2000 - WAEC Mathematics Past Questions and Answers - page 2
Given that the logarithm of a number is \(\bar{1}.8732\), find, correct to 2 significant figures the square root of the number.
Let 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
A car moves at an average speed of 30kmh\(^{-1}\), how long does it take to cover 200 meters?
Speed = 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
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.
Price of television = N25,000
Paid N10, 000
Remainder = N(25,000 - 10,000)
= N15,000
Eight equal installments = \(\frac{15,000}{8}\)
= N1,875 per installment.
A tree is 8km due south of a building. Kofi is standing 8km west of the tree. How far is Kofi from the building?
\(x^2 = 8^2 + 8^2\)
\(x^2 = 64 + 64 = 128\)
\(x = \sqrt{128}\)
= \(8\sqrt{2}\) km
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
Which of the following bearings is equivalent to S50°W?
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
In triangle ABC, \(\frac{BC}{AC} = \cos \theta\)
\(\implies \cos \theta = \frac{5}{10} = 0.5\)
\(\theta = \cos^{-1} (0.5) = 60°\)
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
How many students scored 4 marks and above?
Students that scored 4 and above = 5 + 3 + 3 + 0 + 2 + 2 + 2
= 17
How many students took the test?
Calculate the standard deviation of the following marks; 2, 3, 6, 2, 5, 0, 4, 2
| x | 2 | 3 | 6 | 2 | 5 | 0 | 4 | 2 | 24 |
| x - \(\bar{x}\) | -1 | 0 | 3 | -1 | 2 | -3 | 1 | -1 | |
| \((x - \bar{x})^2\) | 1 | 0 | 9 | 1 | 4 | 9 | 1 | 1 | 26 |
Mean = \(\frac{24}{8} = 3\)
Standard deviation = \(\sqrt{\frac{26}{8}}\)
= \(\sqrt{3.25}\)
= 1.802 \(\approeq\) 1.8