Year :
1995
Title :
Mathematics (Core)
Exam :
WASSCE/WAEC MAY/JUNE
Paper 1 | Objectives
11 - 20 of 48 Questions
| # | Question | Ans |
|---|---|---|
| 11. |
Solve the inequality: \(\frac{1}{3}(2x - 1) < 5\) A. x < - 5 B. X<-6 C. X<7 D. x <8 E. x < 16 Detailed Solution\(\frac{1}{3}(2x - 1) < 5\)\(2x - 1 < 15\) \(2x < 16\) \(x < 8\) |
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| 12. |
What is the smaller value of x for which x\(^2\) - 3x + 2= 0? A. 1 B. 2 C. 3 D. 4 E. 5 Detailed Solutionx\(^2\) - 3x + 2 = 0x\(^2\) - 2x - x + 2 = 0 x(x - 2) - 1(x - 2) = 0 (x - 2)(x - 1) = 0 x = 1 or 2. The smaller value of x = 1. |
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| 13. |
Factorize the expression x(a - c) + y(c - a) A. (a - c)(y - x) B. (a - c)(x - y) C. (a + c)(x - y) D. (a + c)(x + y) E. (a - c)(x + y) Detailed Solutionx(a - c) + y(c - a)= x(a - c) - y(a - c) = (x - y)(a - c) |
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| 14. |
If \(y \propto \frac{1}{x^2}\) and x = 3 when y = 4, find y when x = 2. A. 1 B. 3 C. 9 D. 18 E. 21 Detailed Solution\(y \propto \frac{1}{x^2}\)\(y = \frac{k}{x^2}\) \(4 = \frac{k}{3^2}\) \(k = 4 \times 3^2 = 36\) \(y = \frac{36}{x^2}\) When x = 2, \(y = \frac{36}{2^2} = 9\) |
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| 15. |
Solve the equation 3x\(^2\) + 25x -18 = 0 A. -3,2 B. -2,3 C. -2,9 D. -9,2/3 E. 2/3, 9. Detailed Solution3x\(^2\) + 25x - 18 = 03x\(^2\) + 27x - 2x - 18 = 0 3x(x + 9) - 2(x + 9) = 0 (3x - 2)(x + 9) = 0 x = \(\frac{2}{3}\) or x = -9. |
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| 16. |
Solve the equation (x +2)(x - 7) = 0 A. x = 1 or 8 B. x = -2 or 7 C. x = -4 or 5 D. x = -3 or 6 E. x= -5 or -2 Detailed Solution(x + 2)(x - 7) = 0x + 2 = 0 \(\implies\) x = -2 x - 7 = 0 \(\implies\) x = 7 x = -2 or 7 |
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| 17. |
Solve the following simultaneous equations: x+ y = 3/2; x - y = 5/2 and use your result to find the value of 2y + x A. -2 B. -1 C. 1/2 D. 1 E. 31/2 Detailed Solutionx + y = \(\frac{3}{2}\) ... (i)x - y = \(\frac{5}{2}\) ... (ii) (i) - (ii): 2y = \(\frac{-2}{2}\) = -1 y = \(-\frac{1}{2}\) x + y = \(\frac{3}{2}\) x - \(\frac{1}{2}\) = \(\frac{3}{2}\) x = \(\frac{3}{2} + \frac{1}{2}\) = \(\frac{4}{2}\) x = 2 \(\therefore\) 2y + x = 2(\(-\frac{1}{2}\)) + 2 = -1 + 2 = 1. |
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| 18. |
 Use the graph of y = 3x\(^2\) + x - 7 above to answer the questionWhat is the minimum value of y? A. -10 B. -7 C. -4 D. -1 E. 2 |
B |
| 19. |
 Use the graph of y = 3x\(^2\) + x - 7 above to answer the questionFind the roots of the equation y=3x\(^2\) + x - 7 A. 1.0 And -1.2 B. 1.1and-1.3 C. 1.4and-1.7 D. 2.0 and -1.9 E. 2.4 and -2.0 |
C |
| 20. |
 In the diagram above. |AB| = 12cm, |AE| = 8cm, |DCl = 9cm and AB||DC. Calculate |EC| A. 10cm B. 9cm C. 8cm D. 7cm E. 6cm Detailed Solution|AB| = 12 cm; |DC| = 9 cm; |AE| = 8 cm.\(\Delta\) ABE and \(\Delta\) EDC are similar triangles. Hence, \(\frac{|AB|}{|DC|} = \frac{|AE|}{|EC|}\) \(\frac{12}{9} = \frac{8}{|EC|}\) \(\therefore |EC| = \frac{9 \times 8}{12}\) = 6 cm |
| 11. |
Solve the inequality: \(\frac{1}{3}(2x - 1) < 5\) A. x < - 5 B. X<-6 C. X<7 D. x <8 E. x < 16 Detailed Solution\(\frac{1}{3}(2x - 1) < 5\)\(2x - 1 < 15\) \(2x < 16\) \(x < 8\) |
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| 12. |
What is the smaller value of x for which x\(^2\) - 3x + 2= 0? A. 1 B. 2 C. 3 D. 4 E. 5 Detailed Solutionx\(^2\) - 3x + 2 = 0x\(^2\) - 2x - x + 2 = 0 x(x - 2) - 1(x - 2) = 0 (x - 2)(x - 1) = 0 x = 1 or 2. The smaller value of x = 1. |
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| 13. |
Factorize the expression x(a - c) + y(c - a) A. (a - c)(y - x) B. (a - c)(x - y) C. (a + c)(x - y) D. (a + c)(x + y) E. (a - c)(x + y) Detailed Solutionx(a - c) + y(c - a)= x(a - c) - y(a - c) = (x - y)(a - c) |
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| 14. |
If \(y \propto \frac{1}{x^2}\) and x = 3 when y = 4, find y when x = 2. A. 1 B. 3 C. 9 D. 18 E. 21 Detailed Solution\(y \propto \frac{1}{x^2}\)\(y = \frac{k}{x^2}\) \(4 = \frac{k}{3^2}\) \(k = 4 \times 3^2 = 36\) \(y = \frac{36}{x^2}\) When x = 2, \(y = \frac{36}{2^2} = 9\) |
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| 15. |
Solve the equation 3x\(^2\) + 25x -18 = 0 A. -3,2 B. -2,3 C. -2,9 D. -9,2/3 E. 2/3, 9. Detailed Solution3x\(^2\) + 25x - 18 = 03x\(^2\) + 27x - 2x - 18 = 0 3x(x + 9) - 2(x + 9) = 0 (3x - 2)(x + 9) = 0 x = \(\frac{2}{3}\) or x = -9. |
| 16. |
Solve the equation (x +2)(x - 7) = 0 A. x = 1 or 8 B. x = -2 or 7 C. x = -4 or 5 D. x = -3 or 6 E. x= -5 or -2 Detailed Solution(x + 2)(x - 7) = 0x + 2 = 0 \(\implies\) x = -2 x - 7 = 0 \(\implies\) x = 7 x = -2 or 7 |
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| 17. |
Solve the following simultaneous equations: x+ y = 3/2; x - y = 5/2 and use your result to find the value of 2y + x A. -2 B. -1 C. 1/2 D. 1 E. 31/2 Detailed Solutionx + y = \(\frac{3}{2}\) ... (i)x - y = \(\frac{5}{2}\) ... (ii) (i) - (ii): 2y = \(\frac{-2}{2}\) = -1 y = \(-\frac{1}{2}\) x + y = \(\frac{3}{2}\) x - \(\frac{1}{2}\) = \(\frac{3}{2}\) x = \(\frac{3}{2} + \frac{1}{2}\) = \(\frac{4}{2}\) x = 2 \(\therefore\) 2y + x = 2(\(-\frac{1}{2}\)) + 2 = -1 + 2 = 1. |
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| 18. |
Use the graph of y = 3x\(^2\) + x - 7 above to answer the questionWhat is the minimum value of y? A. -10 B. -7 C. -4 D. -1 E. 2 |
B |
| 19. |
Use the graph of y = 3x\(^2\) + x - 7 above to answer the questionFind the roots of the equation y=3x\(^2\) + x - 7 A. 1.0 And -1.2 B. 1.1and-1.3 C. 1.4and-1.7 D. 2.0 and -1.9 E. 2.4 and -2.0 |
C |
| 20. |
In the diagram above. |AB| = 12cm, |AE| = 8cm, |DCl = 9cm and AB||DC. Calculate |EC| A. 10cm B. 9cm C. 8cm D. 7cm E. 6cm Detailed Solution|AB| = 12 cm; |DC| = 9 cm; |AE| = 8 cm.\(\Delta\) ABE and \(\Delta\) EDC are similar triangles. Hence, \(\frac{|AB|}{|DC|} = \frac{|AE|}{|EC|}\) \(\frac{12}{9} = \frac{8}{|EC|}\) \(\therefore |EC| = \frac{9 \times 8}{12}\) = 6 cm |