Year :
2008
Title :
Mathematics (Core)
Exam :
WASSCE/WAEC MAY/JUNE
Paper 1 | Objectives
11 - 20 of 45 Questions
| # | Question | Ans |
|---|---|---|
| 11. |
If 4y is 9 greater than the sum of y and 3x, by how much is greater than x? A. 3 B. 6 C. 9 D. 12 Detailed Solution4y - 9 > y + 3x; 4y - y > 3x + 93y > 3(x + 3); y = > \(\frac{3(x + 3)}{3}\) y > x + 3; y - 3 > x y is greater than x |
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| 12. |
If p = \(\frac{3}{5} \sqrt{\frac{q}{r}}\), express q in terms of p and r A. \(\frac{9}{25} pr^2\) B. \(\frac{9}{25} p^2r\) C. \(\frac{25}{9} p^2r\) D. \(\frac{25}{9} pr^2\) Detailed Solutionp = \(\frac{3}{5} \sqrt{\frac{q}{r}}; \frac{5p}{3} = \sqrt{\frac{q}{r}}\)= (\(\frac{5}{3}p\))2 = \(\frac{q}{r}\) = \(\frac{25p^2}{9} = \frac{q}{r}\) q = \(\frac{25}{9} p^2 r\) |
|
| 13. |
Simplify: \(\frac{2x^2 - 5x - 12}{4x^2 - 9}\) A. \(\frac{x + 4}{2x + 3}\) B. \(\frac{x + 4}{2x - 3}\) C. \(\frac{x - 4}{2x + 3}\) D. \(\frac{x - 4}{2x - 3}\) Detailed Solution\(\frac{2x^2 - 5x - 12}{4x^2 - 9}\) = \(\frac{3x^2 - 8x + 3x - 12}{(2x)^2 - 3^2}\)= \(\frac{32(x - 4) + 3(x - 4)}{(2x - 3)(2x + 3)} - \frac{(x - 4) + (2x + 3)}{(2x - 3) (2x + 3)}\) = \(\frac{x - 4}{2x - 3}\) |
|
| 14. |
PQR is a sector of a circle centre O, radius 4cm. If PQR = 30o, find, correct to 3 significant figures, the area of sector PQR. [Take \(\pi = \frac{22}{7}\)] A. 4.19cm2 B. 8.38cm2 C. 10.5cm2 D. 20.9cm2 Detailed SolutionGiven q2 = 25 - r2.....(1)from pythagora's theorem P2 = q2 + r2.......(2) put (1) into (2) p2 = 25 - p2 + x2 p2 = 25 p = \(\sqrt{25}\) = 5 |
|
| 15. |
If the volume of a cube is 343cm3, find the length of its side A. 3cm B. 6cm C. 7cm D. 96cm Detailed SolutionVolume of a cube = (side)3= (side)3 (side)3 = 343 side = 3\(\sqrt{343}\) side = 7cm |
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| 16. |
The angles of a quadrilateral are (x + 10)o, 2yo, 90o and (100 - y)o, Find y in terms of x A. y = 160 + x B. y = 100 + x C. y = 160 - x D. y = x - 100 Detailed SolutionSum of the angles in a quadrilateral = 360o(x + 10o) + 2y + 90o + 100 - y = 360o x + y + 200 = 360o y = 360 - 200 - x y = 160 - x |
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| 17. |
If xo is obtuse, which of the following is true? A. 90 B. 180 < x < 270 C. x < 90 D. 90 < x < 180 Detailed Solutionobtuse angle \(\to\) 90o < x < 180o |
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| 18. |
If tan x = 1, evaluate sin x + cos x, leaving your answer in the surd form A. 2\(\sqrt{2}\) B. \(\frac{1}{2} \sqrt{2}\) C. \(\sqrt{2}\) D. 2 Detailed Solutiontan x = 1; x = tan-1(10 = 45osin x + cos x = sin 45o + cos 45o = \(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\) = \(\frac{\sqrt{2} + \sqrt{2}}{2}\) = \(\frac{2\sqrt{2}}{2}\) = \(\sqrt{2}\) |
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| 19. |
If cos (x + 25)o = sin 45o, find the value of x A. 20 B. 30 C. 45 D. 60 Detailed Solutioncos(x + 25o) = sin 45ousing cos \(\theta\) = sin(90 - \(\theta\)) cos(x + 25o) = cos(90 - 45) cos(x + 25o) = cos 45 x + 25 = 45 x = 45 - 25 x = 20o |
|
| 20. |
If 2n = 128, find the value of (2n - 1)(5n - 1) A. 5(106) B. 2(106) C. 5(105) D. 2(105) Detailed Solution2n = 1282n = 27 n = 7 (2n - 1)(5n - 2) = (2n - 2.2)(5n - 2) put n = 7 (2n - 1)(5n - 2) = 2(2n - 2 x 5n - 2) = 2(2 x 5)n - 2 = 2(10n - 2) put n = 7 (2n-1)(5n-2) = 2(105) |
| 11. |
If 4y is 9 greater than the sum of y and 3x, by how much is greater than x? A. 3 B. 6 C. 9 D. 12 Detailed Solution4y - 9 > y + 3x; 4y - y > 3x + 93y > 3(x + 3); y = > \(\frac{3(x + 3)}{3}\) y > x + 3; y - 3 > x y is greater than x |
|
| 12. |
If p = \(\frac{3}{5} \sqrt{\frac{q}{r}}\), express q in terms of p and r A. \(\frac{9}{25} pr^2\) B. \(\frac{9}{25} p^2r\) C. \(\frac{25}{9} p^2r\) D. \(\frac{25}{9} pr^2\) Detailed Solutionp = \(\frac{3}{5} \sqrt{\frac{q}{r}}; \frac{5p}{3} = \sqrt{\frac{q}{r}}\)= (\(\frac{5}{3}p\))2 = \(\frac{q}{r}\) = \(\frac{25p^2}{9} = \frac{q}{r}\) q = \(\frac{25}{9} p^2 r\) |
|
| 13. |
Simplify: \(\frac{2x^2 - 5x - 12}{4x^2 - 9}\) A. \(\frac{x + 4}{2x + 3}\) B. \(\frac{x + 4}{2x - 3}\) C. \(\frac{x - 4}{2x + 3}\) D. \(\frac{x - 4}{2x - 3}\) Detailed Solution\(\frac{2x^2 - 5x - 12}{4x^2 - 9}\) = \(\frac{3x^2 - 8x + 3x - 12}{(2x)^2 - 3^2}\)= \(\frac{32(x - 4) + 3(x - 4)}{(2x - 3)(2x + 3)} - \frac{(x - 4) + (2x + 3)}{(2x - 3) (2x + 3)}\) = \(\frac{x - 4}{2x - 3}\) |
|
| 14. |
PQR is a sector of a circle centre O, radius 4cm. If PQR = 30o, find, correct to 3 significant figures, the area of sector PQR. [Take \(\pi = \frac{22}{7}\)] A. 4.19cm2 B. 8.38cm2 C. 10.5cm2 D. 20.9cm2 Detailed SolutionGiven q2 = 25 - r2.....(1)from pythagora's theorem P2 = q2 + r2.......(2) put (1) into (2) p2 = 25 - p2 + x2 p2 = 25 p = \(\sqrt{25}\) = 5 |
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| 15. |
If the volume of a cube is 343cm3, find the length of its side A. 3cm B. 6cm C. 7cm D. 96cm Detailed SolutionVolume of a cube = (side)3= (side)3 (side)3 = 343 side = 3\(\sqrt{343}\) side = 7cm |
| 16. |
The angles of a quadrilateral are (x + 10)o, 2yo, 90o and (100 - y)o, Find y in terms of x A. y = 160 + x B. y = 100 + x C. y = 160 - x D. y = x - 100 Detailed SolutionSum of the angles in a quadrilateral = 360o(x + 10o) + 2y + 90o + 100 - y = 360o x + y + 200 = 360o y = 360 - 200 - x y = 160 - x |
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| 17. |
If xo is obtuse, which of the following is true? A. 90 B. 180 < x < 270 C. x < 90 D. 90 < x < 180 Detailed Solutionobtuse angle \(\to\) 90o < x < 180o |
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| 18. |
If tan x = 1, evaluate sin x + cos x, leaving your answer in the surd form A. 2\(\sqrt{2}\) B. \(\frac{1}{2} \sqrt{2}\) C. \(\sqrt{2}\) D. 2 Detailed Solutiontan x = 1; x = tan-1(10 = 45osin x + cos x = sin 45o + cos 45o = \(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\) = \(\frac{\sqrt{2} + \sqrt{2}}{2}\) = \(\frac{2\sqrt{2}}{2}\) = \(\sqrt{2}\) |
|
| 19. |
If cos (x + 25)o = sin 45o, find the value of x A. 20 B. 30 C. 45 D. 60 Detailed Solutioncos(x + 25o) = sin 45ousing cos \(\theta\) = sin(90 - \(\theta\)) cos(x + 25o) = cos(90 - 45) cos(x + 25o) = cos 45 x + 25 = 45 x = 45 - 25 x = 20o |
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| 20. |
If 2n = 128, find the value of (2n - 1)(5n - 1) A. 5(106) B. 2(106) C. 5(105) D. 2(105) Detailed Solution2n = 1282n = 27 n = 7 (2n - 1)(5n - 2) = (2n - 2.2)(5n - 2) put n = 7 (2n - 1)(5n - 2) = 2(2n - 2 x 5n - 2) = 2(2 x 5)n - 2 = 2(10n - 2) put n = 7 (2n-1)(5n-2) = 2(105) |