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.
12
9
8
7
Correct answer is D
Excellent = 360° - (120° + 90° + 80°)
= 70°
Number of excellent students = \(\frac{70}{360} \times 36\)
= 7 students
Find the area of the trapezium above.
91 cm2
78 cm2
60 cm2
19 cm2
Correct answer is C
Area of trapezium = \(\frac{1}{2} (a + b) \times h\)
= \(\frac{1}{2} (7 + 13) \times 6\)
= \(60 cm^{2}\)
In the diagram above, PQR is a circle centre O. If < QPR is x°, find < QRP.
x°
(90 – x)°
(90 + x)°
(180 – x)°
Correct answer is B
< PQR = 90° (angle in a semi-circle) < QRP = (90 - x)°
In diagram above, QR//TU, < PQR = 80° and < PSU = 95°. Calculate < SUT.
15o
25o
30o
80o
Correct answer is A
< PQR = < PTU = 80°
< TSU = 85°
x = 180° - (80° + 85°)
= 15°
The shaded region above is represented by
the equation
y ≤ 4x + 2
y ≥ 4x + 2
y ≤ -4x + 4
y ≤ 4x + 4
Correct answer is C
Equation of the line
\(\frac{y - 4}{x - 0} = \frac{0 - 4}{1 - 0}\)
\(\frac{y - 4}{x} = \frac{-4}{1}\)
\(\therefore -4x = y - 4\)
\(y = -4x + 4\)
\(\therefore \text{The shaded portion = } y \leq -4x + 4\)