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.
The chord ST of a circle is equal to the radius r of the circle. Find the length of arc ST
\(\frac{\pi r}{3}\)
\(\frac{\pi r}{2}\)
\(\frac{\pi r}{12}\)
\(\frac{\pi r}{6}\)
Correct answer is A
\(\frac{ \frac{r}{2}}{r}\) Sin \(\theta\) = \(\frac{1}{2}\)
\(\theta\) = sin\(^{-1}\) (\(\frac{1}{2}\)) = 30\(^o\) = 60\(^o\)
Length of arc (minor)
ST = \(\frac{\theta}{360}\) x 2\(\pi r\)
\(\frac{60}{360} \times 2 \pi \times r = \frac{\pi}{3}\)
A binary operation x is defined by a x b = a\(^b\). If a x 2 = 2 - a, find the possible values of a?
1, -2
2, -1
2, -2
1, -1
Correct answer is A
a = b = a\(^2\)
a + 2 = a\(^2\).....(i)
a + 2 = 2 - a..............(ii)
a\(^2\) = 2 - a
a\(^2\)+ a - 2 = a\(^2\) + a - 2 = 0
= (a + 2)(a - 1) = 0
a = 1 or - 2
Find the value of x if \(\frac{\sqrt{2}}{x + \sqrt{2}}\) = \(\frac{1}{x - \sqrt{2}}\)
3\(\sqrt{2}\) + 4
3\(\sqrt{2}\) - 4
3 - 2\(\sqrt{2}\)
4 + 2\(\sqrt{2}\)
Correct answer is A
\(\frac{\sqrt{2}}{x + 2}\) = x - \(\frac{1}{\sqrt{2}}\)
x\(\sqrt{2}\) (x - \(\sqrt{2}\)) = x + \(\sqrt{2}\) (cross multiply)
x\(\sqrt{2}\) - 2 = x + \(\sqrt{2}\)
= x\(\sqrt{2}\) - x
= 2 + \(\sqrt{2}\)
x (\(\sqrt{2}\) - 1) = 2 + \(\sqrt{2}\)
= \(\frac{2 + \sqrt{2}}{\sqrt{2} - 1} \times \frac{\sqrt{2} + 1}{\sqrt{2} + 1}\)
x = \(\frac{2 \sqrt{2} + 2 + 2 + \sqrt{2}}{2 - 1}\)
= 3\(\sqrt{2}\) + 4
x > -\(\frac{1}{6}\)
x > 0
0 < x < 6
0 < x <\(\frac{1}{6}\)
Correct answer is A
\(\frac{1}{3x}\) + \(\frac{1}{2}\)x = \(\frac{2 + 3x}{6x}\) > \(\frac{1}{4x}\)
= 4(2 + 3x) > 6x = 12x\(^2\) - 2x = 0
= 2x(6x - 1) > 0 = x(6x - 1) > 0
Case 1 (-, -) = x < 0, 6x - 1 > 0
= x < 0, x < \(\frac{1}{6}\) (solution)
Case 2 (+, +) = x > 0, 6x - 1 > 0 = x > 0
x > \(\frac{1}{6}\)
Combining solutions in cases (1) and (2)
= x > 0, x < \(\frac{1}{6}\) = 0 < x < \(\frac{1}{6}\)
N2,050
N2,600
N3,100
N3,450
Correct answer is A
360\(^o\) - (60\(^o\) + 60\(^o\) + 67 + 50 = 237\(^o\))
360\(^o\) - 237 = 130\(^o\)
B. Salary = \(\frac{123}{360} X \frac{N6000}{1}\)
= N2,050