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.
15
27
50
52
54
Correct answer is E
Total students = 120
grade = 18
Zo = \(\frac{18}{120}\) x \(\frac{360}{1}\)
= 54o
16667km
28850km
8333km
2233km
Correct answer is A
Length of an area = \(\frac{\theta}{360}\) x 2\(\pi\)r
Longitude difference = 147 + 153 = 30NoN
distance between xy = \(\frac{\theta}{360}\) x 2\(\pi\)R cos60o
= \(\frac{300}{360}\) x 4 x 104 x \(\frac{1}{2}\)
= 1.667 x 104km(1667 km)
\(\frac{2}{3}\)
\(\frac{2}{15}\)
\(\frac{1}{2}\)
\(\frac{1}{3}\)
Correct answer is D
P(R1) = \(\frac{6}{10}\)
= \(\frac{2}{3}\)
P(R1 n R11) = P(both red)
\(\frac{3}{5}\) x \(\frac{5}{9}\)
= \(\frac{1}{3}\)
7.0
6.0
5.2
5.0
Correct answer is B
Area of trapezium = 14cm2
Area of trapezium = \(\frac{1}{2}\)(a + b)h
14 = \(\frac{1}{2}\)(4 + 3)h
14 = \(\frac{7}{2}\)h
h = \(\frac{14 \times 2}{7}\)
= 4
Area of SQR = \(\frac{1}{2}\)(3 x 4)
= \(\frac{12}{2}\)
= 6.0
5kg
16kg
19kg
6kg
Correct answer is A
\(\frac{1\sqrt{3}}{(\frac{1}{2})^2}\)
= \(\frac{4}{\sqrt{3}}\)
= \(\frac{\sqrt{3}}{\sqrt{3}}\)
= \(\frac{4\sqrt{3}}{\sqrt{3}}\)
m = 64kg, V = \(\frac{4\pi r^3}{3}\)
= \(\frac{4\pi(4)^3}{3}\)
= \(\frac{256\pi}{3}\) x 10-6m3
density(P) = \(\frac{\text{Mass}}{\text{Volume}}\)
= \(\frac{64}{\frac{256\pi}{3 \times 10^{-6}}}\)
= \(\frac{64 \times 3 \times 10^{-6}}{256}\)
= \(\frac{3}{4 \times 10^{-6}}\)
m = PV = \(\frac{3}{4 \pi \times 10^{-6}}\) x \(\frac{4}{3}\) \(\pi\)[32 - 22] x 10-6
\(\frac{3}{4 \times 10^{-6}}\) x \(\frac{4}{3}\) x 5 x 10-6
= 5kg