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.
\(\frac{x}{sin 40^o}\)
\(\frac{x}{cos 40^o}\)
\(\frac{x}{2 sin 40^o}\)
\(\frac{x}{2 cos 40^o}\)
Correct answer is C
No explanation has been provided for this answer.
14.9 o
15.5 o
74.5 o
75.1 o
Correct answer is A
\(\tan x = \frac{20}{75} = 0.267\)
\(x = \tan^{-1} 0.267 = 14.93°\)
\(\approxeq\) 14.9°
If \(tan x = \frac{1}{\sqrt{3}}\), find cos x - sin x such that \(0^o \leq x \leq 90^o\)<
\(\frac{\sqrt{3}+1}{2}\)
\(\frac{2}{\sqrt{3}+1}\)
\(\frac{\sqrt{3}-1}{2}\)
\(\frac{2}{\sqrt{3}-1}\)
Correct answer is C
\(\cos x = \frac{\sqrt{3}}{2}\)
\(\sin x = \frac{1}{2}\)
\(\cos x - \sin x = \frac{\sqrt{3} - 1}{2}\)
If y varies inversely as x\(^2\), how does x vary with y?
x varies inversely as y2
x varies inversely as √y
x varies directly as y2
x varies directly as
Correct answer is B
\(y \propto \frac{1}{x^2}\)
\(y = \frac{k}{x^2}\)
\(x^2 = \frac{k}{y}\)
\(x = \frac{\sqrt{k}}{\sqrt{y}}\)
Since k is a constant, then \(\sqrt{k}\) is also a constant.
\(\therefore x \propto \frac{1}{\sqrt{y}}\)
Simplify \(\frac{4}{x+1}-\frac{3}{x-1}\)
\(\frac{x+7}{x^2 - 1}\)
\(\frac{x-7}{x^2 + 1}\)
\(\frac{x-7}{x^2 - 1}\)
\(\frac{x-11}{x^2 - 1}\)
Correct answer is C
\(\frac{4}{x+1}-\frac{3}{x-1}\)
=\(\frac{4x - 4 - 3x - 3}{(x+1)(x-1)}=\frac{x-7}{x^2 - 1}\)