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.
An arc of a circle of radius 6cm is 8cm long. Find the area of the sector
5\(\frac{1}{3}\)cm2
24cm2
36cm2
84cm2
Correct answer is B
Radius of the circle r = 6cm, Length of the arc = 8cm
Area of sector = \(\frac{\theta}{360}\) x \(\pi\)r2........(i)
Length of arc = \(\frac{\theta}{360}\) x 2\(\pi\)r........(ii)
from eqn. (ii) \(\theta\) = \(\frac{240}{\pi}\), subt. for \(\theta\) in eqn (i)
Area x \(\frac{240}{1}\) x \(\frac{1}{360}\) x \(\frac{\pi 6}{1}\)
= 24cm\(^2\)
69m
57m
51m
21m
Correct answer is C
QY = 452 + 242 = 2025 + 576
= 2601
QY = \(\sqrt{2601}\)
= 51
A regular polygon of n sides has 160o as the size of each interior angle. Find n
18
16
14
12
Correct answer is A
No explanation has been provided for this answer.
If cos\(\theta\) = \(\frac{a}{b}\), find 1 + tan2\(\theta\)
\(\frac{b^2}{a^2}\)
\(\frac{a^2}{b^2}\)
\(\frac{a^2 + b^2}{b^2 - a^2}\)
\(\frac{2a^2 + b^2}{a^2 + b^2}\)
Correct answer is A
cos\(\theta\) = \(\frac{a}{b}\), Sin\(\theta\) = \(\sqrt{\frac{b^2 - a^2}{a}}\)
Tan\(\theta\) = \(\sqrt{\frac{b^2 - a^2}{a^2}}\), Tan 2 = \(\sqrt{\frac{b^2 - a^2}{a^2}}\)
1 + tan2\(\theta\) = 1 + \(\frac{b^2 - a^2}{a^2}\)
= \(\frac{a^2 + b^2 - a^2}{a^2}\)
= \(\frac{b^2}{a^2}\)
At what points does the straight line y = 2x + 1 intersect the curve y = 2x2 + 5x - 1?
(-2, -3) and(\(\frac{1}{2}\), 2)
(1, 0), (1, 3)
(4, 0) and (0,1)
(2, 0) and (0,1)
Correct answer is A
No explanation has been provided for this answer.