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 the surface area of a sphere increased by 44%, find the percentage increase in diameter
44
30
22
20
Correct answer is D
Surface Area of Sphere A = 4\(\pi r^2\)
∴ A = 4\(\pi\)\(\frac{(D)^2}{2}\)
= \(\frac{(D)^2}{2}\)
= \(\pi\)D2
When increased by 44% A = \(\frac{144 \pi D^2}{100}\)
\(\pi\)\(\frac{(12D)^2}{10}\) = \(\pi\)\(\frac{(6D)^2}{5}\)
Increase in diameter = \(\frac{6D}{5}\) - D = \(\frac{1}{5}\)D
Percentage increase = \(\frac{1}{5}\) x \(\frac{1}{100}\)%
= 20%
2\(\frac{2}{3}\)
4
5\(\frac{1}{3}\)
8
Correct answer is B
P = N800 (Principal), r = 12\(\frac{1}{2}\)% or 0.125
After sometimes, A = 1.5 x 800 = N1,200
A = P(1 + Tr) = 1200
= 800(1 + 0.125T)
= 1200
= 800(1 + 0.125T)
1200 = 800(1 + 0.125T)
0.125T = 0.5
T = \(\frac{0.5}{0.125}\)
= 4
If a : b = 5 : 8, x : y = 25 : 16; evaluate \(\frac{a}{x}\) : \(\frac{b}{y}\)
125 : 128
3 : 5
3 : 4
2 : 5
Correct answer is D
a : b = 5 : 8 = 2.5 : 40
x : y = 25 : 16
\(\frac{a}{x}\) : \(\frac{b}{y}\) = \(\frac{25}{25}\) : \(\frac{40}{16}\)
= 1 : \(\frac{40}{16}\)
= 16 : 40
= 2 : 5
10-5
7 x 10-4
8 x 10-5
10-6
Correct answer is A
0.007685 = 0.00769 (three significant figures)
0.007685 = 0.0077(4d.p)
the difference = 0.0077 - 0.00769
= 0.00001
= 1.0 x 10-5
Simplify \(\sqrt[3]{(64r^{-6})^{\frac{1}{2}}}\)
\(\frac{r}{2}\)
2r
\(\frac{1}{2r}\)
\(\frac{2}{r}\)
Correct answer is B
\(\sqrt[3]{(64r^{-6})^{\frac{1}{2}}}\)
= \(((64r^{-6})^{1/_2})^{1/_3}\)
=\((64r^{-6})^{1/_6}\)
=\(64)^{1/_6}(r^{-6})^{1/_6}\)
=2/r