How good are you with figures and formulas? Find out with these Mathematics past questions and answers. This Test is useful for both job aptitude test candidates and students preparing for JAMB, WAEC, NECO or Post UTME.
If \(\sin x° = \frac{a}{b}\), what is \(\sin (90 - x)°\)?
\(\frac{\sqrt{b^2 - a^2}}{b}\)
1\(\frac{-a}{b}\)
\(\frac{b^2 - a^2}{b}\)
\(\frac{a^2 - b^2}{b}\)
\(\sqrt{b^2 - a^3}\)
Correct answer is A
\(\sin x = \frac{a}{b}\)
\(\sin^{2} x + \cos^{2} x = 1\)
\(\sin^{2} x = \frac{a^{2}}{b^{2}}\)
\(\cos^{2} x = 1 - \frac{a^{2}}{b^{2}} = \frac{b^{2} - a^{2}}{b^{2}}\)
\(\therefore \cos x = \frac{\sqrt{b^{2} - a^{2}}}{b}\)
\(\sin (90 - x) = \sin 90 \cos x - \cos 90 \sin x\)
= \((1 \times \frac{\sqrt{b^{2} - a^{2}}}{b}) - (0 \times \frac{a}{b})\)
= \(\frac{\sqrt{b^{2} - a^{2}}}{b}\)
6
5
8
7
10
Correct answer is D
| x | 5 | 6 | 7 | 8 | 9 |
| f | 6 | 4 | 8 | 7 | 5 |
Mode is the No having the highest frequency = 7
6
5
8
7
10
Correct answer is B
Add all the No.s together to give 100
\(\sum\)xf = 100
N = 20
x = \(\frac{\sum xf}{N}\)
= \(\frac{100}{20}\)
= 5
What is the area between two concentric circles of diameters 26cm and 20cm?
100\(\pi\)
169\(\pi\)
69\(\pi\)
9\(\pi\)
269\(\pi\)
Correct answer is C
Area of circle 1 with diameter 26cm:
\(\pi r^{2} = \pi \times (\frac{26}{2})^{2} \)
= \(169 \pi cm^{2}\)
Area of circle 2 with diameter 20 cm:
\(\pi R^{2} = \pi \times (\frac{20}{2})^{2}\)
= \(100 \pi cm^{2}\)
Area between the two circles = \((169 - 100) \pi cm^{2}\)
= \(69 \pi cm^{2}\)
Evaluate correct to 4 decimal places 827.51 x 0.015
8.8415
12.4127
124.1265
12.4120
114.1265
Correct answer is B
827.51 x 0.015 By normal multiplication or use of four figure table, 827.51 x 0.015 = 12.4127 (to 4 decimal places).