frac{\\sum f|d|}{\\sum f} = \\frac{69}{50} \\)\n= \\(1.38\\)\n(b) Let E denote the event of getting a score of at least 4.\n\\(n(E) = 11 + 9 + 10 = 30\\)\n\\(p(E) = \\frac{n(E)}{n(S)} = \\frac{30}{50}\\)\n= \\(0.6\\)\n\n', '(a)\n\n\n\nScores (x)\nFrequency (f)\n\\(fx\\)\n\n\n1\n2\n2\n\n\n2\n5\n10\n\n\n3\n13\n39\n\n\n4\n11\n44\n\n\n5\n9\n45\n\n\n6\n10\n60\n\n\n\n\\(\\sum f = 50\\)\n\\(\\sum fx = 200\\)\n\n\n\n\\(Mean (\\bar{x}) = \\frac{\\sum fx}{\\sum f} = \\frac{200}{50} = 4\\)\n\n\n\n\\(d = x - 4\\)\n\\(|d|\\)\nf\n\\(f|d|\\)\n\n\n-3\n3\n2\n6\n\n\n-2\n2\n5\n10\n\n\n-1\n1\n13\n13\n\n\n0\n0\n11\n0\n\n\n1\n1\n9\n9\n\n\n2\n2\n10\n20\n\n\n\n\n\n\\(\\sum f|d| = 69\\)\n\n\n\nHence, Mean Deviation = \\(\\frac{\\sum f|d|}{\\sum f} = \\frac{69}{50} \\)\n= \\(1.38\\)~(b) Let E denote the event of getting a score of at least 4.\n\\(n(E) = 11 + 9 + 10 = 30\\)\n\\(p(E) = \\frac{n(E)}{n(S)} = \\frac{30}{50}\\)\n= \\(0.6\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(380, '2020', '\n\nQuestion\n\n(a) Given that \\(5 \\cos (x + 8.5)° - 1 = 0, 0° \\leq x \\leq 90°\\), calculate, correct to the nearest degree, the value of x.\n(b) The bearing of Q from P is 0150° and the bearing of P from R is 015°. If Q and R are 24km and 32km respectively from P : (i) represent this information in a diagram;\n(ii) calculate the distance between Q and R, correct to two decimal places ; (iii) find the bearing of R from Q, correct to the nearest degree.\n\n', '', '(a) Given that \\(5 \\cos (x + 8.5)° - 1 = 0, 0° \\leq x \\leq 90°\\), calculate, correct to the nearest degree, the value of x.\n(b) The bearing of Q from P is 0150° and the bearing of P from R is 015°. If Q and R are 24km and 32km respectively from P : (i) represent this information in a diagram;\n(ii) calculate the distance between Q and R, correct to two decimal places ; (iii) find the bearing of R from Q, correct to the nearest degree.', '(a) Given that \\(5 \\cos (x + 8.5)° - 1 = 0, 0° \\leq x \\leq 90°\\), calculate, correct to the nearest degree, the value of x.~(b) The bearing of Q from P is 0150° and the bearing of P from R is 015°. If Q and R are 24km and 32km respectively from P : (i) represent this information in a diagram;\n(ii) calculate the distance between Q and R, correct to two decimal places ; (iii) find the bearing of R from Q, correct to the nearest degree.', NULL, '\n\nExplanation\n(a) \\(5 \\cos (x + 8.5)° - 1 = 0\\)\n\\(5 \\cos (x + 8.5)° = 1\\)\n\\(\\cos (x + 8.5)° = \\frac{1}{5} = 0.2\\)\n\\((x + 8.5)° = \\cos^{-1}(0.2) = 78.463°\\)\n\\(x = 78.463° - 8.5 = 69.963°\\)\n\\(\\approxeq 70°\\) (to the nearest degree).\n(b)(i)\n(ii) By the cosine rule,\n\\(|QR|^{2} = 32^{2} + 24^{2} - 2 \\times 32 \\times 24 \\times \\cos 45\\)\n\\(|QR|^{2} = 1024 + 576 - 1536 \\cos 45\\)\n= \\(1600 - 1086.1056\\)\n\\(|QR|^{2} = 513.8944\\)\n\\(|QR| = \\sqrt{513.8944} = 22.669 km\\)\n\\(\\approxeq 22.67 km\\) ( 2 decimal place)\n(iii) By the sine rule,\n\\(\\frac{32}{\\sin \\alpha} = \\frac{22.67}{\\sin 45}\\)\n\\(\\sin \\alpha = \\frac{32 \\times \\sin 45}{22.67}\\)\n= \\(0.9981\\)\n\\(\\alpha = \\sin^{-1} (0.9981) = 86.4787°\\)\nThe diagram below shows all the angles at Q;\nThe bearing of R from Q is given by the reflex angle NQR. Thus\nreflex < NQR = 360° - (86.47° + 30°) = 360° - 116.47°\n= 243.53°\nHence, the bearing of R from Q = 244° (to the nearest degree).\n\n', '', 'Explanation\n(a) \\(5 \\cos (x + 8.5)° - 1 = 0\\)\n\\(5 \\cos (x + 8.5)° = 1\\)\n\\(\\cos (x + 8.5)° = \\frac{1}{5} = 0.2\\)\n\\((x + 8.5)° = \\cos^{-1}(0.2) = 78.463°\\)\n\\(x = 78.463° - 8.5 = 69.963°\\)\n\\(\\approxeq 70°\\) (to the nearest degree).\n(b)(i)\n(ii) By the cosine rule,\n\\(|QR|^{2} = 32^{2} + 24^{2} - 2 \\times 32 \\times 24 \\times \\cos 45\\)\n\\(|QR|^{2} = 1024 + 576 - 1536 \\cos 45\\)\n= \\(1600 - 1086.1056\\)\n\\(|QR|^{2} = 513.8944\\)\n\\(|QR| = \\sqrt{513.8944} = 22.669 km\\)\n\\(\\approxeq 22.67 km\\) ( 2 decimal place)\n(iii) By the sine rule,\n\\(\\frac{32}{\\sin \\alpha} = \\frac{22.67}{\\sin 45}\\)\n\\(\\sin \\alpha = \\frac{32 \\times \\sin 45}{22.67}\\)\n= \\(0.9981\\)\n\\(\\alpha = \\sin^{-1} (0.9981) = 86.4787°\\)\nThe diagram below shows all the angles at Q;\nThe bearing of R from Q is given by the reflex angle NQR. Thus\nreflex < NQR = 360° - (86.47° + 30°) = 360° - 116.47°\n= 243.53°\nHence, the bearing of R from Q = 244° (to the nearest degree).\n\n', '(a) \\(5 \\cos (x + 8.5)° - 1 = 0\\)\n\\(5 \\cos (x + 8.5)° = 1\\)\n\\(\\cos (x + 8.5)° = \\frac{1}{5} = 0.2\\)\n\\((x + 8.5)° = \\cos^{-1}(0.2) = 78.463°\\)\n\\(x = 78.463° - 8.5 = 69.963°\\)\n\\(\\approxeq 70°\\) (to the nearest degree).~(b)(i)\n(ii) By the cosine rule,\n\\(|QR|^{2} = 32^{2} + 24^{2} - 2 \\times 32 \\times 24 \\times \\cos 45\\)\n\\(|QR|^{2} = 1024 + 576 - 1536 \\cos 45\\)\n= \\(1600 - 1086.1056\\)\n\\(|QR|^{2} = 513.8944\\)\n\\(|QR| = \\sqrt{513.8944} = 22.669 km\\)\n\\(\\approxeq 22.67 km\\) ( 2 decimal place)\n(iii) By the sine rule,\n\\(\\frac{32}{\\sin \\alpha} = \\frac{22.67}{\\sin 45}\\)\n\\(\\sin \\alpha = \\frac{32 \\times \\sin 45}{22.67}\\)\n= \\(0.9981\\)\n\\(\\alpha = \\sin^{-1} (0.9981) = 86.4787°\\)\nThe diagram below shows all the angles at Q;\nThe bearing of R from Q is given by the reflex angle NQR. Thus\nreflex < NQR = 360° - (86.47° + 30°) = 360° - 116.47°\n= 243.53°\nHence, the bearing of R from Q = 244° (to the nearest degree).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(381, '2020', '\n\nQuestion\n\n(a) Two functions, f and g, are defined by \\(f : x \\to 2x^{2} - 1\\) and \\(g : x \\to 3x + 2\\) where x is a real number.\n(i) If \\(f(x - 1) - 7 = 0\\), find the values of x.\n(ii) Evaluate : \\(\\frac{f(-\\frac{1}{2}) . g(3)}{f(4) - g(5)}\\).\n(b) An operation, \\((\\ast)\\) is defined on the set R, of real numbers, by \\(m \\ast n = \\frac{-n}{m^{2} + 1}\\), where \\(m, n \\in R\\). If \\(-3, -10 \\in R\\), show whether or not \\(\\ast\\) is commutative.\n\n', '', '(a) Two functions, f and g, are defined by \\(f : x \\to 2x^{2} - 1\\) and \\(g : x \\to 3x + 2\\) where x is a real number.\n(i) If \\(f(x - 1) - 7 = 0\\), find the values of x.\n(ii) Evaluate : \\(\\frac{f(-\\frac{1}{2}) . g(3)}{f(4) - g(5)}\\).\n(b) An operation, \\((\\ast)\\) is defined on the set R, of real numbers, by \\(m \\ast n = \\frac{-n}{m^{2} + 1}\\), where \\(m, n \\in R\\). If \\(-3, -10 \\in R\\), show whether or not \\(\\ast\\) is commutative.', '(a) Two functions, f and g, are defined by \\(f : x \\to 2x^{2} - 1\\) and \\(g : x \\to 3x + 2\\) where x is a real number.\n(i) If \\(f(x - 1) - 7 = 0\\), find the values of x.\n(ii) Evaluate : \\(\\frac{f(-\\frac{1}{2}) . g(3)}{f(4) - g(5)}\\).~(b) An operation, \\((\\ast)\\) is defined on the set R, of real numbers, by \\(m \\ast n = \\frac{-n}{m^{2} + 1}\\), where \\(m, n \\in R\\). If \\(-3, -10 \\in R\\), show whether or not \\(\\ast\\) is commutative.', NULL, '\n\nExplanation\n(a) \\(f : x \\to 2x^{2} - 1 ; g : x \\to 3x + 2\\)\n(i) \\(f(x - 1) - 7 = 0\\)\n\\(f(x - 1) = 2(x - 1)^{2} - 1 = 2(x^{2} - 2x + 1) - 1\\)\n= \\(2x^{2} - 4x + 2 - 1\\)\n\\(f(x - 1) - 7 = 2x^{2} - 4x + 1- 7 = 2x^{2} - 4x - 6 = 0\\)\n\\(2x^{2} - 6x + 2x - 6 = 0 \\implies 2x(x - 3) + 2(x - 3) = 0\\)\n\\((2x + 2)(x - 3) = 0 \\implies 2x = -2; x = 3\\)\n\\(x = -1 ; 3\\)\n(ii)\\(\\frac{f(-\\frac{1}{2}) . g(3)}{f(4) - g(5)}\\)\n\\(f(-\\frac{1}{2}) = 2(-\\frac{1}{2})^{2} - 1 = \\frac{1}{2} - 1 = -\\frac{1}{2}\\)\n\\(g(3) = 3(3) + 2 = 9 + 2 = 11\\)\n\\(f(4) = 2(4^{2}) - 1 = 32 - 1 = 31\\)\n\\(g(5) = 3(5) + 2 = 15 + 2 = 17\\)\n\\(\\frac{f(-\\frac{1}{2}) . g(3)}{f(4) - g(5)} = \\frac{(-\\frac{1}{2}) . (11)}{31 - 17}\\)\n= \\(\\frac{-\\frac{11}{2}}{14}\\)\n= \\(-\\frac{11}{28}\\)\n(b) \\(m \\ast n = \\frac{-n}{m^{2} + 1}; m, n \\in R\\)\n\\(-3 \\ast -10 = \\frac{-(-10)}{(-3)^{2} + 1}\\)\n= \\(\\frac{10}{10} = 1\\)\n\\(-10 \\ast -3 = \\frac{-(-3)}{(-10)^{2} + 1}\\)\n= \\(\\frac{3}{101}\\)\nHence, \\(\\ast\\) is not commutative as \\(m \\ast n \\neq n \\ast m\\).\n\n', '', 'Explanation\n(a) \\(f : x \\to 2x^{2} - 1 ; g : x \\to 3x + 2\\)\n(i) \\(f(x - 1) - 7 = 0\\)\n\\(f(x - 1) = 2(x - 1)^{2} - 1 = 2(x^{2} - 2x + 1) - 1\\)\n= \\(2x^{2} - 4x + 2 - 1\\)\n\\(f(x - 1) - 7 = 2x^{2} - 4x + 1- 7 = 2x^{2} - 4x - 6 = 0\\)\n\\(2x^{2} - 6x + 2x - 6 = 0 \\implies 2x(x - 3) + 2(x - 3) = 0\\)\n\\((2x + 2)(x - 3) = 0 \\implies 2x = -2; x = 3\\)\n\\(x = -1 ; 3\\)\n(ii)\\(\\frac{f(-\\frac{1}{2}) . g(3)}{f(4) - g(5)}\\)\n\\(f(-\\frac{1}{2}) = 2(-\\frac{1}{2})^{2} - 1 = \\frac{1}{2} - 1 = -\\frac{1}{2}\\)\n\\(g(3) = 3(3) + 2 = 9 + 2 = 11\\)\n\\(f(4) = 2(4^{2}) - 1 = 32 - 1 = 31\\)\n\\(g(5) = 3(5) + 2 = 15 + 2 = 17\\)\n\\(\\frac{f(-\\frac{1}{2}) . g(3)}{f(4) - g(5)} = \\frac{(-\\frac{1}{2}) . (11)}{31 - 17}\\)\n= \\(\\frac{-\\frac{11}{2}}{14}\\)\n= \\(-\\frac{11}{28}\\)\n(b) \\(m \\ast n = \\frac{-n}{m^{2} + 1}; m, n \\in R\\)\n\\(-3 \\ast -10 = \\frac{-(-10)}{(-3)^{2} + 1}\\)\n= \\(\\frac{10}{10} = 1\\)\n\\(-10 \\ast -3 = \\frac{-(-3)}{(-10)^{2} + 1}\\)\n= \\(\\frac{3}{101}\\)\nHence, \\(\\ast\\) is not commutative as \\(m \\ast n \\neq n \\ast m\\).\n\n', '(a) \\(f : x \\to 2x^{2} - 1 ; g : x \\to 3x + 2\\)\n(i) \\(f(x - 1) - 7 = 0\\)\n\\(f(x - 1) = 2(x - 1)^{2} - 1 = 2(x^{2} - 2x + 1) - 1\\)\n= \\(2x^{2} - 4x + 2 - 1\\)\n\\(f(x - 1) - 7 = 2x^{2} - 4x + 1- 7 = 2x^{2} - 4x - 6 = 0\\)\n\\(2x^{2} - 6x + 2x - 6 = 0 \\implies 2x(x - 3) + 2(x - 3) = 0\\)\n\\((2x + 2)(x - 3) = 0 \\implies 2x = -2; x = 3\\)\n\\(x = -1 ; 3\\)\n(ii)\\(\\frac{f(-\\frac{1}{2}) . g(3)}{f(4) - g(5)}\\)\n\\(f(-\\frac{1}{2}) = 2(-\\frac{1}{2})^{2} - 1 = \\frac{1}{2} - 1 = -\\frac{1}{2}\\)\n\\(g(3) = 3(3) + 2 = 9 + 2 = 11\\)\n\\(f(4) = 2(4^{2}) - 1 = 32 - 1 = 31\\)\n\\(g(5) = 3(5) + 2 = 15 + 2 = 17\\)\n\\(\\frac{f(-\\frac{1}{2}) . g(3)}{f(4) - g(5)} = \\frac{(-\\frac{1}{2}) . (11)}{31 - 17}\\)\n= \\(\\frac{-\\frac{11}{2}}{14}\\)\n= \\(-\\frac{11}{28}\\)~(b) \\(m \\ast n = \\frac{-n}{m^{2} + 1}; m, n \\in R\\)\n\\(-3 \\ast -10 = \\frac{-(-10)}{(-3)^{2} + 1}\\)\n= \\(\\frac{10}{10} = 1\\)\n\\(-10 \\ast -3 = \\frac{-(-3)}{(-10)^{2} + 1}\\)\n= \\(\\frac{3}{101}\\)\nHence, \\(\\ast\\) is not commutative as \\(m \\ast n \\neq n \\ast m\\).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(382, '2020', '\n\nQuestion\n\n(a) Simplify, without using tables or calculator : \\(\\frac{\\frac{3}{4}(3\\frac{3}{8} + 1\\frac{5}{8})}{2\\frac{1}{8} - 1\\frac{1}{2}}\\).\n(b) Given that \\(\\log_{10} 2 = 0.3010\\) and \\(\\log_{10} 3 = 0.4771\\), evaluate, correct to 2 significant figures and without using tables or calculator, \\(\\log_{10} 1.125\\).\n\n', '', '(a) Simplify, without using tables or calculator : \\(\\frac{\\frac{3}{4}(3\\frac{3}{8} + 1\\frac{5}{8})}{2\\frac{1}{8} - 1\\frac{1}{2}}\\).\n(b) Given that \\(\\log_{10} 2 = 0.3010\\) and \\(\\log_{10} 3 = 0.4771\\), evaluate, correct to 2 significant figures and without using tables or calculator, \\(\\log_{10} 1.125\\).', '(a) Simplify, without using tables or calculator : \\(\\frac{\\frac{3}{4}(3\\frac{3}{8} + 1\\frac{5}{8})}{2\\frac{1}{8} - 1\\frac{1}{2}}\\).~(b) Given that \\(\\log_{10} 2 = 0.3010\\) and \\(\\log_{10} 3 = 0.4771\\), evaluate, correct to 2 significant figures and without using tables or calculator, \\(\\log_{10} 1.125\\).', NULL, '\n\nExplanation\n(a) \\(\\frac{\\frac{3}{4}(3\\frac{3}{8} + 1\\frac{5}{8})}{2\\frac{1}{8} - 1\\frac{1}{2}}\\)\n\\(\\frac{3}{4}(3\\frac{3}{8} + 1\\frac{5}{8}) = \\frac{3}{4}(\\frac{27}{8} + \\frac{13}{8})\\)\n= \\(\\frac{3}{4}(\\frac{40}{8})\\)\n= \\(\\frac{15}{4}\\)\n\\(2\\frac{1}{8} - 1\\frac{1}{2} = \\frac{17}{8} - \\frac{3}{2}\\)\n= \\(\\frac{17}{8} - \\frac{12}{8}\\)\n= \\(\\frac{5}{8}\\)\n\\(\\therefore \\frac{\\frac{3}{4}(3\\frac{3}{8} + 1\\frac{5}{8})}{2\\frac{1}{8} - 1\\frac{1}{2}} = \\frac{15}{4} \\div \\frac{5}{8}\\)\n= \\(\\frac{15}{4} \\times \\frac{8}{5}\\)\n= \\(6\\)\n(b) \\(\\log_{10} 2 = 0.3010 ; \\log_{10} 3 = 0.4771\\)\n\\(\\log_{10} 1.125 = \\log_{10}(\\frac{1125}{1000})\\)(Dividing through with 125)\n= \\(\\log_{10} (\\frac{9}{8})\\)\n= \\(\\log_{10} 9 - \\log_{10} 8\\)\n\\(\\log_{10} 9 = \\log_{10} 3^{2} = 2\\log_{10} 3 = 2 \\times 0.4771 = 0.9542\\)\n\\(\\log_{10} 8 = \\log_{10} 2^{3} = 3\\log_{10} 2 = 3 \\times 0.3010 = 0.9030\\)\n\\(\\therefore \\log_{10} (\\frac{9}{8}) = 0.9542 - 0.9030 = 0.0512\\)\n\\(\\approxeq 0.051\\) ( 2 significant figures)\n\n', '', 'Explanation\n(a) \\(\\frac{\\frac{3}{4}(3\\frac{3}{8} + 1\\frac{5}{8})}{2\\frac{1}{8} - 1\\frac{1}{2}}\\)\n\\(\\frac{3}{4}(3\\frac{3}{8} + 1\\frac{5}{8}) = \\frac{3}{4}(\\frac{27}{8} + \\frac{13}{8})\\)\n= \\(\\frac{3}{4}(\\frac{40}{8})\\)\n= \\(\\frac{15}{4}\\)\n\\(2\\frac{1}{8} - 1\\frac{1}{2} = \\frac{17}{8} - \\frac{3}{2}\\)\n= \\(\\frac{17}{8} - \\frac{12}{8}\\)\n= \\(\\frac{5}{8}\\)\n\\(\\therefore \\frac{\\frac{3}{4}(3\\frac{3}{8} + 1\\frac{5}{8})}{2\\frac{1}{8} - 1\\frac{1}{2}} = \\frac{15}{4} \\div \\frac{5}{8}\\)\n= \\(\\frac{15}{4} \\times \\frac{8}{5}\\)\n= \\(6\\)\n(b) \\(\\log_{10} 2 = 0.3010 ; \\log_{10} 3 = 0.4771\\)\n\\(\\log_{10} 1.125 = \\log_{10}(\\frac{1125}{1000})\\)(Dividing through with 125)\n= \\(\\log_{10} (\\frac{9}{8})\\)\n= \\(\\log_{10} 9 - \\log_{10} 8\\)\n\\(\\log_{10} 9 = \\log_{10} 3^{2} = 2\\log_{10} 3 = 2 \\times 0.4771 = 0.9542\\)\n\\(\\log_{10} 8 = \\log_{10} 2^{3} = 3\\log_{10} 2 = 3 \\times 0.3010 = 0.9030\\)\n\\(\\therefore \\log_{10} (\\frac{9}{8}) = 0.9542 - 0.9030 = 0.0512\\)\n\\(\\approxeq 0.051\\) ( 2 significant figures)\n\n', '(a) \\(\\frac{\\frac{3}{4}(3\\frac{3}{8} + 1\\frac{5}{8})}{2\\frac{1}{8} - 1\\frac{1}{2}}\\)\n\\(\\frac{3}{4}(3\\frac{3}{8} + 1\\frac{5}{8}) = \\frac{3}{4}(\\frac{27}{8} + \\frac{13}{8})\\)\n= \\(\\frac{3}{4}(\\frac{40}{8})\\)\n= \\(\\frac{15}{4}\\)\n\\(2\\frac{1}{8} - 1\\frac{1}{2} = \\frac{17}{8} - \\frac{3}{2}\\)\n= \\(\\frac{17}{8} - \\frac{12}{8}\\)\n= \\(\\frac{5}{8}\\)\n\\(\\therefore \\frac{\\frac{3}{4}(3\\frac{3}{8} + 1\\frac{5}{8})}{2\\frac{1}{8} - 1\\frac{1}{2}} = \\frac{15}{4} \\div \\frac{5}{8}\\)\n= \\(\\frac{15}{4} \\times \\frac{8}{5}\\)\n= \\(6\\)~(b) \\(\\log_{10} 2 = 0.3010 ; \\log_{10} 3 = 0.4771\\)\n\\(\\log_{10} 1.125 = \\log_{10}(\\frac{1125}{1000})\\)(Dividing through with 125)\n= \\(\\log_{10} (\\frac{9}{8})\\)\n= \\(\\log_{10} 9 - \\log_{10} 8\\)\n\\(\\log_{10} 9 = \\log_{10} 3^{2} = 2\\log_{10} 3 = 2 \\times 0.4771 = 0.9542\\)\n\\(\\log_{10} 8 = \\log_{10} 2^{3} = 3\\log_{10} 2 = 3 \\times 0.3010 = 0.9030\\)\n\\(\\therefore \\log_{10} (\\frac{9}{8}) = 0.9542 - 0.9030 = 0.0512\\)\n\\(\\approxeq 0.051\\) ( 2 significant figures)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(383, '2020', '\n\nQuestion\n\n(a) Solve : \\(7x + 4 < \\frac{1}{2}(4x + 3)\\).\n(b) Salem, Sunday and Shaka shared a sum of N1,100.00. For every N2.00 that Salem gets, Sunday gets 50 kobo and for every N4.00 Sunday gets, Shaka gets N2.00. Find Shaka\'s share.\n\n', '', '(a) Solve : \\(7x + 4 < \\frac{1}{2}(4x + 3)\\).\n(b) Salem, Sunday and Shaka shared a sum of N1,100.00. For every N2.00 that Salem gets, Sunday gets 50 kobo and for every N4.00 Sunday gets, Shaka gets N2.00. Find Shaka\'s share.', '(a) Solve : \\(7x + 4 < \\frac{1}{2}(4x + 3)\\).~(b) Salem, Sunday and Shaka shared a sum of N1,100.00. For every N2.00 that Salem gets, Sunday gets 50 kobo and for every N4.00 Sunday gets, Shaka gets N2.00. Find Shaka\'s share.', NULL, '\n\nExplanation\n(a) \\(7x + 4 < \\frac{1}{2}(4x + 3)\\)\n\\(7x + 4 < 2x + \\frac{3}{2}\\)\n\\(7x - 2x < \\frac{3}{2} - 4\\)\n\\(5x < -\\frac{5}{2}\\)\n\\(\\frac{5x}{5} < \\frac{-\\frac{5}{2}}{5}\\)\n\\(x < -\\frac{1}{2}\\).\n(b) When Salem gets N2 ------> Sunday gets 50 kobo\n(Recall N1 = 100 kobo)\nSalemgets 200 kobo ------> Sunday gets 50 kobo.\nSunday gets N4 -----> Shaka gets N2.\n====> Sunday gets 400 kobo ------> Shakagets 200 kobo.\nWhen Sunday gets 400 kobo ------> Salem gets 1600 kobo.\n\\(\\therefore\\) Salem : Sunday : Shaka = 1600 : 400 : 200\n= 8 : 2 : 1\n\\(\\therefore \\text{Shaka\'s share} = \\frac{1}{11} \\times N1,100 = N100\\)\n\n', '', 'Explanation\n(a) \\(7x + 4 < \\frac{1}{2}(4x + 3)\\)\n\\(7x + 4 < 2x + \\frac{3}{2}\\)\n\\(7x - 2x < \\frac{3}{2} - 4\\)\n\\(5x < -\\frac{5}{2}\\)\n\\(\\frac{5x}{5} < \\frac{-\\frac{5}{2}}{5}\\)\n\\(x < -\\frac{1}{2}\\).\n(b) When Salem gets N2 ------> Sunday gets 50 kobo\n(Recall N1 = 100 kobo)\nSalemgets 200 kobo ------> Sunday gets 50 kobo.\nSunday gets N4 -----> Shaka gets N2.\n====> Sunday gets 400 kobo ------> Shakagets 200 kobo.\nWhen Sunday gets 400 kobo ------> Salem gets 1600 kobo.\n\\(\\therefore\\) Salem : Sunday : Shaka = 1600 : 400 : 200\n= 8 : 2 : 1\n\\(\\therefore \\text{Shaka\'s share} = \\frac{1}{11} \\times N1,100 = N100\\)\n\n', '(a) \\(7x + 4 < \\frac{1}{2}(4x + 3)\\)\n\\(7x + 4 < 2x + \\frac{3}{2}\\)\n\\(7x - 2x < \\frac{3}{2} - 4\\)\n\\(5x < -\\frac{5}{2}\\)\n\\(\\frac{5x}{5} < \\frac{-\\frac{5}{2}}{5}\\)\n\\(x < -\\frac{1}{2}\\).~(b) When Salem gets N2 ------> Sunday gets 50 kobo\n(Recall N1 = 100 kobo)\nSalemgets 200 kobo ------> Sunday gets 50 kobo.\nSunday gets N4 -----> Shaka gets N2.\n====> Sunday gets 400 kobo ------> Shakagets 200 kobo.\nWhen Sunday gets 400 kobo ------> Salem gets 1600 kobo.\n\\(\\therefore\\) Salem : Sunday : Shaka = 1600 : 400 : 200\n= 8 : 2 : 1\n\\(\\therefore \\text{Shaka\'s share} = \\frac{1}{11} \\times N1,100 = N100\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(384, '2020', '\n\nQuestion\n\n(a) The present ages of a father and his son are in the ratio 10 : 3. If the son is 15 years old now, in how many years will the ratio of their ages be 2 : 1?\n(b) The arithmetic mean of x, y and z is 6 while that ofx, y, z, l, u, v and w is 9. Calculate the arithmetic mean of l, u, v and w.\n\n', '', '(a) The present ages of a father and his son are in the ratio 10 : 3. If the son is 15 years old now, in how many years will the ratio of their ages be 2 : 1?\n(b) The arithmetic mean of x, y and z is 6 while that ofx, y, z, l, u, v and w is 9. Calculate the arithmetic mean of l, u, v and w.', '(a) The present ages of a father and his son are in the ratio 10 : 3. If the son is 15 years old now, in how many years will the ratio of their ages be 2 : 1?~(b) The arithmetic mean of x, y and z is 6 while that ofx, y, z, l, u, v and w is 9. Calculate the arithmetic mean of l, u, v and w.', NULL, '\n\nExplanation\n(a) Present ages of a father and his son are in the ratio 10 : 3.\nLet the sum of their ages be x.\nThe son\'s age = 15 years = \\(\\frac{3}{13} \\times x = 15\\)\n\\(x = \\frac{15 \\times 13}{3} = 65\\)\n\\(\\therefore\\) The Father\'s present age = 65 - 15 = 50 years.\nIn z years time, the ratio of their ages = 2 : 1\n\\(\\frac{50 + z}{15 + z} = \\frac{2}{1}\\)\n\\(\\implies 50 + z = 2(15 + z)\\)\n\\(50 + z = 30 + 2z \\implies 50 - 30 = 2z - z\\)\n\\(20 = z\\)\nTherefore, in 20 years\' time, the ratio of their age will be 2 : 1.\n(b) \\(\\frac{x + y + z}{3} = 6 \\implies x + y + z = 18 .... (1)\\)\n\\(\\frac{x + y + z + l + u + v + w}{7} = 9 \\implies x + y + z + l + u + v + w = 63 .... (2)\\)\nPutting (1) into (2), we have\n\\(18 + l + u + v + w = 63 \\)\n\\(\\therefore l + u + v + w = 63 - 18 = 45\\)\nMean of l, u, v and w = \\(\\frac{45}{4} = 11.25\\).\n\n', '', 'Explanation\n(a) Present ages of a father and his son are in the ratio 10 : 3.\nLet the sum of their ages be x.\nThe son\'s age = 15 years = \\(\\frac{3}{13} \\times x = 15\\)\n\\(x = \\frac{15 \\times 13}{3} = 65\\)\n\\(\\therefore\\) The Father\'s present age = 65 - 15 = 50 years.\nIn z years time, the ratio of their ages = 2 : 1\n\\(\\frac{50 + z}{15 + z} = \\frac{2}{1}\\)\n\\(\\implies 50 + z = 2(15 + z)\\)\n\\(50 + z = 30 + 2z \\implies 50 - 30 = 2z - z\\)\n\\(20 = z\\)\nTherefore, in 20 years\' time, the ratio of their age will be 2 : 1.\n(b) \\(\\frac{x + y + z}{3} = 6 \\implies x + y + z = 18 .... (1)\\)\n\\(\\frac{x + y + z + l + u + v + w}{7} = 9 \\implies x + y + z + l + u + v + w = 63 .... (2)\\)\nPutting (1) into (2), we have\n\\(18 + l + u + v + w = 63 \\)\n\\(\\therefore l + u + v + w = 63 - 18 = 45\\)\nMean of l, u, v and w = \\(\\frac{45}{4} = 11.25\\).\n\n', '(a) Present ages of a father and his son are in the ratio 10 : 3.\nLet the sum of their ages be x.\nThe son\'s age = 15 years = \\(\\frac{3}{13} \\times x = 15\\)\n\\(x = \\frac{15 \\times 13}{3} = 65\\)\n\\(\\therefore\\) The Father\'s present age = 65 - 15 = 50 years.\nIn z years time, the ratio of their ages = 2 : 1\n\\(\\frac{50 + z}{15 + z} = \\frac{2}{1}\\)\n\\(\\implies 50 + z = 2(15 + z)\\)\n\\(50 + z = 30 + 2z \\implies 50 - 30 = 2z - z\\)\n\\(20 = z\\)\nTherefore, in 20 years\' time, the ratio of their age will be 2 : 1.~(b) \\(\\frac{x + y + z}{3} = 6 \\implies x + y + z = 18 .... (1)\\)\n\\(\\frac{x + y + z + l + u + v + w}{7} = 9 \\implies x + y + z + l + u + v + w = 63 .... (2)\\)\nPutting (1) into (2), we have\n\\(18 + l + u + v + w = 63 \\)\n\\(\\therefore l + u + v + w = 63 - 18 = 45\\)\nMean of l, u, v and w = \\(\\frac{45}{4} = 11.25\\).\n', 'WAEC-PK-1', 'WAEC-MATH', 1);
INSERT INTO `exam_simulation2` (`id`, `examYear`, `questions`, `parent_que`, `child_que`, `gen_child_que_desc`, `imageDir`, `explanation`, `parent_que_expl`, `child_que_exp`, `gen_child_que_expl`, `package_id`, `subject_id`, `status`) VALUES
(385, '2020', '\n\nQuestion\n\nThe area of a circle is \\(154cm^{2}\\). It is divided into three sectors such that two of the sectors are equal in size and the third sector is three times the size of the other two put together. Calculate the perimeter of the third sector. [Take \\(\\pi = \\frac{22}{7}\\)].\n\n', 'The area of a circle is \\(154cm^{2}\\). It is divided into three sectors such that two of the sectors are equal in size and the third sector is three times the size of the other two put together. Calculate the perimeter of the third sector. [Take \\(\\pi = \\frac{22}{7}\\)].', '', '', NULL, '\n\nExplanation\nSince it is divide into 3 sectors\n then \\(8x° = 360°\\)\n\\(x° = 45°\\)\nThe 3rd sector = 3 times the size of the other two sectors put together\n= \\(3(45 + 45) = 3(90) = 270°\\)\nPerimeter of a sector = \\(2r + l\\)\nwhere r = radius of the circle ; l = length of arc = \\(\\frac{\\theta}{360°} \\times 2\\pi r\\)\nArea of circle = \\(\\frac{22}{7} \\times r^{2} = 154 cm^{2}\\)\n\\(r^{2} = \\frac{154 \\times 7}{22} = 49\\)\n\\(r = 7 cm\\)\n\\(l = \\frac{270}{360} \\times 2 \\times \\frac{22}{7} \\times 7 = 33 cm\\)\nPerimeter of sector = \\(2(7) + 33 = 14 + 33 = 47 cm\\).\n\n', 'Since it is divide into 3 sectors\n then \\(8x° = 360°\\)\n\\(x° = 45°\\)\nThe 3rd sector = 3 times the size of the other two sectors put together\n= \\(3(45 + 45) = 3(90) = 270°\\)\nPerimeter of a sector = \\(2r + l\\)\nwhere r = radius of the circle ; l = length of arc = \\(\\frac{\\theta}{360°} \\times 2\\pi r\\)\nArea of circle = \\(\\frac{22}{7} \\times r^{2} = 154 cm^{2}\\)\n\\(r^{2} = \\frac{154 \\times 7}{22} = 49\\)\n\\(r = 7 cm\\)\n\\(l = \\frac{270}{360} \\times 2 \\times \\frac{22}{7} \\times 7 = 33 cm\\)\nPerimeter of sector = \\(2(7) + 33 = 14 + 33 = 47 cm\\).\n\n', '', '', 'WAEC-PK-1', 'WAEC-MATH', 1),
(386, '2020', '\n\nQuestion\n\nA boy 1.2m tall, stands 6m away from the foot of a vertical lamp pole 4.2m long. If the lamp is at the tip of the pole,\n(a) represent this information in a diagram ;\n(b) calculate the (i) length of the shadow of the boy cast by the lamp ; (ii) angle of elevation of the lamp from the boy, correct to the nearest degree.\n\n', '', 'A boy 1.2m tall, stands 6m away from the foot of a vertical lamp pole 4.2m long. If the lamp is at the tip of the pole,\n(a) represent this information in a diagram ;\n(b) calculate the (i) length of the shadow of the boy cast by the lamp ; (ii) angle of elevation of the lamp from the boy, correct to the nearest degree.', 'A boy 1.2m tall, stands 6m away from the foot of a vertical lamp pole 4.2m long. If the lamp is at the tip of the pole,\n(a) represent this information in a diagram ;~(b) calculate the (i) length of the shadow of the boy cast by the lamp ; (ii) angle of elevation of the lamp from the boy, correct to the nearest degree.', NULL, '\n\nExplanation\n(a)\n(b) \\(\\tan \\theta = \\frac{4.2}{6 + x} ; \\tan \\theta = \\frac{1.2}{x}\\)\n\\(\\tan \\theta = \\tan \\theta\\)\n\\(\\implies \\frac{4.2}{6 + x} = \\frac{1.2}{x}\\)\n\\(4.2 x = 1.2(6 + x) \\)\n\\(4.2 x = 7.2 + 1.2 x \\implies 4.2x - 1.2x = 7.2\\)\n\\(3x = 7.2 \\implies x = 2.4 m\\) (length of the shadow cast by the lamp)\n(ii) \\(\\tan \\theta = \\frac{1.2}{2.4} = 0.5\\)\n\\(\\theta = \\tan^{-1} (0.5) \\)\n= \\(27°\\) (angle of elevation)\n\n', '', 'Explanation\n(a)\n(b) \\(\\tan \\theta = \\frac{4.2}{6 + x} ; \\tan \\theta = \\frac{1.2}{x}\\)\n\\(\\tan \\theta = \\tan \\theta\\)\n\\(\\implies \\frac{4.2}{6 + x} = \\frac{1.2}{x}\\)\n\\(4.2 x = 1.2(6 + x) \\)\n\\(4.2 x = 7.2 + 1.2 x \\implies 4.2x - 1.2x = 7.2\\)\n\\(3x = 7.2 \\implies x = 2.4 m\\) (length of the shadow cast by the lamp)\n(ii) \\(\\tan \\theta = \\frac{1.2}{2.4} = 0.5\\)\n\\(\\theta = \\tan^{-1} (0.5) \\)\n= \\(27°\\) (angle of elevation)\n\n', '~(b) \\(\\tan \\theta = \\frac{4.2}{6 + x} ; \\tan \\theta = \\frac{1.2}{x}\\)\n\\(\\tan \\theta = \\tan \\theta\\)\n\\(\\implies \\frac{4.2}{6 + x} = \\frac{1.2}{x}\\)\n\\(4.2 x = 1.2(6 + x) \\)\n\\(4.2 x = 7.2 + 1.2 x \\implies 4.2x - 1.2x = 7.2\\)\n\\(3x = 7.2 \\implies x = 2.4 m\\) (length of the shadow cast by the lamp)\n(ii) \\(\\tan \\theta = \\frac{1.2}{2.4} = 0.5\\)\n\\(\\theta = \\tan^{-1} (0.5) \\)\n= \\(27°\\) (angle of elevation)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(387, '2020', '\n\nQuestion\n\n(a) Two positive whole numbers p and q are such that p is greater than q and their sum is equal to three times their difference;\n(i) Express p in terms of q ; (ii) Hence, evaluate\\(\\frac{p^{2} + q^{2}}{pq}\\).\n(b) A man sold 100 articles at 25 for N66.00 and made a gain of 32%. Calculate his gain or loss percent if he sold them at 20 for N50.00.\n\n', '', '(a) Two positive whole numbers p and q are such that p is greater than q and their sum is equal to three times their difference;\n(i) Express p in terms of q ; (ii) Hence, evaluate\\(\\frac{p^{2} + q^{2}}{pq}\\).\n(b) A man sold 100 articles at 25 for N66.00 and made a gain of 32%. Calculate his gain or loss percent if he sold them at 20 for N50.00.', '(a) Two positive whole numbers p and q are such that p is greater than q and their sum is equal to three times their difference;\n(i) Express p in terms of q ; (ii) Hence, evaluate\\(\\frac{p^{2} + q^{2}}{pq}\\).~(b) A man sold 100 articles at 25 for N66.00 and made a gain of 32%. Calculate his gain or loss percent if he sold them at 20 for N50.00.', NULL, '\n\nExplanation\n(a)(i) \\(p > q .... (1)\\)\n\\(p + q = 3(p - q) ...... (2)\\)\nFrom (2), \\(p + q = 3p - 3q\\)\n\\(p - 3p = - 3q - q \\implies - 2p = - 4q\\)\n\\(p = 2q\\)\n(ii) \\(\\frac{p^{2} + q^{2}}{pq}\\)\n= \\(\\frac{(2q)^{2} + q^{2}}{2q \\times q}\\)\n\\(\\frac{4q^{2} + q^{2}}{2q^{2}}\\)\n= \\(\\frac{5q^{2}}{2q^{2}}\\)\n= \\(\\frac{5}{2}\\)\n(b) Selling price = \\(\\frac{100 \\times 66}{25}\\)\n= \\(N264.00\\)\nUsing \\(\\frac{SP- CP}{CP} = % gain\\)\n\\(\\frac{264 - CP}{CP} = \\frac{32}{100}\\)\n\\(100(264 - CP) = 32CP\\)\n\\(26400 = 32CP + 100CP = 132CP\\)\n\\(CP = \\frac{26400}{132}\\)\n= \\(N200\\)\nWhen he sells 20 for N50, Selling price = \\(\\frac{100 \\times 50}{20} = N250.00\\)\nHence, he made gain.\n\\(% gain = \\frac{250 - 200}{200} \\times 100%\\)\n\\(\\frac{1}{4} \\times 100%\\)\n= \\(25% gain\\)\n\n', '', 'Explanation\n(a)(i) \\(p > q .... (1)\\)\n\\(p + q = 3(p - q) ...... (2)\\)\nFrom (2), \\(p + q = 3p - 3q\\)\n\\(p - 3p = - 3q - q \\implies - 2p = - 4q\\)\n\\(p = 2q\\)\n(ii) \\(\\frac{p^{2} + q^{2}}{pq}\\)\n= \\(\\frac{(2q)^{2} + q^{2}}{2q \\times q}\\)\n\\(\\frac{4q^{2} + q^{2}}{2q^{2}}\\)\n= \\(\\frac{5q^{2}}{2q^{2}}\\)\n= \\(\\frac{5}{2}\\)\n(b) Selling price = \\(\\frac{100 \\times 66}{25}\\)\n= \\(N264.00\\)\nUsing \\(\\frac{SP- CP}{CP} = % gain\\)\n\\(\\frac{264 - CP}{CP} = \\frac{32}{100}\\)\n\\(100(264 - CP) = 32CP\\)\n\\(26400 = 32CP + 100CP = 132CP\\)\n\\(CP = \\frac{26400}{132}\\)\n= \\(N200\\)\nWhen he sells 20 for N50, Selling price = \\(\\frac{100 \\times 50}{20} = N250.00\\)\nHence, he made gain.\n\\(% gain = \\frac{250 - 200}{200} \\times 100%\\)\n\\(\\frac{1}{4} \\times 100%\\)\n= \\(25% gain\\)\n\n', '(a)(i) \\(p > q .... (1)\\)\n\\(p + q = 3(p - q) ...... (2)\\)\nFrom (2), \\(p + q = 3p - 3q\\)\n\\(p - 3p = - 3q - q \\implies - 2p = - 4q\\)\n\\(p = 2q\\)\n(ii) \\(\\frac{p^{2} + q^{2}}{pq}\\)\n= \\(\\frac{(2q)^{2} + q^{2}}{2q \\times q}\\)\n\\(\\frac{4q^{2} + q^{2}}{2q^{2}}\\)\n= \\(\\frac{5q^{2}}{2q^{2}}\\)\n= \\(\\frac{5}{2}\\)~(b) Selling price = \\(\\frac{100 \\times 66}{25}\\)\n= \\(N264.00\\)\nUsing \\(\\frac{SP- CP}{CP} = % gain\\)\n\\(\\frac{264 - CP}{CP} = \\frac{32}{100}\\)\n\\(100(264 - CP) = 32CP\\)\n\\(26400 = 32CP + 100CP = 132CP\\)\n\\(CP = \\frac{26400}{132}\\)\n= \\(N200\\)\nWhen he sells 20 for N50, Selling price = \\(\\frac{100 \\times 50}{20} = N250.00\\)\nHence, he made gain.\n\\(% gain = \\frac{250 - 200}{200} \\times 100%\\)\n\\(\\frac{1}{4} \\times 100%\\)\n= \\(25% gain\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(388, '2020', '\n\nQuestion\n\n(a) Copy and complete the table of values for the relation \\(y = 3x^{2} - 5x - 7\\).\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\ny\n35\n\n\n-7\n-9\n\n5\n\n\n\n\n(b) Using scales 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 = 3x^{2} - 5x - 7, -3 \\leq x \\leq 4\\).\n(c) From the graph : (i) find the roots of the equation \\(3x^{2} - 5x - 7 = 0\\) ; (ii) estimate the minimum value of y ; (iii) calculate the gradient of the curve at the point x = 2.\n\n\n(a) Copy and complete the table of values for the relation \\(y = 3x^{2} - 5x - 7\\).\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\ny\n35\n\n\n-7\n-9\n\n5\n\n\n\n', '', '(a) Copy and complete the table of values for the relation \\(y = 3x^{2} - 5x - 7\\).\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\ny\n35\n\n\n-7\n-9\n\n5\n\n\n\n\n(b) Using scales 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 = 3x^{2} - 5x - 7, -3 \\leq x \\leq 4\\).\n(c) From the graph : (i) find the roots of the equation \\(3x^{2} - 5x - 7 = 0\\) ; (ii) estimate the minimum value of y ; (iii) calculate the gradient of the curve at the point x = 2.\n\n\n(a) Copy and complete the table of values for the relation \\(y = 3x^{2} - 5x - 7\\).\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\ny\n35\n\n\n-7\n-9\n\n5\n\n', '(a) Copy and complete the table of values for the relation \\(y = 3x^{2} - 5x - 7\\).\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\ny\n35\n\n\n-7\n-9\n\n5\n\n\n\n~(b) Using scales 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 = 3x^{2} - 5x - 7, -3 \\leq x \\leq 4\\).\n(c) From the graph : (i) find the roots of the equation \\(3x^{2} - 5x - 7 = 0\\) ; (ii) estimate the minimum value of y ; (iii) calculate the gradient of the curve at the point x = 2.\n\n\n(a) Copy and complete the table of values for the relation \\(y = 3x^{2} - 5x - 7\\).\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\ny\n35\n\n\n-7\n-9\n\n5\n\n~(c) From the graph : (i) find the roots of the equation \\(3x^{2} - 5x - 7 = 0\\) ; (ii) estimate the minimum value of y ; (iii) calculate the gradient of the curve at the point x = 2.\n\n\n(a) Copy and complete the table of values for the relation \\(y = 3x^{2} - 5x - 7\\).\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\ny\n35\n\n\n-7\n-9\n\n5\n\n', NULL, '\n\nExplanation\n(a)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\ny\n35\n15\n1\n-7\n-9\n-5\n5\n21\n\n\n\n(b)\n(c) (i) The roots of the equation is where the graph cuts the x- axis = -0.9, 2.6.\n(ii) Minimum value of y = -9.21\n(iii) At x = 2,\ngradient = \\(\\frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\)\n= \\(\\frac{4 - (-8.5)}{3.3 - 1.5}\\)\n= \\(\\frac{12.5}{1.8}\\)\n= 6.9444\n\\(\\approxeq\\) 6.9\n\n', '', 'Explanation\n(a)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\ny\n35\n15\n1\n-7\n-9\n-5\n5\n21\n\n\n\n(b)\n(c) (i) The roots of the equation is where the graph cuts the x- axis = -0.9, 2.6.\n(ii) Minimum value of y = -9.21\n(iii) At x = 2,\ngradient = \\(\\frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\)\n= \\(\\frac{4 - (-8.5)}{3.3 - 1.5}\\)\n= \\(\\frac{12.5}{1.8}\\)\n= 6.9444\n\\(\\approxeq\\) 6.9\n\n', '(a)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n4\n\n\ny\n35\n15\n1\n-7\n-9\n-5\n5\n21\n\n\n~(b)\n(c) (i) The roots of the equation is where the graph cuts the x- axis = -0.9, 2.6.\n(ii) Minimum value of y = -9.21\n(iii) At x = 2,\ngradient = \\(\\frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\)\n= \\(\\frac{4 - (-8.5)}{3.3 - 1.5}\\)\n= \\(\\frac{12.5}{1.8}\\)\n= 6.9444\n\\(\\approxeq\\) 6.9\n~(c) (i) The roots of the equation is where the graph cuts the x- axis = -0.9, 2.6.\n(ii) Minimum value of y = -9.21\n(iii) At x = 2,\ngradient = \\(\\frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\)\n= \\(\\frac{4 - (-8.5)}{3.3 - 1.5}\\)\n= \\(\\frac{12.5}{1.8}\\)\n= 6.9444\n\\(\\approxeq\\) 6.9\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(389, '2020', '\n\nQuestion\n\n(a) If (3 - x), 6, (7 - 5x) are consecutive terms of a geometric progression (GP) with constant ratio r > 0, find the :\n(i) values of x ; (ii) constant ratio.\n(b)In the diagram, |AB| = 3 cm, |BC| = 4 cm, |CD| = 6 cm and |DA| = 7 cm. Calculate <ADC, correct to the nearest degree.\n\n', '', '(a) If (3 - x), 6, (7 - 5x) are consecutive terms of a geometric progression (GP) with constant ratio r > 0, find the :\n(i) values of x ; (ii) constant ratio.\n(b)In the diagram, |AB| = 3 cm, |BC| = 4 cm, |CD| = 6 cm and |DA| = 7 cm. Calculate <ADC, correct to the nearest degree.', '(a) If (3 - x), 6, (7 - 5x) are consecutive terms of a geometric progression (GP) with constant ratio r > 0, find the :\n(i) values of x ; (ii) constant ratio.~(b)In the diagram, |AB| = 3 cm, |BC| = 4 cm, |CD| = 6 cm and |DA| = 7 cm. Calculate <ADC, correct to the nearest degree.', NULL, '\n\nExplanation\n(a) (i) (3 - x), 6, (7 - 5x).\nFirst term = a = (3 - x)\ncommon ratio = \\(\\frac{6}{3 - x} = \\frac{7 - 5x}{6}\\)\n\\(36 = (3 - x)(7 - 5x)\\) \n\\(36 = 21 - 15x - 7x + 5x^{2}\\)\n\\(36 = 21 - 22x + 5x^{2}\\)\n\\(5x^{2} - 22x + 21 - 36 = 0\\)\n\\(5x^{2} - 22x - 15 = 0\\)\n\\(5x^{2} - 25x + 3x - 15 = 0 \\implies 5x(x - 5) + 3(x - 5) = 0\\)\n\\((x - 5)(5x + 3) = 0\\)\n\\(x = -\\frac{3}{5} \\text{ or = } 5\\)\n(ii) Constant ratio = \\(\\frac{6}{3 - x} = \\frac{7 - 5x}{6}\\)\nUsing x = 5,\n\\(r = \\frac{7 - 5(5)}{6} = \\frac{-18}{6}\\)\n= \\(-3\\)\n(b)Considering \\(\\Delta\\) AOC\n\\(b^{2} = 4^{2} + 3^{2}\\)\n\\(b^{2} = 16 + 9 = 25\\)\n\\(b = \\sqrt{25} = 5 cm\\)\nConsidering \\(\\Delta\\) ACD,\nUsing cosine rule,\n\\(\\cos D = \\frac{a^{2} + b^{2} - c^{2}}{2ab} = \\frac{a^{2} + c^{2} - b^{2}}{2ac}\\)\n= \\(\\frac{6^{2} + 7^{2} - 5^{2}}{2 (6)(7)}\\)\n= \\(\\frac{36 + 49 - 25}{84}\\)\n\\(\\cos D = \\frac{60}{84}\\)\n\\(\\cos D = 0.714\\)\n\\(D = \\cos^{-1} (0.714)\\)\n\\(D = 44.4° \\approxeq 44°\\) (to the nearest degree)\n\n', '', 'Explanation\n(a) (i) (3 - x), 6, (7 - 5x).\nFirst term = a = (3 - x)\ncommon ratio = \\(\\frac{6}{3 - x} = \\frac{7 - 5x}{6}\\)\n\\(36 = (3 - x)(7 - 5x)\\) \n\\(36 = 21 - 15x - 7x + 5x^{2}\\)\n\\(36 = 21 - 22x + 5x^{2}\\)\n\\(5x^{2} - 22x + 21 - 36 = 0\\)\n\\(5x^{2} - 22x - 15 = 0\\)\n\\(5x^{2} - 25x + 3x - 15 = 0 \\implies 5x(x - 5) + 3(x - 5) = 0\\)\n\\((x - 5)(5x + 3) = 0\\)\n\\(x = -\\frac{3}{5} \\text{ or = } 5\\)\n(ii) Constant ratio = \\(\\frac{6}{3 - x} = \\frac{7 - 5x}{6}\\)\nUsing x = 5,\n\\(r = \\frac{7 - 5(5)}{6} = \\frac{-18}{6}\\)\n= \\(-3\\)\n(b)Considering \\(\\Delta\\) AOC\n\\(b^{2} = 4^{2} + 3^{2}\\)\n\\(b^{2} = 16 + 9 = 25\\)\n\\(b = \\sqrt{25} = 5 cm\\)\nConsidering \\(\\Delta\\) ACD,\nUsing cosine rule,\n\\(\\cos D = \\frac{a^{2} + b^{2} - c^{2}}{2ab} = \\frac{a^{2} + c^{2} - b^{2}}{2ac}\\)\n= \\(\\frac{6^{2} + 7^{2} - 5^{2}}{2 (6)(7)}\\)\n= \\(\\frac{36 + 49 - 25}{84}\\)\n\\(\\cos D = \\frac{60}{84}\\)\n\\(\\cos D = 0.714\\)\n\\(D = \\cos^{-1} (0.714)\\)\n\\(D = 44.4° \\approxeq 44°\\) (to the nearest degree)\n\n', '(a) (i) (3 - x), 6, (7 - 5x).\nFirst term = a = (3 - x)\ncommon ratio = \\(\\frac{6}{3 - x} = \\frac{7 - 5x}{6}\\)\n\\(36 = (3 - x)(7 - 5x)\\) \n\\(36 = 21 - 15x - 7x + 5x^{2}\\)\n\\(36 = 21 - 22x + 5x^{2}\\)\n\\(5x^{2} - 22x + 21 - 36 = 0\\)\n\\(5x^{2} - 22x - 15 = 0\\)\n\\(5x^{2} - 25x + 3x - 15 = 0 \\implies 5x(x - 5) + 3(x - 5) = 0\\)\n\\((x - 5)(5x + 3) = 0\\)\n\\(x = -\\frac{3}{5} \\text{ or = } 5\\)\n(ii) Constant ratio = \\(\\frac{6}{3 - x} = \\frac{7 - 5x}{6}\\)\nUsing x = 5,\n\\(r = \\frac{7 - 5(5)}{6} = \\frac{-18}{6}\\)\n= \\(-3\\)~(b)Considering \\(\\Delta\\) AOC\n\\(b^{2} = 4^{2} + 3^{2}\\)\n\\(b^{2} = 16 + 9 = 25\\)\n\\(b = \\sqrt{25} = 5 cm\\)\nConsidering \\(\\Delta\\) ACD,\nUsing cosine rule,\n\\(\\cos D = \\frac{a^{2} + b^{2} - c^{2}}{2ab} = \\frac{a^{2} + c^{2} - b^{2}}{2ac}\\)\n= \\(\\frac{6^{2} + 7^{2} - 5^{2}}{2 (6)(7)}\\)\n= \\(\\frac{36 + 49 - 25}{84}\\)\n\\(\\cos D = \\frac{60}{84}\\)\n\\(\\cos D = 0.714\\)\n\\(D = \\cos^{-1} (0.714)\\)\n\\(D = 44.4° \\approxeq 44°\\) (to the nearest degree)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(390, '2020', '\n\nQuestion\n\n(a) Using ruler and a pair of compasses only, construct : (i) a trapezium WXYZ such that |WX| = 10.2 cm, |XY| = 5.6 cm, |YZ| = 5.8 cm, < WXY = 60° and WX is parallel to YZ (ii) a perpendicular from Z to meet \\(\\overline{WX}\\) at N.\n(b) Measure : (i) |WZ| ; (ii) |ZN| .\n\n', '', '(a) Using ruler and a pair of compasses only, construct : (i) a trapezium WXYZ such that |WX| = 10.2 cm, |XY| = 5.6 cm, |YZ| = 5.8 cm, < WXY = 60° and WX is parallel to YZ (ii) a perpendicular from Z to meet \\(\\overline{WX}\\) at N.\n(b) Measure : (i) |WZ| ; (ii) |ZN| .', '(a) Using ruler and a pair of compasses only, construct : (i) a trapezium WXYZ such that |WX| = 10.2 cm, |XY| = 5.6 cm, |YZ| = 5.8 cm, < WXY = 60° and WX is parallel to YZ (ii) a perpendicular from Z to meet \\(\\overline{WX}\\) at N.~(b) Measure : (i) |WZ| ; (ii) |ZN| .', NULL, '\n\nExplanation\n(a)\n(b) From the figure above,\n(i) |WZ| = 5.25 cm ; (ii) |ZN| = 4.86 cm.\n\n', '', 'Explanation\n(a)\n(b) From the figure above,\n(i) |WZ| = 5.25 cm ; (ii) |ZN| = 4.86 cm.\n\n', '~(b) From the figure above,\n(i) |WZ| = 5.25 cm ; (ii) |ZN| = 4.86 cm.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(391, '2020', '\n\nQuestion\n\n(a)\nA segment of a circle is cut off from a rectangular board as shown in the diagram. If the radius of the circle is \\(1\\frac{1}{2}\\) times the length of the chord;calculate, correct to 2 decimal places, the perimeter of the remaining portion. [Take \\(\\pi = \\frac{22}{7}\\)]\n(b) Evaluate without using calculators or tables, \\(\\frac{3}{\\sqrt{3}}(\\frac{2}{\\sqrt{3}} - \\frac{\\sqrt{12}}{6})\\).\n\n', '', '(a)\nA segment of a circle is cut off from a rectangular board as shown in the diagram. If the radius of the circle is \\(1\\frac{1}{2}\\) times the length of the chord;calculate, correct to 2 decimal places, the perimeter of the remaining portion. [Take \\(\\pi = \\frac{22}{7}\\)]\n(b) Evaluate without using calculators or tables, \\(\\frac{3}{\\sqrt{3}}(\\frac{2}{\\sqrt{3}} - \\frac{\\sqrt{12}}{6})\\).', '(a)\nA segment of a circle is cut off from a rectangular board as shown in the diagram. If the radius of the circle is \\(1\\frac{1}{2}\\) times the length of the chord;calculate, correct to 2 decimal places, the perimeter of the remaining portion. [Take \\(\\pi = \\frac{22}{7}\\)]~(b) Evaluate without using calculators or tables, \\(\\frac{3}{\\sqrt{3}}(\\frac{2}{\\sqrt{3}} - \\frac{\\sqrt{12}}{6})\\).', NULL, '\n\nExplanation\n(a)Length of the chord = 14 cm\nRadius = \\(\\frac{3}{2} \\times 14 cm = 21 cm\\)\nThe chord AB = \\(2r \\sin \\frac{\\theta}{2}\\)\n\\(14 = 2(21) \\sin \\frac{\\theta}{2}\\)\n\\(\\sin \\frac{\\theta}{2} = \\frac{14}{42} = 0.333\\)\n\\(\\frac{\\theta}{2} = \\sin^{-1} (0.333)\\)\n\\(\\frac{\\theta}{2} = 19.469°\\)\n\\(\\theta = 19.469° \\times 2 = 38.938°\\)\nLength of the arc = \\(\\frac{\\theta}{360°} \\times 2\\pi r\\)\n\\(\\frac{38.938}{360} \\times 2 \\times \\frac{22}{7} \\times 21\\)\n= \\(14.277 cm\\)\nPerimeter of the remaining portion = 22 + 12 + 12 + 5 + 3 + 14.277\n= 68.277 cm\n\\(\\approxeq\\) 68.28 cm (2 decimal places).\n(b) \\(\\frac{3}{\\sqrt{3}}(\\frac{2}{\\sqrt{3}} - \\frac{\\sqrt{12}}{6}\\)\n= \\(\\frac{3}{\\sqrt{3}}(\\frac{12 - \\sqrt{36}}{6\\sqrt{3}}\\)\n= \\(\\frac{3}{\\sqrt{3}}(\\frac{12 - 6}{6\\sqrt{3}}\\)\n= \\(\\frac{3}{\\sqrt{3}}(\\frac{6}{6\\sqrt{3}}\\)\n= \\(\\frac{18}{6\\sqrt{9}}\\)\n= \\(\\frac{18}{18}\\)\n= 1.\n\n', '', 'Explanation\n(a)Length of the chord = 14 cm\nRadius = \\(\\frac{3}{2} \\times 14 cm = 21 cm\\)\nThe chord AB = \\(2r \\sin \\frac{\\theta}{2}\\)\n\\(14 = 2(21) \\sin \\frac{\\theta}{2}\\)\n\\(\\sin \\frac{\\theta}{2} = \\frac{14}{42} = 0.333\\)\n\\(\\frac{\\theta}{2} = \\sin^{-1} (0.333)\\)\n\\(\\frac{\\theta}{2} = 19.469°\\)\n\\(\\theta = 19.469° \\times 2 = 38.938°\\)\nLength of the arc = \\(\\frac{\\theta}{360°} \\times 2\\pi r\\)\n\\(\\frac{38.938}{360} \\times 2 \\times \\frac{22}{7} \\times 21\\)\n= \\(14.277 cm\\)\nPerimeter of the remaining portion = 22 + 12 + 12 + 5 + 3 + 14.277\n= 68.277 cm\n\\(\\approxeq\\) 68.28 cm (2 decimal places).\n(b) \\(\\frac{3}{\\sqrt{3}}(\\frac{2}{\\sqrt{3}} - \\frac{\\sqrt{12}}{6}\\)\n= \\(\\frac{3}{\\sqrt{3}}(\\frac{12 - \\sqrt{36}}{6\\sqrt{3}}\\)\n= \\(\\frac{3}{\\sqrt{3}}(\\frac{12 - 6}{6\\sqrt{3}}\\)\n= \\(\\frac{3}{\\sqrt{3}}(\\frac{6}{6\\sqrt{3}}\\)\n= \\(\\frac{18}{6\\sqrt{9}}\\)\n= \\(\\frac{18}{18}\\)\n= 1.\n\n', '(a)Length of the chord = 14 cm\nRadius = \\(\\frac{3}{2} \\times 14 cm = 21 cm\\)\nThe chord AB = \\(2r \\sin \\frac{\\theta}{2}\\)\n\\(14 = 2(21) \\sin \\frac{\\theta}{2}\\)\n\\(\\sin \\frac{\\theta}{2} = \\frac{14}{42} = 0.333\\)\n\\(\\frac{\\theta}{2} = \\sin^{-1} (0.333)\\)\n\\(\\frac{\\theta}{2} = 19.469°\\)\n\\(\\theta = 19.469° \\times 2 = 38.938°\\)\nLength of the arc = \\(\\frac{\\theta}{360°} \\times 2\\pi r\\)\n\\(\\frac{38.938}{360} \\times 2 \\times \\frac{22}{7} \\times 21\\)\n= \\(14.277 cm\\)\nPerimeter of the remaining portion = 22 + 12 + 12 + 5 + 3 + 14.277\n= 68.277 cm\n\\(\\approxeq\\) 68.28 cm (2 decimal places).~(b) \\(\\frac{3}{\\sqrt{3}}(\\frac{2}{\\sqrt{3}} - \\frac{\\sqrt{12}}{6}\\)\n= \\(\\frac{3}{\\sqrt{3}}(\\frac{12 - \\sqrt{36}}{6\\sqrt{3}}\\)\n= \\(\\frac{3}{\\sqrt{3}}(\\frac{12 - 6}{6\\sqrt{3}}\\)\n= \\(\\frac{3}{\\sqrt{3}}(\\frac{6}{6\\sqrt{3}}\\)\n= \\(\\frac{18}{6\\sqrt{9}}\\)\n= \\(\\frac{18}{18}\\)\n= 1.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(392, '2020', '\n\nQuestion\n\nThe frequency distribution table shows the marks obtained by 100 students in a Mathematics test.\n\n\n\n\nMarks (%)\n\n1-10\n11-20\n21-30\n31-40\n41-50\n51-60\n61-70\n71-80\n81-90\n91-100\n\n\nFrequency\n2\n3\n5\n13\n19\n31\n13\n9\n4\n1\n\n\n\n(a) Draw the cumulative curve for the distribution.\n(b) Use the graph to find the : (i) 60th percentile ; (ii) probability that a student passed the test if the pass mark was fixed at 35%.\n\n\nThe frequency distribution table shows the marks obtained by 100 students in a Mathematics test.\n\n\n\n\nMarks (%)\n\n1-10\n11-20\n21-30\n31-40\n41-50\n51-60\n61-70\n71-80\n81-90\n91-100\n\n\nFrequency\n2\n3\n5\n13\n19\n31\n13\n9\n4\n1\n\n\n', '', 'The frequency distribution table shows the marks obtained by 100 students in a Mathematics test.\n\n\n\n\nMarks (%)\n\n1-10\n11-20\n21-30\n31-40\n41-50\n51-60\n61-70\n71-80\n81-90\n91-100\n\n\nFrequency\n2\n3\n5\n13\n19\n31\n13\n9\n4\n1\n\n\n\n(a) Draw the cumulative curve for the distribution.\n(b) Use the graph to find the : (i) 60th percentile ; (ii) probability that a student passed the test if the pass mark was fixed at 35%.\n\n\nThe frequency distribution table shows the marks obtained by 100 students in a Mathematics test.\n\n\n\n\nMarks (%)\n\n1-10\n11-20\n21-30\n31-40\n41-50\n51-60\n61-70\n71-80\n81-90\n91-100\n\n\nFrequency\n2\n3\n5\n13\n19\n31\n13\n9\n4\n1\n', 'The frequency distribution table shows the marks obtained by 100 students in a Mathematics test.\n\n\n\n\nMarks (%)\n\n1-10\n11-20\n21-30\n31-40\n41-50\n51-60\n61-70\n71-80\n81-90\n91-100\n\n\nFrequency\n2\n3\n5\n13\n19\n31\n13\n9\n4\n1\n\n\n\n(a) Draw the cumulative curve for the distribution.~(b) Use the graph to find the : (i) 60th percentile ; (ii) probability that a student passed the test if the pass mark was fixed at 35%.\n\n\nThe frequency distribution table shows the marks obtained by 100 students in a Mathematics test.\n\n\n\n\nMarks (%)\n\n1-10\n11-20\n21-30\n31-40\n41-50\n51-60\n61-70\n71-80\n81-90\n91-100\n\n\nFrequency\n2\n3\n5\n13\n19\n31\n13\n9\n4\n1\n', NULL, '\n\nExplanation\nMarks\nfrequency\ncum freq\nlower bound\nupper bound\n\n\n1 - 10\n2\n2\n0.5\n10.5\n\n\n11 - 20\n3\n5\n10.5\n20.5\n\n\n21 - 30\n5\n10\n20.5\n30.5\n\n\n31 - 40\n13\n23\n30.5\n40.5\n\n\n41 - 50\n19\n42\n40.5\n50.5\n\n\n51 - 60\n31\n73\n50.5\n60.5\n\n\n61 - 70\n13\n86\n60.5\n70.5\n\n\n71 - 80\n9\n95\n70.5\n80.5\n\n\n81 - 90\n4\n99\n80.5\n90.5\n\n\n91 - 100\n1\n100\n90.5\n100.5\n\n\n\n(a)\n(b) (i) 60th percentile is where the line from the cumulative frequency of 60 touches the curve i.e. 60th percentile = 58.5\n(ii) If the pass mark was fixed at 35%, then the frequency will be 13 with the class mark 31 - 40.\n\n', '', 'Explanation\nMarks\nfrequency\ncum freq\nlower bound\nupper bound\n\n\n1 - 10\n2\n2\n0.5\n10.5\n\n\n11 - 20\n3\n5\n10.5\n20.5\n\n\n21 - 30\n5\n10\n20.5\n30.5\n\n\n31 - 40\n13\n23\n30.5\n40.5\n\n\n41 - 50\n19\n42\n40.5\n50.5\n\n\n51 - 60\n31\n73\n50.5\n60.5\n\n\n61 - 70\n13\n86\n60.5\n70.5\n\n\n71 - 80\n9\n95\n70.5\n80.5\n\n\n81 - 90\n4\n99\n80.5\n90.5\n\n\n91 - 100\n1\n100\n90.5\n100.5\n\n\n\n(a)\n(b) (i) 60th percentile is where the line from the cumulative frequency of 60 touches the curve i.e. 60th percentile = 58.5\n(ii) If the pass mark was fixed at 35%, then the frequency will be 13 with the class mark 31 - 40.\n\n', '~(b) (i) 60th percentile is where the line from the cumulative frequency of 60 touches the curve i.e. 60th percentile = 58.5\n(ii) If the pass mark was fixed at 35%, then the frequency will be 13 with the class mark 31 - 40.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(393, '2020', '\n\nQuestion\n\nAn aeroplane flies due North from a town T on the equator at a speed of 950km per hour for 4 hours to another town P. It then flies eastwards to town Q on longitude 65°E. If the longitude of T is 15°E,\n(a) represent this information in a diagram ;\n(b) calculate the : (i) latitude of P, correct to the nearest degree ; (ii) distance between P and Q, correct to four significant figures. [Take \\(\\pi = \\frac{22}{7}\\); Radius of the earth = 6400km].\n\n', '', 'An aeroplane flies due North from a town T on the equator at a speed of 950km per hour for 4 hours to another town P. It then flies eastwards to town Q on longitude 65°E. If the longitude of T is 15°E,\n(a) represent this information in a diagram ;\n(b) calculate the : (i) latitude of P, correct to the nearest degree ; (ii) distance between P and Q, correct to four significant figures. [Take \\(\\pi = \\frac{22}{7}\\); Radius of the earth = 6400km].', 'An aeroplane flies due North from a town T on the equator at a speed of 950km per hour for 4 hours to another town P. It then flies eastwards to town Q on longitude 65°E. If the longitude of T is 15°E,\n(a) represent this information in a diagram ;~(b) calculate the : (i) latitude of P, correct to the nearest degree ; (ii) distance between P and Q, correct to four significant figures. [Take \\(\\pi = \\frac{22}{7}\\); Radius of the earth = 6400km].', NULL, '\n\nExplanation\n(a)   \n(b)(i) \\(Speed = \\frac{distance}{time}\\)\n\\(950 = \\frac{d}{4} \\implies d = 950 \\times 4\\)\n\\(d = 3800km\\)\n\\(d = \\frac{\\theta}{360} \\times 2\\pi r\\)\n\\(3800 = \\frac{\\theta}{360} \\times 2 \\times \\frac{22}{7} \\times 6400\\)\n\\(3800 = \\frac{\\theta \\times 281600}{2520}\\)\n\\(\\theta = \\frac{2520 \\times 3800}{281600}\\)\n\\(\\theta = 34.01°\\)\n\\(\\theta = 34°N\\) (to the nearest degree)\n(ii) Distance between P and Q, correct to four significant figures.\nLongitude difference = 65° - 15° = 50°\nUsing, \\(d = \\frac{\\theta}{360} \\times 2 \\pi r\\)\nwhere \\(r = R \\cos \\theta\\)\n\\(d = \\frac{\\theta}{360} \\times 2 \\pi R \\cos \\theta\\)\n= \\(\\frac{50}{360} \\times 2 \\times \\frac{22}{7} \\times 6400 \\cos 34.01°\\)\n= \\(4631.53 km\\)\n\\(\\approxeq 4632km\\) (to 4 significant figures).\n\n', '', 'Explanation\n(a)   \n(b)(i) \\(Speed = \\frac{distance}{time}\\)\n\\(950 = \\frac{d}{4} \\implies d = 950 \\times 4\\)\n\\(d = 3800km\\)\n\\(d = \\frac{\\theta}{360} \\times 2\\pi r\\)\n\\(3800 = \\frac{\\theta}{360} \\times 2 \\times \\frac{22}{7} \\times 6400\\)\n\\(3800 = \\frac{\\theta \\times 281600}{2520}\\)\n\\(\\theta = \\frac{2520 \\times 3800}{281600}\\)\n\\(\\theta = 34.01°\\)\n\\(\\theta = 34°N\\) (to the nearest degree)\n(ii) Distance between P and Q, correct to four significant figures.\nLongitude difference = 65° - 15° = 50°\nUsing, \\(d = \\frac{\\theta}{360} \\times 2 \\pi r\\)\nwhere \\(r = R \\cos \\theta\\)\n\\(d = \\frac{\\theta}{360} \\times 2 \\pi R \\cos \\theta\\)\n= \\(\\frac{50}{360} \\times 2 \\times \\frac{22}{7} \\times 6400 \\cos 34.01°\\)\n= \\(4631.53 km\\)\n\\(\\approxeq 4632km\\) (to 4 significant figures).\n\n', '(a)   ~(b)(i) \\(Speed = \\frac{distance}{time}\\)\n\\(950 = \\frac{d}{4} \\implies d = 950 \\times 4\\)\n\\(d = 3800km\\)\n\\(d = \\frac{\\theta}{360} \\times 2\\pi r\\)\n\\(3800 = \\frac{\\theta}{360} \\times 2 \\times \\frac{22}{7} \\times 6400\\)\n\\(3800 = \\frac{\\theta \\times 281600}{2520}\\)\n\\(\\theta = \\frac{2520 \\times 3800}{281600}\\)\n\\(\\theta = 34.01°\\)\n\\(\\theta = 34°N\\) (to the nearest degree)\n(ii) Distance between P and Q, correct to four significant figures.\nLongitude difference = 65° - 15° = 50°\nUsing, \\(d = \\frac{\\theta}{360} \\times 2 \\pi r\\)\nwhere \\(r = R \\cos \\theta\\)\n\\(d = \\frac{\\theta}{360} \\times 2 \\pi R \\cos \\theta\\)\n= \\(\\frac{50}{360} \\times 2 \\times \\frac{22}{7} \\times 6400 \\cos 34.01°\\)\n= \\(4631.53 km\\)\n\\(\\approxeq 4632km\\) (to 4 significant figures).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(394, '2020', '\n\nQuestion\n\nWhen one end of a ladder, LM, is placed against a vertical wall at a point 5 metres above the ground, the ladder makes an angle of 37° with the horizontal ground.\n(a) Represent this information in a diagram ;\n(b) Calculate, correct to 3 significant figures, the length of the ladder ;\n(c) If the foot of the ladder is pushed towards the wall by 2 metres, calculate,correct to the nearest degree, the angle which the ladder nows makes with the ground.\n\n', '', 'When one end of a ladder, LM, is placed against a vertical wall at a point 5 metres above the ground, the ladder makes an angle of 37° with the horizontal ground.\n(a) Represent this information in a diagram ;\n(b) Calculate, correct to 3 significant figures, the length of the ladder ;\n(c) If the foot of the ladder is pushed towards the wall by 2 metres, calculate,correct to the nearest degree, the angle which the ladder nows makes with the ground.', 'When one end of a ladder, LM, is placed against a vertical wall at a point 5 metres above the ground, the ladder makes an angle of 37° with the horizontal ground.\n(a) Represent this information in a diagram ;~(b) Calculate, correct to 3 significant figures, the length of the ladder ;\n(c) If the foot of the ladder is pushed towards the wall by 2 metres, calculate,correct to the nearest degree, the angle which the ladder nows makes with the ground.~(c) If the foot of the ladder is pushed towards the wall by 2 metres, calculate,correct to the nearest degree, the angle which the ladder nows makes with the ground.', NULL, '\n\nExplanation\n(a)\n(b) From the above diagram,\n\\(\\sin 37° = \\frac{5}{|LM|}\\)\n\\(|LM| = \\frac{5}{\\sin 37}\\)\n\\(|LM| = \\frac{5}{0.6018}\\)\n= \\(8.308 m\\)\n\\(\\approxeq 8.31 m\\)\n\\(\\cos 37 = \\frac{|AM|}{|LM|}\\)\n\\(|AM| = 8.308 \\times \\cos 37\\)\n\\(|AM| = 8.308 \\times 0.7986\\)\n\\(|AM| = 6.6348 m\\)\n(c) The new |AM| = 6.6348 m - 2 = 4.6348 m\n\\(\\cos y = \\frac{4.6348}{8.308}\\)\n\\(\\cos y = 0.5579\\)\n\\(y = \\cos^{-1} (0.5579)\\)\n\\(y = 56.089°\\)\n\\(y = 56°\\) (to the nearest degree).\n\n', '', 'Explanation\n(a)\n(b) From the above diagram,\n\\(\\sin 37° = \\frac{5}{|LM|}\\)\n\\(|LM| = \\frac{5}{\\sin 37}\\)\n\\(|LM| = \\frac{5}{0.6018}\\)\n= \\(8.308 m\\)\n\\(\\approxeq 8.31 m\\)\n\\(\\cos 37 = \\frac{|AM|}{|LM|}\\)\n\\(|AM| = 8.308 \\times \\cos 37\\)\n\\(|AM| = 8.308 \\times 0.7986\\)\n\\(|AM| = 6.6348 m\\)\n(c) The new |AM| = 6.6348 m - 2 = 4.6348 m\n\\(\\cos y = \\frac{4.6348}{8.308}\\)\n\\(\\cos y = 0.5579\\)\n\\(y = \\cos^{-1} (0.5579)\\)\n\\(y = 56.089°\\)\n\\(y = 56°\\) (to the nearest degree).\n\n', '~(b) From the above diagram,\n\\(\\sin 37° = \\frac{5}{|LM|}\\)\n\\(|LM| = \\frac{5}{\\sin 37}\\)\n\\(|LM| = \\frac{5}{0.6018}\\)\n= \\(8.308 m\\)\n\\(\\approxeq 8.31 m\\)\n\\(\\cos 37 = \\frac{|AM|}{|LM|}\\)\n\\(|AM| = 8.308 \\times \\cos 37\\)\n\\(|AM| = 8.308 \\times 0.7986\\)\n\\(|AM| = 6.6348 m\\)\n(c) The new |AM| = 6.6348 m - 2 = 4.6348 m\n\\(\\cos y = \\frac{4.6348}{8.308}\\)\n\\(\\cos y = 0.5579\\)\n\\(y = \\cos^{-1} (0.5579)\\)\n\\(y = 56.089°\\)\n\\(y = 56°\\) (to the nearest degree).\n~(c) The new |AM| = 6.6348 m - 2 = 4.6348 m\n\\(\\cos y = \\frac{4.6348}{8.308}\\)\n\\(\\cos y = 0.5579\\)\n\\(y = \\cos^{-1} (0.5579)\\)\n\\(y = 56.089°\\)\n\\(y = 56°\\) (to the nearest degree).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(395, '2020', '\n\nQuestion\n\n(a) Simplify : \\(\\frac{1\\frac{1}{4} + \\frac{7}{9}}{1\\frac{4}{9} - 2\\frac{2}{3} \\times \\frac{9}{64}}\\)\n(b) Given that \\(\\sin x = \\frac{2}{3}\\), evaluate, leaving your answer in surd form and without using tables or calculator, \\(\\tan x - \\cos x\\).\n\n', '', '(a) Simplify : \\(\\frac{1\\frac{1}{4} + \\frac{7}{9}}{1\\frac{4}{9} - 2\\frac{2}{3} \\times \\frac{9}{64}}\\)\n(b) Given that \\(\\sin x = \\frac{2}{3}\\), evaluate, leaving your answer in surd form and without using tables or calculator, \\(\\tan x - \\cos x\\).', '(a) Simplify : \\(\\frac{1\\frac{1}{4} + \\frac{7}{9}}{1\\frac{4}{9} - 2\\frac{2}{3} \\times \\frac{9}{64}}\\)~(b) Given that \\(\\sin x = \\frac{2}{3}\\), evaluate, leaving your answer in surd form and without using tables or calculator, \\(\\tan x - \\cos x\\).', NULL, '\n\nExplanation\n(a) \\(\\frac{1\\frac{1}{4} + \\frac{7}{9}}{1\\frac{4}{9} - 2\\frac{2}{3} \\times \\frac{9}{64}}\\)\n\\(1\\frac{1}{4} + \\frac{7}{9} = \\frac{5}{4} + \\frac{7}{9}\\)\n= \\(\\frac{45 + 28}{36}\\)\n= \\(\\frac{73}{36}\\)\n\\(1\\frac{4}{9} - 2\\frac{2}{3} \\times \\frac{9}{64} = \\frac{13}{9} - (\\frac{8}{3} \\times \\frac{9}{64})\\)\n= \\(\\frac{13}{9} - \\frac{3}{8}\\)\n= \\(\\frac{104 - 27}{72}\\)\n= \\(\\frac{77}{72}\\)\n\\(\\therefore \\frac{1\\frac{1}{4} + \\frac{7}{9}}{1\\frac{4}{9} - 2\\frac{2}{3} \\times \\frac{9}{64}} = \\frac{73}{36} \\div \\frac{77}{72}\\)\n= \\(\\frac{73}{36} \\times \\frac{72}{77}\\)\n= \\(\\frac{146}{77}\\)\n= \\(1\\frac{69}{77}\\)\n(b) \\(\\sin x = \\frac{2}{3}\\)\n In \\(\\Delta ABC, 3^{2} = 2^{2} + |BC|^{2}\\)\n\\(\\therefore |BC|^{2} = 9 - 4 = 5\\)\n\\(|BC| = \\sqrt{5}\\)\n\\(\\tan x = \\frac{2}{\\sqrt{5}} ; \\cos x = \\frac{\\sqrt{5}}{3}\\)\n\\(\\therefore \\tan x - \\cos x = \\frac{2}{\\sqrt{5}} - \\frac{\\sqrt{5}}{3}\\)\n= \\(\\frac{6 - 5}{3\\sqrt{5}}\\)\n= \\(\\frac{1}{3\\sqrt{5}}\\)\nRationalizing, we have\n\\(\\frac{1}{3\\sqrt{5}} \\times \\frac{\\sqrt{5}}{\\sqrt{5}} = \\frac{\\sqrt{5}}{15}\\)\n\n', '', 'Explanation\n(a) \\(\\frac{1\\frac{1}{4} + \\frac{7}{9}}{1\\frac{4}{9} - 2\\frac{2}{3} \\times \\frac{9}{64}}\\)\n\\(1\\frac{1}{4} + \\frac{7}{9} = \\frac{5}{4} + \\frac{7}{9}\\)\n= \\(\\frac{45 + 28}{36}\\)\n= \\(\\frac{73}{36}\\)\n\\(1\\frac{4}{9} - 2\\frac{2}{3} \\times \\frac{9}{64} = \\frac{13}{9} - (\\frac{8}{3} \\times \\frac{9}{64})\\)\n= \\(\\frac{13}{9} - \\frac{3}{8}\\)\n= \\(\\frac{104 - 27}{72}\\)\n= \\(\\frac{77}{72}\\)\n\\(\\therefore \\frac{1\\frac{1}{4} + \\frac{7}{9}}{1\\frac{4}{9} - 2\\frac{2}{3} \\times \\frac{9}{64}} = \\frac{73}{36} \\div \\frac{77}{72}\\)\n= \\(\\frac{73}{36} \\times \\frac{72}{77}\\)\n= \\(\\frac{146}{77}\\)\n= \\(1\\frac{69}{77}\\)\n(b) \\(\\sin x = \\frac{2}{3}\\)\n In \\(\\Delta ABC, 3^{2} = 2^{2} + |BC|^{2}\\)\n\\(\\therefore |BC|^{2} = 9 - 4 = 5\\)\n\\(|BC| = \\sqrt{5}\\)\n\\(\\tan x = \\frac{2}{\\sqrt{5}} ; \\cos x = \\frac{\\sqrt{5}}{3}\\)\n\\(\\therefore \\tan x - \\cos x = \\frac{2}{\\sqrt{5}} - \\frac{\\sqrt{5}}{3}\\)\n= \\(\\frac{6 - 5}{3\\sqrt{5}}\\)\n= \\(\\frac{1}{3\\sqrt{5}}\\)\nRationalizing, we have\n\\(\\frac{1}{3\\sqrt{5}} \\times \\frac{\\sqrt{5}}{\\sqrt{5}} = \\frac{\\sqrt{5}}{15}\\)\n\n', '(a) \\(\\frac{1\\frac{1}{4} + \\frac{7}{9}}{1\\frac{4}{9} - 2\\frac{2}{3} \\times \\frac{9}{64}}\\)\n\\(1\\frac{1}{4} + \\frac{7}{9} = \\frac{5}{4} + \\frac{7}{9}\\)\n= \\(\\frac{45 + 28}{36}\\)\n= \\(\\frac{73}{36}\\)\n\\(1\\frac{4}{9} - 2\\frac{2}{3} \\times \\frac{9}{64} = \\frac{13}{9} - (\\frac{8}{3} \\times \\frac{9}{64})\\)\n= \\(\\frac{13}{9} - \\frac{3}{8}\\)\n= \\(\\frac{104 - 27}{72}\\)\n= \\(\\frac{77}{72}\\)\n\\(\\therefore \\frac{1\\frac{1}{4} + \\frac{7}{9}}{1\\frac{4}{9} - 2\\frac{2}{3} \\times \\frac{9}{64}} = \\frac{73}{36} \\div \\frac{77}{72}\\)\n= \\(\\frac{73}{36} \\times \\frac{72}{77}\\)\n= \\(\\frac{146}{77}\\)\n= \\(1\\frac{69}{77}\\)~(b) \\(\\sin x = \\frac{2}{3}\\)\n In \\(\\Delta ABC, 3^{2} = 2^{2} + |BC|^{2}\\)\n\\(\\therefore |BC|^{2} = 9 - 4 = 5\\)\n\\(|BC| = \\sqrt{5}\\)\n\\(\\tan x = \\frac{2}{\\sqrt{5}} ; \\cos x = \\frac{\\sqrt{5}}{3}\\)\n\\(\\therefore \\tan x - \\cos x = \\frac{2}{\\sqrt{5}} - \\frac{\\sqrt{5}}{3}\\)\n= \\(\\frac{6 - 5}{3\\sqrt{5}}\\)\n= \\(\\frac{1}{3\\sqrt{5}}\\)\nRationalizing, we have\n\\(\\frac{1}{3\\sqrt{5}} \\times \\frac{\\sqrt{5}}{\\sqrt{5}} = \\frac{\\sqrt{5}}{15}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(396, '2020', '\n\nQuestion\n\nSonny is twice as old as Wale. Four years ago, he was four times as old as Wale. When will the sum of their ages be 66?\n\n', 'Sonny is twice as old as Wale. Four years ago, he was four times as old as Wale. When will the sum of their ages be 66?', '', '', NULL, '\n\nExplanation\nLet Wale\'s present age = x.\nThen Sonny\'s age = 2x\nFour years ago, Wale = x - 4\nSonny = 4(x - 4) = 2x - 4\n\\(4x - 16 = 2x - 4 \\implies 4x - 2x = -4 + 16\\)\n\\(2x = 12 \\implies x = 6\\)\nSonny = 2x = 2(6) = 12 years\nIn z years\' time, the sum of their ages = 66\nWale = 6 + z\nSonny = 12 + z\n\\(6 + z + 12 + z = 66 \\implies 18 + 2z = 66\\)\n\\(2z = 66 - 18 = 48 \\implies z = 24\\)\nTherefore, in 24 years time, the sum of their ages will be 66.\n\n', 'Let Wale\'s present age = x.\nThen Sonny\'s age = 2x\nFour years ago, Wale = x - 4\nSonny = 4(x - 4) = 2x - 4\n\\(4x - 16 = 2x - 4 \\implies 4x - 2x = -4 + 16\\)\n\\(2x = 12 \\implies x = 6\\)\nSonny = 2x = 2(6) = 12 years\nIn z years\' time, the sum of their ages = 66\nWale = 6 + z\nSonny = 12 + z\n\\(6 + z + 12 + z = 66 \\implies 18 + 2z = 66\\)\n\\(2z = 66 - 18 = 48 \\implies z = 24\\)\nTherefore, in 24 years time, the sum of their ages will be 66.\n\n', '', '', 'WAEC-PK-1', 'WAEC-MATH', 1),
(397, '2020', '\n\nQuestion\n\n(a)In the diagram, TU is tangent to the circle. < RVU = 100° and < URS = 36°. Calculate the value of angle STU.\n(b) In triangle XYZ, |XY| = 5 cm, |YZ| = 8 cm and |XZ| = 6 cm. P is a point on the side XY such that |XP| = 2 cm and the line through P, parallel to YZ meets XZ at Q. Calculate |QZ|.\n\n', '', '(a)In the diagram, TU is tangent to the circle. < RVU = 100° and < URS = 36°. Calculate the value of angle STU.\n(b) In triangle XYZ, |XY| = 5 cm, |YZ| = 8 cm and |XZ| = 6 cm. P is a point on the side XY such that |XP| = 2 cm and the line through P, parallel to YZ meets XZ at Q. Calculate |QZ|.', '(a)In the diagram, TU is tangent to the circle. < RVU = 100° and < URS = 36°. Calculate the value of angle STU.~(b) In triangle XYZ, |XY| = 5 cm, |YZ| = 8 cm and |XZ| = 6 cm. P is a point on the side XY such that |XP| = 2 cm and the line through P, parallel to YZ meets XZ at Q. Calculate |QZ|.', NULL, '\n\nExplanation\n(a) < UST = 100° (exterior angle of a cyclic quad)\n< SUT = 36° (angle in alternate segment)\n\\(\\therefore < UST + < SUT + < STU = 180°\\) (sum of angles in a triangle)\n\\(\\therefore 100° + 36° + < STU = 180°\\)\n\\(< STU = 180° - 136° = 44°\\)\n(b)\n\\(\\frac{y}{6} = \\frac{2}{5}\\)\n\\(\\therefore 5y = 2(6) = 12\\)\n\\(\\therefore y = \\frac{12}{5} = 2.4 cm\\)\n\\(\\therefore XQ + QZ = XZ\\)\n\\(QZ = XZ - XQ\\)\n= \\(6 cm - 2.4 cm = 3.6 cm\\)\n\\(\\therefore QZ = 3.6 cm\\).\n\n', '', 'Explanation\n(a) < UST = 100° (exterior angle of a cyclic quad)\n< SUT = 36° (angle in alternate segment)\n\\(\\therefore < UST + < SUT + < STU = 180°\\) (sum of angles in a triangle)\n\\(\\therefore 100° + 36° + < STU = 180°\\)\n\\(< STU = 180° - 136° = 44°\\)\n(b)\n\\(\\frac{y}{6} = \\frac{2}{5}\\)\n\\(\\therefore 5y = 2(6) = 12\\)\n\\(\\therefore y = \\frac{12}{5} = 2.4 cm\\)\n\\(\\therefore XQ + QZ = XZ\\)\n\\(QZ = XZ - XQ\\)\n= \\(6 cm - 2.4 cm = 3.6 cm\\)\n\\(\\therefore QZ = 3.6 cm\\).\n\n', '(a) < UST = 100° (exterior angle of a cyclic quad)\n< SUT = 36° (angle in alternate segment)\n\\(\\therefore < UST + < SUT + < STU = 180°\\) (sum of angles in a triangle)\n\\(\\therefore 100° + 36° + < STU = 180°\\)\n\\(< STU = 180° - 136° = 44°\\)~(b)\n\\(\\frac{y}{6} = \\frac{2}{5}\\)\n\\(\\therefore 5y = 2(6) = 12\\)\n\\(\\therefore y = \\frac{12}{5} = 2.4 cm\\)\n\\(\\therefore XQ + QZ = XZ\\)\n\\(QZ = XZ - XQ\\)\n= \\(6 cm - 2.4 cm = 3.6 cm\\)\n\\(\\therefore QZ = 3.6 cm\\).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(398, '2020', '\n\nQuestion\n\n(a) A box contains 40 identical discs which are either red or white. If the probability of picking a red disc is \\(\\frac{1}{4}\\); Calculate the number of (i) white discs ; (ii) red discs that should be added such that the probability of picking a red disc will be \\(\\frac{1}{3}\\).\n(b) A salesman bought some plates at N50.00 each. If he sold all of them for N600.00 and made a profit of 20% on the transaction, how many plates did he buy?\n\n', '', '(a) A box contains 40 identical discs which are either red or white. If the probability of picking a red disc is \\(\\frac{1}{4}\\); Calculate the number of (i) white discs ; (ii) red discs that should be added such that the probability of picking a red disc will be \\(\\frac{1}{3}\\).\n(b) A salesman bought some plates at N50.00 each. If he sold all of them for N600.00 and made a profit of 20% on the transaction, how many plates did he buy?', '(a) A box contains 40 identical discs which are either red or white. If the probability of picking a red disc is \\(\\frac{1}{4}\\); Calculate the number of (i) white discs ; (ii) red discs that should be added such that the probability of picking a red disc will be \\(\\frac{1}{3}\\).~(b) A salesman bought some plates at N50.00 each. If he sold all of them for N600.00 and made a profit of 20% on the transaction, how many plates did he buy?', NULL, '\n\nExplanation\n(a) (i) No of red discs = \\(\\frac{1}{4} \\times 40 = 10\\)\nTotal no of discs = 40\n\\(\\therefore\\) No of white balls = 40 - 10 = 30\n(ii) Let the number of red balls to be added = z\n\\(\\therefore \\frac{10 + z}{40 + z} = \\frac{1}{3}\\)\n\\(3(10 + z) = 40 + z\\)\n\\(30 + 3z = 40 + z \\implies 3z - z = 40 - 30\\)\n\\(2z = 10 \\implies z = 5\\)\nTherefore, adding 5 red discs, the probability of picking a red disc = \\(\\frac{1}{3}\\).\n(b) CP of a plate = N50.00\nLet the number of plates be x\n\\(\\therefore \\text{CP of x plates} = N50x\\)\n\\(\\text{SP of all plates} = N600\\)\n\\(\\frac{100 + 20}{100} \\times N50x = N600 \\implies N60x = N600\\)\n\\(x = \\frac{600}{60} = 10 plates\\)\n\n', '', 'Explanation\n(a) (i) No of red discs = \\(\\frac{1}{4} \\times 40 = 10\\)\nTotal no of discs = 40\n\\(\\therefore\\) No of white balls = 40 - 10 = 30\n(ii) Let the number of red balls to be added = z\n\\(\\therefore \\frac{10 + z}{40 + z} = \\frac{1}{3}\\)\n\\(3(10 + z) = 40 + z\\)\n\\(30 + 3z = 40 + z \\implies 3z - z = 40 - 30\\)\n\\(2z = 10 \\implies z = 5\\)\nTherefore, adding 5 red discs, the probability of picking a red disc = \\(\\frac{1}{3}\\).\n(b) CP of a plate = N50.00\nLet the number of plates be x\n\\(\\therefore \\text{CP of x plates} = N50x\\)\n\\(\\text{SP of all plates} = N600\\)\n\\(\\frac{100 + 20}{100} \\times N50x = N600 \\implies N60x = N600\\)\n\\(x = \\frac{600}{60} = 10 plates\\)\n\n', '(a) (i) No of red discs = \\(\\frac{1}{4} \\times 40 = 10\\)\nTotal no of discs = 40\n\\(\\therefore\\) No of white balls = 40 - 10 = 30\n(ii) Let the number of red balls to be added = z\n\\(\\therefore \\frac{10 + z}{40 + z} = \\frac{1}{3}\\)\n\\(3(10 + z) = 40 + z\\)\n\\(30 + 3z = 40 + z \\implies 3z - z = 40 - 30\\)\n\\(2z = 10 \\implies z = 5\\)\nTherefore, adding 5 red discs, the probability of picking a red disc = \\(\\frac{1}{3}\\).~(b) CP of a plate = N50.00\nLet the number of plates be x\n\\(\\therefore \\text{CP of x plates} = N50x\\)\n\\(\\text{SP of all plates} = N600\\)\n\\(\\frac{100 + 20}{100} \\times N50x = N600 \\implies N60x = N600\\)\n\\(x = \\frac{600}{60} = 10 plates\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(399, '2020', '\n\nQuestion\n\nIn the diagram, O is the centre of the circleand XY is a chord. If the radius is 5 cm and /XY/ = 6 cm, calculate, correct to 2 decimal places, the :\n(a) angle which XY subtends at the centre O ;\n(b) area of the shaded portion.\n\n\n', '', 'In the diagram, O is the centre of the circleand XY is a chord. If the radius is 5 cm and /XY/ = 6 cm, calculate, correct to 2 decimal places, the :\n(a) angle which XY subtends at the centre O ;\n(b) area of the shaded portion.\n', 'In the diagram, O is the centre of the circleand XY is a chord. If the radius is 5 cm and /XY/ = 6 cm, calculate, correct to 2 decimal places, the :\n(a) angle which XY subtends at the centre O ;~(b) area of the shaded portion.\n', NULL, '\n\nExplanation\n(a) Let the angle subtended at the centre by XY = \\(\\theta\\)\nLength of chord = \\(2r \\sin \\frac{\\theta}{2}\\)\n\\(\\implies 6 cm = 2 \\times 5 \\times \\frac{\\theta}{2}\\)\n\\(\\sin \\frac{\\theta}{2} = \\frac{6}{10} = 0.6\\)\n\\(\\frac{\\theta}{2} = \\sin^{-1} (0.6) = 36.87°\\)\n\\(\\theta = 2 \\times 36.87° = 73.74°\\)\n(b) Area of sector = \\(\\frac{\\theta}{360°} \\times \\pi r^{2}\\)\n= \\(\\frac{73.74}{360} \\times \\frac{22}{7} \\times 5^{2}\\)\n= \\(16.094 cm^{2}\\)\nArea of triangle = \\(\\sqrt{s(s - a)(s - b)(s - c)}\\)\nwhere \\(s = \\frac{a + b + c}{2}\\)\n\\( a = 5 cm ; b = 5 cm ; c = 6 cm \\therefore s = \\frac{5 + 5 + 6}{2} = \\frac{16}{2} = 8\\)\n\\(\\therefore Area = \\sqrt{8(8 - 5)(8 - 5)(8 - 6)}\\)\n= \\(\\sqrt{8(3)(3)(2)}\\)\n= \\(\\sqrt{144}\\)\n= \\(12 cm^{2}\\)\nArea of shaded portion = \\(16.094 cm^{2} - 12 cm^{2} = 4.094 cm^{2}\\)\n\n', '', 'Explanation\n(a) Let the angle subtended at the centre by XY = \\(\\theta\\)\nLength of chord = \\(2r \\sin \\frac{\\theta}{2}\\)\n\\(\\implies 6 cm = 2 \\times 5 \\times \\frac{\\theta}{2}\\)\n\\(\\sin \\frac{\\theta}{2} = \\frac{6}{10} = 0.6\\)\n\\(\\frac{\\theta}{2} = \\sin^{-1} (0.6) = 36.87°\\)\n\\(\\theta = 2 \\times 36.87° = 73.74°\\)\n(b) Area of sector = \\(\\frac{\\theta}{360°} \\times \\pi r^{2}\\)\n= \\(\\frac{73.74}{360} \\times \\frac{22}{7} \\times 5^{2}\\)\n= \\(16.094 cm^{2}\\)\nArea of triangle = \\(\\sqrt{s(s - a)(s - b)(s - c)}\\)\nwhere \\(s = \\frac{a + b + c}{2}\\)\n\\( a = 5 cm ; b = 5 cm ; c = 6 cm \\therefore s = \\frac{5 + 5 + 6}{2} = \\frac{16}{2} = 8\\)\n\\(\\therefore Area = \\sqrt{8(8 - 5)(8 - 5)(8 - 6)}\\)\n= \\(\\sqrt{8(3)(3)(2)}\\)\n= \\(\\sqrt{144}\\)\n= \\(12 cm^{2}\\)\nArea of shaded portion = \\(16.094 cm^{2} - 12 cm^{2} = 4.094 cm^{2}\\)\n\n', '(a) Let the angle subtended at the centre by XY = \\(\\theta\\)\nLength of chord = \\(2r \\sin \\frac{\\theta}{2}\\)\n\\(\\implies 6 cm = 2 \\times 5 \\times \\frac{\\theta}{2}\\)\n\\(\\sin \\frac{\\theta}{2} = \\frac{6}{10} = 0.6\\)\n\\(\\frac{\\theta}{2} = \\sin^{-1} (0.6) = 36.87°\\)\n\\(\\theta = 2 \\times 36.87° = 73.74°\\)~(b) Area of sector = \\(\\frac{\\theta}{360°} \\times \\pi r^{2}\\)\n= \\(\\frac{73.74}{360} \\times \\frac{22}{7} \\times 5^{2}\\)\n= \\(16.094 cm^{2}\\)\nArea of triangle = \\(\\sqrt{s(s - a)(s - b)(s - c)}\\)\nwhere \\(s = \\frac{a + b + c}{2}\\)\n\\( a = 5 cm ; b = 5 cm ; c = 6 cm \\therefore s = \\frac{5 + 5 + 6}{2} = \\frac{16}{2} = 8\\)\n\\(\\therefore Area = \\sqrt{8(8 - 5)(8 - 5)(8 - 6)}\\)\n= \\(\\sqrt{8(3)(3)(2)}\\)\n= \\(\\sqrt{144}\\)\n= \\(12 cm^{2}\\)\nArea of shaded portion = \\(16.094 cm^{2} - 12 cm^{2} = 4.094 cm^{2}\\)\n~c)}\\)\nwhere \\(s = \\frac{a + b + c}{2}\\)\n\\( a = 5 cm ; b = 5 cm ; c = 6 cm \\therefore s = \\frac{5 + 5 + 6}{2} = \\frac{16}{2} = 8\\)\n\\(\\therefore Area = \\sqrt{8(8 - 5)(8 - 5)(8 - 6)}\\)\n= \\(\\sqrt{8(3)(3)(2)}\\)\n= \\(\\sqrt{144}\\)\n= \\(12 cm^{2}\\)\nArea of shaded portion = \\(16.094 cm^{2} - 12 cm^{2} = 4.094 cm^{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
(400, '2020', '\n\nQuestion\n\n(a) A boy had M Dalasis (D). He spent D15 and shared the remainder equally with his sister. If the sister\'s share was equal to \\(\\frac{1}{3}\\) of M, find the value of M.\n(b) A number of tourists were interviewed on their choice of means of travel. Two- thirds said that they travelled by road, \\(\\frac{13}{30}\\) by air and \\(\\frac{4}{15}\\) by both air and road. If 20 tourists did not travel by either air or road ; (i) represent the information on a Venn diagram ; (ii) how many tourists (1) were interviewed ; (2) travelled by air only?\n\n', '', '(a) A boy had M Dalasis (D). He spent D15 and shared the remainder equally with his sister. If the sister\'s share was equal to \\(\\frac{1}{3}\\) of M, find the value of M.\n(b) A number of tourists were interviewed on their choice of means of travel. Two- thirds said that they travelled by road, \\(\\frac{13}{30}\\) by air and \\(\\frac{4}{15}\\) by both air and road. If 20 tourists did not travel by either air or road ; (i) represent the information on a Venn diagram ; (ii) how many tourists (1) were interviewed ; (2) travelled by air only?', '(a) A boy had M Dalasis (D). He spent D15 and shared the remainder equally with his sister. If the sister\'s share was equal to \\(\\frac{1}{3}\\) of M, find the value of M.~(b) A number of tourists were interviewed on their choice of means of travel. Two- thirds said that they travelled by road, \\(\\frac{13}{30}\\) by air and \\(\\frac{4}{15}\\) by both air and road. If 20 tourists did not travel by either air or road ; (i) represent the information on a Venn diagram ; (ii) how many tourists (1) were interviewed ; (2) travelled by air only?', NULL, '\n\nExplanation\n(a) Amount shared = D(M - 15)\nSister\'s share = \\(\\frac{D(M - 15)}{2}\\)\n\\(\\implies \\frac{D(M - 15)}{2} = \\frac{1}{3}M\\)\n\\(2M = 3(M - 15) \\implies 3M - 2M = 45\\)\n\\(M = 45.00\\)\n(b) (i)\n(ii) (1) Total no interviewed = \\(\\frac{2}{5}x + \\frac{4}{15}x + \\frac{1}{6}x + 20 = x\\)\n\\(\\therefore 12x + 8x + 5x + 600 = 30x\\) (Multiplying through with their LCM which is 30)\n\\(600 = 30x - 12x - 8x - 5x = 5x \\implies 600 = 5x\\)\n\\(x = \\frac{600}{5} = 120\\)\n(2) Those that travelled by air only = \\((\\frac{13}{30} \\times 120) - (\\frac{4}{15} \\times 120)\\)\n= \\(52 - 32 = 20\\)\n\n', '', 'Explanation\n(a) Amount shared = D(M - 15)\nSister\'s share = \\(\\frac{D(M - 15)}{2}\\)\n\\(\\implies \\frac{D(M - 15)}{2} = \\frac{1}{3}M\\)\n\\(2M = 3(M - 15) \\implies 3M - 2M = 45\\)\n\\(M = 45.00\\)\n(b) (i)\n(ii) (1) Total no interviewed = \\(\\frac{2}{5}x + \\frac{4}{15}x + \\frac{1}{6}x + 20 = x\\)\n\\(\\therefore 12x + 8x + 5x + 600 = 30x\\) (Multiplying through with their LCM which is 30)\n\\(600 = 30x - 12x - 8x - 5x = 5x \\implies 600 = 5x\\)\n\\(x = \\frac{600}{5} = 120\\)\n(2) Those that travelled by air only = \\((\\frac{13}{30} \\times 120) - (\\frac{4}{15} \\times 120)\\)\n= \\(52 - 32 = 20\\)\n\n', '(a) Amount shared = D(M - 15)\nSister\'s share = \\(\\frac{D(M - 15)}{2}\\)\n\\(\\implies \\frac{D(M - 15)}{2} = \\frac{1}{3}M\\)\n\\(2M = 3(M - 15) \\implies 3M - 2M = 45\\)\n\\(M = 45.00\\)~(b) (i)\n(ii) (1) Total no interviewed = \\(\\frac{2}{5}x + \\frac{4}{15}x + \\frac{1}{6}x + 20 = x\\)\n\\(\\therefore 12x + 8x + 5x + 600 = 30x\\) (Multiplying through with their LCM which is 30)\n\\(600 = 30x - 12x - 8x - 5x = 5x \\implies 600 = 5x\\)\n\\(x = \\frac{600}{5} = 120\\)\n(2) Those that travelled by air only = \\((\\frac{13}{30} \\times 120) - (\\frac{4}{15} \\times 120)\\)\n= \\(52 - 32 = 20\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(401, '2020', '\n\nQuestion\n\n(a) (i) Using a scale of 2 cm to 1 unit on both axes, on the same graph sheet, draw the graphs of \\(y - \\frac{3x}{4} = 3\\) and \\(y + 2x = 6\\).\n(ii) From your graph, find the coordinates of the point of intersection of the two graphs.\n(iii) Show, on the graph sheet, the region satisfied by the inequality \\(y - \\frac{3}{4}x \\geq 3\\).\n(b) Given that \\(x^{2} + bx + 18\\) is factorized as \\((x + 2)(x + c)\\). Find the values of c and b.\n\n', '', '(a) (i) Using a scale of 2 cm to 1 unit on both axes, on the same graph sheet, draw the graphs of \\(y - \\frac{3x}{4} = 3\\) and \\(y + 2x = 6\\).\n(ii) From your graph, find the coordinates of the point of intersection of the two graphs.\n(iii) Show, on the graph sheet, the region satisfied by the inequality \\(y - \\frac{3}{4}x \\geq 3\\).\n(b) Given that \\(x^{2} + bx + 18\\) is factorized as \\((x + 2)(x + c)\\). Find the values of c and b.', '(a) (i) Using a scale of 2 cm to 1 unit on both axes, on the same graph sheet, draw the graphs of \\(y - \\frac{3x}{4} = 3\\) and \\(y + 2x = 6\\).\n(ii) From your graph, find the coordinates of the point of intersection of the two graphs.\n(iii) Show, on the graph sheet, the region satisfied by the inequality \\(y - \\frac{3}{4}x \\geq 3\\).~(b) Given that \\(x^{2} + bx + 18\\) is factorized as \\((x + 2)(x + c)\\). Find the values of c and b.', NULL, '\n\nExplanation\n(a)(i)R is the required region.\n(iii) Coordinates of the point of intersection of the two graphs = (1.1, 3.8)\n(b) Assuming \\(y - \\frac{3x}{4} = 3\\)\nIf x = 0, \\(y - \\frac{3}{4}(0) = 3 \\implies y = 3\\)\nIf y = 0, \\(0 - \\frac{3}{4}x = 3 \\implies -3x = 12 ; x = - 4\\)\n\n\n\nx\n0\n-4\n\n\ny\n3\n0\n\n\n\nAssuming \\(y + 2x = 6\\)\nIf x = 0, \\(y + 2(0) = 6 \\implies y = 6\\)\nIf y = 0, \\(0 + 2x = 6 \\implies 2x = 6 ; x = 3\\)\n\n\n\nx\n0\n3\n\n\ny\n6\n0\n\n\n\n(b) \\((x + 2)(x + c) = x^{2} + 2x + cx + 2c\\)\n\\(x^{2} + (2 + c)x + 2c = x^{2} + bx + 18\\)\n\\(2c = 18 ; c = 9\\)\n\\(b = 2 + c = 2 + 9 = 11\\)\n(b, c) = (11, 9).\n\n', '', 'Explanation\n(a)(i)R is the required region.\n(iii) Coordinates of the point of intersection of the two graphs = (1.1, 3.8)\n(b) Assuming \\(y - \\frac{3x}{4} = 3\\)\nIf x = 0, \\(y - \\frac{3}{4}(0) = 3 \\implies y = 3\\)\nIf y = 0, \\(0 - \\frac{3}{4}x = 3 \\implies -3x = 12 ; x = - 4\\)\n\n\n\nx\n0\n-4\n\n\ny\n3\n0\n\n\n\nAssuming \\(y + 2x = 6\\)\nIf x = 0, \\(y + 2(0) = 6 \\implies y = 6\\)\nIf y = 0, \\(0 + 2x = 6 \\implies 2x = 6 ; x = 3\\)\n\n\n\nx\n0\n3\n\n\ny\n6\n0\n\n\n\n(b) \\((x + 2)(x + c) = x^{2} + 2x + cx + 2c\\)\n\\(x^{2} + (2 + c)x + 2c = x^{2} + bx + 18\\)\n\\(2c = 18 ; c = 9\\)\n\\(b = 2 + c = 2 + 9 = 11\\)\n(b, c) = (11, 9).\n\n', '(a)(i)R is the required region.\n(iii) Coordinates of the point of intersection of the two graphs = (1.1, 3.8)\n(b) Assuming \\(y - \\frac{3x}{4} = 3\\)\nIf x = 0, \\(y - \\frac{3}{4}(0) = 3 \\implies y = 3\\)\nIf y = 0, \\(0 - \\frac{3}{4}x = 3 \\implies -3x = 12 ; x = - 4\\)\n\n\n\nx\n0\n-4\n\n\ny\n3\n0\n\n\n\nAssuming \\(y + 2x = 6\\)\nIf x = 0, \\(y + 2(0) = 6 \\implies y = 6\\)\nIf y = 0, \\(0 + 2x = 6 \\implies 2x = 6 ; x = 3\\)\n\n\n\nx\n0\n3\n\n\ny\n6\n0\n\n\n~(b) Assuming \\(y - \\frac{3x}{4} = 3\\)\nIf x = 0, \\(y - \\frac{3}{4}(0) = 3 \\implies y = 3\\)\nIf y = 0, \\(0 - \\frac{3}{4}x = 3 \\implies -3x = 12 ; x = - 4\\)\n\n\n\nx\n0\n-4\n\n\ny\n3\n0\n\n\n\nAssuming \\(y + 2x = 6\\)\nIf x = 0, \\(y + 2(0) = 6 \\implies y = 6\\)\nIf y = 0, \\(0 + 2x = 6 \\implies 2x = 6 ; x = 3\\)\n\n\n\nx\n0\n3\n\n\ny\n6\n0\n\n\n\n(b) \\((x + 2)(x + c) = x^{2} + 2x + cx + 2c\\)\n\\(x^{2} + (2 + c)x + 2c = x^{2} + bx + 18\\)\n\\(2c = 18 ; c = 9\\)\n\\(b = 2 + c = 2 + 9 = 11\\)\n(b, c) = (11, 9).\n~c) = x^{2} + 2x + cx + 2c\\)\n\\(x^{2} + (2 + c)x + 2c = x^{2} + bx + 18\\)\n\\(2c = 18 ; c = 9\\)\n\\(b = 2 + c = 2 + 9 = 11\\)\n(b, c) = (11, 9).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(402, '2020', '\n\nQuestion\n\nA point H is 20 m away from the foot of a tower on the same horizontal ground. From the point H, the angle of elevation of the point P on the tower and the top (T) of the tower are 30° and 50° respectively. Calculate, correct to 3 significant figures :\n(a) /PT/; (b) the distance between H and the top of the tower\n(c) The position of H if theangle of depression of H from the top of the tower is to be 40°.\n\n', '', 'A point H is 20 m away from the foot of a tower on the same horizontal ground. From the point H, the angle of elevation of the point P on the tower and the top (T) of the tower are 30° and 50° respectively. Calculate, correct to 3 significant figures :\n(a) /PT/; (b) the distance between H and the top of the tower\n(c) The position of H if theangle of depression of H from the top of the tower is to be 40°.', '~(b) the distance between H and the top of the tower\n(c) The position of H if theangle of depression of H from the top of the tower is to be 40°.~(c) The position of H if theangle of depression of H from the top of the tower is to be 40°.', NULL, '\n\nExplanation\n(a) \\(\\tan 50° = \\frac{|TF|}{20} \\implies |TF| = 20 \\tan 50\\)\n= \\(20 \\times 1.1918\\)\n= \\(23.836 m\\)\n\\(\\tan 30 = \\frac{|PF|}{20} \\implies |PF| = 20 \\tan 30\\)\n= \\(20 \\times 0.5774\\)\n= \\(11.548 m\\)\n\\(\\therefore |PT| = 23.836 m - 11.548m\\)\n= \\(12.288 m \\approxeq 12.3 m\\)\n(b) \\(|TH| = \\frac{20}{\\cos 50}\\)\n= \\(\\frac{20}{0.6428}\\)\n= \\(31.11 m\\)\n(c) \\(|HF| = \\frac{20 \\tan 50°}{\\tan 40°}\\)\n= \\(\\frac{20 \\times 1.1918}{0.8391}\\)\n= \\(28.41 m\\)\n\n', '', 'Explanation\n(a) \\(\\tan 50° = \\frac{|TF|}{20} \\implies |TF| = 20 \\tan 50\\)\n= \\(20 \\times 1.1918\\)\n= \\(23.836 m\\)\n\\(\\tan 30 = \\frac{|PF|}{20} \\implies |PF| = 20 \\tan 30\\)\n= \\(20 \\times 0.5774\\)\n= \\(11.548 m\\)\n\\(\\therefore |PT| = 23.836 m - 11.548m\\)\n= \\(12.288 m \\approxeq 12.3 m\\)\n(b) \\(|TH| = \\frac{20}{\\cos 50}\\)\n= \\(\\frac{20}{0.6428}\\)\n= \\(31.11 m\\)\n(c) \\(|HF| = \\frac{20 \\tan 50°}{\\tan 40°}\\)\n= \\(\\frac{20 \\times 1.1918}{0.8391}\\)\n= \\(28.41 m\\)\n\n', '(a) \\(\\tan 50° = \\frac{|TF|}{20} \\implies |TF| = 20 \\tan 50\\)\n= \\(20 \\times 1.1918\\)\n= \\(23.836 m\\)\n\\(\\tan 30 = \\frac{|PF|}{20} \\implies |PF| = 20 \\tan 30\\)\n= \\(20 \\times 0.5774\\)\n= \\(11.548 m\\)\n\\(\\therefore |PT| = 23.836 m - 11.548m\\)\n= \\(12.288 m \\approxeq 12.3 m\\)~(b) \\(|TH| = \\frac{20}{\\cos 50}\\)\n= \\(\\frac{20}{0.6428}\\)\n= \\(31.11 m\\)\n(c) \\(|HF| = \\frac{20 \\tan 50°}{\\tan 40°}\\)\n= \\(\\frac{20 \\times 1.1918}{0.8391}\\)\n= \\(28.41 m\\)\n~(c) \\(|HF| = \\frac{20 \\tan 50°}{\\tan 40°}\\)\n= \\(\\frac{20 \\times 1.1918}{0.8391}\\)\n= \\(28.41 m\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(403, '2020', '\n\nQuestion\n\nThree towns X, Y and Z are such that Y is 20 km from X and 22 km from Z. Town X is 18 km from Z. A health centre is to be built by the government to serve the three towns. The centre is to be located such that patients from X and Y travel equal distance to access the health centre while patients from Z will travel exactly 10 km to reach the Health centre.\n(a) Using a scale of 1 cm to 2 km, find the construction, using a pair of compasses and ruler only, the possible positions the Health centre can be located.\n(b) In how many possible locations can the Health centre be built?\n(c) Measure and record the distances of the location from town X.\n(d) Which of these locations would be convenient for all three towns?\n\n', '', 'Three towns X, Y and Z are such that Y is 20 km from X and 22 km from Z. Town X is 18 km from Z. A health centre is to be built by the government to serve the three towns. The centre is to be located such that patients from X and Y travel equal distance to access the health centre while patients from Z will travel exactly 10 km to reach the Health centre.\n(a) Using a scale of 1 cm to 2 km, find the construction, using a pair of compasses and ruler only, the possible positions the Health centre can be located.\n(b) In how many possible locations can the Health centre be built?\n(c) Measure and record the distances of the location from town X.\n(d) Which of these locations would be convenient for all three towns?', 'Three towns X, Y and Z are such that Y is 20 km from X and 22 km from Z. Town X is 18 km from Z. A health centre is to be built by the government to serve the three towns. The centre is to be located such that patients from X and Y travel equal distance to access the health centre while patients from Z will travel exactly 10 km to reach the Health centre.\n(a) Using a scale of 1 cm to 2 km, find the construction, using a pair of compasses and ruler only, the possible positions the Health centre can be located.~(b) In how many possible locations can the Health centre be built?\n(c) Measure and record the distances of the location from town X.\n(d) Which of these locations would be convenient for all three towns?~(c) Measure and record the distances of the location from town X.\n(d) Which of these locations would be convenient for all three towns?~(d) Which of these locations would be convenient for all three towns?', NULL, '\n\nExplanation\n(b) Two possible ways\n(c) The distances of the locations from town X = 27.6 km and 12.4km.\n(d) 12.4 km\n\n', '', 'Explanation\n(b) Two possible ways\n(c) The distances of the locations from town X = 27.6 km and 12.4km.\n(d) 12.4 km\n\n', '~(b) Two possible ways\n(c) The distances of the locations from town X = 27.6 km and 12.4km.\n(d) 12.4 km\n~(c) The distances of the locations from town X = 27.6 km and 12.4km.\n(d) 12.4 km\n~(d) 12.4 km\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(404, '2020', '\n\nQuestion\n\n\n\n\nClass\nInterval\n\nFrequency\n\n\n60 - 64\n2\n\n\n65 - 69\n3\n\n\n70 - 74\n6\n\n\n75 - 79\n11\n\n\n80 - 84\n8\n\n\n85 - 89\n7\n\n\n90 - 94\n2\n\n\n95 - 99\n1\n\n\n\nThe table shows the distribution of marks scored by students in an examination. Calculate, correct to 2 decimal places, the\n(a) mean ; (b) standard deviation of the distribution.\n\n\n\n\n\n\nClass\nInterval\n\nFrequency\n\n\n60 - 64\n2\n\n\n65 - 69\n3\n\n\n70 - 74\n6\n\n\n75 - 79\n11\n\n\n80 - 84\n8\n\n\n85 - 89\n7\n\n\n90 - 94\n2\n\n\n95 - 99\n1\n\n\n', '', 'Class\nInterval\n\nFrequency\n\n\n60 - 64\n2\n\n\n65 - 69\n3\n\n\n70 - 74\n6\n\n\n75 - 79\n11\n\n\n80 - 84\n8\n\n\n85 - 89\n7\n\n\n90 - 94\n2\n\n\n95 - 99\n1\n\n\n\nThe table shows the distribution of marks scored by students in an examination. Calculate, correct to 2 decimal places, the\n(a) mean ; (b) standard deviation of the distribution.\n\n\n\n\n\n\nClass\nInterval\n\nFrequency\n\n\n60 - 64\n2\n\n\n65 - 69\n3\n\n\n70 - 74\n6\n\n\n75 - 79\n11\n\n\n80 - 84\n8\n\n\n85 - 89\n7\n\n\n90 - 94\n2\n\n\n95 - 99\n1\n', '~(b) standard deviation of the distribution.\n\n\n\n\n\n\nClass\nInterval\n\nFrequency\n\n\n60 - 64\n2\n\n\n65 - 69\n3\n\n\n70 - 74\n6\n\n\n75 - 79\n11\n\n\n80 - 84\n8\n\n\n85 - 89\n7\n\n\n90 - 94\n2\n\n\n95 - 99\n1\n', NULL, '\n\nExplanation\nClass\nInterval\n\n\nFreq\n(f)\n\n\nMid-point\n(x)\n\n\\(fx\\)\n\\(x^{2}\\)\n\\(fx^{2}\\)\n\n\n60 - 64\n2\n62\n124\n3844\n7688\n\n\n65 - 69\n3\n67\n201\n4489\n13487\n\n\n70 - 74\n6\n72\n432\n5184\n31104\n\n\n75 - 79\n11\n77\n847\n5929\n65219\n\n\n80 - 84\n8\n82\n656\n6724\n53792\n\n\n85 - 89\n7\n87\n609\n7569\n52983\n\n\n90 - 94\n2\n92\n184\n8464\n16928\n\n\n95 - 99\n1\n97\n97\n9409\n9409\n\n\n\\(\\sum\\)\n40\n\n3150\n\n250590\n\n\n\n(a) Mean \\(\\bar{x} = \\frac{\\sum fx}{\\sum f}\\)\n= \\(\\frac{3150}{40}\\)\n= 78.75\n(b) \\(SD = \\sqrt{\\frac{\\sum fx^{2}}{\\sum f} - (\\frac{\\sum fx}{\\sum f})^{2}}\\)\n= \\(\\sqrt{\\frac{250590}{40} - (\\frac{3150}{40})^{2}}\\)\n= \\(\\sqrt{6264.75 - (78.75)^{2}}\\)\n= \\(\\sqrt{6264.75 - 6201.5625}\\)\n= \\(\\sqrt{63.1875}\\)\n= 7.95.\n\n\n', '', 'Explanation\nClass\nInterval\n\n\nFreq\n(f)\n\n\nMid-point\n(x)\n\n\\(fx\\)\n\\(x^{2}\\)\n\\(fx^{2}\\)\n\n\n60 - 64\n2\n62\n124\n3844\n7688\n\n\n65 - 69\n3\n67\n201\n4489\n13487\n\n\n70 - 74\n6\n72\n432\n5184\n31104\n\n\n75 - 79\n11\n77\n847\n5929\n65219\n\n\n80 - 84\n8\n82\n656\n6724\n53792\n\n\n85 - 89\n7\n87\n609\n7569\n52983\n\n\n90 - 94\n2\n92\n184\n8464\n16928\n\n\n95 - 99\n1\n97\n97\n9409\n9409\n\n\n\\(\\sum\\)\n40\n\n3150\n\n250590\n\n\n\n(a) Mean \\(\\bar{x} = \\frac{\\sum fx}{\\sum f}\\)\n= \\(\\frac{3150}{40}\\)\n= 78.75\n(b) \\(SD = \\sqrt{\\frac{\\sum fx^{2}}{\\sum f} - (\\frac{\\sum fx}{\\sum f})^{2}}\\)\n= \\(\\sqrt{\\frac{250590}{40} - (\\frac{3150}{40})^{2}}\\)\n= \\(\\sqrt{6264.75 - (78.75)^{2}}\\)\n= \\(\\sqrt{6264.75 - 6201.5625}\\)\n= \\(\\sqrt{63.1875}\\)\n= 7.95.\n\n\n', '(a) Mean \\(\\bar{x} = \\frac{\\sum fx}{\\sum f}\\)\n= \\(\\frac{3150}{40}\\)\n= 78.75~(b) \\(SD = \\sqrt{\\frac{\\sum fx^{2}}{\\sum f} - (\\frac{\\sum fx}{\\sum f})^{2}}\\)\n= \\(\\sqrt{\\frac{250590}{40} - (\\frac{3150}{40})^{2}}\\)\n= \\(\\sqrt{6264.75 - (78.75)^{2}}\\)\n= \\(\\sqrt{6264.75 - 6201.5625}\\)\n= \\(\\sqrt{63.1875}\\)\n= 7.95.\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(405, '2020', '\n\nQuestion\n\n(a)\nIn the diagram, ABCD is a rectangular garden (3n - 1)m long and (2n + 1)m wide. A wire mesh 135m long is used to mark its boundary and to divide it into 8 equal plots. Find the value of n.\n(b) A cylinder with base radius 14 cm has the same volume as a cube of side 22 cm. Calculate the ratio of the total surface area of the cylinder to that of the cube. [Take \\(\\pi = \\frac{22}{7}\\)]\n\n', '', '(a)\nIn the diagram, ABCD is a rectangular garden (3n - 1)m long and (2n + 1)m wide. A wire mesh 135m long is used to mark its boundary and to divide it into 8 equal plots. Find the value of n.\n(b) A cylinder with base radius 14 cm has the same volume as a cube of side 22 cm. Calculate the ratio of the total surface area of the cylinder to that of the cube. [Take \\(\\pi = \\frac{22}{7}\\)]', '(a)\nIn the diagram, ABCD is a rectangular garden (3n - 1)m long and (2n + 1)m wide. A wire mesh 135m long is used to mark its boundary and to divide it into 8 equal plots. Find the value of n.~(b) A cylinder with base radius 14 cm has the same volume as a cube of side 22 cm. Calculate the ratio of the total surface area of the cylinder to that of the cube. [Take \\(\\pi = \\frac{22}{7}\\)]', NULL, '\n\nExplanation\n(a) \\(3(3n - 1) + 5(2n + 1) = 135\\)\n\\(9n - 3 + 10n + 5 = 135\\)\n\\(19n + 2 = 135\\)\n\\(19n = 135 - 2 = 133\\)\n\\(n = \\frac{133}{19} = 7\\)\n(b) Radius of a cylinder = 14 cm\nside of a cube = 22 cm\n\\(\\therefore \\frac{22}{7} \\times 14 \\times 14 \\times h = 22 \\times 22 \\times 22\\)\n\\(44 \\times 14 \\times h = 22 \\times 22 \\times 22\\)\n\\(h = \\frac{22 \\times 22 \\times 22}{44 \\times 14}\\)\n= \\(\\frac{121}{7} cm\\)\nTSA of a cylinder = \\(2\\pi r (r + h) = 2 \\times \\frac{22}{7} \\times 14 (14 + \\frac{121}{7})\\)\n= \\(88(\\frac{98 + 121}{7}) = \\frac{88 \\times 219}{7}\\)\n= \\(\\frac{19272}{7} cm^{2}\\)\nTSA of a cube = \\(6s^{2}\\)\n= \\(6 \\times 22 \\times 22\\)\n= \\(2904 cm^{2}\\)\nRatio = \\(\\frac{19272}{7} : 2904\\)\n= \\(73 : 77\\)\n\n', '', 'Explanation\n(a) \\(3(3n - 1) + 5(2n + 1) = 135\\)\n\\(9n - 3 + 10n + 5 = 135\\)\n\\(19n + 2 = 135\\)\n\\(19n = 135 - 2 = 133\\)\n\\(n = \\frac{133}{19} = 7\\)\n(b) Radius of a cylinder = 14 cm\nside of a cube = 22 cm\n\\(\\therefore \\frac{22}{7} \\times 14 \\times 14 \\times h = 22 \\times 22 \\times 22\\)\n\\(44 \\times 14 \\times h = 22 \\times 22 \\times 22\\)\n\\(h = \\frac{22 \\times 22 \\times 22}{44 \\times 14}\\)\n= \\(\\frac{121}{7} cm\\)\nTSA of a cylinder = \\(2\\pi r (r + h) = 2 \\times \\frac{22}{7} \\times 14 (14 + \\frac{121}{7})\\)\n= \\(88(\\frac{98 + 121}{7}) = \\frac{88 \\times 219}{7}\\)\n= \\(\\frac{19272}{7} cm^{2}\\)\nTSA of a cube = \\(6s^{2}\\)\n= \\(6 \\times 22 \\times 22\\)\n= \\(2904 cm^{2}\\)\nRatio = \\(\\frac{19272}{7} : 2904\\)\n= \\(73 : 77\\)\n\n', '(a) \\(3(3n - 1) + 5(2n + 1) = 135\\)\n\\(9n - 3 + 10n + 5 = 135\\)\n\\(19n + 2 = 135\\)\n\\(19n = 135 - 2 = 133\\)\n\\(n = \\frac{133}{19} = 7\\)~(b) Radius of a cylinder = 14 cm\nside of a cube = 22 cm\n\\(\\therefore \\frac{22}{7} \\times 14 \\times 14 \\times h = 22 \\times 22 \\times 22\\)\n\\(44 \\times 14 \\times h = 22 \\times 22 \\times 22\\)\n\\(h = \\frac{22 \\times 22 \\times 22}{44 \\times 14}\\)\n= \\(\\frac{121}{7} cm\\)\nTSA of a cylinder = \\(2\\pi r (r + h) = 2 \\times \\frac{22}{7} \\times 14 (14 + \\frac{121}{7})\\)\n= \\(88(\\frac{98 + 121}{7}) = \\frac{88 \\times 219}{7}\\)\n= \\(\\frac{19272}{7} cm^{2}\\)\nTSA of a cube = \\(6s^{2}\\)\n= \\(6 \\times 22 \\times 22\\)\n= \\(2904 cm^{2}\\)\nRatio = \\(\\frac{19272}{7} : 2904\\)\n= \\(73 : 77\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(406, '2020', '\n\nQuestion\n\n(a) Copy and complete the table of values for \\(y = 1 - 4\\cos x\\).\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n240°\n270°\n\n300°\n\n\n\ny\n-3.0\n\n\n1.0\n\n\n\n4.5\n\n\n-1.0\n\n\n\n(b) Using a scale of 2cm to 30° on the x- axis and 2cm to 1 unit on the y- axis, draw the graph of \\(y = 1 - 4\\cos x\\) for \\(0° \\leq x \\leq 360°\\).\n(c) Use the graph to : (i) solve the equation \\(1 - 4\\cos x = 0\\) ; (ii) find the value of y when x = 105° ; (iii) find xwhen y = 1.5.\n\n\n(a) Copy and complete the table of values for \\(y = 1 - 4\\cos x\\).\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n240°\n270°\n\n300°\n\n\n\ny\n-3.0\n\n\n1.0\n\n\n\n4.5\n\n\n-1.0\n\n\n', '', '(a) Copy and complete the table of values for \\(y = 1 - 4\\cos x\\).\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n240°\n270°\n\n300°\n\n\n\ny\n-3.0\n\n\n1.0\n\n\n\n4.5\n\n\n-1.0\n\n\n\n(b) Using a scale of 2cm to 30° on the x- axis and 2cm to 1 unit on the y- axis, draw the graph of \\(y = 1 - 4\\cos x\\) for \\(0° \\leq x \\leq 360°\\).\n(c) Use the graph to : (i) solve the equation \\(1 - 4\\cos x = 0\\) ; (ii) find the value of y when x = 105° ; (iii) find xwhen y = 1.5.\n\n\n(a) Copy and complete the table of values for \\(y = 1 - 4\\cos x\\).\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n240°\n270°\n\n300°\n\n\n\ny\n-3.0\n\n\n1.0\n\n\n\n4.5\n\n\n-1.0\n', '(a) Copy and complete the table of values for \\(y = 1 - 4\\cos x\\).\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n240°\n270°\n\n300°\n\n\n\ny\n-3.0\n\n\n1.0\n\n\n\n4.5\n\n\n-1.0\n\n\n~(b) Using a scale of 2cm to 30° on the x- axis and 2cm to 1 unit on the y- axis, draw the graph of \\(y = 1 - 4\\cos x\\) for \\(0° \\leq x \\leq 360°\\).\n(c) Use the graph to : (i) solve the equation \\(1 - 4\\cos x = 0\\) ; (ii) find the value of y when x = 105° ; (iii) find xwhen y = 1.5.\n\n\n(a) Copy and complete the table of values for \\(y = 1 - 4\\cos x\\).\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n240°\n270°\n\n300°\n\n\n\ny\n-3.0\n\n\n1.0\n\n\n\n4.5\n\n\n-1.0\n~(c) Use the graph to : (i) solve the equation \\(1 - 4\\cos x = 0\\) ; (ii) find the value of y when x = 105° ; (iii) find xwhen y = 1.5.\n\n\n(a) Copy and complete the table of values for \\(y = 1 - 4\\cos x\\).\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n240°\n270°\n\n300°\n\n\n\ny\n-3.0\n\n\n1.0\n\n\n\n4.5\n\n\n-1.0\n', NULL, '\n\nExplanation\n(a)\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n240°\n270°\n\n300°\n\n\n\ny\n-3.0\n-2.5\n-1.0\n1.0\n3.0\n4.5\n5.0\n4.5\n3.0\n1.0\n-1.0\n\n\n\n(b)\n(c)(i) x = 75° or x = 285°\n(ii) When x = 105°, y = 2.\n(iii) When y = 1.5, x = 96° or x = 264°.\n\n', '', 'Explanation\n(a)\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n240°\n270°\n\n300°\n\n\n\ny\n-3.0\n-2.5\n-1.0\n1.0\n3.0\n4.5\n5.0\n4.5\n3.0\n1.0\n-1.0\n\n\n\n(b)\n(c)(i) x = 75° or x = 285°\n(ii) When x = 105°, y = 2.\n(iii) When y = 1.5, x = 96° or x = 264°.\n\n', '(a)\n\n\n\nx\n0°\n30°\n60°\n90°\n120°\n150°\n180°\n210°\n240°\n270°\n\n300°\n\n\n\ny\n-3.0\n-2.5\n-1.0\n1.0\n3.0\n4.5\n5.0\n4.5\n3.0\n1.0\n-1.0\n\n\n~(b)\n(c)(i) x = 75° or x = 285°\n(ii) When x = 105°, y = 2.\n(iii) When y = 1.5, x = 96° or x = 264°.\n~(c)(i) x = 75° or x = 285°\n(ii) When x = 105°, y = 2.\n(iii) When y = 1.5, x = 96° or x = 264°.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(407, '2020', '\n\nQuestion\n\n(a)In the diagram, /PQ/ = 6 cm, /QR/ = 13 cm, /RS/ = 5 cm and < RSQ is a right- angled triangle. Calculate, correct to one decimal place, /PS/.\n(b)The diagram show a wooden structure in the form of a cone mounted on a hemispherical base. The vertical height of the cone is 24 cm and the base radius 7 cm. Calculate, correct to 3 significant figures, the surface area of the structure. [Take \\(\\pi = \\frac{22}{7}\\)].\n\n', '', '(a)In the diagram, /PQ/ = 6 cm, /QR/ = 13 cm, /RS/ = 5 cm and < RSQ is a right- angled triangle. Calculate, correct to one decimal place, /PS/.\n(b)The diagram show a wooden structure in the form of a cone mounted on a hemispherical base. The vertical height of the cone is 24 cm and the base radius 7 cm. Calculate, correct to 3 significant figures, the surface area of the structure. [Take \\(\\pi = \\frac{22}{7}\\)].', '(a)In the diagram, /PQ/ = 6 cm, /QR/ = 13 cm, /RS/ = 5 cm and < RSQ is a right- angled triangle. Calculate, correct to one decimal place, /PS/.~(b)The diagram show a wooden structure in the form of a cone mounted on a hemispherical base. The vertical height of the cone is 24 cm and the base radius 7 cm. Calculate, correct to 3 significant figures, the surface area of the structure. [Take \\(\\pi = \\frac{22}{7}\\)].', NULL, '\n\nExplanation\n(a)\n\\(\\cos R = \\frac{5}{13} = 0.3846\\)\n\\(R = \\cos^{-1} (0.3846) = 67.38°\\)\n\\(/PS/^{2} = 5^{2} + 19^{2} - 2(5)(19)\\cos 67.38\\)\n= \\(25 + 361 - 190(0.3872)\\)\n= \\(386 - 73.568\\)\n\\(/PS/^{2} = 312.432\\)\n\\(/PS/ = \\sqrt{312.432}\\)\n= \\(17.68 \\approxeq 17.7 cm\\)\n(b) Slant height of a cone = \\(l = \\sqrt{7^{2} + 24^{2}}\\)\n\\(\\sqrt{49 + 576} = \\sqrt{625}\\)\n= \\(25 cm\\)\nC.S.A of a cone = \\(\\pi rl\\)\n= \\(\\frac{22}{7} \\times 7 \\times 25\\)\n= \\(550 cm^{2}\\)\nArea of hemisphere = \\(\\frac{1}{2} \\times 4\\pi r^{2} = 2 \\pi r^{2}\\)\n\\(2 \\times \\frac{22}{7} \\times 7 \\times 7 = 308 cm^{2}\\)\nTSA of the structure = \\(550 cm^{2} + 308 cm^{2} = 858 cm^{2}\\)\n\n', '', 'Explanation\n(a)\n\\(\\cos R = \\frac{5}{13} = 0.3846\\)\n\\(R = \\cos^{-1} (0.3846) = 67.38°\\)\n\\(/PS/^{2} = 5^{2} + 19^{2} - 2(5)(19)\\cos 67.38\\)\n= \\(25 + 361 - 190(0.3872)\\)\n= \\(386 - 73.568\\)\n\\(/PS/^{2} = 312.432\\)\n\\(/PS/ = \\sqrt{312.432}\\)\n= \\(17.68 \\approxeq 17.7 cm\\)\n(b) Slant height of a cone = \\(l = \\sqrt{7^{2} + 24^{2}}\\)\n\\(\\sqrt{49 + 576} = \\sqrt{625}\\)\n= \\(25 cm\\)\nC.S.A of a cone = \\(\\pi rl\\)\n= \\(\\frac{22}{7} \\times 7 \\times 25\\)\n= \\(550 cm^{2}\\)\nArea of hemisphere = \\(\\frac{1}{2} \\times 4\\pi r^{2} = 2 \\pi r^{2}\\)\n\\(2 \\times \\frac{22}{7} \\times 7 \\times 7 = 308 cm^{2}\\)\nTSA of the structure = \\(550 cm^{2} + 308 cm^{2} = 858 cm^{2}\\)\n\n', '(a)\n\\(\\cos R = \\frac{5}{13} = 0.3846\\)\n\\(R = \\cos^{-1} (0.3846) = 67.38°\\)\n\\(/PS/^{2} = 5^{2} + 19^{2} - 2(5)(19)\\cos 67.38\\)\n= \\(25 + 361 - 190(0.3872)\\)\n= \\(386 - 73.568\\)\n\\(/PS/^{2} = 312.432\\)\n\\(/PS/ = \\sqrt{312.432}\\)\n= \\(17.68 \\approxeq 17.7 cm\\)~(b) Slant height of a cone = \\(l = \\sqrt{7^{2} + 24^{2}}\\)\n\\(\\sqrt{49 + 576} = \\sqrt{625}\\)\n= \\(25 cm\\)\nC.S.A of a cone = \\(\\pi rl\\)\n= \\(\\frac{22}{7} \\times 7 \\times 25\\)\n= \\(550 cm^{2}\\)\nArea of hemisphere = \\(\\frac{1}{2} \\times 4\\pi r^{2} = 2 \\pi r^{2}\\)\n\\(2 \\times \\frac{22}{7} \\times 7 \\times 7 = 308 cm^{2}\\)\nTSA of the structure = \\(550 cm^{2} + 308 cm^{2} = 858 cm^{2}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(408, '2020', '\n\nQuestion\n\n(a) Simplify : \\(\\frac{\\frac{1}{2} of \\frac{1}{4} \\div \\frac{1}{3}}{\\frac{1}{6} - \\frac{3}{4} + \\frac{1}{2}}\\).\n(b) Given that \\(\\sqrt{x} = 10^{\\bar{1}.6741}\\), without using calculators, find the value of x.\n\n', '', '(a) Simplify : \\(\\frac{\\frac{1}{2} of \\frac{1}{4} \\div \\frac{1}{3}}{\\frac{1}{6} - \\frac{3}{4} + \\frac{1}{2}}\\).\n(b) Given that \\(\\sqrt{x} = 10^{\\bar{1}.6741}\\), without using calculators, find the value of x.', '(a) Simplify : \\(\\frac{\\frac{1}{2} of \\frac{1}{4} \\div \\frac{1}{3}}{\\frac{1}{6} - \\frac{3}{4} + \\frac{1}{2}}\\).~(b) Given that \\(\\sqrt{x} = 10^{\\bar{1}.6741}\\), without using calculators, find the value of x.', NULL, '\n\nExplanation\n(a) \\(\\frac{\\frac{1}{2} of \\frac{1}{4} \\div \\frac{1}{3}}{\\frac{1}{6} - \\frac{3}{4} + \\frac{1}{2}}\\)\nNumerator - \\(\\frac{1}{2} \\times \\frac{1}{4} \\div \\frac{1}{3} = (\\frac{1}{2} \\times \\frac{1}{4}) \\div \\frac{1}{3}\\)\n\\(\\frac{1}{8} \\times 3 = \\frac{3}{8}\\)\nDenominator - \\(\\frac{1}{6} - \\frac{3}{4} + \\frac{1}{2} = \\frac{1}{6} - (\\frac{3}{4} -\\frac{1}{2})\\)\n\\(\\frac{1}{6} - (\\frac{3 -2}{4}) = \\frac{1}{6} - \\frac{1}{4}\\)\n\\(\\frac{2 - 3}{12} = \\frac{-1}{12}\\)\n\\(\\therefore \\frac{\\frac{1}{2} \\times \\frac{1}{4} \\div \\frac{1}{3}}{\\frac{1}{6} - \\frac{3}{4} + \\frac{1}{2}} = \\frac{3}{8} \\div \\frac{-1}{12}\\)\n\\(\\frac{3}{8} \\times -12 = \\frac{-9}{2}\\)\n= \\(-4.5\\)\n(b) \\(\\sqrt{x} = 10^{\\bar{1}.6741}}\\)\nSquaring both sides, we have\n\\(x = (10^{\\bar{1}.6741}})^{2}\\)\n\\(x = 10^{\\bar{1}.3482}\\) (checking antilogarithm)\n\\(x = 0.2229\\)\n\n', '', 'Explanation\n(a) \\(\\frac{\\frac{1}{2} of \\frac{1}{4} \\div \\frac{1}{3}}{\\frac{1}{6} - \\frac{3}{4} + \\frac{1}{2}}\\)\nNumerator - \\(\\frac{1}{2} \\times \\frac{1}{4} \\div \\frac{1}{3} = (\\frac{1}{2} \\times \\frac{1}{4}) \\div \\frac{1}{3}\\)\n\\(\\frac{1}{8} \\times 3 = \\frac{3}{8}\\)\nDenominator - \\(\\frac{1}{6} - \\frac{3}{4} + \\frac{1}{2} = \\frac{1}{6} - (\\frac{3}{4} -\\frac{1}{2})\\)\n\\(\\frac{1}{6} - (\\frac{3 -2}{4}) = \\frac{1}{6} - \\frac{1}{4}\\)\n\\(\\frac{2 - 3}{12} = \\frac{-1}{12}\\)\n\\(\\therefore \\frac{\\frac{1}{2} \\times \\frac{1}{4} \\div \\frac{1}{3}}{\\frac{1}{6} - \\frac{3}{4} + \\frac{1}{2}} = \\frac{3}{8} \\div \\frac{-1}{12}\\)\n\\(\\frac{3}{8} \\times -12 = \\frac{-9}{2}\\)\n= \\(-4.5\\)\n(b) \\(\\sqrt{x} = 10^{\\bar{1}.6741}}\\)\nSquaring both sides, we have\n\\(x = (10^{\\bar{1}.6741}})^{2}\\)\n\\(x = 10^{\\bar{1}.3482}\\) (checking antilogarithm)\n\\(x = 0.2229\\)\n\n', '(a) \\(\\frac{\\frac{1}{2} of \\frac{1}{4} \\div \\frac{1}{3}}{\\frac{1}{6} - \\frac{3}{4} + \\frac{1}{2}}\\)\nNumerator - \\(\\frac{1}{2} \\times \\frac{1}{4} \\div \\frac{1}{3} = (\\frac{1}{2} \\times \\frac{1}{4}) \\div \\frac{1}{3}\\)\n\\(\\frac{1}{8} \\times 3 = \\frac{3}{8}\\)\nDenominator - \\(\\frac{1}{6} - \\frac{3}{4} + \\frac{1}{2} = \\frac{1}{6} - (\\frac{3}{4} -\\frac{1}{2})\\)\n\\(\\frac{1}{6} - (\\frac{3 -2}{4}) = \\frac{1}{6} - \\frac{1}{4}\\)\n\\(\\frac{2 - 3}{12} = \\frac{-1}{12}\\)\n\\(\\therefore \\frac{\\frac{1}{2} \\times \\frac{1}{4} \\div \\frac{1}{3}}{\\frac{1}{6} - \\frac{3}{4} + \\frac{1}{2}} = \\frac{3}{8} \\div \\frac{-1}{12}\\)\n\\(\\frac{3}{8} \\times -12 = \\frac{-9}{2}\\)\n= \\(-4.5\\)~(b) \\(\\sqrt{x} = 10^{\\bar{1}.6741}}\\)\nSquaring both sides, we have\n\\(x = (10^{\\bar{1}.6741}})^{2}\\)\n\\(x = 10^{\\bar{1}.3482}\\) (checking antilogarithm)\n\\(x = 0.2229\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(409, '2020', '\n\nQuestion\n\n(a) Make q the subject of the relation \\(t = \\sqrt{\\frac{pq}{r} - r^{2}q}\\).\n(b) If \\(9^{(1 - x)} = 27^{y}\\) and \\(x - y = -1\\frac{1}{2}\\), find the value of x and y.\n\n', '', '(a) Make q the subject of the relation \\(t = \\sqrt{\\frac{pq}{r} - r^{2}q}\\).\n(b) If \\(9^{(1 - x)} = 27^{y}\\) and \\(x - y = -1\\frac{1}{2}\\), find the value of x and y.', '(a) Make q the subject of the relation \\(t = \\sqrt{\\frac{pq}{r} - r^{2}q}\\).~(b) If \\(9^{(1 - x)} = 27^{y}\\) and \\(x - y = -1\\frac{1}{2}\\), find the value of x and y.', NULL, '\n\nExplanation\n(a) \\(t = \\sqrt{\\frac{pq}{r} - r^{2}q}\\)\nSquaring both sides,\n\\(t^{2} = \\frac{pq}{r} - r^{2}q\\)\n\\(t^{2} = \\frac{pq - r^{3}q}{r}\\)\n\\(t^{2}r = pq - r^{3}q \\implies t^{2}r = q(p - r^{3})\\)\n\\(q = \\frac{t^{2}r}{p - r^{3}}\\).\n(b) \\(9^{(1 - x)} = 27^{y}\\)\n\\(\\implies 3^{2(1 - x)} = 3^{3y}\\)\n\\(2 - 2x = 3y ..... (1)\\)\n\\(x -y = -1\\frac{1}{2} ....... (2)\\)\nFrom (2), \\(x = -1\\frac{1}{2} +y\\)\n\\(\\therefore 2 - 2(-1\\frac{1}{2} +y) = 3y\\)\n\\(2 + 3 -2y = 3y \\implies 5 = 3y +2y = 5y\\)\n\\(\\implies y = 1\\)\n\\(x = -1\\frac{1}{2} + 1= -\\frac{1}{2}\\)\n\\(\\therefore x + y = -\\frac{1}{2} + 1 = \\frac{1}{2}\\)\n\n', '', 'Explanation\n(a) \\(t = \\sqrt{\\frac{pq}{r} - r^{2}q}\\)\nSquaring both sides,\n\\(t^{2} = \\frac{pq}{r} - r^{2}q\\)\n\\(t^{2} = \\frac{pq - r^{3}q}{r}\\)\n\\(t^{2}r = pq - r^{3}q \\implies t^{2}r = q(p - r^{3})\\)\n\\(q = \\frac{t^{2}r}{p - r^{3}}\\).\n(b) \\(9^{(1 - x)} = 27^{y}\\)\n\\(\\implies 3^{2(1 - x)} = 3^{3y}\\)\n\\(2 - 2x = 3y ..... (1)\\)\n\\(x -y = -1\\frac{1}{2} ....... (2)\\)\nFrom (2), \\(x = -1\\frac{1}{2} +y\\)\n\\(\\therefore 2 - 2(-1\\frac{1}{2} +y) = 3y\\)\n\\(2 + 3 -2y = 3y \\implies 5 = 3y +2y = 5y\\)\n\\(\\implies y = 1\\)\n\\(x = -1\\frac{1}{2} + 1= -\\frac{1}{2}\\)\n\\(\\therefore x + y = -\\frac{1}{2} + 1 = \\frac{1}{2}\\)\n\n', '(a) \\(t = \\sqrt{\\frac{pq}{r} - r^{2}q}\\)\nSquaring both sides,\n\\(t^{2} = \\frac{pq}{r} - r^{2}q\\)\n\\(t^{2} = \\frac{pq - r^{3}q}{r}\\)\n\\(t^{2}r = pq - r^{3}q \\implies t^{2}r = q(p - r^{3})\\)\n\\(q = \\frac{t^{2}r}{p - r^{3}}\\).~(b) \\(9^{(1 - x)} = 27^{y}\\)\n\\(\\implies 3^{2(1 - x)} = 3^{3y}\\)\n\\(2 - 2x = 3y ..... (1)\\)\n\\(x -y = -1\\frac{1}{2} ....... (2)\\)\nFrom (2), \\(x = -1\\frac{1}{2} +y\\)\n\\(\\therefore 2 - 2(-1\\frac{1}{2} +y) = 3y\\)\n\\(2 + 3 -2y = 3y \\implies 5 = 3y +2y = 5y\\)\n\\(\\implies y = 1\\)\n\\(x = -1\\frac{1}{2} + 1= -\\frac{1}{2}\\)\n\\(\\therefore x + y = -\\frac{1}{2} + 1 = \\frac{1}{2}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(410, '2020', '\n\nQuestion\n\nA sector of a circle with radius 21 cm has an area of 280\\(cm^{2}\\).\n(a) Calculate, correct to 1 decimal place, the perimeter of the sector.\n(b) If the sector is bent such that its straight edges coincide to form a cone, calculate, correct to the nearest degree, the vertical angle of the cone. [Take \\(\\pi = \\frac{22}{7}\\)].\n\n', '', 'A sector of a circle with radius 21 cm has an area of 280\\(cm^{2}\\).\n(a) Calculate, correct to 1 decimal place, the perimeter of the sector.\n(b) If the sector is bent such that its straight edges coincide to form a cone, calculate, correct to the nearest degree, the vertical angle of the cone. [Take \\(\\pi = \\frac{22}{7}\\)].', 'A sector of a circle with radius 21 cm has an area of 280\\(cm^{2}\\).\n(a) Calculate, correct to 1 decimal place, the perimeter of the sector.~(b) If the sector is bent such that its straight edges coincide to form a cone, calculate, correct to the nearest degree, the vertical angle of the cone. [Take \\(\\pi = \\frac{22}{7}\\)].', NULL, '\n\nExplanation\n(a) Area of sector = \\(\\frac{\\theta}{360} \\times \\pi r^{2}\\)\n\\(280 = \\frac{\\theta}{360} \\times \\frac{22}{7} \\times 21 \\times 21\\)\n\\(280 = \\frac{1386 \\theta}{360}\\)\n\\(\\theta = \\frac{280 \\times 360}{1386}\\)\n\\(\\theta = 72.72°\\)\nPerimeter of sector = \\(2r + \\frac{\\theta}{360} \\times 2\\pi r\\)\n= \\(2(21) + \\frac{72.72}{360} \\times 2 \\times \\frac{22}{7} \\times 21\\)\n= \\(42 + (0.202 \\times 2 \\times 66)\\)\n= \\(42 + 26.667\\)\n= \\(68.667 cm\\)\n\\(\\approxeq 68.7 cm\\)\n(b)When the sector is bent to form a cone, its radius becomes the slant height of the cone.\nThe radius of the base of the cone is obtained from the relation \\(r = \\frac{R \\theta}{360}\\),\nwhere r = radius of the base of the cone, R = radius of the sector, θ = angle of the sector.\nTherefore,r = \\(\\frac{21 \\times 800}{11 \\times 360}\\)\n= \\(\\frac{140}{33}\\)\nIf y is the vertical angle of the cone, then \\(sin \\frac{y}{2} = \\frac{r}{l}\\)\n= \\(\\frac{140}{33 \\times 21}\\)\n= 0.2020\nHence, required angle =y = 2x sin\\(^{-1}\\) (0.2020)= 23\\(^o\\).\n\n', '', 'Explanation\n(a) Area of sector = \\(\\frac{\\theta}{360} \\times \\pi r^{2}\\)\n\\(280 = \\frac{\\theta}{360} \\times \\frac{22}{7} \\times 21 \\times 21\\)\n\\(280 = \\frac{1386 \\theta}{360}\\)\n\\(\\theta = \\frac{280 \\times 360}{1386}\\)\n\\(\\theta = 72.72°\\)\nPerimeter of sector = \\(2r + \\frac{\\theta}{360} \\times 2\\pi r\\)\n= \\(2(21) + \\frac{72.72}{360} \\times 2 \\times \\frac{22}{7} \\times 21\\)\n= \\(42 + (0.202 \\times 2 \\times 66)\\)\n= \\(42 + 26.667\\)\n= \\(68.667 cm\\)\n\\(\\approxeq 68.7 cm\\)\n(b)When the sector is bent to form a cone, its radius becomes the slant height of the cone.\nThe radius of the base of the cone is obtained from the relation \\(r = \\frac{R \\theta}{360}\\),\nwhere r = radius of the base of the cone, R = radius of the sector, θ = angle of the sector.\nTherefore,r = \\(\\frac{21 \\times 800}{11 \\times 360}\\)\n= \\(\\frac{140}{33}\\)\nIf y is the vertical angle of the cone, then \\(sin \\frac{y}{2} = \\frac{r}{l}\\)\n= \\(\\frac{140}{33 \\times 21}\\)\n= 0.2020\nHence, required angle =y = 2x sin\\(^{-1}\\) (0.2020)= 23\\(^o\\).\n\n', '(a) Area of sector = \\(\\frac{\\theta}{360} \\times \\pi r^{2}\\)\n\\(280 = \\frac{\\theta}{360} \\times \\frac{22}{7} \\times 21 \\times 21\\)\n\\(280 = \\frac{1386 \\theta}{360}\\)\n\\(\\theta = \\frac{280 \\times 360}{1386}\\)\n\\(\\theta = 72.72°\\)\nPerimeter of sector = \\(2r + \\frac{\\theta}{360} \\times 2\\pi r\\)\n= \\(2(21) + \\frac{72.72}{360} \\times 2 \\times \\frac{22}{7} \\times 21\\)\n= \\(42 + (0.202 \\times 2 \\times 66)\\)\n= \\(42 + 26.667\\)\n= \\(68.667 cm\\)\n\\(\\approxeq 68.7 cm\\)~(b)When the sector is bent to form a cone, its radius becomes the slant height of the cone.\nThe radius of the base of the cone is obtained from the relation \\(r = \\frac{R \\theta}{360}\\),\nwhere r = radius of the base of the cone, R = radius of the sector, θ = angle of the sector.\nTherefore,r = \\(\\frac{21 \\times 800}{11 \\times 360}\\)\n= \\(\\frac{140}{33}\\)\nIf y is the vertical angle of the cone, then \\(sin \\frac{y}{2} = \\frac{r}{l}\\)\n= \\(\\frac{140}{33 \\times 21}\\)\n= 0.2020\nHence, required angle =y = 2x sin\\(^{-1}\\) (0.2020)= 23\\(^o\\).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(411, '2020', '\n\nQuestion\n\n(a)\nIn the diagram, PQRST is a quadrilateral. PT // QS, < PTQ = 42°, < TSQ = 38° and < QSR = 30°. If < QTS = x and < POT = y, find: (i) x ; (ii) y.\n(b)\nIn the diagram, PQRS is a circle centre O. If POQ = 150°, < QSR = 40° and < SQP = 45°, calculate < RQS.\n\n', '', '(a)\nIn the diagram, PQRST is a quadrilateral. PT // QS, < PTQ = 42°, < TSQ = 38° and < QSR = 30°. If < QTS = x and < POT = y, find: (i) x ; (ii) y.\n(b)\nIn the diagram, PQRS is a circle centre O. If POQ = 150°, < QSR = 40° and < SQP = 45°, calculate < RQS.', '(a)\nIn the diagram, PQRST is a quadrilateral. PT // QS, < PTQ = 42°, < TSQ = 38° and < QSR = 30°. If < QTS = x and < POT = y, find: (i) x ; (ii) y.~(b)\nIn the diagram, PQRS is a circle centre O. If POQ = 150°, < QSR = 40° and < SQP = 45°, calculate < RQS.', NULL, '\n\nExplanation\n(a)(i) < PTS = 38° (alternate angle)\n42° + x + 38° = 180° (angles on a straight line)\n80° + x = 180°\nx = 180° - 80° = 100°\n(ii) < SQT = 42° (alternate angle)\n< SQR = 60° (sum of angles in a triangle)\ny + 42° + 60° = 180° (angle on a straight line)\ny + 102° = 180°\ny = 180° - 102° = 78°\n(b) \\(< QSP = \\frac{150}{2}\\) (angle at the centre is twicw angle at the circumference)\n\\(< QSP = 75°\\)\n\\(\\hat{R} + \\hat{P} = 180°\\) (opposite angles in a cyclic quad are supplementary)\n\\(\\hat{P} + 75° + 45° = 180°\\) (sum of angle in a triangle)\n\\(\\hat{P} = 180° - 120°\\)\n\\(\\hat{P} = 60°\\)\n\\(\\hat{R} + \\hat{P} = 180°\\)\n\\(\\hat{R} = 180° - 60°\\)\n\\(\\hat{R} = 120°\\)\n\\(< RQS + < QRS + < QSR = 180°\\) (sum of angle in a triangle)\n\\(< RQS = 180° - (40° + 120°)\\)\n= \\(180° - 160°\\)\n\\(< RQS = 20°\\)\n\n', '', 'Explanation\n(a)(i) < PTS = 38° (alternate angle)\n42° + x + 38° = 180° (angles on a straight line)\n80° + x = 180°\nx = 180° - 80° = 100°\n(ii) < SQT = 42° (alternate angle)\n< SQR = 60° (sum of angles in a triangle)\ny + 42° + 60° = 180° (angle on a straight line)\ny + 102° = 180°\ny = 180° - 102° = 78°\n(b) \\(< QSP = \\frac{150}{2}\\) (angle at the centre is twicw angle at the circumference)\n\\(< QSP = 75°\\)\n\\(\\hat{R} + \\hat{P} = 180°\\) (opposite angles in a cyclic quad are supplementary)\n\\(\\hat{P} + 75° + 45° = 180°\\) (sum of angle in a triangle)\n\\(\\hat{P} = 180° - 120°\\)\n\\(\\hat{P} = 60°\\)\n\\(\\hat{R} + \\hat{P} = 180°\\)\n\\(\\hat{R} = 180° - 60°\\)\n\\(\\hat{R} = 120°\\)\n\\(< RQS + < QRS + < QSR = 180°\\) (sum of angle in a triangle)\n\\(< RQS = 180° - (40° + 120°)\\)\n= \\(180° - 160°\\)\n\\(< RQS = 20°\\)\n\n', '(a)(i) < PTS = 38° (alternate angle)\n42° + x + 38° = 180° (angles on a straight line)\n80° + x = 180°\nx = 180° - 80° = 100°\n(ii) < SQT = 42° (alternate angle)\n< SQR = 60° (sum of angles in a triangle)\ny + 42° + 60° = 180° (angle on a straight line)\ny + 102° = 180°\ny = 180° - 102° = 78°~(b) \\(< QSP = \\frac{150}{2}\\) (angle at the centre is twicw angle at the circumference)\n\\(< QSP = 75°\\)\n\\(\\hat{R} + \\hat{P} = 180°\\) (opposite angles in a cyclic quad are supplementary)\n\\(\\hat{P} + 75° + 45° = 180°\\) (sum of angle in a triangle)\n\\(\\hat{P} = 180° - 120°\\)\n\\(\\hat{P} = 60°\\)\n\\(\\hat{R} + \\hat{P} = 180°\\)\n\\(\\hat{R} = 180° - 60°\\)\n\\(\\hat{R} = 120°\\)\n\\(< RQS + < QRS + < QSR = 180°\\) (sum of angle in a triangle)\n\\(< RQS = 180° - (40° + 120°)\\)\n= \\(180° - 160°\\)\n\\(< RQS = 20°\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(412, '2020', '\n\nQuestion\n\nA library received$1,300 grant. It spends 10% of the grant on magazine subscriptions, 35% on new books, 15% to repair damaged books, 30% to buy new furniture and 10% to train library staff.\n(a) Represent this information on a pie chart.\n(b) Calculate, correct to the nearest whole number, the percentage increase of the amount for buying books over that of new furniture.\n\n', '', 'A library received$1,300 grant. It spends 10% of the grant on magazine subscriptions, 35% on new books, 15% to repair damaged books, 30% to buy new furniture and 10% to train library staff.\n(a) Represent this information on a pie chart.\n(b) Calculate, correct to the nearest whole number, the percentage increase of the amount for buying books over that of new furniture.', 'A library received$1,300 grant. It spends 10% of the grant on magazine subscriptions, 35% on new books, 15% to repair damaged books, 30% to buy new furniture and 10% to train library staff.\n(a) Represent this information on a pie chart.~(b) Calculate, correct to the nearest whole number, the percentage increase of the amount for buying books over that of new furniture.', NULL, '\n\nExplanation\nIncome =$1,300 grant\nGrant on different aspects:\nMagazine = \\(\\frac{10}{100} \\times 1300 =$130\\)\nNew books = \\(\\frac{35}{100} \\times 1300 =$455\\)\nDamaged books = \\(\\frac{15}{100} \\times 1300 =$195\\)\nNew furniture = \\(\\frac{30}{100} \\times 1300 =$390\\)\nLibrary staff = \\(\\frac{10}{100} \\times 1300 =$130\\)\nFor pie chart,\nMagazine = \\(\\frac{130}{1300} \\times 360° = 36°\\)\nNew books = \\(\\frac{455}{1300} \\times 360° = 126°\\)\nDamaged books = \\(\\frac{195}{1300} \\times 360° = 54°\\)\nNew furniture = \\(\\frac{390}{1300} \\times 360° = 108°\\)\nLibrary staff = \\(\\frac{130}{1300} \\times 360° = 36°\\)\n\n(b) Percentage increase = \\(\\frac{455 - 390}{390} \\times 100% =16.67%\\)\n\n', '', 'Explanation\nIncome =$1,300 grant\nGrant on different aspects:\nMagazine = \\(\\frac{10}{100} \\times 1300 =$130\\)\nNew books = \\(\\frac{35}{100} \\times 1300 =$455\\)\nDamaged books = \\(\\frac{15}{100} \\times 1300 =$195\\)\nNew furniture = \\(\\frac{30}{100} \\times 1300 =$390\\)\nLibrary staff = \\(\\frac{10}{100} \\times 1300 =$130\\)\nFor pie chart,\nMagazine = \\(\\frac{130}{1300} \\times 360° = 36°\\)\nNew books = \\(\\frac{455}{1300} \\times 360° = 126°\\)\nDamaged books = \\(\\frac{195}{1300} \\times 360° = 54°\\)\nNew furniture = \\(\\frac{390}{1300} \\times 360° = 108°\\)\nLibrary staff = \\(\\frac{130}{1300} \\times 360° = 36°\\)\n\n(b) Percentage increase = \\(\\frac{455 - 390}{390} \\times 100% =16.67%\\)\n\n', '~(b) Percentage increase = \\(\\frac{455 - 390}{390} \\times 100% =16.67%\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(413, '2020', '\n\nQuestion\n\nIn a class of 40 students, 18 passed Mathematics, 19 passed Accounts, 16 passed Economics, 5 passed Mathematics and Accounts only, 6 Mathematics only, 9 Accounts only, 2 Accounts and Economics only. If each student offered at least one of the subjects,\n(a) how many students failed in all subjects?\n(b) find the percentage number that failed in at least one of Economics and Mathematics\n(c) calculate the probability that a student picked at random failed in Accounts?\n\n', '', 'In a class of 40 students, 18 passed Mathematics, 19 passed Accounts, 16 passed Economics, 5 passed Mathematics and Accounts only, 6 Mathematics only, 9 Accounts only, 2 Accounts and Economics only. If each student offered at least one of the subjects,\n(a) how many students failed in all subjects?\n(b) find the percentage number that failed in at least one of Economics and Mathematics\n(c) calculate the probability that a student picked at random failed in Accounts?', 'In a class of 40 students, 18 passed Mathematics, 19 passed Accounts, 16 passed Economics, 5 passed Mathematics and Accounts only, 6 Mathematics only, 9 Accounts only, 2 Accounts and Economics only. If each student offered at least one of the subjects,\n(a) how many students failed in all subjects?~(b) find the percentage number that failed in at least one of Economics and Mathematics\n(c) calculate the probability that a student picked at random failed in Accounts?~(c) calculate the probability that a student picked at random failed in Accounts?', NULL, '\n\nExplanation\n(a) \\(n(U) = 40\\)\n\\(n(M) = 18\\)                \\(n(E) = 16\\)\n\\(n(A) = 19\\)                 \\(n(M \\cap E) = 5\\)\n\\(n(A \\cap E) = 2\\)             \\(n(M) only = 6\\)\n\\(n(A) only = 9\\)\n\nAccounts:\nx + 5 + 2 + 9 = 19\nx + 16 = 19\nx = 19 - 16 = 3.\nMathematics :\ny + 6 + x + 5 = 18\ny + 6 + 3 + 5 = 18\ny + 14 = 18\ny = 18 - 14 = 4.\nEconomics :\nz + y + x + 2 = 16\nz + 4 + 3 + 2 = 16\nz + 9 = 16\nz = 16 - 9 = 7.\nLet the number of students that failed be r.\n40 - r = 6 + 5 + 9 + 2 + 3 + 4 + 7\n40 - r = 36\nr = 40 - 36 = 4.\n(b) Percentage that failed at least one of Economics and Mathematics\n= 100% - (% of people that passed at least one of Economics and Mathemics)\nNumber that passed at least one of Economics and Mathematics = n(E) + n(M) + n(E and M)\n= 6 + 7 + 4 = 17 students passed at least one of Economics and Mathematics\n% passed = \\(\\frac{17}{40} \\times 100% = 42.5%\\)\nPercentage that failed at least one of Economics and Mathematics = 100% - 42.5% = 57.5%\n(c) Probability of failed in Accounts = 1 = Probability of passed in Accounts.\nP(passed Accounts) = \\(\\frac{n(A)}{n(U)}\\)\n= \\(\\frac{19}{40}\\)\nP(failed accounts) = \\(1 - \\frac{19}{40}\\)\n= \\(\\frac{21}{40} = 0.525\\)\n\n', '', 'Explanation\n(a) \\(n(U) = 40\\)\n\\(n(M) = 18\\)                \\(n(E) = 16\\)\n\\(n(A) = 19\\)                 \\(n(M \\cap E) = 5\\)\n\\(n(A \\cap E) = 2\\)             \\(n(M) only = 6\\)\n\\(n(A) only = 9\\)\n\nAccounts:\nx + 5 + 2 + 9 = 19\nx + 16 = 19\nx = 19 - 16 = 3.\nMathematics :\ny + 6 + x + 5 = 18\ny + 6 + 3 + 5 = 18\ny + 14 = 18\ny = 18 - 14 = 4.\nEconomics :\nz + y + x + 2 = 16\nz + 4 + 3 + 2 = 16\nz + 9 = 16\nz = 16 - 9 = 7.\nLet the number of students that failed be r.\n40 - r = 6 + 5 + 9 + 2 + 3 + 4 + 7\n40 - r = 36\nr = 40 - 36 = 4.\n(b) Percentage that failed at least one of Economics and Mathematics\n= 100% - (% of people that passed at least one of Economics and Mathemics)\nNumber that passed at least one of Economics and Mathematics = n(E) + n(M) + n(E and M)\n= 6 + 7 + 4 = 17 students passed at least one of Economics and Mathematics\n% passed = \\(\\frac{17}{40} \\times 100% = 42.5%\\)\nPercentage that failed at least one of Economics and Mathematics = 100% - 42.5% = 57.5%\n(c) Probability of failed in Accounts = 1 = Probability of passed in Accounts.\nP(passed Accounts) = \\(\\frac{n(A)}{n(U)}\\)\n= \\(\\frac{19}{40}\\)\nP(failed accounts) = \\(1 - \\frac{19}{40}\\)\n= \\(\\frac{21}{40} = 0.525\\)\n\n', '(a) \\(n(U) = 40\\)\n\\(n(M) = 18\\)                \\(n(E) = 16\\)\n\\(n(A) = 19\\)                 \\(n(M \\cap E) = 5\\)\n\\(n(A \\cap E) = 2\\)             \\(n(M) only = 6\\)\n\\(n(A) only = 9\\)\n\nAccounts:\nx + 5 + 2 + 9 = 19\nx + 16 = 19\nx = 19 - 16 = 3.\nMathematics :\ny + 6 + x + 5 = 18\ny + 6 + 3 + 5 = 18\ny + 14 = 18\ny = 18 - 14 = 4.\nEconomics :\nz + y + x + 2 = 16\nz + 4 + 3 + 2 = 16\nz + 9 = 16\nz = 16 - 9 = 7.\nLet the number of students that failed be r.\n40 - r = 6 + 5 + 9 + 2 + 3 + 4 + 7\n40 - r = 36\nr = 40 - 36 = 4.~(b) Percentage that failed at least one of Economics and Mathematics\n= 100% - (% of people that passed at least one of Economics and Mathemics)\nNumber that passed at least one of Economics and Mathematics = n(E) + n(M) + n(E and M)\n= 6 + 7 + 4 = 17 students passed at least one of Economics and Mathematics\n% passed = \\(\\frac{17}{40} \\times 100% = 42.5%\\)\nPercentage that failed at least one of Economics and Mathematics = 100% - 42.5% = 57.5%\n(c) Probability of failed in Accounts = 1 = Probability of passed in Accounts.\nP(passed Accounts) = \\(\\frac{n(A)}{n(U)}\\)\n= \\(\\frac{19}{40}\\)\nP(failed accounts) = \\(1 - \\frac{19}{40}\\)\n= \\(\\frac{21}{40} = 0.525\\)\n~(c) Probability of failed in Accounts = 1 = Probability of passed in Accounts.\nP(passed Accounts) = \\(\\frac{n(A)}{n(U)}\\)\n= \\(\\frac{19}{40}\\)\nP(failed accounts) = \\(1 - \\frac{19}{40}\\)\n= \\(\\frac{21}{40} = 0.525\\)\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
(414, '2020', '\n\nQuestion\n\n(a) Divide \\(\\frac{x^{2} - 4}{x^{2} + x}\\) by \\(\\frac{x^{2} - 4x + 4}{x + 1}\\).\n(b) The diagram below shows the graphs of \\(y = ax^{2} + bx + c\\) and \\(y = mx + k\\) where a, b, c and m are constants. Use the graph(s) to :\n(i) find the roots of the equation \\(ax^{2} + bx + c = mx + k\\);\n(ii) determine the values of a, b and c using the coordinates of points L, M and N and hence write down the equation of the curve;\n(iii) determine the line of symmetry of the curve \\(y = ax^{2} + bx + c\\).\n\n\n', '', '(a) Divide \\(\\frac{x^{2} - 4}{x^{2} + x}\\) by \\(\\frac{x^{2} - 4x + 4}{x + 1}\\).\n(b) The diagram below shows the graphs of \\(y = ax^{2} + bx + c\\) and \\(y = mx + k\\) where a, b, c and m are constants. Use the graph(s) to :\n(i) find the roots of the equation \\(ax^{2} + bx + c = mx + k\\);\n(ii) determine the values of a, b and c using the coordinates of points L, M and N and hence write down the equation of the curve;\n(iii) determine the line of symmetry of the curve \\(y = ax^{2} + bx + c\\).\n', '(a) Divide \\(\\frac{x^{2} - 4}{x^{2} + x}\\) by \\(\\frac{x^{2} - 4x + 4}{x + 1}\\).~(b) The diagram below shows the graphs of \\(y = ax^{2} + bx + c\\) and \\(y = mx + k\\) where a, b, c and m are constants. Use the graph(s) to :\n(i) find the roots of the equation \\(ax^{2} + bx + c = mx + k\\);\n(ii) determine the values of a, b and c using the coordinates of points L, M and N and hence write down the equation of the curve;\n(iii) determine the line of symmetry of the curve \\(y = ax^{2} + bx + c\\).\n', NULL, '\n\nExplanation\n(a) \\(\\frac{x^{2} - 4}{x^{2} + x} \\div \\frac{x^{2} - 4x + 4}{x + 1}\\)\n= \\(\\frac{x^{2} - 4}{x^{2} + x} \\times \\frac{x + 1}{x^{2} - 4x + 4}\\)\n= \\(\\frac{(x - 2)(x + 2)}{x(x + 1)} \\times \\frac{x + 1}{x^{2} - 2x - 2x + 4}\\) (Using difference of two squares)\n= \\(\\frac{(x - 2)(x + 2)}{x} \\times \\frac{1}{(x - 2)^{2}}\\)\n= \\(\\frac{(x + 2)}{x(x - 2)}.\n(b) Roots of \\(ax^{2} +bx + c = mx + k\\)\nFor graph \\(y = ax^{2} + bx + c\\),\n\\(x = -1\\) and \\(x = 2\\)\n\\(x + 1 = 0\\) and \\(x - 2 = 0\\)\n\\((x + 1)(x - 2) = 0\\)\n\\(x^{2} - 2x + x - 2 = 0\\)\n\\(x^{2} - x - 2 = 0 ..... (i)\\)\n\\(-x^{2} + x + 2 = 0 ...... (ii)\\)\nFor graph \\(y = mx + k\\)\n\\(m = \\frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \\frac{1 - (-1)}{1.3 - (-1.3)}\\)\n\\(m = \\frac{2}{2.6} = 0.769\\)\n\\(k = 0.25\\)\n\\(x^{2} - x - 2 = 0.769x + 0.25\\)\n\\(x^{2} - x - 0.769x - 2 - 0.25 = 0\\)\n\\(x^{2} - 1.769x - 2.25 = 0\\)\nUsing formula method, \\(x = \\frac{-b \\pm \\sqrt{b^{2} - 4ac}}{2a}\\)\n\\(x = \\frac{-(-1.769) \\pm \\sqrt{(-1.769)^{2} - 4(1)(-2.25)}}{2(1)}\\)\n\\(x = \\frac{1.769 \\pm \\sqrt{3.1294 + 9}}{2}\\)\n\\(x = \\frac{1.769 \\pm 3.483}{2}\\)\n\\(x = \\frac{1.769 + 3.483}{2}\\) or \\(x = \\frac{1.769 - 3.483}{2}\\)\n\\(x = \\frac{5.252}{2} \\approxeq 2.626\\) or \\(x = \\frac{-1.714}{2} \\approxeq-0.857\\)\n(ii) The coordinates of points L, M and N are (-1, 0), (0, 2) and (2, 0) respectively.\nUsing coordinate for L:\n\\(0 = a(-1^{2}) + b(-1) + c \\implies 0 = a - b + c ..... (1)\\)\nUsing M :\n\\(2 = a(0^{2}) + b(0) + c \\implies c = 2 ...... (2)\\)\nUsing N :\n\\(0 = a(2^{2}) + b(2) + c \\implies 0 = 4a + 2b + c .... (3)\\)\nFrom (2), c = 2\n\\(\\therefore (1) : a - b + 2 = 0 \\implies a - b = -2 .... (1a)\\)\n\\(\\therefore (3) : 4a + 2b + 2 = 0 \\implies 4a + 2b = -2 ..... (3a)\\)\nSolving equations (1a) and (3a) simultaneously, we get\n\\(a = -1 , b = 1 , c = 2\\)\n\\(\\therefore\\) Equation of the curve : \\(-x^{2} + x + 2 = 0\\)\n(iii) Line of symmetry = \\(x = \\frac{-b}{2a}\\)\n\\(y = -x^{2} + x + 2 \\)\nLine of symmetry = \\(\\frac{-1}{2(-1)}\\)\n\\(x = \\frac{1}{2}\\)\n\n', '', 'Explanation\n(a) \\(\\frac{x^{2} - 4}{x^{2} + x} \\div \\frac{x^{2} - 4x + 4}{x + 1}\\)\n= \\(\\frac{x^{2} - 4}{x^{2} + x} \\times \\frac{x + 1}{x^{2} - 4x + 4}\\)\n= \\(\\frac{(x - 2)(x + 2)}{x(x + 1)} \\times \\frac{x + 1}{x^{2} - 2x - 2x + 4}\\) (Using difference of two squares)\n= \\(\\frac{(x - 2)(x + 2)}{x} \\times \\frac{1}{(x - 2)^{2}}\\)\n= \\(\\frac{(x + 2)}{x(x - 2)}.\n(b) Roots of \\(ax^{2} +bx + c = mx + k\\)\nFor graph \\(y = ax^{2} + bx + c\\),\n\\(x = -1\\) and \\(x = 2\\)\n\\(x + 1 = 0\\) and \\(x - 2 = 0\\)\n\\((x + 1)(x - 2) = 0\\)\n\\(x^{2} - 2x + x - 2 = 0\\)\n\\(x^{2} - x - 2 = 0 ..... (i)\\)\n\\(-x^{2} + x + 2 = 0 ...... (ii)\\)\nFor graph \\(y = mx + k\\)\n\\(m = \\frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \\frac{1 - (-1)}{1.3 - (-1.3)}\\)\n\\(m = \\frac{2}{2.6} = 0.769\\)\n\\(k = 0.25\\)\n\\(x^{2} - x - 2 = 0.769x + 0.25\\)\n\\(x^{2} - x - 0.769x - 2 - 0.25 = 0\\)\n\\(x^{2} - 1.769x - 2.25 = 0\\)\nUsing formula method, \\(x = \\frac{-b \\pm \\sqrt{b^{2} - 4ac}}{2a}\\)\n\\(x = \\frac{-(-1.769) \\pm \\sqrt{(-1.769)^{2} - 4(1)(-2.25)}}{2(1)}\\)\n\\(x = \\frac{1.769 \\pm \\sqrt{3.1294 + 9}}{2}\\)\n\\(x = \\frac{1.769 \\pm 3.483}{2}\\)\n\\(x = \\frac{1.769 + 3.483}{2}\\) or \\(x = \\frac{1.769 - 3.483}{2}\\)\n\\(x = \\frac{5.252}{2} \\approxeq 2.626\\) or \\(x = \\frac{-1.714}{2} \\approxeq-0.857\\)\n(ii) The coordinates of points L, M and N are (-1, 0), (0, 2) and (2, 0) respectively.\nUsing coordinate for L:\n\\(0 = a(-1^{2}) + b(-1) + c \\implies 0 = a - b + c ..... (1)\\)\nUsing M :\n\\(2 = a(0^{2}) + b(0) + c \\implies c = 2 ...... (2)\\)\nUsing N :\n\\(0 = a(2^{2}) + b(2) + c \\implies 0 = 4a + 2b + c .... (3)\\)\nFrom (2), c = 2\n\\(\\therefore (1) : a - b + 2 = 0 \\implies a - b = -2 .... (1a)\\)\n\\(\\therefore (3) : 4a + 2b + 2 = 0 \\implies 4a + 2b = -2 ..... (3a)\\)\nSolving equations (1a) and (3a) simultaneously, we get\n\\(a = -1 , b = 1 , c = 2\\)\n\\(\\therefore\\) Equation of the curve : \\(-x^{2} + x + 2 = 0\\)\n(iii) Line of symmetry = \\(x = \\frac{-b}{2a}\\)\n\\(y = -x^{2} + x + 2 \\)\nLine of symmetry = \\(\\frac{-1}{2(-1)}\\)\n\\(x = \\frac{1}{2}\\)\n\n', '(a) \\(\\frac{x^{2} - 4}{x^{2} + x} \\div \\frac{x^{2} - 4x + 4}{x + 1}\\)\n= \\(\\frac{x^{2} - 4}{x^{2} + x} \\times \\frac{x + 1}{x^{2} - 4x + 4}\\)\n= \\(\\frac{(x - 2)(x + 2)}{x(x + 1)} \\times \\frac{x + 1}{x^{2} - 2x - 2x + 4}\\) (Using difference of two squares)\n= \\(\\frac{(x - 2)(x + 2)}{x} \\times \\frac{1}{(x - 2)^{2}}\\)\n= \\(\\frac{(x + 2)}{x(x - 2)}.~(b) Roots of \\(ax^{2} +bx + c = mx + k\\)\nFor graph \\(y = ax^{2} + bx + c\\),\n\\(x = -1\\) and \\(x = 2\\)\n\\(x + 1 = 0\\) and \\(x - 2 = 0\\)\n\\((x + 1)(x - 2) = 0\\)\n\\(x^{2} - 2x + x - 2 = 0\\)\n\\(x^{2} - x - 2 = 0 ..... (i)\\)\n\\(-x^{2} + x + 2 = 0 ...... (ii)\\)\nFor graph \\(y = mx + k\\)\n\\(m = \\frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \\frac{1 - (-1)}{1.3 - (-1.3)}\\)\n\\(m = \\frac{2}{2.6} = 0.769\\)\n\\(k = 0.25\\)\n\\(x^{2} - x - 2 = 0.769x + 0.25\\)\n\\(x^{2} - x - 0.769x - 2 - 0.25 = 0\\)\n\\(x^{2} - 1.769x - 2.25 = 0\\)\nUsing formula method, \\(x = \\frac{-b \\pm \\sqrt{b^{2} - 4ac}}{2a}\\)\n\\(x = \\frac{-(-1.769) \\pm \\sqrt{(-1.769)^{2} - 4(1)(-2.25)}}{2(1)}\\)\n\\(x = \\frac{1.769 \\pm \\sqrt{3.1294 + 9}}{2}\\)\n\\(x = \\frac{1.769 \\pm 3.483}{2}\\)\n\\(x = \\frac{1.769 + 3.483}{2}\\) or \\(x = \\frac{1.769 - 3.483}{2}\\)\n\\(x = \\frac{5.252}{2} \\approxeq 2.626\\) or \\(x = \\frac{-1.714}{2} \\approxeq-0.857\\)\n(ii) The coordinates of points L, M and N are (-1, 0), (0, 2) and (2, 0) respectively.\nUsing coordinate for L:\n\\(0 = a(-1^{2}) + b(-1) + c \\implies 0 = a - b + c ..... (1)\\)\nUsing M :\n\\(2 = a(0^{2}) + b(0) + c \\implies c = 2 ...... (2)\\)\nUsing N :\n\\(0 = a(2^{2}) + b(2) + c \\implies 0 = 4a + 2b + c .... (3)\\)\nFrom (2), c = 2\n\\(\\therefore (1) : a - b + 2 = 0 \\implies a - b = -2 .... (1a)\\)\n\\(\\therefore (3) : 4a + 2b + 2 = 0 \\implies 4a + 2b = -2 ..... (3a)\\)\nSolving equations (1a) and (3a) simultaneously, we get\n\\(a = -1 , b = 1 , c = 2\\)\n\\(\\therefore\\) Equation of the curve : \\(-x^{2} + x + 2 = 0\\)\n(iii) Line of symmetry = \\(x = \\frac{-b}{2a}\\)\n\\(y = -x^{2} + x + 2 \\)\nLine of symmetry = \\(\\frac{-1}{2(-1)}\\)\n\\(x = \\frac{1}{2}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(415, '2020', '\n\nQuestion\n\n(a) Given that \\(\\sin x = 0.6, 0° \\leq x \\leq 90°\\), evaluate \\(2\\cos x + 3\\sin x\\), leaving your answer in the form \\(\\frac{m}{n}\\), where m and n are integers.\n(b)\nIn the diagram, a semi-circle WXYZ with centre O is inscribed in an isosceles triangle ABC. If /AC/= /BC/, |OC| = 30 cm and < ACB = 130°, calculate, correct to one decimal place, the (i) radius of the circle ; (ii) area oc the shaded portion. [Take \\(\\pi = \\frac{22}{7}\\)].\n\n', '', '(a) Given that \\(\\sin x = 0.6, 0° \\leq x \\leq 90°\\), evaluate \\(2\\cos x + 3\\sin x\\), leaving your answer in the form \\(\\frac{m}{n}\\), where m and n are integers.\n(b)\nIn the diagram, a semi-circle WXYZ with centre O is inscribed in an isosceles triangle ABC. If /AC/= /BC/, |OC| = 30 cm and < ACB = 130°, calculate, correct to one decimal place, the (i) radius of the circle ; (ii) area oc the shaded portion. [Take \\(\\pi = \\frac{22}{7}\\)].', '(a) Given that \\(\\sin x = 0.6, 0° \\leq x \\leq 90°\\), evaluate \\(2\\cos x + 3\\sin x\\), leaving your answer in the form \\(\\frac{m}{n}\\), where m and n are integers.~(b)\nIn the diagram, a semi-circle WXYZ with centre O is inscribed in an isosceles triangle ABC. If /AC/= /BC/, |OC| = 30 cm and < ACB = 130°, calculate, correct to one decimal place, the (i) radius of the circle ; (ii) area oc the shaded portion. [Take \\(\\pi = \\frac{22}{7}\\)].', NULL, '\n\nExplanation\nGiven \\(\\sin x = 0.6 = \\frac{3}{5}\\)\n\\(\\therefore Opp = 3; Hyp = 5\\)\nUsing Pythagoras theorem,\n\\(5^{2} = 3^{2} + Adj^{2}\\)\n\\(Adj = \\sqrt{5^{2} - 3^{2}}\\)\n\\(Adj = \\sqrt{25 - 9} = \\sqrt{16} = 4\\)\n\\(\\therefore \\cos x = \\frac{4}{5} = 0.8\\)\n\\(2\\cos x + 3\\sin x = 2(\\frac{4}{5}) + 3(\\frac{3}{5})\\)\n= \\(\\frac{8}{5} + \\frac{9}{5}\\)\n= \\(\\frac{17}{5}\\).\n(b)\nJoin O to X and O to Y.\nConsidering either \\(\\Delta OXC\\) or \\(\\Delta OYC\\),\n\\(\\sin 65 = \\frac{r}{30}\\)\n\\(r = 30 \\sin 65\\)\n\\(r = 30 \\times 0.9063\\)\n\\(r = 27.189 cm \\approxeq 27.2 cm\\) (to one decimal place)\n(b) Area of shaded portion = Area of triangle ABC - Area of semi-circle WXYZ.\n= \\(\\frac{1}{2} bh - \\frac{1}{2} \\pi r^{2}\\)\n= \\((\\frac{1}{2} \\times 2c \\times 30) - (\\frac{1}{2} \\times \\frac{22}{7} \\times (27.189)^{2})\\)\nFrom \\(\\Delta COA, \\tan 65 = \\frac{c}{30}\\)\n\\(\\implies c = 30 \\tan 65 \\approxeq 64.34 cm\\)\n\\(\\therefore\\) Area of shaded portion = \\(30(64.34) - \\frac{8131.659}{7}\\)\n= \\(1930.056- 1161.666\\)\n= \\(768.39 cm^{2}\\)\n\n', '', 'Explanation\nGiven \\(\\sin x = 0.6 = \\frac{3}{5}\\)\n\\(\\therefore Opp = 3; Hyp = 5\\)\nUsing Pythagoras theorem,\n\\(5^{2} = 3^{2} + Adj^{2}\\)\n\\(Adj = \\sqrt{5^{2} - 3^{2}}\\)\n\\(Adj = \\sqrt{25 - 9} = \\sqrt{16} = 4\\)\n\\(\\therefore \\cos x = \\frac{4}{5} = 0.8\\)\n\\(2\\cos x + 3\\sin x = 2(\\frac{4}{5}) + 3(\\frac{3}{5})\\)\n= \\(\\frac{8}{5} + \\frac{9}{5}\\)\n= \\(\\frac{17}{5}\\).\n(b)\nJoin O to X and O to Y.\nConsidering either \\(\\Delta OXC\\) or \\(\\Delta OYC\\),\n\\(\\sin 65 = \\frac{r}{30}\\)\n\\(r = 30 \\sin 65\\)\n\\(r = 30 \\times 0.9063\\)\n\\(r = 27.189 cm \\approxeq 27.2 cm\\) (to one decimal place)\n(b) Area of shaded portion = Area of triangle ABC - Area of semi-circle WXYZ.\n= \\(\\frac{1}{2} bh - \\frac{1}{2} \\pi r^{2}\\)\n= \\((\\frac{1}{2} \\times 2c \\times 30) - (\\frac{1}{2} \\times \\frac{22}{7} \\times (27.189)^{2})\\)\nFrom \\(\\Delta COA, \\tan 65 = \\frac{c}{30}\\)\n\\(\\implies c = 30 \\tan 65 \\approxeq 64.34 cm\\)\n\\(\\therefore\\) Area of shaded portion = \\(30(64.34) - \\frac{8131.659}{7}\\)\n= \\(1930.056- 1161.666\\)\n= \\(768.39 cm^{2}\\)\n\n', '~(b)\nJoin O to X and O to Y.\nConsidering either \\(\\Delta OXC\\) or \\(\\Delta OYC\\),\n\\(\\sin 65 = \\frac{r}{30}\\)\n\\(r = 30 \\sin 65\\)\n\\(r = 30 \\times 0.9063\\)\n\\(r = 27.189 cm \\approxeq 27.2 cm\\) (to one decimal place)\n(b) Area of shaded portion = Area of triangle ABC - Area of semi-circle WXYZ.\n= \\(\\frac{1}{2} bh - \\frac{1}{2} \\pi r^{2}\\)\n= \\((\\frac{1}{2} \\times 2c \\times 30) - (\\frac{1}{2} \\times \\frac{22}{7} \\times (27.189)^{2})\\)\nFrom \\(\\Delta COA, \\tan 65 = \\frac{c}{30}\\)\n\\(\\implies c = 30 \\tan 65 \\approxeq 64.34 cm\\)\n\\(\\therefore\\) Area of shaded portion = \\(30(64.34) - \\frac{8131.659}{7}\\)\n= \\(1930.056- 1161.666\\)\n= \\(768.39 cm^{2}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(416, '2020', '\n\nQuestion\n\nUsing ruler and a pair of compasses only,\n(a) construct a rhombus PQRS of side 7 cm and < PQR = 60°;\n(b) locate point X such that X lies on the locus of points equidistant from PQ and QR and also equidistant from Q and R ;\n(c) measure |XR|.\n\n', '', 'Using ruler and a pair of compasses only,\n(a) construct a rhombus PQRS of side 7 cm and < PQR = 60°;\n(b) locate point X such that X lies on the locus of points equidistant from PQ and QR and also equidistant from Q and R ;\n(c) measure |XR|.', 'Using ruler and a pair of compasses only,\n(a) construct a rhombus PQRS of side 7 cm and < PQR = 60°;~(b) locate point X such that X lies on the locus of points equidistant from PQ and QR and also equidistant from Q and R ;\n(c) measure |XR|.~(c) measure |XR|.', NULL, '\n\nExplanation\n(c) |XP| = 3.8 cm.\n\n', '(c) |XP| = 3.8 cm.\n\n', '', '', 'WAEC-PK-1', 'WAEC-MATH', 1),
(417, '2020', '\n\nQuestion\n\n(a) The total surface area of two spheres are in the ratio 9 : 49. If the radius of the smaller sphere is 12 cm, find, correct to the nearest \\(cm^{3}\\), the volume of the bigger sphere.\n(b) A cyclist starts from a point X and rides 3 km due West to a point Y. At Y, he changes direction and rides 5 km North- West to a point Z.\n(i) How far is he from the starting point, correct to the nearest km? ; (ii) Find the bearing of Z from X, to the nearest degree.\n\n', '', '(a) The total surface area of two spheres are in the ratio 9 : 49. If the radius of the smaller sphere is 12 cm, find, correct to the nearest \\(cm^{3}\\), the volume of the bigger sphere.\n(b) A cyclist starts from a point X and rides 3 km due West to a point Y. At Y, he changes direction and rides 5 km North- West to a point Z.\n(i) How far is he from the starting point, correct to the nearest km? ; (ii) Find the bearing of Z from X, to the nearest degree.', '(a) The total surface area of two spheres are in the ratio 9 : 49. If the radius of the smaller sphere is 12 cm, find, correct to the nearest \\(cm^{3}\\), the volume of the bigger sphere.~(b) A cyclist starts from a point X and rides 3 km due West to a point Y. At Y, he changes direction and rides 5 km North- West to a point Z.\n(i) How far is he from the starting point, correct to the nearest km? ; (ii) Find the bearing of Z from X, to the nearest degree.', NULL, '\n\nExplanation\n(a) Total surface area of a sphere = \\(4\\pi r^{2}\\)\nLet the TSA of the smaller sphere be \\(S_{1}\\) with radius r and the bigger sphere be \\(S_{2}\\) with radius R.\n\\(\\frac{S_{1}}{S_{2}} = \\frac{9}{49}\\)\n\\(\\frac{9}{49} = \\frac{4\\pi r^{2}}{4\\pi R^{2}}\\)\n\\(\\implies \\frac{9}{49} = \\frac{12^{2}}{R^{2}}\\)\n\\(\\frac{49 \\times 144}{9} = R^{2}\\)\n\\(R = \\sqrt{49 \\times 16} = 28 cm\\)\nVolume of bigger sphere = \\(\\frac{4}{3} \\pi r^{3}\\)\n= \\(\\frac{4}{3} \\times \\frac{22}{7} \\times 28 \\times 28 \\times 28\\)\n= \\(\\frac{275,968}{3}\\)\n= \\(91989.33 cm^{3}\\)\n\\(\\approxeq 91989 cm^{3}\\) (nearest whole number).\n(b)\n\\(|ZX|^{2} = |YZ|^{2} + |YX|^{2} - 2|YZ||YX| \\cos Y\\)\n= \\(5^{2} + 3^{2} - 2(5)(3) \\cos 135\\)\n= \\(25 + 9 - 30(-0.7071)\\)\n= \\(34+ 21.213\\)\n\\(|ZX|^{2} = 55.213\\)\n\\(|ZX| = 7.43km \\approxeq 7 km\\)\n(ii) Using sine rule,\n\\(\\frac{\\sin \\theta}{5} = \\frac{\\sin 135}{7.43}\\)\n\\(\\sin \\theta = \\frac{5 \\times \\sin 135}{7.43}\\)\n= \\(\\frac{3.5355}{7.43}\\)\n\\(\\sin \\theta = 0.4758\\)\n\\(\\theta = \\sin^{-1} (0.4758)\\)\n= \\(28.41° \\approxeq 28°\\) (to the nearest degree).\nBearing of Z from X = 270° + 28.41° = 298.41°\n\\(\\approxeq\\) 298°.\n\n', '', 'Explanation\n(a) Total surface area of a sphere = \\(4\\pi r^{2}\\)\nLet the TSA of the smaller sphere be \\(S_{1}\\) with radius r and the bigger sphere be \\(S_{2}\\) with radius R.\n\\(\\frac{S_{1}}{S_{2}} = \\frac{9}{49}\\)\n\\(\\frac{9}{49} = \\frac{4\\pi r^{2}}{4\\pi R^{2}}\\)\n\\(\\implies \\frac{9}{49} = \\frac{12^{2}}{R^{2}}\\)\n\\(\\frac{49 \\times 144}{9} = R^{2}\\)\n\\(R = \\sqrt{49 \\times 16} = 28 cm\\)\nVolume of bigger sphere = \\(\\frac{4}{3} \\pi r^{3}\\)\n= \\(\\frac{4}{3} \\times \\frac{22}{7} \\times 28 \\times 28 \\times 28\\)\n= \\(\\frac{275,968}{3}\\)\n= \\(91989.33 cm^{3}\\)\n\\(\\approxeq 91989 cm^{3}\\) (nearest whole number).\n(b)\n\\(|ZX|^{2} = |YZ|^{2} + |YX|^{2} - 2|YZ||YX| \\cos Y\\)\n= \\(5^{2} + 3^{2} - 2(5)(3) \\cos 135\\)\n= \\(25 + 9 - 30(-0.7071)\\)\n= \\(34+ 21.213\\)\n\\(|ZX|^{2} = 55.213\\)\n\\(|ZX| = 7.43km \\approxeq 7 km\\)\n(ii) Using sine rule,\n\\(\\frac{\\sin \\theta}{5} = \\frac{\\sin 135}{7.43}\\)\n\\(\\sin \\theta = \\frac{5 \\times \\sin 135}{7.43}\\)\n= \\(\\frac{3.5355}{7.43}\\)\n\\(\\sin \\theta = 0.4758\\)\n\\(\\theta = \\sin^{-1} (0.4758)\\)\n= \\(28.41° \\approxeq 28°\\) (to the nearest degree).\nBearing of Z from X = 270° + 28.41° = 298.41°\n\\(\\approxeq\\) 298°.\n\n', '(a) Total surface area of a sphere = \\(4\\pi r^{2}\\)\nLet the TSA of the smaller sphere be \\(S_{1}\\) with radius r and the bigger sphere be \\(S_{2}\\) with radius R.\n\\(\\frac{S_{1}}{S_{2}} = \\frac{9}{49}\\)\n\\(\\frac{9}{49} = \\frac{4\\pi r^{2}}{4\\pi R^{2}}\\)\n\\(\\implies \\frac{9}{49} = \\frac{12^{2}}{R^{2}}\\)\n\\(\\frac{49 \\times 144}{9} = R^{2}\\)\n\\(R = \\sqrt{49 \\times 16} = 28 cm\\)\nVolume of bigger sphere = \\(\\frac{4}{3} \\pi r^{3}\\)\n= \\(\\frac{4}{3} \\times \\frac{22}{7} \\times 28 \\times 28 \\times 28\\)\n= \\(\\frac{275,968}{3}\\)\n= \\(91989.33 cm^{3}\\)\n\\(\\approxeq 91989 cm^{3}\\) (nearest whole number).~(b)\n\\(|ZX|^{2} = |YZ|^{2} + |YX|^{2} - 2|YZ||YX| \\cos Y\\)\n= \\(5^{2} + 3^{2} - 2(5)(3) \\cos 135\\)\n= \\(25 + 9 - 30(-0.7071)\\)\n= \\(34+ 21.213\\)\n\\(|ZX|^{2} = 55.213\\)\n\\(|ZX| = 7.43km \\approxeq 7 km\\)\n(ii) Using sine rule,\n\\(\\frac{\\sin \\theta}{5} = \\frac{\\sin 135}{7.43}\\)\n\\(\\sin \\theta = \\frac{5 \\times \\sin 135}{7.43}\\)\n= \\(\\frac{3.5355}{7.43}\\)\n\\(\\sin \\theta = 0.4758\\)\n\\(\\theta = \\sin^{-1} (0.4758)\\)\n= \\(28.41° \\approxeq 28°\\) (to the nearest degree).\nBearing of Z from X = 270° + 28.41° = 298.41°\n\\(\\approxeq\\) 298°.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(418, '2020', '\n\nQuestion\n\nThe table shows the scores obtained when a fair die was thrown a number of times.\n\n\n\nScore\n1\n2\n3\n4\n5\n6\n\n\nFrequency\n2\n5\nx\n11\n9\n10\n\n\n\nIf the probability of obtaining a 3 is 0.26, find the (a) median\n(b) standard deviation of the distribution.\n\n\nThe table shows the scores obtained when a fair die was thrown a number of times.\n\n\n\nScore\n1\n2\n3\n4\n5\n6\n\n\nFrequency\n2\n5\nx\n11\n9\n10\n\n\n', '', 'The table shows the scores obtained when a fair die was thrown a number of times.\n\n\n\nScore\n1\n2\n3\n4\n5\n6\n\n\nFrequency\n2\n5\nx\n11\n9\n10\n\n\n\nIf the probability of obtaining a 3 is 0.26, find the (a) median\n(b) standard deviation of the distribution.\n\n\nThe table shows the scores obtained when a fair die was thrown a number of times.\n\n\n\nScore\n1\n2\n3\n4\n5\n6\n\n\nFrequency\n2\n5\nx\n11\n9\n10\n', 'The table shows the scores obtained when a fair die was thrown a number of times.\n\n\n\nScore\n1\n2\n3\n4\n5\n6\n\n\nFrequency\n2\n5\nx\n11\n9\n10\n\n\n\nIf the probability of obtaining a 3 is 0.26, find the (a) median~(b) standard deviation of the distribution.\n\n\nThe table shows the scores obtained when a fair die was thrown a number of times.\n\n\n\nScore\n1\n2\n3\n4\n5\n6\n\n\nFrequency\n2\n5\nx\n11\n9\n10\n', NULL, '\n\nExplanation\nScore\n1\n2\n3\n4\n5\n6\n\n\nFrequency\n2\n5\nx\n11\n9\n10\n\n\n\nP(3) = 0.26\n\\(\\frac{x}{37 + x} = 0.26\\)\n\\(x = 0.26(37 + x) \\implies x = 9.62 + 0.26x\\)\n\\(x - 0.26x = 9.62 \\implies 0.74x = 9.62\\)\n\\(x = \\frac{9.62}{0.74} = 13\\)\nTotal toss frequency = 37 + x\n= 37 + 13\n= 50.\n(a) Median = \\(\\frac{n + 1}{2}\\)\n= \\(\\frac{50 + 1}{2}\\)\n= 25.5\nFrom the table, 25.5 occurs at 4 which is the median.\n(b) Standard deviation = \\(\\sqrt{\\frac{\\sum fd^{2}}{\\sum f}}\\)\n\n\n\n\\(x\\)\n\\(f\\)\n\\(\\fx\\)\n\\(d = |x - \\bar{x}|\\)\n\\(d^{2}\\)\n\\(fd^{2}\\)\n\n\n1\n2\n2\n-3\n9\n18\n\n\n2\n5\n10\n-2\n4\n20\n\n\n3\n13\n39\n-1\n1\n13\n\n\n4\n11\n44\n0\n0\n0\n\n\n5\n9\n45\n1\n1\n9\n\n\n6\n10\n60\n2\n4\n40\n\n\n\\(\\sum\\)\n50\n200\n\n\n100\n\n\n\n\\(Mean(\\bar{x}) = \\frac{\\sum fx}{\\sum f}\\)\n= \\(\\frac{200}{50}\n= 4\nS.D = \\(\\sqrt{\\frac{100}{50}}\\)\n= \\(\\sqrt{2}\\)\n= 1.4142\n\n', '', 'Explanation\nScore\n1\n2\n3\n4\n5\n6\n\n\nFrequency\n2\n5\nx\n11\n9\n10\n\n\n\nP(3) = 0.26\n\\(\\frac{x}{37 + x} = 0.26\\)\n\\(x = 0.26(37 + x) \\implies x = 9.62 + 0.26x\\)\n\\(x - 0.26x = 9.62 \\implies 0.74x = 9.62\\)\n\\(x = \\frac{9.62}{0.74} = 13\\)\nTotal toss frequency = 37 + x\n= 37 + 13\n= 50.\n(a) Median = \\(\\frac{n + 1}{2}\\)\n= \\(\\frac{50 + 1}{2}\\)\n= 25.5\nFrom the table, 25.5 occurs at 4 which is the median.\n(b) Standard deviation = \\(\\sqrt{\\frac{\\sum fd^{2}}{\\sum f}}\\)\n\n\n\n\\(x\\)\n\\(f\\)\n\\(\\fx\\)\n\\(d = |x - \\bar{x}|\\)\n\\(d^{2}\\)\n\\(fd^{2}\\)\n\n\n1\n2\n2\n-3\n9\n18\n\n\n2\n5\n10\n-2\n4\n20\n\n\n3\n13\n39\n-1\n1\n13\n\n\n4\n11\n44\n0\n0\n0\n\n\n5\n9\n45\n1\n1\n9\n\n\n6\n10\n60\n2\n4\n40\n\n\n\\(\\sum\\)\n50\n200\n\n\n100\n\n\n\n\\(Mean(\\bar{x}) = \\frac{\\sum fx}{\\sum f}\\)\n= \\(\\frac{200}{50}\n= 4\nS.D = \\(\\sqrt{\\frac{100}{50}}\\)\n= \\(\\sqrt{2}\\)\n= 1.4142\n\n', '(a) Median = \\(\\frac{n + 1}{2}\\)\n= \\(\\frac{50 + 1}{2}\\)\n= 25.5\nFrom the table, 25.5 occurs at 4 which is the median.~(b) Standard deviation = \\(\\sqrt{\\frac{\\sum fd^{2}}{\\sum f}}\\)\n\n\n\n\\(x\\)\n\\(f\\)\n\\(\\fx\\)\n\\(d = |x - \\bar{x}|\\)\n\\(d^{2}\\)\n\\(fd^{2}\\)\n\n\n1\n2\n2\n-3\n9\n18\n\n\n2\n5\n10\n-2\n4\n20\n\n\n3\n13\n39\n-1\n1\n13\n\n\n4\n11\n44\n0\n0\n0\n\n\n5\n9\n45\n1\n1\n9\n\n\n6\n10\n60\n2\n4\n40\n\n\n\\(\\sum\\)\n50\n200\n\n\n100\n\n\n\n\\(Mean(\\bar{x}) = \\frac{\\sum fx}{\\sum f}\\)\n= \\(\\frac{200}{50}\n= 4\nS.D = \\(\\sqrt{\\frac{100}{50}}\\)\n= \\(\\sqrt{2}\\)\n= 1.4142\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(419, '2020', '\n\nQuestion\n\n(a) The area of trapezium PQRS is 60\\(cm^{2}\\). PQ // RS, /PQ/ = 15 cm, /RS/ = 25 cm and < PSR = 60°. Calculate the : (i) perpendicular height of PQRS ; (ii) |PS|.\n(b) Ade received \\(\\frac{3}{5}\\) of a sum of money, Nelly \\(\\frac{1}{3}\\) of the remainder while Austin took the rest. If Austin\'s share is greater than Nelly\'s share by N3,000, how much did Ade get?\n\n', '', '(a) The area of trapezium PQRS is 60\\(cm^{2}\\). PQ // RS, /PQ/ = 15 cm, /RS/ = 25 cm and < PSR = 60°. Calculate the : (i) perpendicular height of PQRS ; (ii) |PS|.\n(b) Ade received \\(\\frac{3}{5}\\) of a sum of money, Nelly \\(\\frac{1}{3}\\) of the remainder while Austin took the rest. If Austin\'s share is greater than Nelly\'s share by N3,000, how much did Ade get?', '(a) The area of trapezium PQRS is 60\\(cm^{2}\\). PQ // RS, /PQ/ = 15 cm, /RS/ = 25 cm and < PSR = 60°. Calculate the : (i) perpendicular height of PQRS ; (ii) |PS|.~(b) Ade received \\(\\frac{3}{5}\\) of a sum of money, Nelly \\(\\frac{1}{3}\\) of the remainder while Austin took the rest. If Austin\'s share is greater than Nelly\'s share by N3,000, how much did Ade get?', NULL, '\n\nExplanation\n(a)\n\\(\\theta = \\frac{60°}{2} = 30°\\)\nArea of trapezium = \\(\\frac{a + b}{2} h\\)\n\\(60 = \\frac{15 + 25}{2} h \\implies 60 = 20h\\)\n\\(h = \\frac{60}{20} = 3 cm\\)\n(ii) Considering \\(\\Delta PSR\\),\n\\(\\sin \\theta = \\frac{Opp}{Hyp}\\)\n\\(\\sin 30 = \\frac{3}{|PS|}\\)\n\\(|PS| = \\frac{3}{\\sin 30}\\)\n= \\(\\frac{3}{0.5} = 6 cm\\)\n(b) Let the total amount to be shared = k.\nAde\'s share = \\(\\frac{3}{5} k \\)\nRemainder = \\(k - \\frac{3}{5}k \\)\n= \\(\\frac{2}{5}k\\)\nNelly\'s share : \\(\\frac{1}{3} \\times \\frac{2}{5}k\\)\n= \\(\\frac{2}{15}k\\)\nAustin\'s share : \\(\\frac{2}{5}k - \\frac{2}{15}k\\)\n= \\(\\frac{4}{15}k\\)\n\\(\\frac{4}{15}k - \\frac{2}{15}k = \\frac{2}{15}k\\)\n\\(\\frac{2}{15}k = N3,000\\)\n\\(\\implies k = \\frac{3,000 \\times 15}{2}\\)\n= \\(N22,500\\)\nAde\'s share = \\(\\frac{3}{5}k\\)\n= \\(\\frac{3}{5} \\times N22,500 = N13,500\\).\n\n\n\n', '', 'Explanation\n(a)\n\\(\\theta = \\frac{60°}{2} = 30°\\)\nArea of trapezium = \\(\\frac{a + b}{2} h\\)\n\\(60 = \\frac{15 + 25}{2} h \\implies 60 = 20h\\)\n\\(h = \\frac{60}{20} = 3 cm\\)\n(ii) Considering \\(\\Delta PSR\\),\n\\(\\sin \\theta = \\frac{Opp}{Hyp}\\)\n\\(\\sin 30 = \\frac{3}{|PS|}\\)\n\\(|PS| = \\frac{3}{\\sin 30}\\)\n= \\(\\frac{3}{0.5} = 6 cm\\)\n(b) Let the total amount to be shared = k.\nAde\'s share = \\(\\frac{3}{5} k \\)\nRemainder = \\(k - \\frac{3}{5}k \\)\n= \\(\\frac{2}{5}k\\)\nNelly\'s share : \\(\\frac{1}{3} \\times \\frac{2}{5}k\\)\n= \\(\\frac{2}{15}k\\)\nAustin\'s share : \\(\\frac{2}{5}k - \\frac{2}{15}k\\)\n= \\(\\frac{4}{15}k\\)\n\\(\\frac{4}{15}k - \\frac{2}{15}k = \\frac{2}{15}k\\)\n\\(\\frac{2}{15}k = N3,000\\)\n\\(\\implies k = \\frac{3,000 \\times 15}{2}\\)\n= \\(N22,500\\)\nAde\'s share = \\(\\frac{3}{5}k\\)\n= \\(\\frac{3}{5} \\times N22,500 = N13,500\\).\n\n\n\n', '(a)\n\\(\\theta = \\frac{60°}{2} = 30°\\)\nArea of trapezium = \\(\\frac{a + b}{2} h\\)\n\\(60 = \\frac{15 + 25}{2} h \\implies 60 = 20h\\)\n\\(h = \\frac{60}{20} = 3 cm\\)\n(ii) Considering \\(\\Delta PSR\\),\n\\(\\sin \\theta = \\frac{Opp}{Hyp}\\)\n\\(\\sin 30 = \\frac{3}{|PS|}\\)\n\\(|PS| = \\frac{3}{\\sin 30}\\)\n= \\(\\frac{3}{0.5} = 6 cm\\)~(b) Let the total amount to be shared = k.\nAde\'s share = \\(\\frac{3}{5} k \\)\nRemainder = \\(k - \\frac{3}{5}k \\)\n= \\(\\frac{2}{5}k\\)\nNelly\'s share : \\(\\frac{1}{3} \\times \\frac{2}{5}k\\)\n= \\(\\frac{2}{15}k\\)\nAustin\'s share : \\(\\frac{2}{5}k - \\frac{2}{15}k\\)\n= \\(\\frac{4}{15}k\\)\n\\(\\frac{4}{15}k - \\frac{2}{15}k = \\frac{2}{15}k\\)\n\\(\\frac{2}{15}k = N3,000\\)\n\\(\\implies k = \\frac{3,000 \\times 15}{2}\\)\n= \\(N22,500\\)\nAde\'s share = \\(\\frac{3}{5}k\\)\n= \\(\\frac{3}{5} \\times N22,500 = N13,500\\).\n\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(420, '2020', '\n\nQuestion\n\n(a) P varies directly as Q and inversely as the square of R. If P = 1 when Q = 8 and R = 2, find the value of Q when P = 3 and R = 5.\n(b) An aeroplane flies from town A(20°N, 60°E) to town B(20°N, 20°E). (i) if the journey takes 6 hours, calculate, correct to 3 significant figures, the average speed of the aeroplane. (ii) if it then flies due North from town B to town C, 420 km away, calculate correct to the nearest degree, the latitude of town C. [Take radius of the earth = 6400 km and \\(\\pi\\) = 3.142].\n\n', '', '(a) P varies directly as Q and inversely as the square of R. If P = 1 when Q = 8 and R = 2, find the value of Q when P = 3 and R = 5.\n(b) An aeroplane flies from town A(20°N, 60°E) to town B(20°N, 20°E). (i) if the journey takes 6 hours, calculate, correct to 3 significant figures, the average speed of the aeroplane. (ii) if it then flies due North from town B to town C, 420 km away, calculate correct to the nearest degree, the latitude of town C. [Take radius of the earth = 6400 km and \\(\\pi\\) = 3.142].', '(a) P varies directly as Q and inversely as the square of R. If P = 1 when Q = 8 and R = 2, find the value of Q when P = 3 and R = 5.~(b) An aeroplane flies from town A(20°N, 60°E) to town B(20°N, 20°E). (i) if the journey takes 6 hours, calculate, correct to 3 significant figures, the average speed of the aeroplane. (ii) if it then flies due North from town B to town C, 420 km away, calculate correct to the nearest degree, the latitude of town C. [Take radius of the earth = 6400 km and \\(\\pi\\) = 3.142].', NULL, '\n\nExplanation\n(a) \\(P \\propto Q\\) and \\(P \\propto \\frac{1}{R^{2}}\\).\n\\(P \\propto \\frac{Q}{R^{2}} \\implies P = \\frac{kQ}{R^{2}}\\)\nWhen P = 1, Q = 8, R = 2\n\\(1 = \\frac{8k}{2^{2}} \\implies 8k = 4\\)\n\\(k = \\frac{4}{8} = 0.5\\)\n\\(P = \\frac{Q}{2R^{2}}\\)\nWhen P = 3, R = 5, Q = ?\n\\(3 = \\frac{Q}{2(5^{2})}\\)\n\\(3 = \\frac{Q}{50} \\implies Q = 150\\).\n(b)\nAngular difference : 60° - 20° = 40°\n\\(\\frac{\\theta}{360} \\times 2\\pi r \\)\n\\(r = R\\cos \\theta\\)\n\\(\\frac{40}{360} \\times 2 \\times 3.142 \\times 6400 \\cos 20 = \\frac{1511687.278}{360}\\)\n= \\(4199.13 km\\)\n\\(Speed = \\frac{Distance}{Time}\\)\n= \\(\\frac{4199.13}{6}\\)\n= \\(699.85 km/hr\\)\n\\(\\approxeq 700 km/hr\\) (to 3 significant figures)\n(ii)\\(D= \\frac{\\theta}{360} \\times 2\\pi r\\)\n\\(420 = \\frac{\\theta}{360} \\times 2 \\times 3.142 \\times 6400\\)\n\\(\\theta = \\frac{420 \\times 360}{2 \\times 3.142 \\times 6400}\\)\n\\(\\theta = \\frac{151200}{40217.6} = 3.76°\\)\nLatitude of C : 20° + 3.76° = 23.76°\nNote that due North implies same longitude but different latitude.\n\n', '', 'Explanation\n(a) \\(P \\propto Q\\) and \\(P \\propto \\frac{1}{R^{2}}\\).\n\\(P \\propto \\frac{Q}{R^{2}} \\implies P = \\frac{kQ}{R^{2}}\\)\nWhen P = 1, Q = 8, R = 2\n\\(1 = \\frac{8k}{2^{2}} \\implies 8k = 4\\)\n\\(k = \\frac{4}{8} = 0.5\\)\n\\(P = \\frac{Q}{2R^{2}}\\)\nWhen P = 3, R = 5, Q = ?\n\\(3 = \\frac{Q}{2(5^{2})}\\)\n\\(3 = \\frac{Q}{50} \\implies Q = 150\\).\n(b)\nAngular difference : 60° - 20° = 40°\n\\(\\frac{\\theta}{360} \\times 2\\pi r \\)\n\\(r = R\\cos \\theta\\)\n\\(\\frac{40}{360} \\times 2 \\times 3.142 \\times 6400 \\cos 20 = \\frac{1511687.278}{360}\\)\n= \\(4199.13 km\\)\n\\(Speed = \\frac{Distance}{Time}\\)\n= \\(\\frac{4199.13}{6}\\)\n= \\(699.85 km/hr\\)\n\\(\\approxeq 700 km/hr\\) (to 3 significant figures)\n(ii)\\(D= \\frac{\\theta}{360} \\times 2\\pi r\\)\n\\(420 = \\frac{\\theta}{360} \\times 2 \\times 3.142 \\times 6400\\)\n\\(\\theta = \\frac{420 \\times 360}{2 \\times 3.142 \\times 6400}\\)\n\\(\\theta = \\frac{151200}{40217.6} = 3.76°\\)\nLatitude of C : 20° + 3.76° = 23.76°\nNote that due North implies same longitude but different latitude.\n\n', '(a) \\(P \\propto Q\\) and \\(P \\propto \\frac{1}{R^{2}}\\).\n\\(P \\propto \\frac{Q}{R^{2}} \\implies P = \\frac{kQ}{R^{2}}\\)\nWhen P = 1, Q = 8, R = 2\n\\(1 = \\frac{8k}{2^{2}} \\implies 8k = 4\\)\n\\(k = \\frac{4}{8} = 0.5\\)\n\\(P = \\frac{Q}{2R^{2}}\\)\nWhen P = 3, R = 5, Q = ?\n\\(3 = \\frac{Q}{2(5^{2})}\\)\n\\(3 = \\frac{Q}{50} \\implies Q = 150\\).~(b)\nAngular difference : 60° - 20° = 40°\n\\(\\frac{\\theta}{360} \\times 2\\pi r \\)\n\\(r = R\\cos \\theta\\)\n\\(\\frac{40}{360} \\times 2 \\times 3.142 \\times 6400 \\cos 20 = \\frac{1511687.278}{360}\\)\n= \\(4199.13 km\\)\n\\(Speed = \\frac{Distance}{Time}\\)\n= \\(\\frac{4199.13}{6}\\)\n= \\(699.85 km/hr\\)\n\\(\\approxeq 700 km/hr\\) (to 3 significant figures)\n(ii)\\(D= \\frac{\\theta}{360} \\times 2\\pi r\\)\n\\(420 = \\frac{\\theta}{360} \\times 2 \\times 3.142 \\times 6400\\)\n\\(\\theta = \\frac{420 \\times 360}{2 \\times 3.142 \\times 6400}\\)\n\\(\\theta = \\frac{151200}{40217.6} = 3.76°\\)\nLatitude of C : 20° + 3.76° = 23.76°\nNote that due North implies same longitude but different latitude.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(421, '2020', '\n\nQuestion\n\nA = {2, 4, 6, 8}, B = {2, 3, 7, 9} and C = {x : 3 < x < 9} are subsets of the universal set U = {2, 3, 4, 5, 6, 7, 8, 9}. Find\n(a) \\(A \\cap (B\' \\cap C\')\\) ;\n(b) \\((A \\cup B) \\cap (B \\cup C)\\).\n\n', '', 'A = {2, 4, 6, 8}, B = {2, 3, 7, 9} and C = {x : 3 < x < 9} are subsets of the universal set U = {2, 3, 4, 5, 6, 7, 8, 9}. Find\n(a) \\(A \\cap (B\' \\cap C\')\\) ;\n(b) \\((A \\cup B) \\cap (B \\cup C)\\).', 'A = {2, 4, 6, 8}, B = {2, 3, 7, 9} and C = {x : 3 < x < 9} are subsets of the universal set U = {2, 3, 4, 5, 6, 7, 8, 9}. Find\n(a) \\(A \\cap (B\' \\cap C\')\\) ;~(b) \\((A \\cup B) \\cap (B \\cup C)\\).', NULL, '\n\nExplanation\n\\(U = {2, 3, 4, 5, 6, 7, 8, 9}\\)\n\\(A = {2, 4, 6, 8}\\)\n\\(B = {2, 3, 7, 9}\\)\n\\(C = {4, 5, 6, 7, 8}\\)\n(a) \\(A \\cap (B\' \\cap C\')\\) = \\({2, 4, 6, 8} \\cap ({4, 5, 6, 8} \\cap {2, 3, 9})\\)\n= \\({2, 4, 6, 8} \\cap { }\\)\n= \\({ }\\)\n(b) \\((A \\cup B) \\cap (B \\cup C) = {2, 3, 4, 6, 7, 8, 9} \\cap {2, 3, 4, 5, 6, 7, 8, 9}\\)\n= \\({2, 3, 4, 6, 7, 8, 9}\\)\n\n', '', 'Explanation\n\\(U = {2, 3, 4, 5, 6, 7, 8, 9}\\)\n\\(A = {2, 4, 6, 8}\\)\n\\(B = {2, 3, 7, 9}\\)\n\\(C = {4, 5, 6, 7, 8}\\)\n(a) \\(A \\cap (B\' \\cap C\')\\) = \\({2, 4, 6, 8} \\cap ({4, 5, 6, 8} \\cap {2, 3, 9})\\)\n= \\({2, 4, 6, 8} \\cap { }\\)\n= \\({ }\\)\n(b) \\((A \\cup B) \\cap (B \\cup C) = {2, 3, 4, 6, 7, 8, 9} \\cap {2, 3, 4, 5, 6, 7, 8, 9}\\)\n= \\({2, 3, 4, 6, 7, 8, 9}\\)\n\n', '(a) \\(A \\cap (B\' \\cap C\')\\) = \\({2, 4, 6, 8} \\cap ({4, 5, 6, 8} \\cap {2, 3, 9})\\)\n= \\({2, 4, 6, 8} \\cap { }\\)\n= \\({ }\\)~(b) \\((A \\cup B) \\cap (B \\cup C) = {2, 3, 4, 6, 7, 8, 9} \\cap {2, 3, 4, 5, 6, 7, 8, 9}\\)\n= \\({2, 3, 4, 6, 7, 8, 9}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(422, '2020', '\n\nQuestion\n\n(a) The angle of depression of a boat from the mid-point of a vertical cliff is 35°. If the boat is 120m from the foot of the cliff, calculate the height of the cliff.\n(b) Towns P and Q are x km apart. Two motorists set out at the same time from P to Q at steady speeds of 60 km/h and 80 km/h. The faster motorist got to Q 30 minutes earlier than the other. Find the value of x.\n\n', '', '(a) The angle of depression of a boat from the mid-point of a vertical cliff is 35°. If the boat is 120m from the foot of the cliff, calculate the height of the cliff.\n(b) Towns P and Q are x km apart. Two motorists set out at the same time from P to Q at steady speeds of 60 km/h and 80 km/h. The faster motorist got to Q 30 minutes earlier than the other. Find the value of x.', '(a) The angle of depression of a boat from the mid-point of a vertical cliff is 35°. If the boat is 120m from the foot of the cliff, calculate the height of the cliff.~(b) Towns P and Q are x km apart. Two motorists set out at the same time from P to Q at steady speeds of 60 km/h and 80 km/h. The faster motorist got to Q 30 minutes earlier than the other. Find the value of x.', NULL, '\n\nExplanation\n(a)\n\\(\\frac{FM}{FB} = \\tan 35\\)\n\\(\\therefore \\frac{FM}{120} = \\tan 35\\)\n\\(FM = 120 \\tan 35 \\)\n= \\(120 \\times 0.7002\\)\n= \\(84.024 m\\)\nTherefore, the height of the cliff = \\(2 \\times 84.024 = 168.048m\\)\n(b) Distance between P and Q = x km\nSpeed for 1st motorist= 60 km/h\n\\(\\therefore \\text{1st motorist\'s time} = \\frac{x}{60} hrs\\)\nSpeed for 2nd motorist= 80 km/h\n\\(\\therefore \\text{2nd motorist\'s time} = \\frac{x}{80} hrs\\)\n\\(\\frac{x}{60} - \\frac{x}{80} = \\frac{30}{60}\\)\n\\(\\frac{4x - 3x}{240} = \\frac{1}{2}\\)\n\\(\\frac{x}{240} = \\frac{1}{2} \\implies 2x = 240\\)\n\\(x = 120 km\\).\n\n', '', 'Explanation\n(a)\n\\(\\frac{FM}{FB} = \\tan 35\\)\n\\(\\therefore \\frac{FM}{120} = \\tan 35\\)\n\\(FM = 120 \\tan 35 \\)\n= \\(120 \\times 0.7002\\)\n= \\(84.024 m\\)\nTherefore, the height of the cliff = \\(2 \\times 84.024 = 168.048m\\)\n(b) Distance between P and Q = x km\nSpeed for 1st motorist= 60 km/h\n\\(\\therefore \\text{1st motorist\'s time} = \\frac{x}{60} hrs\\)\nSpeed for 2nd motorist= 80 km/h\n\\(\\therefore \\text{2nd motorist\'s time} = \\frac{x}{80} hrs\\)\n\\(\\frac{x}{60} - \\frac{x}{80} = \\frac{30}{60}\\)\n\\(\\frac{4x - 3x}{240} = \\frac{1}{2}\\)\n\\(\\frac{x}{240} = \\frac{1}{2} \\implies 2x = 240\\)\n\\(x = 120 km\\).\n\n', '(a)\n\\(\\frac{FM}{FB} = \\tan 35\\)\n\\(\\therefore \\frac{FM}{120} = \\tan 35\\)\n\\(FM = 120 \\tan 35 \\)\n= \\(120 \\times 0.7002\\)\n= \\(84.024 m\\)\nTherefore, the height of the cliff = \\(2 \\times 84.024 = 168.048m\\)~(b) Distance between P and Q = x km\nSpeed for 1st motorist= 60 km/h\n\\(\\therefore \\text{1st motorist\'s time} = \\frac{x}{60} hrs\\)\nSpeed for 2nd motorist= 80 km/h\n\\(\\therefore \\text{2nd motorist\'s time} = \\frac{x}{80} hrs\\)\n\\(\\frac{x}{60} - \\frac{x}{80} = \\frac{30}{60}\\)\n\\(\\frac{4x - 3x}{240} = \\frac{1}{2}\\)\n\\(\\frac{x}{240} = \\frac{1}{2} \\implies 2x = 240\\)\n\\(x = 120 km\\).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(423, '2020', '\n\nQuestion\n\n(a)\nIn the diagram, < PQR = 125°, < QRS = r, < RST = 80° and < STU = 44°. Calculate the value of r.\n(b) In the diagram TS is a tangent to the circle at A. AB // CE, < AEC = 5x°, < ADB = 60° and < TAE = x. Find the value of x.\n\n', '', '(a)\nIn the diagram, < PQR = 125°, < QRS = r, < RST = 80° and < STU = 44°. Calculate the value of r.\n(b) In the diagram TS is a tangent to the circle at A. AB // CE, < AEC = 5x°, < ADB = 60° and < TAE = x. Find the value of x.', '(a)\nIn the diagram, < PQR = 125°, < QRS = r, < RST = 80° and < STU = 44°. Calculate the value of r.~(b) In the diagram TS is a tangent to the circle at A. AB // CE, < AEC = 5x°, < ADB = 60° and < TAE = x. Find the value of x.', NULL, '\n\nExplanation\n(a)\\(\\therefore r = 36° + 55° = 91°\\)\n(b)\n\\(< BAS = 60°\\) (angles in alternate segments)\n\\(< BAE = 180° - 5x°\\) (adjacent angles on a transversal)\n\\(\\therefore < BAS + < BAE + < EAT = 180°\\)\n\\(\\therefore 60° + 180° - 5x + x = 180°\\)\n\\(\\implies 240° - 180° = 4x\\)\n\\(60° = 4x \\implies x = \\frac{60}{4} = 15°\\)\n\n\n', '', 'Explanation\n(a)\\(\\therefore r = 36° + 55° = 91°\\)\n(b)\n\\(< BAS = 60°\\) (angles in alternate segments)\n\\(< BAE = 180° - 5x°\\) (adjacent angles on a transversal)\n\\(\\therefore < BAS + < BAE + < EAT = 180°\\)\n\\(\\therefore 60° + 180° - 5x + x = 180°\\)\n\\(\\implies 240° - 180° = 4x\\)\n\\(60° = 4x \\implies x = \\frac{60}{4} = 15°\\)\n\n\n', '(a)\\(\\therefore r = 36° + 55° = 91°\\)~(b)\n\\(< BAS = 60°\\) (angles in alternate segments)\n\\(< BAE = 180° - 5x°\\) (adjacent angles on a transversal)\n\\(\\therefore < BAS + < BAE + < EAT = 180°\\)\n\\(\\therefore 60° + 180° - 5x + x = 180°\\)\n\\(\\implies 240° - 180° = 4x\\)\n\\(60° = 4x \\implies x = \\frac{60}{4} = 15°\\)\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(424, '2020', '\n\nQuestion\n\n(a)\nCurved Surface Area = \\(\\pi rl\\)\n\\(115.5 = \\frac{22}{7} \\times r \\times 10.5\\)\n\\(115.5 = 33r\\)\n\\(r = \\frac{115.5}{33} = 3.5 cm\\)\n(b)\n\\(\\therefore h^{2} +(3.50)^{2} = (10.5)^{2}\\)\n\\(h^{2} = 10.5^{2} - 3.5^{2}\\)\n\\(h^{2} = 98 \\implies h = \\sqrt{98}\\)\n\\(h = 9.8994 cm \\approxeq 9.90 cm\\)\n(c) Volume of a cone = \\(\\frac{1}{3} \\pi r^{2} h\\)\n= \\(\\frac{1}{3} \\times \\frac{22}{7} \\times \\frac{7}{2} \\times \\frac{7}{2} \\times 9.90\\)\n= \\(\\frac{23.1 \\times 11}{2}\\)\n= \\(127.05 cm^{3} \\approxeq 127 cm^{3}\\)\n\n', '', '(a)\nCurved Surface Area = \\(\\pi rl\\)\n\\(115.5 = \\frac{22}{7} \\times r \\times 10.5\\)\n\\(115.5 = 33r\\)\n\\(r = \\frac{115.5}{33} = 3.5 cm\\)\n(b)\n\\(\\therefore h^{2} +(3.50)^{2} = (10.5)^{2}\\)\n\\(h^{2} = 10.5^{2} - 3.5^{2}\\)\n\\(h^{2} = 98 \\implies h = \\sqrt{98}\\)\n\\(h = 9.8994 cm \\approxeq 9.90 cm\\)\n(c) Volume of a cone = \\(\\frac{1}{3} \\pi r^{2} h\\)\n= \\(\\frac{1}{3} \\times \\frac{22}{7} \\times \\frac{7}{2} \\times \\frac{7}{2} \\times 9.90\\)\n= \\(\\frac{23.1 \\times 11}{2}\\)\n= \\(127.05 cm^{3} \\approxeq 127 cm^{3}\\)', '(a)\nCurved Surface Area = \\(\\pi rl\\)\n\\(115.5 = \\frac{22}{7} \\times r \\times 10.5\\)\n\\(115.5 = 33r\\)\n\\(r = \\frac{115.5}{33} = 3.5 cm\\)~(b)\n\\(\\therefore h^{2} +(3.50)^{2} = (10.5)^{2}\\)\n\\(h^{2} = 10.5^{2} - 3.5^{2}\\)\n\\(h^{2} = 98 \\implies h = \\sqrt{98}\\)\n\\(h = 9.8994 cm \\approxeq 9.90 cm\\)\n(c) Volume of a cone = \\(\\frac{1}{3} \\pi r^{2} h\\)\n= \\(\\frac{1}{3} \\times \\frac{22}{7} \\times \\frac{7}{2} \\times \\frac{7}{2} \\times 9.90\\)\n= \\(\\frac{23.1 \\times 11}{2}\\)\n= \\(127.05 cm^{3} \\approxeq 127 cm^{3}\\)~(c) Volume of a cone = \\(\\frac{1}{3} \\pi r^{2} h\\)\n= \\(\\frac{1}{3} \\times \\frac{22}{7} \\times \\frac{7}{2} \\times \\frac{7}{2} \\times 9.90\\)\n= \\(\\frac{23.1 \\times 11}{2}\\)\n= \\(127.05 cm^{3} \\approxeq 127 cm^{3}\\)', NULL, '\n\nExplanation\nNo official explanation is available for this question at this time. Please check contributions posted by others below. If you can provide an explanation to help other student learn, please click here\n\n', 'No official explanation is available for this question at this time. Please check contributions posted by others below. If you can provide an explanation to help other student learn, please click here\n\n', '', '', 'WAEC-PK-1', 'WAEC-MATH', 1),
(425, '2020', '\n\nQuestion\n\nTwo fair die are thrown. M is the event described by \"The sum of the scores is 10\" and N is the event described by \"The difference between the scores is 3\".\n(a) Write out the elements of M and N.\n(b) Find the probability of M or N.\n(c) Are M and N mutually exclusive? Give reasons.\n\n', '', 'Two fair die are thrown. M is the event described by \"The sum of the scores is 10\" and N is the event described by \"The difference between the scores is 3\".\n(a) Write out the elements of M and N.\n(b) Find the probability of M or N.\n(c) Are M and N mutually exclusive? Give reasons.', 'Two fair die are thrown. M is the event described by \"The sum of the scores is 10\" and N is the event described by \"The difference between the scores is 3\".\n(a) Write out the elements of M and N.~(b) Find the probability of M or N.\n(c) Are M and N mutually exclusive? Give reasons.~(c) Are M and N mutually exclusive? Give reasons.', NULL, '\n\nExplanation\n(a) M = {(4, 6), (5, 5), (6,4)}\nN = {(1, 4), (2, 5), (3, 6), (4, 1), (5, 2), (6, 3)}\n(b) Pr(M or N) = Pr(M) + Pr(N)\nTotal sample = 36.\n\\(Pr(M) = \\frac{3}{36} = \\frac{1}{12}\\)\n\\(Pr(N) = \\frac{6}{36} = \\frac{1}{6}\\)\n\\(Pr(\\text{M or N}) = \\frac{1}{12} + \\frac{1}{6}\\)\n= \\(\\frac{2 + 1}{12}\\)\n= \\(\\frac{3}{12} = \\frac{1}{4}\\)\n(c) Yes, they are mutually exclusive since \\(M \\cap N = \\emptyset\\).\n\n', '', 'Explanation\n(a) M = {(4, 6), (5, 5), (6,4)}\nN = {(1, 4), (2, 5), (3, 6), (4, 1), (5, 2), (6, 3)}\n(b) Pr(M or N) = Pr(M) + Pr(N)\nTotal sample = 36.\n\\(Pr(M) = \\frac{3}{36} = \\frac{1}{12}\\)\n\\(Pr(N) = \\frac{6}{36} = \\frac{1}{6}\\)\n\\(Pr(\\text{M or N}) = \\frac{1}{12} + \\frac{1}{6}\\)\n= \\(\\frac{2 + 1}{12}\\)\n= \\(\\frac{3}{12} = \\frac{1}{4}\\)\n(c) Yes, they are mutually exclusive since \\(M \\cap N = \\emptyset\\).\n\n', '(a) M = {(4, 6), (5, 5), (6,4)}\nN = {(1, 4), (2, 5), (3, 6), (4, 1), (5, 2), (6, 3)}~(b) Pr(M or N) = Pr(M) + Pr(N)\nTotal sample = 36.\n\\(Pr(M) = \\frac{3}{36} = \\frac{1}{12}\\)\n\\(Pr(N) = \\frac{6}{36} = \\frac{1}{6}\\)\n\\(Pr(\\text{M or N}) = \\frac{1}{12} + \\frac{1}{6}\\)\n= \\(\\frac{2 + 1}{12}\\)\n= \\(\\frac{3}{12} = \\frac{1}{4}\\)\n(c) Yes, they are mutually exclusive since \\(M \\cap N = \\emptyset\\).\n~(c) Yes, they are mutually exclusive since \\(M \\cap N = \\emptyset\\).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(426, '2020', '\n\nQuestion\n\n(a) The scale of a map is 1 : 20,000. Calculate the area, in square centimetres, on the map of a forest reserve which covers 85\\(km^{2}\\).\n(b) A rectangular playing field is 18m wide. It is surrounded by a path 6m wide such that its area is equal to the perimeterof the path. Calculate the length of the field.\n(c) The diagram shows a circle centre O. If < POQ = x°, the diameter of the circle is 7 cm and the area of the shaded portion is 27.5\\(cm^{2}\\). Find, correct to the nearest degree, the value of x. [Take \\(\\pi = \\frac{22}{7}\\)].\n\n', '', '(a) The scale of a map is 1 : 20,000. Calculate the area, in square centimetres, on the map of a forest reserve which covers 85\\(km^{2}\\).\n(b) A rectangular playing field is 18m wide. It is surrounded by a path 6m wide such that its area is equal to the perimeterof the path. Calculate the length of the field.\n(c) The diagram shows a circle centre O. If < POQ = x°, the diameter of the circle is 7 cm and the area of the shaded portion is 27.5\\(cm^{2}\\). Find, correct to the nearest degree, the value of x. [Take \\(\\pi = \\frac{22}{7}\\)].', '(a) The scale of a map is 1 : 20,000. Calculate the area, in square centimetres, on the map of a forest reserve which covers 85\\(km^{2}\\).~(b) A rectangular playing field is 18m wide. It is surrounded by a path 6m wide such that its area is equal to the perimeterof the path. Calculate the length of the field.\n(c) The diagram shows a circle centre O. If < POQ = x°, the diameter of the circle is 7 cm and the area of the shaded portion is 27.5\\(cm^{2}\\). Find, correct to the nearest degree, the value of x. [Take \\(\\pi = \\frac{22}{7}\\)].~(c) The diagram shows a circle centre O. If < POQ = x°, the diameter of the circle is 7 cm and the area of the shaded portion is 27.5\\(cm^{2}\\). Find, correct to the nearest degree, the value of x. [Take \\(\\pi = \\frac{22}{7}\\)].', NULL, '\n\nExplanation\n(a) 1km = 100,000 cm.\nArea = \\(\\frac{85 \\times 100,000 \\times 100,000}{20,000 \\times 20,000}\\)\n= \\(85 \\times 5 \\times 5\\)\n= \\(2125 cm^{2}\\)\n(b)\nPerimeter of path = \\(2(length + width)\\)\n= \\(2(30 + (12 + a)) = 2(42 + a)\\)\n= \\(84 + 2a\\)\nArea of field = \\(length \\times width\\)\n= \\(18 \\times a\\)\n= 18a\n\\(\\therefore 18a = 84 + 2a\\)\n\\(18a - 2a = 84 \\implies 16a = 84\\)\n\\(a = \\frac{84}{16}\\)\n= \\(5.25 cm\\)\n(c)Area of major sector = \\(\\frac{360 - x}{360} \\times \\pi r^{2}\\)\n\\(27.5 = \\frac{360 - x}{360} \\times \\frac{22}{7} \\times \\frac{7}{2} \\times \\frac{7}{2}\\)\n\\(27.5 = \\frac{360 - x}{360} \\times \\frac{77}{2}\\)\n\\(360 - x = \\frac{27.5 \\times 360 \\times 2}{77}\\)\n\\(360 - x = 257.14\\)\n\\(x = 360° - 257.14° = 102.86°\\)\n\\(x \\approxeq 103°\\) (to the nearest degree).\n\n', '', 'Explanation\n(a) 1km = 100,000 cm.\nArea = \\(\\frac{85 \\times 100,000 \\times 100,000}{20,000 \\times 20,000}\\)\n= \\(85 \\times 5 \\times 5\\)\n= \\(2125 cm^{2}\\)\n(b)\nPerimeter of path = \\(2(length + width)\\)\n= \\(2(30 + (12 + a)) = 2(42 + a)\\)\n= \\(84 + 2a\\)\nArea of field = \\(length \\times width\\)\n= \\(18 \\times a\\)\n= 18a\n\\(\\therefore 18a = 84 + 2a\\)\n\\(18a - 2a = 84 \\implies 16a = 84\\)\n\\(a = \\frac{84}{16}\\)\n= \\(5.25 cm\\)\n(c)Area of major sector = \\(\\frac{360 - x}{360} \\times \\pi r^{2}\\)\n\\(27.5 = \\frac{360 - x}{360} \\times \\frac{22}{7} \\times \\frac{7}{2} \\times \\frac{7}{2}\\)\n\\(27.5 = \\frac{360 - x}{360} \\times \\frac{77}{2}\\)\n\\(360 - x = \\frac{27.5 \\times 360 \\times 2}{77}\\)\n\\(360 - x = 257.14\\)\n\\(x = 360° - 257.14° = 102.86°\\)\n\\(x \\approxeq 103°\\) (to the nearest degree).\n\n', '(a) 1km = 100,000 cm.\nArea = \\(\\frac{85 \\times 100,000 \\times 100,000}{20,000 \\times 20,000}\\)\n= \\(85 \\times 5 \\times 5\\)\n= \\(2125 cm^{2}\\)~(b)\nPerimeter of path = \\(2(length + width)\\)\n= \\(2(30 + (12 + a)) = 2(42 + a)\\)\n= \\(84 + 2a\\)\nArea of field = \\(length \\times width\\)\n= \\(18 \\times a\\)\n= 18a\n\\(\\therefore 18a = 84 + 2a\\)\n\\(18a - 2a = 84 \\implies 16a = 84\\)\n\\(a = \\frac{84}{16}\\)\n= \\(5.25 cm\\)\n(c)Area of major sector = \\(\\frac{360 - x}{360} \\times \\pi r^{2}\\)\n\\(27.5 = \\frac{360 - x}{360} \\times \\frac{22}{7} \\times \\frac{7}{2} \\times \\frac{7}{2}\\)\n\\(27.5 = \\frac{360 - x}{360} \\times \\frac{77}{2}\\)\n\\(360 - x = \\frac{27.5 \\times 360 \\times 2}{77}\\)\n\\(360 - x = 257.14\\)\n\\(x = 360° - 257.14° = 102.86°\\)\n\\(x \\approxeq 103°\\) (to the nearest degree).\n~(c)Area of major sector = \\(\\frac{360 - x}{360} \\times \\pi r^{2}\\)\n\\(27.5 = \\frac{360 - x}{360} \\times \\frac{22}{7} \\times \\frac{7}{2} \\times \\frac{7}{2}\\)\n\\(27.5 = \\frac{360 - x}{360} \\times \\frac{77}{2}\\)\n\\(360 - x = \\frac{27.5 \\times 360 \\times 2}{77}\\)\n\\(360 - x = 257.14\\)\n\\(x = 360° - 257.14° = 102.86°\\)\n\\(x \\approxeq 103°\\) (to the nearest degree).\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
(427, '2020', '\n\nQuestion\n\n(a) Madam Kwakyewaa imported a quantity of frozen fish costing GH¢ 400.00. The goods attracted an import duty of 15% of its cost. She also paid a sales taxof 10% of the total cost of the goods including the import duty and then sold the goods for GH¢ 660.00. Calculate the percentage profit.\n(b) In a school, there are 1000 boys and a number of girls. The 48% of the total number of students that were successful in an examination was made up of 50%of the boys and 40% of the girls. Find the number of girls in the school.\n\n', '', '(a) Madam Kwakyewaa imported a quantity of frozen fish costing GH¢ 400.00. The goods attracted an import duty of 15% of its cost. She also paid a sales taxof 10% of the total cost of the goods including the import duty and then sold the goods for GH¢ 660.00. Calculate the percentage profit.\n(b) In a school, there are 1000 boys and a number of girls. The 48% of the total number of students that were successful in an examination was made up of 50%of the boys and 40% of the girls. Find the number of girls in the school.', '(a) Madam Kwakyewaa imported a quantity of frozen fish costing GH¢ 400.00. The goods attracted an import duty of 15% of its cost. She also paid a sales taxof 10% of the total cost of the goods including the import duty and then sold the goods for GH¢ 660.00. Calculate the percentage profit.~(b) In a school, there are 1000 boys and a number of girls. The 48% of the total number of students that were successful in an examination was made up of 50%of the boys and 40% of the girls. Find the number of girls in the school.', NULL, '\n\nExplanation\n(a) Import duty :\\(\\frac{15}{100} \\times 400 = GH¢ 60.00\\)\nNew cost price :GH¢400.00 +GH¢ 60 =GH¢ 460.00\nSales tax : \\(\\frac{10}{100} \\times 460 =GH¢ 46.00\\)\nTotal cost price : \\(GH¢(460 + 46) =GH¢ 506.00\\)\nProfit = GH¢(660 - 506)\n= GH¢ 154.00\n\\(% profit = \\frac{154}{506} \\times 100%\\)\n= \\(30.43%\\).\n(b) Let the number of girls be t.\nTherefore the total number of students = (1000 + t) students.\nThe number that passed the exam = \\(\\frac{48}{100} \\times (1000 + t)\\)\n= \\(0.48(1000 + t)\\)\n\\(\\implies 0.48(1000 + t) = (\\frac{50}{100} \\times 1000 + \\frac{40}{100} \\times t)\\)\n\\(0.48(1000 + t) = 500 + 0.4t\\)\n\\(480 + 0.48t = 500 + 0.4t \\implies 0.48t - 0.4t = 500 - 480\\)\n\\(0.08t = 20\\)\n\\(t = \\frac{20}{0.08} = 250\\)\nThere are 250 girls in the school.\n\n', '', 'Explanation\n(a) Import duty :\\(\\frac{15}{100} \\times 400 = GH¢ 60.00\\)\nNew cost price :GH¢400.00 +GH¢ 60 =GH¢ 460.00\nSales tax : \\(\\frac{10}{100} \\times 460 =GH¢ 46.00\\)\nTotal cost price : \\(GH¢(460 + 46) =GH¢ 506.00\\)\nProfit = GH¢(660 - 506)\n= GH¢ 154.00\n\\(% profit = \\frac{154}{506} \\times 100%\\)\n= \\(30.43%\\).\n(b) Let the number of girls be t.\nTherefore the total number of students = (1000 + t) students.\nThe number that passed the exam = \\(\\frac{48}{100} \\times (1000 + t)\\)\n= \\(0.48(1000 + t)\\)\n\\(\\implies 0.48(1000 + t) = (\\frac{50}{100} \\times 1000 + \\frac{40}{100} \\times t)\\)\n\\(0.48(1000 + t) = 500 + 0.4t\\)\n\\(480 + 0.48t = 500 + 0.4t \\implies 0.48t - 0.4t = 500 - 480\\)\n\\(0.08t = 20\\)\n\\(t = \\frac{20}{0.08} = 250\\)\nThere are 250 girls in the school.\n\n', '(a) Import duty :\\(\\frac{15}{100} \\times 400 = GH¢ 60.00\\)\nNew cost price :GH¢400.00 +GH¢ 60 =GH¢ 460.00\nSales tax : \\(\\frac{10}{100} \\times 460 =GH¢ 46.00\\)\nTotal cost price : \\(GH¢(460 + 46) =GH¢ 506.00\\)\nProfit = GH¢(660 - 506)\n= GH¢ 154.00\n\\(% profit = \\frac{154}{506} \\times 100%\\)\n= \\(30.43%\\).~(b) Let the number of girls be t.\nTherefore the total number of students = (1000 + t) students.\nThe number that passed the exam = \\(\\frac{48}{100} \\times (1000 + t)\\)\n= \\(0.48(1000 + t)\\)\n\\(\\implies 0.48(1000 + t) = (\\frac{50}{100} \\times 1000 + \\frac{40}{100} \\times t)\\)\n\\(0.48(1000 + t) = 500 + 0.4t\\)\n\\(480 + 0.48t = 500 + 0.4t \\implies 0.48t - 0.4t = 500 - 480\\)\n\\(0.08t = 20\\)\n\\(t = \\frac{20}{0.08} = 250\\)\nThere are 250 girls in the school.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(428, '2020', '\n\nQuestion\n\nUsing ruler and a pair of compasses only,\n(a) construct (i) a quadrilateral PQRS with /PS/ = 6 cm, < RSP = 9 cm, /QR/ = 8.4 cm and /PQ/ = 5.4 cm; (ii) the bisectors of < RSP and < SPQ to meet at X ;            (iii) the perpendicular XT to meet PS at T.\n(b) Measure /XT/.\n\n', '', 'Using ruler and a pair of compasses only,\n(a) construct (i) a quadrilateral PQRS with /PS/ = 6 cm, < RSP = 9 cm, /QR/ = 8.4 cm and /PQ/ = 5.4 cm; (ii) the bisectors of < RSP and < SPQ to meet at X ;            (iii) the perpendicular XT to meet PS at T.\n(b) Measure /XT/.', 'Using ruler and a pair of compasses only,\n(a) construct (i) a quadrilateral PQRS with /PS/ = 6 cm, < RSP = 9 cm, /QR/ = 8.4 cm and /PQ/ = 5.4 cm; (ii) the bisectors of < RSP and < SPQ to meet at X ;            (iii) the perpendicular XT to meet PS at T.~(b) Measure /XT/.', NULL, '\n\nExplanation\n(a)\n(b) /XT/ = 3.4 cm.\n\n', '', 'Explanation\n(a)\n(b) /XT/ = 3.4 cm.\n\n', '~(b) /XT/ = 3.4 cm.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(429, '2020', '\n\nQuestion\n\nIn the diagram, /AB/ = 8 km, /BC/ = 13 km, the bearing of A from B is 310° and the bearing of B from C is 230°. Calculate, correct to 3 significant figures,\n(a) the distance AC ;\n(b) the bearing of C from A ;\n(c) how far east of B, C is.\n\n', '', 'In the diagram, /AB/ = 8 km, /BC/ = 13 km, the bearing of A from B is 310° and the bearing of B from C is 230°. Calculate, correct to 3 significant figures,\n(a) the distance AC ;\n(b) the bearing of C from A ;\n(c) how far east of B, C is.', 'In the diagram, /AB/ = 8 km, /BC/ = 13 km, the bearing of A from B is 310° and the bearing of B from C is 230°. Calculate, correct to 3 significant figures,\n(a) the distance AC ;~(b) the bearing of C from A ;\n(c) how far east of B, C is.~(c) how far east of B, C is.', NULL, '\n\nExplanation\n(a)\n< ABC = 100°\n\\(\\therefore b^{2} = 8^{2} + 13^{2} - 2(13)(8) \\cos 100°\\)\n= \\(64 + 169 - (208 \\times - 0.1736)\\)\n= \\(233 + 36.12\\)\n\\(b^{2} = 269.12 \\implies b = \\sqrt{269.12} = 16.405 km\\)\n\\(\\approxeq 16.4 km\\) (3 significant figures)\n(b) \\(\\frac{\\sin B}{b} = \\frac{\\sin A}{a}\\)\n\\(\\frac{\\sin 100}{16.4} = \\frac{\\sin A}{13}\\)\n\\(\\sin A = \\frac{13 \\times \\sin 100}{16.4}\\)\n\\(\\sin A = \\frac{12.803}{16.4}\\)\n\\(\\sin A = 0.7806\\)\n\\(A = \\sin^{-1} (0.7806) = 51.32°\\)\n\\(\\therefore\\) Bearing ofC from A = 180° - (50° + 51.32°)\n= 180° - 101.32°\n= 78.68° \\(\\approxeq\\) 78.7° (3 significant figure).\n(c) \\(\\cos 40 = \\frac{BD}{13}\\)\n\\(BD = 13 \\cos 40\\)\n= \\(13 \\times 0.7660\\)\n= 9.959 km\n\\(\\approxeq\\) 9.96 km east of B.\n\n', '', 'Explanation\n(a)\n< ABC = 100°\n\\(\\therefore b^{2} = 8^{2} + 13^{2} - 2(13)(8) \\cos 100°\\)\n= \\(64 + 169 - (208 \\times - 0.1736)\\)\n= \\(233 + 36.12\\)\n\\(b^{2} = 269.12 \\implies b = \\sqrt{269.12} = 16.405 km\\)\n\\(\\approxeq 16.4 km\\) (3 significant figures)\n(b) \\(\\frac{\\sin B}{b} = \\frac{\\sin A}{a}\\)\n\\(\\frac{\\sin 100}{16.4} = \\frac{\\sin A}{13}\\)\n\\(\\sin A = \\frac{13 \\times \\sin 100}{16.4}\\)\n\\(\\sin A = \\frac{12.803}{16.4}\\)\n\\(\\sin A = 0.7806\\)\n\\(A = \\sin^{-1} (0.7806) = 51.32°\\)\n\\(\\therefore\\) Bearing ofC from A = 180° - (50° + 51.32°)\n= 180° - 101.32°\n= 78.68° \\(\\approxeq\\) 78.7° (3 significant figure).\n(c) \\(\\cos 40 = \\frac{BD}{13}\\)\n\\(BD = 13 \\cos 40\\)\n= \\(13 \\times 0.7660\\)\n= 9.959 km\n\\(\\approxeq\\) 9.96 km east of B.\n\n', '(a)\n< ABC = 100°\n\\(\\therefore b^{2} = 8^{2} + 13^{2} - 2(13)(8) \\cos 100°\\)\n= \\(64 + 169 - (208 \\times - 0.1736)\\)\n= \\(233 + 36.12\\)\n\\(b^{2} = 269.12 \\implies b = \\sqrt{269.12} = 16.405 km\\)\n\\(\\approxeq 16.4 km\\) (3 significant figures)~(b) \\(\\frac{\\sin B}{b} = \\frac{\\sin A}{a}\\)\n\\(\\frac{\\sin 100}{16.4} = \\frac{\\sin A}{13}\\)\n\\(\\sin A = \\frac{13 \\times \\sin 100}{16.4}\\)\n\\(\\sin A = \\frac{12.803}{16.4}\\)\n\\(\\sin A = 0.7806\\)\n\\(A = \\sin^{-1} (0.7806) = 51.32°\\)\n\\(\\therefore\\) Bearing ofC from A = 180° - (50° + 51.32°)\n= 180° - 101.32°\n= 78.68° \\(\\approxeq\\) 78.7° (3 significant figure).\n(c) \\(\\cos 40 = \\frac{BD}{13}\\)\n\\(BD = 13 \\cos 40\\)\n= \\(13 \\times 0.7660\\)\n= 9.959 km\n\\(\\approxeq\\) 9.96 km east of B.\n~(c) \\(\\cos 40 = \\frac{BD}{13}\\)\n\\(BD = 13 \\cos 40\\)\n= \\(13 \\times 0.7660\\)\n= 9.959 km\n\\(\\approxeq\\) 9.96 km east of B.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(430, '2020', '\n\nQuestion\n\n(a) Copy and complete the table of values for the relation \\(y = -x^{2} + x + 2; -3 \\leq x \\leq 3\\).\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n\n\ny\n\n-4\n\n2\n\n\n-4\n\n\n\n(b) Using scales of 2 cm to 1 unit on the x- axis and 2 cm to 2 units on the y- axis, draw a graph of the relation \\(y = -x^{2} + x + 2\\).\n(c) From the graph, find the : (i) minimum value of y ; (ii) roots of equation \\(x^{2} - x - 2 = 0\\) ; (iii) gradient of the curve at x = -0.5.\n\n\n(a) Copy and complete the table of values for the relation \\(y = -x^{2} + x + 2; -3 \\leq x \\leq 3\\).\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n\n\ny\n\n-4\n\n2\n\n\n-4\n\n\n', '', '(a) Copy and complete the table of values for the relation \\(y = -x^{2} + x + 2; -3 \\leq x \\leq 3\\).\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n\n\ny\n\n-4\n\n2\n\n\n-4\n\n\n\n(b) Using scales of 2 cm to 1 unit on the x- axis and 2 cm to 2 units on the y- axis, draw a graph of the relation \\(y = -x^{2} + x + 2\\).\n(c) From the graph, find the : (i) minimum value of y ; (ii) roots of equation \\(x^{2} - x - 2 = 0\\) ; (iii) gradient of the curve at x = -0.5.\n\n\n(a) Copy and complete the table of values for the relation \\(y = -x^{2} + x + 2; -3 \\leq x \\leq 3\\).\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n\n\ny\n\n-4\n\n2\n\n\n-4\n', '(a) Copy and complete the table of values for the relation \\(y = -x^{2} + x + 2; -3 \\leq x \\leq 3\\).\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n\n\ny\n\n-4\n\n2\n\n\n-4\n\n\n~(b) Using scales of 2 cm to 1 unit on the x- axis and 2 cm to 2 units on the y- axis, draw a graph of the relation \\(y = -x^{2} + x + 2\\).\n(c) From the graph, find the : (i) minimum value of y ; (ii) roots of equation \\(x^{2} - x - 2 = 0\\) ; (iii) gradient of the curve at x = -0.5.\n\n\n(a) Copy and complete the table of values for the relation \\(y = -x^{2} + x + 2; -3 \\leq x \\leq 3\\).\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n\n\ny\n\n-4\n\n2\n\n\n-4\n~(c) From the graph, find the : (i) minimum value of y ; (ii) roots of equation \\(x^{2} - x - 2 = 0\\) ; (iii) gradient of the curve at x = -0.5.\n\n\n(a) Copy and complete the table of values for the relation \\(y = -x^{2} + x + 2; -3 \\leq x \\leq 3\\).\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n\n\ny\n\n-4\n\n2\n\n\n-4\n', NULL, '\n\nExplanation\n(a)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n\n\ny\n-10\n-4\n0\n2\n2\n0\n-4\n\n\n\n(b)\n(c) (ii) roots of the equation : \\(x^{2} - x - 2 = 0\\) are x = -1 or x = 2.\n(iii) gradient of the curve at x = -0.5 is 2.1.\n\n', '', 'Explanation\n(a)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n\n\ny\n-10\n-4\n0\n2\n2\n0\n-4\n\n\n\n(b)\n(c) (ii) roots of the equation : \\(x^{2} - x - 2 = 0\\) are x = -1 or x = 2.\n(iii) gradient of the curve at x = -0.5 is 2.1.\n\n', '(a)\n\n\n\nx\n-3\n-2\n-1\n0\n1\n2\n3\n\n\ny\n-10\n-4\n0\n2\n2\n0\n-4\n\n\n~(b)\n(c) (ii) roots of the equation : \\(x^{2} - x - 2 = 0\\) are x = -1 or x = 2.\n(iii) gradient of the curve at x = -0.5 is 2.1.\n~(c) (ii) roots of the equation : \\(x^{2} - x - 2 = 0\\) are x = -1 or x = 2.\n(iii) gradient of the curve at x = -0.5 is 2.1.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(431, '2020', '\n\nQuestion\n\n(a)\nIn the diagram, < PTQ = < PSR = 90°, /PQ/ = 10 cm, /PS/ = 14.4 cm and /TQ/ = 6 cm. Calculate the area of the quadrilateral QRST.\n(b) Two opposite sides of a square are each decreased by 10% while the other two are each increased by 15% to form a rectangle. Find the ratio of the area of the rectangleto that of the square.\n\n', '', '(a)\nIn the diagram, < PTQ = < PSR = 90°, /PQ/ = 10 cm, /PS/ = 14.4 cm and /TQ/ = 6 cm. Calculate the area of the quadrilateral QRST.\n(b) Two opposite sides of a square are each decreased by 10% while the other two are each increased by 15% to form a rectangle. Find the ratio of the area of the rectangleto that of the square.', '(a)\nIn the diagram, < PTQ = < PSR = 90°, /PQ/ = 10 cm, /PS/ = 14.4 cm and /TQ/ = 6 cm. Calculate the area of the quadrilateral QRST.~(b) Two opposite sides of a square are each decreased by 10% while the other two are each increased by 15% to form a rectangle. Find the ratio of the area of the rectangleto that of the square.', NULL, '\n\nExplanation\n(a) \\((PT)^{2} + (TQ)^{2} = (PQ)^{2}\\)\n\\((PT)^{2} + 6^{2} = 10^{2}\\)\n\\((PT)^{2} = 100 - 36 = 64\\)\n\\(\\therefore PT = 8 cm\\)\n\\(\\therefore \\frac{8}{6} = \\frac{14.4}{SR}\\)\n\\(SR = \\frac{6 \\times 14.4}{8}\\)\n\\(SR = 10.8 cm\\)\nArea of the quadrilateral QRST = \\(\\frac{1}{2} (6 + 10.8) \\times 6.4\\)\n= \\(53.76 cm^{2}\\)\n(b) Opp sides of a square = \\((\\frac{(100 - 10)x}{100} =0.9x\\)\nother two sides of the square = \\(\\frac{(100 + 15)x}{100} = 1.15x\\)\nArea of the square :\\(x \\times x = x^{2}\\)\nArea of the rectangle : \\(0.9x \\times 1.15x = 1.035x^{2}\\)\n\\(\\therefore \\text{The ratio of the area of the rectangle to the square} = 1.035x^{2} : x^{2}\\)\n= \\(1.035 : 1\\)\n\n', '', 'Explanation\n(a) \\((PT)^{2} + (TQ)^{2} = (PQ)^{2}\\)\n\\((PT)^{2} + 6^{2} = 10^{2}\\)\n\\((PT)^{2} = 100 - 36 = 64\\)\n\\(\\therefore PT = 8 cm\\)\n\\(\\therefore \\frac{8}{6} = \\frac{14.4}{SR}\\)\n\\(SR = \\frac{6 \\times 14.4}{8}\\)\n\\(SR = 10.8 cm\\)\nArea of the quadrilateral QRST = \\(\\frac{1}{2} (6 + 10.8) \\times 6.4\\)\n= \\(53.76 cm^{2}\\)\n(b) Opp sides of a square = \\((\\frac{(100 - 10)x}{100} =0.9x\\)\nother two sides of the square = \\(\\frac{(100 + 15)x}{100} = 1.15x\\)\nArea of the square :\\(x \\times x = x^{2}\\)\nArea of the rectangle : \\(0.9x \\times 1.15x = 1.035x^{2}\\)\n\\(\\therefore \\text{The ratio of the area of the rectangle to the square} = 1.035x^{2} : x^{2}\\)\n= \\(1.035 : 1\\)\n\n', '(a) \\((PT)^{2} + (TQ)^{2} = (PQ)^{2}\\)\n\\((PT)^{2} + 6^{2} = 10^{2}\\)\n\\((PT)^{2} = 100 - 36 = 64\\)\n\\(\\therefore PT = 8 cm\\)\n\\(\\therefore \\frac{8}{6} = \\frac{14.4}{SR}\\)\n\\(SR = \\frac{6 \\times 14.4}{8}\\)\n\\(SR = 10.8 cm\\)\nArea of the quadrilateral QRST = \\(\\frac{1}{2} (6 + 10.8) \\times 6.4\\)\n= \\(53.76 cm^{2}\\)~(b) Opp sides of a square = \\((\\frac{(100 - 10)x}{100} =0.9x\\)\nother two sides of the square = \\(\\frac{(100 + 15)x}{100} = 1.15x\\)\nArea of the square :\\(x \\times x = x^{2}\\)\nArea of the rectangle : \\(0.9x \\times 1.15x = 1.035x^{2}\\)\n\\(\\therefore \\text{The ratio of the area of the rectangle to the square} = 1.035x^{2} : x^{2}\\)\n= \\(1.035 : 1\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(432, '2020', '\n\nQuestion\n\nThe frequency distribution of the weight of 100 participants in a high jump competition is as shown below :\n\n\n\nWeight (kg)\n20 - 29\n30 - 39\n40 - 49\n50 - 59\n60 - 69\n70 - 79\n\n\n\nNumber of\nParticipants\n\n10\n18\n22\n25\n16\n9\n\n\n\n(a) Construct the cumulative frequency table.\n(b) Draw the cumulative frequency curve.\n(c) From the curve, estimate the : (i) median ; (ii) semi- interquartile range ; (iii) probability that a participant chosen at random weighs at least 60 kg.\n\n\n\nThe frequency distribution of the weight of 100 participants in a high jump competition is as shown below :\n\n\n\nWeight (kg)\n20 - 29\n30 - 39\n40 - 49\n50 - 59\n60 - 69\n70 - 79\n\n\n\nNumber of\nParticipants\n\n10\n18\n22\n25\n16\n9\n\n\n', '', 'The frequency distribution of the weight of 100 participants in a high jump competition is as shown below :\n\n\n\nWeight (kg)\n20 - 29\n30 - 39\n40 - 49\n50 - 59\n60 - 69\n70 - 79\n\n\n\nNumber of\nParticipants\n\n10\n18\n22\n25\n16\n9\n\n\n\n(a) Construct the cumulative frequency table.\n(b) Draw the cumulative frequency curve.\n(c) From the curve, estimate the : (i) median ; (ii) semi- interquartile range ; (iii) probability that a participant chosen at random weighs at least 60 kg.\n\n\n\nThe frequency distribution of the weight of 100 participants in a high jump competition is as shown below :\n\n\n\nWeight (kg)\n20 - 29\n30 - 39\n40 - 49\n50 - 59\n60 - 69\n70 - 79\n\n\n\nNumber of\nParticipants\n\n10\n18\n22\n25\n16\n9\n', 'The frequency distribution of the weight of 100 participants in a high jump competition is as shown below :\n\n\n\nWeight (kg)\n20 - 29\n30 - 39\n40 - 49\n50 - 59\n60 - 69\n70 - 79\n\n\n\nNumber of\nParticipants\n\n10\n18\n22\n25\n16\n9\n\n\n\n(a) Construct the cumulative frequency table.~(b) Draw the cumulative frequency curve.\n(c) From the curve, estimate the : (i) median ; (ii) semi- interquartile range ; (iii) probability that a participant chosen at random weighs at least 60 kg.\n\n\n\nThe frequency distribution of the weight of 100 participants in a high jump competition is as shown below :\n\n\n\nWeight (kg)\n20 - 29\n30 - 39\n40 - 49\n50 - 59\n60 - 69\n70 - 79\n\n\n\nNumber of\nParticipants\n\n10\n18\n22\n25\n16\n9\n~(c) From the curve, estimate the : (i) median ; (ii) semi- interquartile range ; (iii) probability that a participant chosen at random weighs at least 60 kg.\n\n\n\nThe frequency distribution of the weight of 100 participants in a high jump competition is as shown below :\n\n\n\nWeight (kg)\n20 - 29\n30 - 39\n40 - 49\n50 - 59\n60 - 69\n70 - 79\n\n\n\nNumber of\nParticipants\n\n10\n18\n22\n25\n16\n9\n', NULL, '\n\nExplanation\n(a)\n\n\n\n\nClass\nIntervals\n\nfreq\n\ncum\nfreq\n\n\nclass\nboundaries\n\n\n\n20 - 29\n10\n10\n19.5 - 29.5\n\n\n30 - 39\n18\n28\n29.5 - 39.5\n\n\n40 - 49\n22\n50\n39.5 - 49.5\n\n\n50 - 59\n25\n75\n49.5 - 59.5\n\n\n60 - 69\n16\n91\n59.5 - 69.5\n\n\n70 - 79\n9\n100\n69.5 - 79.5\n\n\n\n(b)\n(c) (i) Median (\\(Q_{2}\\)) = 49.5\n(ii) Semi- Interquartile range = \\(\\frac{1}{2} (Q_{3} - Q_{1})\\)\n\\(Q_{1} = 38.5\\); \\(Q_{3} = 59.5\\).\nSemi- Interquartile range = \\(\\frac{1}{2} (59.5 - 38.5)\\)\n= \\(\\frac{1}{2} (21)\\)\n= 10.5.\n(iii) Pr(\\(\\geq 60kg = \\frac{25}{100} = \\frac{1}{4}\\)\n\n', '', 'Explanation\n(a)\n\n\n\n\nClass\nIntervals\n\nfreq\n\ncum\nfreq\n\n\nclass\nboundaries\n\n\n\n20 - 29\n10\n10\n19.5 - 29.5\n\n\n30 - 39\n18\n28\n29.5 - 39.5\n\n\n40 - 49\n22\n50\n39.5 - 49.5\n\n\n50 - 59\n25\n75\n49.5 - 59.5\n\n\n60 - 69\n16\n91\n59.5 - 69.5\n\n\n70 - 79\n9\n100\n69.5 - 79.5\n\n\n\n(b)\n(c) (i) Median (\\(Q_{2}\\)) = 49.5\n(ii) Semi- Interquartile range = \\(\\frac{1}{2} (Q_{3} - Q_{1})\\)\n\\(Q_{1} = 38.5\\); \\(Q_{3} = 59.5\\).\nSemi- Interquartile range = \\(\\frac{1}{2} (59.5 - 38.5)\\)\n= \\(\\frac{1}{2} (21)\\)\n= 10.5.\n(iii) Pr(\\(\\geq 60kg = \\frac{25}{100} = \\frac{1}{4}\\)\n\n', '(a)\n\n\n\n\nClass\nIntervals\n\nfreq\n\ncum\nfreq\n\n\nclass\nboundaries\n\n\n\n20 - 29\n10\n10\n19.5 - 29.5\n\n\n30 - 39\n18\n28\n29.5 - 39.5\n\n\n40 - 49\n22\n50\n39.5 - 49.5\n\n\n50 - 59\n25\n75\n49.5 - 59.5\n\n\n60 - 69\n16\n91\n59.5 - 69.5\n\n\n70 - 79\n9\n100\n69.5 - 79.5\n\n\n~(b)\n(c) (i) Median (\\(Q_{2}\\)) = 49.5\n(ii) Semi- Interquartile range = \\(\\frac{1}{2} (Q_{3} - Q_{1})\\)\n\\(Q_{1} = 38.5\\); \\(Q_{3} = 59.5\\).\nSemi- Interquartile range = \\(\\frac{1}{2} (59.5 - 38.5)\\)\n= \\(\\frac{1}{2} (21)\\)\n= 10.5.\n(iii) Pr(\\(\\geq 60kg = \\frac{25}{100} = \\frac{1}{4}\\)\n~(c) (i) Median (\\(Q_{2}\\)) = 49.5\n(ii) Semi- Interquartile range = \\(\\frac{1}{2} (Q_{3} - Q_{1})\\)\n\\(Q_{1} = 38.5\\); \\(Q_{3} = 59.5\\).\nSemi- Interquartile range = \\(\\frac{1}{2} (59.5 - 38.5)\\)\n= \\(\\frac{1}{2} (21)\\)\n= 10.5.\n(iii) Pr(\\(\\geq 60kg = \\frac{25}{100} = \\frac{1}{4}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(433, '2020', '\n\nQuestion\n\n(a) The third term of a Geometric Progression (G.P) is 24 and its seventh term is \\(4\\frac{20}{27}\\). Find its first term.\n(b) Given that y varies directly as x and inversely as the square of z. If y = 4, when x = 3 and z = 1, find y when x = 3 and z = 2.\n\n', '', '(a) The third term of a Geometric Progression (G.P) is 24 and its seventh term is \\(4\\frac{20}{27}\\). Find its first term.\n(b) Given that y varies directly as x and inversely as the square of z. If y = 4, when x = 3 and z = 1, find y when x = 3 and z = 2.', '(a) The third term of a Geometric Progression (G.P) is 24 and its seventh term is \\(4\\frac{20}{27}\\). Find its first term.~(b) Given that y varies directly as x and inversely as the square of z. If y = 4, when x = 3 and z = 1, find y when x = 3 and z = 2.', NULL, '\n\nExplanation\n\\(T_{n} = ar^{n - 1}\\) (Terms of a G.P)\n\\(T_{3} = ar^{2} =24 .... (1)\\)\n\\(T_{7} = ar^{6} = 4\\frac{20}{27} = \\frac{128}{27} .......... (2)\\)\nDivide (2) by (1),\n\\(\\frac{ar^{6}}{ar^{2}} = \\frac{\\frac{128}{27}}{24}\\)\n\\(r^{4} = \\frac{16}{81}\\)\n\\(r = \\sqrt[4]{\\frac{16}{81}}\\)\n= \\(\\frac{2}{3}\\)\nPutting r in equation (1),\n\\(ar^{2} = a(\\frac{2}{3})^{2} = 24\\)\n\\(\\frac{4}{9} a = 24 \\implies a = \\frac{24 \\times 9}{4}\\)\na = 54.\n(b) \\(y \\propto x \\propto \\frac{1}{z^{2}}\\)\n\\(\\implies y = \\frac{kx}{z^{2}}\\)\n\\(4 = \\frac{3k}{1^{2}} \\implies k = \\frac{4}{3}\\)\n\\(\\therefore y = \\frac{4x}{3z^{2}}\\)\nWhen x = 3, z = 2, y = ?\n\\(y = \\frac{4(3)}{3(2^{2})}\\)\n= \\(\\frac{12}{12} = 1\\).\n\n', '', 'Explanation\n\\(T_{n} = ar^{n - 1}\\) (Terms of a G.P)\n\\(T_{3} = ar^{2} =24 .... (1)\\)\n\\(T_{7} = ar^{6} = 4\\frac{20}{27} = \\frac{128}{27} .......... (2)\\)\nDivide (2) by (1),\n\\(\\frac{ar^{6}}{ar^{2}} = \\frac{\\frac{128}{27}}{24}\\)\n\\(r^{4} = \\frac{16}{81}\\)\n\\(r = \\sqrt[4]{\\frac{16}{81}}\\)\n= \\(\\frac{2}{3}\\)\nPutting r in equation (1),\n\\(ar^{2} = a(\\frac{2}{3})^{2} = 24\\)\n\\(\\frac{4}{9} a = 24 \\implies a = \\frac{24 \\times 9}{4}\\)\na = 54.\n(b) \\(y \\propto x \\propto \\frac{1}{z^{2}}\\)\n\\(\\implies y = \\frac{kx}{z^{2}}\\)\n\\(4 = \\frac{3k}{1^{2}} \\implies k = \\frac{4}{3}\\)\n\\(\\therefore y = \\frac{4x}{3z^{2}}\\)\nWhen x = 3, z = 2, y = ?\n\\(y = \\frac{4(3)}{3(2^{2})}\\)\n= \\(\\frac{12}{12} = 1\\).\n\n', '~(b) \\(y \\propto x \\propto \\frac{1}{z^{2}}\\)\n\\(\\implies y = \\frac{kx}{z^{2}}\\)\n\\(4 = \\frac{3k}{1^{2}} \\implies k = \\frac{4}{3}\\)\n\\(\\therefore y = \\frac{4x}{3z^{2}}\\)\nWhen x = 3, z = 2, y = ?\n\\(y = \\frac{4(3)}{3(2^{2})}\\)\n= \\(\\frac{12}{12} = 1\\).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(434, '2020', '\n\nQuestion\n\n(a) Given that \\((\\sqrt{3} - 5\\sqrt{2})(\\sqrt{3} + \\sqrt{2}) = a + b\\sqrt{6}\\), find a and b.\n(b) If \\(\\frac{2^{1 - y} \\times 2^{y - 1}}{2^{y + 2}} = 8^{2 - 3y}\\), find y.\n\n', '', '(a) Given that \\((\\sqrt{3} - 5\\sqrt{2})(\\sqrt{3} + \\sqrt{2}) = a + b\\sqrt{6}\\), find a and b.\n(b) If \\(\\frac{2^{1 - y} \\times 2^{y - 1}}{2^{y + 2}} = 8^{2 - 3y}\\), find y.', '(a) Given that \\((\\sqrt{3} - 5\\sqrt{2})(\\sqrt{3} + \\sqrt{2}) = a + b\\sqrt{6}\\), find a and b.~(b) If \\(\\frac{2^{1 - y} \\times 2^{y - 1}}{2^{y + 2}} = 8^{2 - 3y}\\), find y.', NULL, '\n\nExplanation\n(a) \\((\\sqrt{3} - 5\\sqrt{2})(\\sqrt{3} + \\sqrt{2}) = \\sqrt{3 \\times 3} + \\sqrt{3 \\times 2} - 5\\sqrt{2 \\times 3} - 5\\sqrt{2 \\times 2}\\)\n= \\(3 + \\sqrt{6} - 5\\sqrt{6} - 5(2)\\)\n= \\(-7 - 4\\sqrt{6}\\)\n\\(\\therefore\\) a = -7 and b = -4.\n(b) \\(\\frac{2^{1 - y} \\times 2^{y - 1}}{2^{y + 2}} = 8^{2 - 3y}\\)\n\\(\\frac{2^{1 - y + y - 1}}{2^{y + 2}} = (2^{3})^{2 - 3y}\\)\n\\(2^{0 - (y + 2)} = 2^{6 - 9y} \\implies 2^{- y - 2} = 2^{6-9y}\\)\n\\(- y - 2 = 6 - 9y\\implies - y + 9y= 6 + 2\\)\n\\(8y= 8 \\implies y = 1\\).\n\n', '', 'Explanation\n(a) \\((\\sqrt{3} - 5\\sqrt{2})(\\sqrt{3} + \\sqrt{2}) = \\sqrt{3 \\times 3} + \\sqrt{3 \\times 2} - 5\\sqrt{2 \\times 3} - 5\\sqrt{2 \\times 2}\\)\n= \\(3 + \\sqrt{6} - 5\\sqrt{6} - 5(2)\\)\n= \\(-7 - 4\\sqrt{6}\\)\n\\(\\therefore\\) a = -7 and b = -4.\n(b) \\(\\frac{2^{1 - y} \\times 2^{y - 1}}{2^{y + 2}} = 8^{2 - 3y}\\)\n\\(\\frac{2^{1 - y + y - 1}}{2^{y + 2}} = (2^{3})^{2 - 3y}\\)\n\\(2^{0 - (y + 2)} = 2^{6 - 9y} \\implies 2^{- y - 2} = 2^{6-9y}\\)\n\\(- y - 2 = 6 - 9y\\implies - y + 9y= 6 + 2\\)\n\\(8y= 8 \\implies y = 1\\).\n\n', '(a) \\((\\sqrt{3} - 5\\sqrt{2})(\\sqrt{3} + \\sqrt{2}) = \\sqrt{3 \\times 3} + \\sqrt{3 \\times 2} - 5\\sqrt{2 \\times 3} - 5\\sqrt{2 \\times 2}\\)\n= \\(3 + \\sqrt{6} - 5\\sqrt{6} - 5(2)\\)\n= \\(-7 - 4\\sqrt{6}\\)\n\\(\\therefore\\) a = -7 and b = -4.~(b) \\(\\frac{2^{1 - y} \\times 2^{y - 1}}{2^{y + 2}} = 8^{2 - 3y}\\)\n\\(\\frac{2^{1 - y + y - 1}}{2^{y + 2}} = (2^{3})^{2 - 3y}\\)\n\\(2^{0 - (y + 2)} = 2^{6 - 9y} \\implies 2^{- y - 2} = 2^{6-9y}\\)\n\\(- y - 2 = 6 - 9y\\implies - y + 9y= 6 + 2\\)\n\\(8y= 8 \\implies y = 1\\).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(435, '2020', '\n\nQuestion\n\n(a) If \\(9 \\cos x - 7 = 1\\) and \\(0° \\leq x \\leq 90°\\), find x.\n(b) Given that x is an integer, find the three greatest values of x which satisfy the inequality \\(7x < 2x - 13\\).\n\n', '', '(a) If \\(9 \\cos x - 7 = 1\\) and \\(0° \\leq x \\leq 90°\\), find x.\n(b) Given that x is an integer, find the three greatest values of x which satisfy the inequality \\(7x < 2x - 13\\).', '(a) If \\(9 \\cos x - 7 = 1\\) and \\(0° \\leq x \\leq 90°\\), find x.~(b) Given that x is an integer, find the three greatest values of x which satisfy the inequality \\(7x < 2x - 13\\).', NULL, '\n\nExplanation\n(a) \\(9 \\cos x - 7 = 1 \\implies 9 \\cos x = 1 + 7 = 8\\)\n\\(\\cos x = \\frac{8}{9}\\)\n\\(x = \\cos^{-1} (0.8889)\\)\n\\(x = 27.26°\\)\n(b) \\(7x < 2x - 13\\)\n\\(7x - 2x < - 13\\)\n\\(5x < - 13\\)\n\\(x < \\frac{-13}{5}\\)\n\\([-2.6] = -3, -4, -5\\)\nNote : The greatest value function rounds upa real number to an integer that is less or equal to the real number.\n\n', '', 'Explanation\n(a) \\(9 \\cos x - 7 = 1 \\implies 9 \\cos x = 1 + 7 = 8\\)\n\\(\\cos x = \\frac{8}{9}\\)\n\\(x = \\cos^{-1} (0.8889)\\)\n\\(x = 27.26°\\)\n(b) \\(7x < 2x - 13\\)\n\\(7x - 2x < - 13\\)\n\\(5x < - 13\\)\n\\(x < \\frac{-13}{5}\\)\n\\([-2.6] = -3, -4, -5\\)\nNote : The greatest value function rounds upa real number to an integer that is less or equal to the real number.\n\n', '(a) \\(9 \\cos x - 7 = 1 \\implies 9 \\cos x = 1 + 7 = 8\\)\n\\(\\cos x = \\frac{8}{9}\\)\n\\(x = \\cos^{-1} (0.8889)\\)\n\\(x = 27.26°\\)~(b) \\(7x < 2x - 13\\)\n\\(7x - 2x < - 13\\)\n\\(5x < - 13\\)\n\\(x < \\frac{-13}{5}\\)\n\\([-2.6] = -3, -4, -5\\)\nNote : The greatest value function rounds upa real number to an integer that is less or equal to the real number.\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(436, '2020', '\n\nQuestion\n\nThe table shows the number of children per family in a community.\n\n\n\nNo of children\n0\n1\n2\n3\n4\n5\n\n\nNo of families\n3\n5\n7\n4\n3\n2\n\n\n\n(a) Find the : (i) mode ; (ii) third quartile ; (iii) probability that a family has at least 2 children.\n(b) If a pie chart were to be drawn for the data, what would be the sectorial angle representing families with one child?\n\n\nThe table shows the number of children per family in a community.\n\n\n\nNo of children\n0\n1\n2\n3\n4\n5\n\n\nNo of families\n3\n5\n7\n4\n3\n2\n\n\n', '', 'The table shows the number of children per family in a community.\n\n\n\nNo of children\n0\n1\n2\n3\n4\n5\n\n\nNo of families\n3\n5\n7\n4\n3\n2\n\n\n\n(a) Find the : (i) mode ; (ii) third quartile ; (iii) probability that a family has at least 2 children.\n(b) If a pie chart were to be drawn for the data, what would be the sectorial angle representing families with one child?\n\n\nThe table shows the number of children per family in a community.\n\n\n\nNo of children\n0\n1\n2\n3\n4\n5\n\n\nNo of families\n3\n5\n7\n4\n3\n2\n', 'The table shows the number of children per family in a community.\n\n\n\nNo of children\n0\n1\n2\n3\n4\n5\n\n\nNo of families\n3\n5\n7\n4\n3\n2\n\n\n\n(a) Find the : (i) mode ; (ii) third quartile ; (iii) probability that a family has at least 2 children.~(b) If a pie chart were to be drawn for the data, what would be the sectorial angle representing families with one child?\n\n\nThe table shows the number of children per family in a community.\n\n\n\nNo of children\n0\n1\n2\n3\n4\n5\n\n\nNo of families\n3\n5\n7\n4\n3\n2\n', NULL, '\n\nExplanation\n(a) (i) Mode = 2\n(ii) Third quartile : Sum of frequency = 24\n\\(Q_{3} = \\frac{3}{4} \\times 24 = 18\\)\nThe 18th position falls at 3.\n(iii) P(at least 2 children) = \\(\\frac{7 + 4 + 3 + 2}{24}\\)\n= \\(\\frac{16}{24}\\)\n= \\(\\frac{2}{3}\\)\n(b) The sector for 1 child = \\(\\frac{5}{24} \\times 360\\)\n= \\(75°\\).\n\n', '', 'Explanation\n(a) (i) Mode = 2\n(ii) Third quartile : Sum of frequency = 24\n\\(Q_{3} = \\frac{3}{4} \\times 24 = 18\\)\nThe 18th position falls at 3.\n(iii) P(at least 2 children) = \\(\\frac{7 + 4 + 3 + 2}{24}\\)\n= \\(\\frac{16}{24}\\)\n= \\(\\frac{2}{3}\\)\n(b) The sector for 1 child = \\(\\frac{5}{24} \\times 360\\)\n= \\(75°\\).\n\n', '(a) (i) Mode = 2\n(ii) Third quartile : Sum of frequency = 24\n\\(Q_{3} = \\frac{3}{4} \\times 24 = 18\\)\nThe 18th position falls at 3.\n(iii) P(at least 2 children) = \\(\\frac{7 + 4 + 3 + 2}{24}\\)\n= \\(\\frac{16}{24}\\)\n= \\(\\frac{2}{3}\\)~(b) The sector for 1 child = \\(\\frac{5}{24} \\times 360\\)\n= \\(75°\\).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(437, '2020', '\n\nQuestion\n\n(a) Out of 30 candidates applying for a post, 17 have degrees, 15 have diplomas and 4 neither degree nor diploma. How many of them have both?\n(b) In triangle PQR, M and N are points on the side PQ and PR respectively such that MN is parallel to QR. If < PRQ = 75°, PN = QN and < PNQ = 125°, determine :\n(i) < NQR ; (ii) < NPM.\n\n', '', '(a) Out of 30 candidates applying for a post, 17 have degrees, 15 have diplomas and 4 neither degree nor diploma. How many of them have both?\n(b) In triangle PQR, M and N are points on the side PQ and PR respectively such that MN is parallel to QR. If < PRQ = 75°, PN = QN and < PNQ = 125°, determine :\n(i) < NQR ; (ii) < NPM.', '(a) Out of 30 candidates applying for a post, 17 have degrees, 15 have diplomas and 4 neither degree nor diploma. How many of them have both?~(b) In triangle PQR, M and N are points on the side PQ and PR respectively such that MN is parallel to QR. If < PRQ = 75°, PN = QN and < PNQ = 125°, determine :\n(i) < NQR ; (ii) < NPM.', NULL, '\n\nExplanation\n(a) Let B represent those with degree.\nLet D represent those with diploma.\nLet x represent those who have both degree and diploma.\n\n(17 - x) + x + (15 - x) + 4 = 30\n36 - x = 30\nx = 36 - 30 = 6.\n(b)\n\\(< PNM = 75°\\) (corresponding angle)\n\\(\\therefore NQR = 125° - 75° \\)\n= \\(50°\\)\n\\(< NPM = \\frac{180° - 125°}{2}\\)\n= \\(\\frac{55°}{2}\\)\n= \\(27.5°\\). (Base angles of an isosceles triangle).\n\n', '', 'Explanation\n(a) Let B represent those with degree.\nLet D represent those with diploma.\nLet x represent those who have both degree and diploma.\n\n(17 - x) + x + (15 - x) + 4 = 30\n36 - x = 30\nx = 36 - 30 = 6.\n(b)\n\\(< PNM = 75°\\) (corresponding angle)\n\\(\\therefore NQR = 125° - 75° \\)\n= \\(50°\\)\n\\(< NPM = \\frac{180° - 125°}{2}\\)\n= \\(\\frac{55°}{2}\\)\n= \\(27.5°\\). (Base angles of an isosceles triangle).\n\n', '(a) Let B represent those with degree.\nLet D represent those with diploma.\nLet x represent those who have both degree and diploma.\n\n(17 - x) + x + (15 - x) + 4 = 30\n36 - x = 30\nx = 36 - 30 = 6.~(b)\n\\(< PNM = 75°\\) (corresponding angle)\n\\(\\therefore NQR = 125° - 75° \\)\n= \\(50°\\)\n\\(< NPM = \\frac{180° - 125°}{2}\\)\n= \\(\\frac{55°}{2}\\)\n= \\(27.5°\\). (Base angles of an isosceles triangle).\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(438, '2014', '\n\nQuestion\n\nIn the diagram, ABCDEF is a triangular prism. < ABC = < DEF = 90°, /AB/ = 24 cm, /BC/ = 7 cm and /CD/ = 40 cm. Calculate :\n(a) /AC/ ;\n(b) the total surface area of the prism.\n\n\n', '', 'In the diagram, ABCDEF is a triangular prism. < ABC = < DEF = 90°, /AB/ = 24 cm, /BC/ = 7 cm and /CD/ = 40 cm. Calculate :\n(a) /AC/ ;\n(b) the total surface area of the prism.\n', 'In the diagram, ABCDEF is a triangular prism. < ABC = < DEF = 90°, /AB/ = 24 cm, /BC/ = 7 cm and /CD/ = 40 cm. Calculate :\n(a) /AC/ ;~(b) the total surface area of the prism.\n', NULL, '\n\nExplanation\n(a) From the diagram,\n\n\\(|AC|^{2} = 24^{2} + 7^{2}\\)\n= \\(576 + 49\\)\n\\(|AC|^{2} = 625 \\implies |AC| = \\sqrt{625} = 25 cm\\)\n\\(Area = \\frac{1}{2} \\times b \\times h\\)\n= \\(\\frac{7 \\times 24}{2}\\)\n= \\(84 cm^{2}\\).\n(b) Total Surface Area of prism\n= Area of two right angles ( ABC and EFD) + Area of 3 rectangles (ACDF, ABEF and BCDE)\n= \\(( 2 \\times 84 cm^{2}) + (25 \\times 40)cm^{2}+ (24 \\times 40)cm^{2}+ (7 \\times 40)cm^{2}\\)\n= \\((168 + 1000 + 960 + 280) cm^{2}\\)\n= \\(2408 cm^{2}\\)\n\n', '', 'Explanation\n(a) From the diagram,\n\n\\(|AC|^{2} = 24^{2} + 7^{2}\\)\n= \\(576 + 49\\)\n\\(|AC|^{2} = 625 \\implies |AC| = \\sqrt{625} = 25 cm\\)\n\\(Area = \\frac{1}{2} \\times b \\times h\\)\n= \\(\\frac{7 \\times 24}{2}\\)\n= \\(84 cm^{2}\\).\n(b) Total Surface Area of prism\n= Area of two right angles ( ABC and EFD) + Area of 3 rectangles (ACDF, ABEF and BCDE)\n= \\(( 2 \\times 84 cm^{2}) + (25 \\times 40)cm^{2}+ (24 \\times 40)cm^{2}+ (7 \\times 40)cm^{2}\\)\n= \\((168 + 1000 + 960 + 280) cm^{2}\\)\n= \\(2408 cm^{2}\\)\n\n', '(a) From the diagram,\n\n\\(|AC|^{2} = 24^{2} + 7^{2}\\)\n= \\(576 + 49\\)\n\\(|AC|^{2} = 625 \\implies |AC| = \\sqrt{625} = 25 cm\\)\n\\(Area = \\frac{1}{2} \\times b \\times h\\)\n= \\(\\frac{7 \\times 24}{2}\\)\n= \\(84 cm^{2}\\).~(b) Total Surface Area of prism\n= Area of two right angles ( ABC and EFD) + Area of 3 rectangles (ACDF, ABEF and BCDE)\n= \\(( 2 \\times 84 cm^{2}) + (25 \\times 40)cm^{2}+ (24 \\times 40)cm^{2}+ (7 \\times 40)cm^{2}\\)\n= \\((168 + 1000 + 960 + 280) cm^{2}\\)\n= \\(2408 cm^{2}\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(439, '2014', '\n\nQuestion\n\n(a) If \\(\\log 5 = 0.6990, \\log 7 = 0.8451\\) and \\(\\log 8 = 0.9031\\), evaluate \\(\\log (\\frac{35 \\times 49}{40 \\div 56})\\).\n(b) For a musical show, x children were present. There were 60 more adults than children. An adult paid D5 and a child D2. If a total of D1280 was collected, calculate the\n(i) value of x ; (ii) ratio of the number of children to the number of adults ; (iii) average amount paid per person ; (iv) percentage gain if the organisers spent D720 on the show.\n\n', '', '(a) If \\(\\log 5 = 0.6990, \\log 7 = 0.8451\\) and \\(\\log 8 = 0.9031\\), evaluate \\(\\log (\\frac{35 \\times 49}{40 \\div 56})\\).\n(b) For a musical show, x children were present. There were 60 more adults than children. An adult paid D5 and a child D2. If a total of D1280 was collected, calculate the\n(i) value of x ; (ii) ratio of the number of children to the number of adults ; (iii) average amount paid per person ; (iv) percentage gain if the organisers spent D720 on the show.', '(a) If \\(\\log 5 = 0.6990, \\log 7 = 0.8451\\) and \\(\\log 8 = 0.9031\\), evaluate \\(\\log (\\frac{35 \\times 49}{40 \\div 56})\\).~(b) For a musical show, x children were present. There were 60 more adults than children. An adult paid D5 and a child D2. If a total of D1280 was collected, calculate the\n(i) value of x ; (ii) ratio of the number of children to the number of adults ; (iii) average amount paid per person ; (iv) percentage gain if the organisers spent D720 on the show.', NULL, '\n\nExplanation\n(a) \\(\\log (\\frac{35 \\times 49}{40 \\div 56})\\)\nGiven \\(\\log 5 = 0.6990 , \\log 7 = 0.8451 , \\log 8 = 0.9031\\)\n\\(\\log (\\frac{35 \\times 49}{40 \\div 56} = \\log (\\frac{(7 \\times 5) \\times 7^{2}}{(8 \\times 5) \\div (8 \\times 7)}\\)\n= \\(\\log (\\frac{7^{3} \\times 5}{5 \\div 7}\\)\n= \\(\\log 7^{3} + \\log 5 - (\\log 5 - \\log 7)\\)\n= \\(3 \\log 7 + \\log 5 - \\log 5 + \\log 7\\)\n= \\(4 \\log 7\\)\n= \\(4 \\times 0.8471\\)\n= \\(3.3804\\)\n(b) (i) Since there are 60 more adults than children, then the number of adults = x + 60\n\\(\\therefore D 5(x + 60) + D x(2) = D 1280\\)\n\\(5x + 300 + 2x = 1280\\)\n\\(7x + 300 = 1280 \\implies 7x = 1280 - 300 = 980\\)\n\\(x = \\frac{980}{7} = 140\\)\nThere were 140 children.\n(ii) Ratio of children to adults = \\(x : x + 60\\)\n= \\(140 : (140 + 60)\\)\n= \\(140 : 200\\)\n= \\(7 : 10\\)\n(iii) Total number of persons at the show : 140 + 200 = 340 persons.\nTotal amount gotten = D 1280\nAverage paid per person = \\( D \\frac{1280}{340}\\)\n= \\(D \\frac{64}{17}\\)\n= D 3.765\n(iv) Percentage profit\nProfit : D (1280 - 720) = D 560\n% profit : \\(\\frac{560}{720} \\times 100% = \\frac{700}{9} %\\)\n= \\(77.78%\\)\n\n', '', 'Explanation\n(a) \\(\\log (\\frac{35 \\times 49}{40 \\div 56})\\)\nGiven \\(\\log 5 = 0.6990 , \\log 7 = 0.8451 , \\log 8 = 0.9031\\)\n\\(\\log (\\frac{35 \\times 49}{40 \\div 56} = \\log (\\frac{(7 \\times 5) \\times 7^{2}}{(8 \\times 5) \\div (8 \\times 7)}\\)\n= \\(\\log (\\frac{7^{3} \\times 5}{5 \\div 7}\\)\n= \\(\\log 7^{3} + \\log 5 - (\\log 5 - \\log 7)\\)\n= \\(3 \\log 7 + \\log 5 - \\log 5 + \\log 7\\)\n= \\(4 \\log 7\\)\n= \\(4 \\times 0.8471\\)\n= \\(3.3804\\)\n(b) (i) Since there are 60 more adults than children, then the number of adults = x + 60\n\\(\\therefore D 5(x + 60) + D x(2) = D 1280\\)\n\\(5x + 300 + 2x = 1280\\)\n\\(7x + 300 = 1280 \\implies 7x = 1280 - 300 = 980\\)\n\\(x = \\frac{980}{7} = 140\\)\nThere were 140 children.\n(ii) Ratio of children to adults = \\(x : x + 60\\)\n= \\(140 : (140 + 60)\\)\n= \\(140 : 200\\)\n= \\(7 : 10\\)\n(iii) Total number of persons at the show : 140 + 200 = 340 persons.\nTotal amount gotten = D 1280\nAverage paid per person = \\( D \\frac{1280}{340}\\)\n= \\(D \\frac{64}{17}\\)\n= D 3.765\n(iv) Percentage profit\nProfit : D (1280 - 720) = D 560\n% profit : \\(\\frac{560}{720} \\times 100% = \\frac{700}{9} %\\)\n= \\(77.78%\\)\n\n', '(a) \\(\\log (\\frac{35 \\times 49}{40 \\div 56})\\)\nGiven \\(\\log 5 = 0.6990 , \\log 7 = 0.8451 , \\log 8 = 0.9031\\)\n\\(\\log (\\frac{35 \\times 49}{40 \\div 56} = \\log (\\frac{(7 \\times 5) \\times 7^{2}}{(8 \\times 5) \\div (8 \\times 7)}\\)\n= \\(\\log (\\frac{7^{3} \\times 5}{5 \\div 7}\\)\n= \\(\\log 7^{3} + \\log 5 - (\\log 5 - \\log 7)\\)\n= \\(3 \\log 7 + \\log 5 - \\log 5 + \\log 7\\)\n= \\(4 \\log 7\\)\n= \\(4 \\times 0.8471\\)\n= \\(3.3804\\)~(b) (i) Since there are 60 more adults than children, then the number of adults = x + 60\n\\(\\therefore D 5(x + 60) + D x(2) = D 1280\\)\n\\(5x + 300 + 2x = 1280\\)\n\\(7x + 300 = 1280 \\implies 7x = 1280 - 300 = 980\\)\n\\(x = \\frac{980}{7} = 140\\)\nThere were 140 children.\n(ii) Ratio of children to adults = \\(x : x + 60\\)\n= \\(140 : (140 + 60)\\)\n= \\(140 : 200\\)\n= \\(7 : 10\\)\n(iii) Total number of persons at the show : 140 + 200 = 340 persons.\nTotal amount gotten = D 1280\nAverage paid per person = \\( D \\frac{1280}{340}\\)\n= \\(D \\frac{64}{17}\\)\n= D 3.765\n(iv) Percentage profit\nProfit : D (1280 - 720) = D 560\n% profit : \\(\\frac{560}{720} \\times 100% = \\frac{700}{9} %\\)\n= \\(77.78%\\)\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(440, '2014', '\n\nQuestion\n\n(a) A woman looking out from the window of a building at a height of 30m, observed that the angle of depression of the top of a flag pole was 44°. If the foot of the pole is 25m from the foot of the building and on the same horizontal ground, find, correct to the nearest whole number, the (i) angle of depression of the foot of the pole from the woman ; (ii) height of the flag pole.\n(b)In the diagram, O is the centre of the circle, < OQR = 32° and < TPQ = 15°. Calculate, (i) < QPR ; (ii) < TQo.\n\n', '', '(a) A woman looking out from the window of a building at a height of 30m, observed that the angle of depression of the top of a flag pole was 44°. If the foot of the pole is 25m from the foot of the building and on the same horizontal ground, find, correct to the nearest whole number, the (i) angle of depression of the foot of the pole from the woman ; (ii) height of the flag pole.\n(b)In the diagram, O is the centre of the circle, < OQR = 32° and < TPQ = 15°. Calculate, (i) < QPR ; (ii) < TQo.', '(a) A woman looking out from the window of a building at a height of 30m, observed that the angle of depression of the top of a flag pole was 44°. If the foot of the pole is 25m from the foot of the building and on the same horizontal ground, find, correct to the nearest whole number, the (i) angle of depression of the foot of the pole from the woman ; (ii) height of the flag pole.~(b)In the diagram, O is the centre of the circle, < OQR = 32° and < TPQ = 15°. Calculate, (i) < QPR ; (ii) < TQo.', NULL, '\n\nExplanation\n(a)\n(i) From \\(\\Delta\\) TFO,\n\\(\\tan x = \\tan \\theta\\) (alternate angles)\n\\(\\tan x = \\frac{25}{30} = 0.8333\\)\n\\(x = \\tan^{-1} (0.8333)\\)\n= \\(39.80° \\approxeq 40°\\)\nAngle of depression of the pole from the woman\n= 90° - 40°\n= 50°.\n(ii) From \\(\\Delta\\) TPQ,\n\\(\\tan 46 = \\frac{25}{TQ}\\)\n\\(TQ = \\frac{25}{\\tan 46}\\)\n\\(TQ = \\frac{25}{1.0355}\\)\n= \\(24.1429 m\\)\nHeight of the pole : \\(TF - TQ\\)\n= \\(30 m - 24.1429 m\\)\n= \\(5.8571 m\\)\n\\(\\approxeq 6m\\)\n\n', '', 'Explanation\n(a)\n(i) From \\(\\Delta\\) TFO,\n\\(\\tan x = \\tan \\theta\\) (alternate angles)\n\\(\\tan x = \\frac{25}{30} = 0.8333\\)\n\\(x = \\tan^{-1} (0.8333)\\)\n= \\(39.80° \\approxeq 40°\\)\nAngle of depression of the pole from the woman\n= 90° - 40°\n= 50°.\n(ii) From \\(\\Delta\\) TPQ,\n\\(\\tan 46 = \\frac{25}{TQ}\\)\n\\(TQ = \\frac{25}{\\tan 46}\\)\n\\(TQ = \\frac{25}{1.0355}\\)\n= \\(24.1429 m\\)\nHeight of the pole : \\(TF - TQ\\)\n= \\(30 m - 24.1429 m\\)\n= \\(5.8571 m\\)\n\\(\\approxeq 6m\\)\n\n', '', 'WAEC-PK-1', 'WAEC-MATH', 1),
(441, '2014', '\n\nQuestion\n\nThe marks scored by 50 students in a Geography examination are as follows :\n60 54 40 67 53 73 37 55 62 43 44 69 39 32 45 58 48 67 39 51 46 59 40 52 61 48 23 60 59 47 65 58 74 47 40 59 68 51 50 50 71 51 26 36 38 70 46 40 51 42.\n(a) Usingclass intervals 21 - 30, 31 - 40, ..., prepare a frequency distribution table.\n(b) Calculate the mean mark of the distribution.\n(c) What percentage of the students scored more than 60%?\n\n', '', 'The marks scored by 50 students in a Geography examination are as follows :\n60 54 40 67 53 73 37 55 62 43 44 69 39 32 45 58 48 67 39 51 46 59 40 52 61 48 23 60 59 47 65 58 74 47 40 59 68 51 50 50 71 51 26 36 38 70 46 40 51 42.\n(a) Usingclass intervals 21 - 30, 31 - 40, ..., prepare a frequency distribution table.\n(b) Calculate the mean mark of the distribution.\n(c) What percentage of the students scored more than 60%?', 'The marks scored by 50 students in a Geography examination are as follows :\n60 54 40 67 53 73 37 55 62 43 44 69 39 32 45 58 48 67 39 51 46 59 40 52 61 48 23 60 59 47 65 58 74 47 40 59 68 51 50 50 71 51 26 36 38 70 46 40 51 42.\n(a) Usingclass intervals 21 - 30, 31 - 40, ..., prepare a frequency distribution table.~(b) Calculate the mean mark of the distribution.\n(c) What percentage of the students scored more than 60%?~(c) What percentage of the students scored more than 60%?', NULL, '\n\nExplanation\n(a)\n\n\n\nMarks\nfrequency\nmid-point\n\\(fx\\)\n\n\n21 - 30\n2\n25.5\n51\n\n\n31 - 40\n10\n35.5\n355\n\n\n41 - 50\n12\n45.5\n546\n\n\n51 - 60\n15\n55.5\n832.5\n\n\n61 - 70\n8\n65.5\n524\n\n\n71 - 80\n3\n75.5\n226.5\n\n\n\\(\\sum\\)\n50\n\n2535\n\n\n\n(b) Mean (x) = \\(\\frac{\\sum fx}{\\sum f}\\)\n= \\(\\frac{2535}{50}\\)\n= \\(50.7\\)\n(c) Number of students that scored above 60% : 8 + 3 = 11\n%age : \\(\\frac{11}{50} \\times 100% = 22%\\)\n\n\n', '', 'Explanation\n(a)\n\n\n\nMarks\nfrequency\nmid-point\n\\(fx\\)\n\n\n21 - 30\n2\n25.5\n51\n\n\n31 - 40\n10\n35.5\n355\n\n\n41 - 50\n12\n45.5\n546\n\n\n51 - 60\n15\n55.5\n832.5\n\n\n61 - 70\n8\n65.5\n524\n\n\n71 - 80\n3\n75.5\n226.5\n\n\n\\(\\sum\\)\n50\n\n2535\n\n\n\n(b) Mean (x) = \\(\\frac{\\sum fx}{\\sum f}\\)\n= \\(\\frac{2535}{50}\\)\n= \\(50.7\\)\n(c) Number of students that scored above 60% : 8 + 3 = 11\n%age : \\(\\frac{11}{50} \\times 100% = 22%\\)\n\n\n', '(a)\n\n\n\nMarks\nfrequency\nmid-point\n\\(fx\\)\n\n\n21 - 30\n2\n25.5\n51\n\n\n31 - 40\n10\n35.5\n355\n\n\n41 - 50\n12\n45.5\n546\n\n\n51 - 60\n15\n55.5\n832.5\n\n\n61 - 70\n8\n65.5\n524\n\n\n71 - 80\n3\n75.5\n226.5\n\n\n\\(\\sum\\)\n50\n\n2535\n\n\n~(b) Mean (x) = \\(\\frac{\\sum fx}{\\sum f}\\)\n= \\(\\frac{2535}{50}\\)\n= \\(50.7\\)\n(c) Number of students that scored above 60% : 8 + 3 = 11\n%age : \\(\\frac{11}{50} \\times 100% = 22%\\)\n\n~(c) Number of students that scored above 60% : 8 + 3 = 11\n%age : \\(\\frac{11}{50} \\times 100% = 22%\\)\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(442, '2014', '\n\nQuestion\n\n(a) Simplify \\(\\frac{x + 2}{x - 2} - \\frac{x + 3}{x - 1}\\)\n(b) The graph of the equation \\(y = Ax^{2} + Bx + C\\) passes through the point (0, 0), (1, 4) and (2, 10). Find the :\n(i) value of C ; (ii) values of A and B ; (iii) co-ordinates of the other point where the graph cuts the x- axis.\n\n', '', '(a) Simplify \\(\\frac{x + 2}{x - 2} - \\frac{x + 3}{x - 1}\\)\n(b) The graph of the equation \\(y = Ax^{2} + Bx + C\\) passes through the point (0, 0), (1, 4) and (2, 10). Find the :\n(i) value of C ; (ii) values of A and B ; (iii) co-ordinates of the other point where the graph cuts the x- axis.', '(a) Simplify \\(\\frac{x + 2}{x - 2} - \\frac{x + 3}{x - 1}\\)~(b) The graph of the equation \\(y = Ax^{2} + Bx + C\\) passes through the point (0, 0), (1, 4) and (2, 10). Find the :\n(i) value of C ; (ii) values of A and B ; (iii) co-ordinates of the other point where the graph cuts the x- axis.', NULL, '\n\nExplanation\n(a) \\(\\frac{x + 2}{x - 2} - \\frac{x + 3}{x - 1}\\)\n= \\(\\frac{(x - 1)(x + 2) - (x - 2)(x + 3)}{(x - 2)(x - 1)}\\)\n= \\(\\frac{x^{2} + 2x - x - 2 - (x^{2} + 3x - 2x - 6)}{(x - 2)(x - 1)}\\)\n= \\(\\frac{x^{2} - x^{2} + x - x - 2 + 6}{(x - 2)(x - 1)}\\)\n= \\(\\frac{4}{(x - 2)(x - 1)}\\)\n(b)\n(1, 4) and (2, 10) are supposed roots of the equation\nWhen x = 1, equation becomes \\(A(1)^{2} + B(1) = 4\\)\nWhen x = 2, \\(A(2^{2}) + B(2) = 10\\)\n\\(\\implies A + B = 4 ...... (1)\\)\n\\(4A + 2B = 10 ....... (2)\\)\nSolving for A and B, we have\nA = 1 and B = 3.\n(iii) \\(y = x^{2} + 3x\\)\n\\(y = x(x + 3)\\)\nx = 0 or x = 3.\n\n\n', '', 'Explanation\n(a) \\(\\frac{x + 2}{x - 2} - \\frac{x + 3}{x - 1}\\)\n= \\(\\frac{(x - 1)(x + 2) - (x - 2)(x + 3)}{(x - 2)(x - 1)}\\)\n= \\(\\frac{x^{2} + 2x - x - 2 - (x^{2} + 3x - 2x - 6)}{(x - 2)(x - 1)}\\)\n= \\(\\frac{x^{2} - x^{2} + x - x - 2 + 6}{(x - 2)(x - 1)}\\)\n= \\(\\frac{4}{(x - 2)(x - 1)}\\)\n(b)\n(1, 4) and (2, 10) are supposed roots of the equation\nWhen x = 1, equation becomes \\(A(1)^{2} + B(1) = 4\\)\nWhen x = 2, \\(A(2^{2}) + B(2) = 10\\)\n\\(\\implies A + B = 4 ...... (1)\\)\n\\(4A + 2B = 10 ....... (2)\\)\nSolving for A and B, we have\nA = 1 and B = 3.\n(iii) \\(y = x^{2} + 3x\\)\n\\(y = x(x + 3)\\)\nx = 0 or x = 3.\n\n\n', '(a) \\(\\frac{x + 2}{x - 2} - \\frac{x + 3}{x - 1}\\)\n= \\(\\frac{(x - 1)(x + 2) - (x - 2)(x + 3)}{(x - 2)(x - 1)}\\)\n= \\(\\frac{x^{2} + 2x - x - 2 - (x^{2} + 3x - 2x - 6)}{(x - 2)(x - 1)}\\)\n= \\(\\frac{x^{2} - x^{2} + x - x - 2 + 6}{(x - 2)(x - 1)}\\)\n= \\(\\frac{4}{(x - 2)(x - 1)}\\)~(b)\n(1, 4) and (2, 10) are supposed roots of the equation\nWhen x = 1, equation becomes \\(A(1)^{2} + B(1) = 4\\)\nWhen x = 2, \\(A(2^{2}) + B(2) = 10\\)\n\\(\\implies A + B = 4 ...... (1)\\)\n\\(4A + 2B = 10 ....... (2)\\)\nSolving for A and B, we have\nA = 1 and B = 3.\n(iii) \\(y = x^{2} + 3x\\)\n\\(y = x(x + 3)\\)\nx = 0 or x = 3.\n\n', 'WAEC-PK-1', 'WAEC-MATH', 1),
(443, '2014', '\n\nQuestion\n\n(a) Using ruler and a pair of compasses only, construct : (i) quadrilateral PQRS such that /PQ/ = 10 cm, /QR/ = 8 cm, /PS/ = 6 cm, < PQR = 60° and < QPS = 75° ;\n(ii) the locus \\(l_{1}\\) of points equidistant from QR and RS ; (iii) locus \\(l_{2}\\) of points equidistant from R and S ;\n(b) Measure /RS/.\n\n', '', '(a) Using ruler and a pair of compasses only, construct : (i) quadrilateral PQRS such that /PQ/ = 10 cm, /QR/ = 8 cm, /PS/ = 6 cm, < PQR = 60° and < QPS = 75° ;\n(ii) the locus \\(l_{1}\\) of points equidistant from QR and RS ; (iii) locus \\(l_{2}\\) of points equidistant from R and S ;\n(b) Measure /RS/.', '(a) Using ruler and a pair of compasses only, construct : (i) quadrilateral PQRS such that /PQ/ = 10 cm, /QR/ = 8 cm, /PS/ = 6 cm, < PQR = 60° and < QPS = 75° ;\n(ii) the locus \\(l_{1}\\) of points equidistant from QR and RS ; (iii) locus \\(l_{2}\\) of points equidistant from R and S ;~(b) Measure /RS/.', NULL, '\n\nExplanation\n(a)\n(b) /RS/ = 4.4 cm.\n\n', '', 'Explanation\n(a)\n(b) /RS/ = 4.4 cm.\n\n', '~(b) /RS/ = 4.4 cm.\n', 'WAEC-PK-1', 'WAEC-MATH', 1);
