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.
11cm
15.4cm
17.64cm
23.10cm
Correct answer is A
\(\frac{θ}{360}\) * 2 * π * r → \(\frac{150 * 2 * 22 * 4.2}{360 x 7}\)
= 11cm
An exterior angle of a regular polygon is 22.5°. Find the number of sides.
13
14
15
16
Correct answer is D
The sum of exterior angles of a polygon is 360°
\(\frac{360}{22.5}\) = 16 sides
Make t the subject of k = \(m \sqrt \frac{t-p}{r}\)
\(\frac{k^2r + p}{m^2}\)
\(\frac{k^2r + pm^2}{m^2}\)
\(\frac{k^2r - p}{m^2}\)
\(\frac{k^2r + p^2}{m^2}\)
Correct answer is B
square both sides to remove the square root
k\(^2\) = m\(^2\) \(\frac{t-p}{r}\)
\(\frac{k^2r}{m^2}\) = t - p
t = \(\frac{k^2r}{m^2}\) + p
t = \(\frac{k^2r + pm^2}{m^2}\)
Find the value of x such that \(\frac{1}{x}\) +\(\frac{4}{3x}\) - \(\frac{5}{6x}\) + 1 = 0
\(\frac{1}{6}\)
\(\frac{1}{4}\)
\(\frac{-3}{2}\)
\(\frac{-7}{6}\)
Correct answer is C
\(\frac{1}{x}\) +\(\frac{4}{3x}\) - \(\frac{5}{6x}\) + 1 = 0
using 6x as lcm
→ \(\frac{6+8-5+6x}{6x}\)
→ \(\frac{9+6x}{6x}\) = 0
9+6x = 0
6x = -9
x = \(\frac{-9}{6}\) or \(\frac{-3}{2}\)
What value of p will make (x\(^2\) - 4x + p) a perfect square?
-2
16
4
-8
Correct answer is C
(x\(^2\) - 4x + p)
Use the coefficient of the middle variable(-4x)
= (\(\frac{-4}{2}\))\(^2\)
= (-2)\(^2\)
= 4