Year :
2000
Title :
Mathematics (Core)
Exam :
WASSCE/WAEC MAY/JUNE
Paper 1 | Objectives
11 - 20 of 48 Questions
| # | Question | Ans |
|---|---|---|
| 11. |
Given that the logarithm of a number is \(\bar{1}.8732\), find, correct to 2 significant figures the square root of the number. A. 0.29 B. 0.75 C. 0.86 D. 0.93 Detailed SolutionLet the number = d.\(\log d = \bar{1}.8732\) \(\log \sqrt{d} = \frac{\bar{1}.8732}{2}\) = \(\frac{\bar{2} + 1.8732}{2}\) = \(\bar{1}.9366\) \(\therefore \sqrt{d} = Antilog (\bar{1}.9366)\) = 0.86 |
|
| 12. |
A car moves at an average speed of 30kmh\(^{-1}\), how long does it take to cover 200 meters? A. 2.4 sec B. 24 sec C. 144 sec D. 240 sec Detailed SolutionSpeed = 30 kmh\(^{-1}\)= \(\frac{30 \times 1000}{3600}\) = \(\frac{25}{3} ms^{-1}\) Time = \(\frac{Distance}{Speed}\) = \(\frac{200}{\frac{25}{3}}\) = \(\frac{200 \times 3}{25}\) = 24 seconds |
|
| 13. |
A man bought a television set on hire purchase for N25,000, out of which he paid N10,000, if he is allowed to pay the balance in eight equal installments, find the value of each installment. A. N1250 B. N1578 C. N1875 D. N3125 Detailed SolutionPrice of television = N25,000Paid N10, 000 Remainder = N(25,000 - 10,000) = N15,000 Eight equal installments = \(\frac{15,000}{8}\) = N1,875 per installment. |
|
| 14. |
A tree is 8km due south of a building. Kofi is standing 8km west of the tree. How far is Kofi from the building? A. 4√2km B. 8km C. 8√2km D. 16km Detailed Solution \(x^2 = 8^2 + 8^2\) \(x^2 = 64 + 64 = 128\) \(x = \sqrt{128}\) = \(8\sqrt{2}\) km |
|
| 15. |
A tree is 8km due south of a building. Kofi is standing 8km west of the tree. Find the bearing of Kofi from the building A. 315o B. 270o C. 225o D. 135o Detailed SolutionThe bearing of Kofi from the building = 180° + 45°= 225° |
|
| 16. |
Which of the following bearings is equivalent to S50°W? A. 040o B. 130o C. 220o D. 230o Detailed Solution S50°W = 180° + 50° = 230° |
|
| 17. |
 In the diagram, AB is a vertical pole and BC is horizontal. If |AC| = 10m and |BC| = 5m, calculate the angle of depression of C from A A. 63o B. 60o C. 45o D. 27o Detailed Solution In triangle ABC, \(\frac{BC}{AC} = \cos \theta\) \(\implies \cos \theta = \frac{5}{10} = 0.5\) \(\theta = \cos^{-1} (0.5) = 60°\) |
|
| 18. |
 The bar chart shows the distribution of marks scored by a group of students in a test. Use the chart to answer the question below A. 15 B. 11 C. 10 D. 17 Detailed SolutionStudents that scored 4 and above = 5 + 3 + 3 + 0 + 2 + 2 + 2= 17 |
|
| 19. |
 The bar chart shows the distribution of marks scored by a group of students in a test. Use the chart to answer the question below A. 38 B. 22 C. 15 D. 11 |
B |
| 20. |
Calculate the standard deviation of the following marks; 2, 3, 6, 2, 5, 0, 4, 2 A. 1.5 B. 1.7 C. 1.8 D. 1.9 Detailed SolutionMean = \(\frac{24}{8} = 3\)Standard deviation = \(\sqrt{\frac{26}{8}}\) = \(\sqrt{3.25}\) = 1.802 \(\approeq\) 1.8 |
| 11. |
Given that the logarithm of a number is \(\bar{1}.8732\), find, correct to 2 significant figures the square root of the number. A. 0.29 B. 0.75 C. 0.86 D. 0.93 Detailed SolutionLet the number = d.\(\log d = \bar{1}.8732\) \(\log \sqrt{d} = \frac{\bar{1}.8732}{2}\) = \(\frac{\bar{2} + 1.8732}{2}\) = \(\bar{1}.9366\) \(\therefore \sqrt{d} = Antilog (\bar{1}.9366)\) = 0.86 |
|
| 12. |
A car moves at an average speed of 30kmh\(^{-1}\), how long does it take to cover 200 meters? A. 2.4 sec B. 24 sec C. 144 sec D. 240 sec Detailed SolutionSpeed = 30 kmh\(^{-1}\)= \(\frac{30 \times 1000}{3600}\) = \(\frac{25}{3} ms^{-1}\) Time = \(\frac{Distance}{Speed}\) = \(\frac{200}{\frac{25}{3}}\) = \(\frac{200 \times 3}{25}\) = 24 seconds |
|
| 13. |
A man bought a television set on hire purchase for N25,000, out of which he paid N10,000, if he is allowed to pay the balance in eight equal installments, find the value of each installment. A. N1250 B. N1578 C. N1875 D. N3125 Detailed SolutionPrice of television = N25,000Paid N10, 000 Remainder = N(25,000 - 10,000) = N15,000 Eight equal installments = \(\frac{15,000}{8}\) = N1,875 per installment. |
|
| 14. |
A tree is 8km due south of a building. Kofi is standing 8km west of the tree. How far is Kofi from the building? A. 4√2km B. 8km C. 8√2km D. 16km Detailed Solution\(x^2 = 8^2 + 8^2\) \(x^2 = 64 + 64 = 128\) \(x = \sqrt{128}\) = \(8\sqrt{2}\) km |
|
| 15. |
A tree is 8km due south of a building. Kofi is standing 8km west of the tree. Find the bearing of Kofi from the building A. 315o B. 270o C. 225o D. 135o Detailed SolutionThe bearing of Kofi from the building = 180° + 45°= 225° |
| 16. |
Which of the following bearings is equivalent to S50°W? A. 040o B. 130o C. 220o D. 230o Detailed SolutionS50°W = 180° + 50° = 230° |
|
| 17. |
In the diagram, AB is a vertical pole and BC is horizontal. If |AC| = 10m and |BC| = 5m, calculate the angle of depression of C from A A. 63o B. 60o C. 45o D. 27o Detailed SolutionIn triangle ABC, \(\frac{BC}{AC} = \cos \theta\) \(\implies \cos \theta = \frac{5}{10} = 0.5\) \(\theta = \cos^{-1} (0.5) = 60°\) |
|
| 18. |
The bar chart shows the distribution of marks scored by a group of students in a test. Use the chart to answer the question below A. 15 B. 11 C. 10 D. 17 Detailed SolutionStudents that scored 4 and above = 5 + 3 + 3 + 0 + 2 + 2 + 2= 17 |
|
| 19. |
The bar chart shows the distribution of marks scored by a group of students in a test. Use the chart to answer the question below A. 38 B. 22 C. 15 D. 11 |
B |
| 20. |
Calculate the standard deviation of the following marks; 2, 3, 6, 2, 5, 0, 4, 2 A. 1.5 B. 1.7 C. 1.8 D. 1.9 Detailed SolutionMean = \(\frac{24}{8} = 3\)Standard deviation = \(\sqrt{\frac{26}{8}}\) = \(\sqrt{3.25}\) = 1.802 \(\approeq\) 1.8 |