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 \(\cos^{2} x (\sec^{2} x + \sec^{2} x \tan^{2} x)\)
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)
If the exterior angles of a pentagon are x°, (x + 5)°, (x + 10)°, (x + 15)° and (x + 20)°, find x
118o
72o
62o
36o
Correct answer is C
The sum of the exterior angles of a regular polygon = 360°.
\(\therefore x + (x + 5) + (x + 10) + (x + 15) + (x + 20) = 360°\)
\(5x + 50 = 360° \implies 5x = 360° - 50° = 310°\)
\(x = \frac{310°}{5} = 62°\)
8\(\sqrt{3}\)
\(\frac{16}{\sqrt{3}}\)
\(\sqrt{3}\)
\(\frac{10}{\sqrt{3}}\)
Correct answer is A
\(\frac{x}{4}\) = \(\frac{\tan 60}{1}\)
x = 4 tan60
= 4\(\sqrt{3}\)
BD = 2x
= 8\(\sqrt{3}\)
The area of a square is 144 sq cm. Find the length of its diagonal
11\(\sqrt{3cm}\)
12cm
12\(\sqrt{2cm}\)
13cm
Correct answer is C
BD = \(\sqrt{x^2 + x^2}\)
= \(\sqrt{12^2 + 12^2}\)
= \(\sqrt{144 + 144}\)
= 2(144)
= 12\(\sqrt{2cm}\)