Year :
1989
Title :
Mathematics (Core)
Exam :
WASSCE/WAEC MAY/JUNE
Paper 1 | Objectives
41 - 48 of 48 Questions
| # | Question | Ans |
|---|---|---|
| 41. |
 In the diagram above, O is the center of the circle. If ∠POR = 114o, calculate ∠PQR A. 123o B. 118.5o C. 117o D. 114o E. 54o Detailed SolutionReflex POR = 2PQRReflex PQR = 360o - 114o = 246o PQR = 246/2 = 123o |
|
| 42. |
The angle of depression of a point on the ground from the top of a building is 20.3o. If the foot of the building is 40m, calculate the height of the building, correct to one decimal place A. 37.5m B. 28.1m C. 27.8m D. 14.8m E. 13.9m Detailed Solution
 h = 40tan 69.7 = 402.703 h = 14.79 = 14.8 |
|
| 43. |
Solve for x: (x2 + 2x + 1) = 25 A. -6, -4 B. 6, -4 C. 6, 4 D. -6, 4 E. 5, 5 Detailed Solutionx2 + 2x + 1 = 25x2 + 6x - 4x -24= 0 x(x + 6) - 4(x + 6) = 0 (x - 4) - (x + 6) = 0 x = 4 or -6 = -6,4 |
|
| 44. |
Given that logp = 2 logx + 3logq, which of the following expresses p in terms of x and q? A. p = 5xq B. p = 6xq C. p = x2 + q3 D. p = 2x + 3q E. p = x2q3 Detailed Solutionlog p = 2log x + 3log qlog p = logx 2 + logq 3 logp = log x 2q 3 p = x 2q 3 |
|
| 45. |
Sinθ = 1/2 and cosθ = -√3/2, what is the value of θ? A. 30 0 B. 60 0 C. 90 o D. 120 o E. 150 o Detailed SolutionIn the 2nd Quadrant; SINE is the only Positive.Where θ is 30 for the first quadrant and 150 in the 2nd quadrant which makes COSINE negative |
|
| 46. |
What is the probability that 3 customers waiting in a bank will be served in the sequence of their arrival at the bank A. 1/6 B. 1/3 C. 1/2 D. 2/3 E. 5/6 Detailed SolutionProbability that 3 customers are attended to in the sequence they arrive= prob of attending to the 1st arrival x prob of attending to the second last x prob of attending to the last = \(\frac{1}{3} \times \frac{1}{2} \times \frac{1}{1}\) = \(\frac{1}{6}\) |
|
| 47. |
 In the diagram, O is the center of the circle, If ∠POQ = 80o and ∠PRQ = 5x, find the value of x. A. 4o B. 8o C. 16o D. 20o E. 32o Detailed Solution80o = 2 x 5x(angle at the center)=80o = 10x; x = 80o/10 = 8o |
|
| 48. |
In an A.P the first term is 2, and the sum of the 1st and the 6th term is 161/2. What is the 4th term A. 12 B. 91/2 C. 8 D. 7 E. 51/2 Detailed Solution1st term a = 2; 6th term = a + (16-1)d = 2 + 5d2 + (2 + 5d) = 161/2; 4 + 5d = 161/2 5d = 12.5 ∴ = 2.5 the 4th term is 2 + (4-1)2.5= 91/2 |
| 41. |
In the diagram above, O is the center of the circle. If ∠POR = 114o, calculate ∠PQR A. 123o B. 118.5o C. 117o D. 114o E. 54o Detailed SolutionReflex POR = 2PQRReflex PQR = 360o - 114o = 246o PQR = 246/2 = 123o |
|
| 42. |
The angle of depression of a point on the ground from the top of a building is 20.3o. If the foot of the building is 40m, calculate the height of the building, correct to one decimal place A. 37.5m B. 28.1m C. 27.8m D. 14.8m E. 13.9m Detailed Solution
h = 40tan 69.7 = 402.703 h = 14.79 = 14.8 |
|
| 43. |
Solve for x: (x2 + 2x + 1) = 25 A. -6, -4 B. 6, -4 C. 6, 4 D. -6, 4 E. 5, 5 Detailed Solutionx2 + 2x + 1 = 25x2 + 6x - 4x -24= 0 x(x + 6) - 4(x + 6) = 0 (x - 4) - (x + 6) = 0 x = 4 or -6 = -6,4 |
|
| 44. |
Given that logp = 2 logx + 3logq, which of the following expresses p in terms of x and q? A. p = 5xq B. p = 6xq C. p = x2 + q3 D. p = 2x + 3q E. p = x2q3 Detailed Solutionlog p = 2log x + 3log qlog p = logx 2 + logq 3 logp = log x 2q 3 p = x 2q 3 |
| 45. |
Sinθ = 1/2 and cosθ = -√3/2, what is the value of θ? A. 30 0 B. 60 0 C. 90 o D. 120 o E. 150 o Detailed SolutionIn the 2nd Quadrant; SINE is the only Positive.Where θ is 30 for the first quadrant and 150 in the 2nd quadrant which makes COSINE negative |
|
| 46. |
What is the probability that 3 customers waiting in a bank will be served in the sequence of their arrival at the bank A. 1/6 B. 1/3 C. 1/2 D. 2/3 E. 5/6 Detailed SolutionProbability that 3 customers are attended to in the sequence they arrive= prob of attending to the 1st arrival x prob of attending to the second last x prob of attending to the last = \(\frac{1}{3} \times \frac{1}{2} \times \frac{1}{1}\) = \(\frac{1}{6}\) |
|
| 47. |
In the diagram, O is the center of the circle, If ∠POQ = 80o and ∠PRQ = 5x, find the value of x. A. 4o B. 8o C. 16o D. 20o E. 32o Detailed Solution80o = 2 x 5x(angle at the center)=80o = 10x; x = 80o/10 = 8o |
|
| 48. |
In an A.P the first term is 2, and the sum of the 1st and the 6th term is 161/2. What is the 4th term A. 12 B. 91/2 C. 8 D. 7 E. 51/2 Detailed Solution1st term a = 2; 6th term = a + (16-1)d = 2 + 5d2 + (2 + 5d) = 161/2; 4 + 5d = 161/2 5d = 12.5 ∴ = 2.5 the 4th term is 2 + (4-1)2.5= 91/2 |