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.
Find the total area of the surface of a solid cylinder whose base radius is 4cm and height is 5cm
56\(\pi\) cm2
72\(\pi\) cm2
96\(\pi\) cm2
192\(\pi\) cm2
Correct answer is B
The total surface area of a cylinder = \(2\pi r (r + h)\)
= \(2 \pi (4) (4 + 5)\)
= \(8 \pi (9) = 72 \pi cm^{2}\)
0.30o
045o
060o
090o
Correct answer is C
tan \(\theta\) = \(\frac{60\sqrt{3}}{60}\) = \(\sqrt{3}\)
\(\theta\) = tan \(\frac{1}{3}\) = 60o
Bearing of y from x = 060o
If cos x = \(\sqrt{\frac{a}{b}}\) find cosec x
\(\frac{b}{\sqrt{b - a}}\)
\(\sqrt{\frac{b}{a}}\)
\(\sqrt{\frac{b}{b - a}}\)
\(\sqrt{\frac{b - a}{a}}\)
Correct answer is C
cosx = \(\sqrt{\frac{a}{b}}\)
y2 + \(\sqrt{(a)^2}\) = \(\sqrt{(b)^2}\) by pythagoras
y2 = b - a
∴ y = b - a
cosec x = \(\frac{1}{sin x}\) = \(\frac{1}{y}\)
\(\frac{b}{y}\) = \(\frac{\sqrt{b}}{\sqrt{b - a}}\)
= \(\sqrt{\frac{b}{b - a}}\)
Simplify \(\cos^{2} x (\sec^{2} x + \sec^{2} x \tan^{2} x)\)
tan x
tanx secx
\(\sec^2 x\)
cosec2x
Correct answer is C
\(\cos^{2} x (\sec^{2} x + \sec^{2} x \tan^{2} x)\)
= \(\cos^{2} x \sec^{2} x + \cos^{2} x \sec^{2} x \tan^{2} x\)
= \(1 + \tan^{2} x\)
= \(1 + \frac{\sin^{2} x}{\cos^{2} x}\)
= \(\frac{\cos^{2} x + \sin^{2} x}{\cos^{2} x}\)
= \(\frac{1}{\cos^{2} x} = \sec^{2} x\)
60 (tan 62o - tan 64o)
60 (cot 64o - cot 62o)
60 (cot 62o - cot 64o)
60 (tan 64o - tan 62o)
Correct answer is D
\(\frac{BC}{60}\) = \(\frac{tan 62}{1}\)
BC = 60 tan 62
\(\frac{AC}{60}\) = \(\frac{tan 62}{1}\)
AC = 60 tan 64
AB = AC - BC
= 60(tan 64o - tan 62o)