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; \(\sqrt{2}(\sqrt{6} + 2\sqrt{2}) - 2\sqrt{3}\)
4
\(\sqrt{3} + 4\)
4 \(\sqrt{2}\)
4\(\sqrt{3} + 4\)
Correct answer is A
\(\sqrt{2}(\sqrt{6} + 2\sqrt{2}) - 2\sqrt{3}\)
\(\sqrt{12}\) + 2 x 2 - 2\(\sqrt{3}\)
2 \(\sqrt{3}\) - 2 \(\sqrt{3}\) + 4
= 4
In what number base was the addition 1 + nn = 100, where n > 0, done?
n - 1
n + 1
n
n + 2
Correct answer is C
No explanation has been provided for this answer.
10.1 years
9.3 years
8.7 years
8 . 3 years
Correct answer is A
x = 10 ; 10 = \(\frac{x}{25}\)
x = 250
x = \(\frac{250 + 12.4}{26}\)
x = 10.09
x = 10.1 years
\(\sqrt{3 - 2}\)
2 - \(\sqrt{3}\)
\(\sqrt{3}\)
-2
Correct answer is C
Tan 60 = 3; Tan 30 = 1
\(\frac{\tan 60^o - 1}{1 - tan 30^o}\) = \(\frac{\sqrt{3 - 1}}{1 - \frac{1}{\sqrt{3}}}\) = \(\frac{\sqrt{3 - 1}}{\frac{3 - 1}{\sqrt{3}}}\)
= \(\frac{\sqrt{3 - 1}}{1} \times \frac{\sqrt{3 - 1}}{\sqrt{3}}\)
= \(\frac{\sqrt{3 - 1}}{1} \times \frac{\sqrt{3}}{\sqrt{3}}\)
= \(\sqrt{3}\)
A bearing of 320\(^o\) expressed as a compass bearing is
N 50\(^o\) W
N 40\(^o\) W
N 50\(^o\) E
N 40\(^o\) E
Correct answer is B
No explanation has been provided for this answer.