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.
Simplify \(\sqrt{50} + \frac{10}{\sqrt{2}}\)
10
10\(\sqrt{2}\)
20
20\(\sqrt{2}\)
Correct answer is B
\(\sqrt{50} + \frac{10}}{\sqrt{2}} = \(\frac{\sqrt{50}}{1} + \sqrt{10}{\sqrt{2}}\)
= \(\frac{\sqrt{50 \times 2} + 10}{\sqrt{2}}\)
= \(\frac{\sqrt{100} + 10}{\sqrt{2}}\)
= \(\frac{10 + 10}{\sqrt{2} = \frac{20}{\sqrt{2}}\)
= \(\frac{20}{\sqrt{2}}\) \times \frac{\sqrt{2}}{\sqrt{2}}\)
= \(\frac{20\sqrt{2}}{2}\)
= 10\(\sqrt{2}\)
Find the product of 0.0409 and 0.0021 leaving your answer in the standard form
8.6 x 10-6
8.6 x 105
8.6 x 10-4
8.6 x 10-5
Correct answer is D
0.0409 x 0.0021 = 409 x 10\(^{-4}\) x 21 x 10\(^{-4}\)
= 409 x 21 x 10\(^{-4 * -4}\)
= 8589 x 10\(^{-8}\)
= 8.589 x 10\(^{3}\) x 10\(^{-8}\)
= 8.6 x 10\(^{3-8}\)
= 8.6 x 10\(^{-5}\)
Evaluate \(\frac{(3.2)^2 - (4.8)^2}{3.2 + 4.8}\)
-0.08
-1.60
-10.24
-12.80
Correct answer is B
\(\frac{(3.2)^2 - (4.8)^2}{3.2 + 1.8} = \frac{(3.2 - 4.8)(3.2 + 4.8)}{(3.2 + 4.8)}\)
= 3.2 - 4.8
= -1.60
If x% of 240 equals 12, find x
x = 1
x = 3
x = 5
x = 7
Correct answer is C
x% of 240 = 12
\(\frac{x}{100} \times 240 = 12\)
x = \(\frac{12 \times 100}{240}\)
x = 5
43.36cm2
32.072
18.212
6.932
Correct answer is C
Area of shaded portion = Area of semicircle
Area of \(\bigtriangleup\) RSO
Area of semicircle = \(\frac {\pi r^2}{2} = \frac{22\times 4 \times 4}{7 \times 2}\)
= 25.14cm2; Area of \(\bigtriangleup\)RSO
=\(\sqrt{s(s - 1)(s - b)(s - c)}\); where
s = \(\frac{a + b + c}{2}\)
s = \(\frac{4 + 4 + 4}{2}\)
= 6cm
= \(\sqrt{6(6 - 4)(6 - 4) (6 - 4)}\)
= \(\sqrt{6(2) (2) (2)}\)
= \(\sqrt{18}\) = 6.93cm2
Area of shaded region
= 25.14 - 6.93
= 18.21cm2