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.
In the diagram, PQRS is a parallelogram. Find the value of < SQR
30o
50o
80o
100o
Correct answer is D
SQR + RQV + VQU = 18o angle on a straight line SP is parallel to QR and PV is parallel to TR
< STP = < RQV = 30o
But SQR + 30o + 50o = 180o
SQR = 180 - 80
= 100o
Find the inequality which represents the shaded portion in the diagram
2x - y - 2 \(\geq\) 0
2x - y - 2 \(\leq\) 0
2x - y - 2 < 0
2x - y - 2 > 0
Correct answer is A
2x - y - 2 \(\geq\) 0 = y \(\leq\) 2x - 2
when x = 0, y = -2, when y = 0, x = 1
y = (x - 3)3
y = (x + 3)3
y = x3 - 27
y = -x3 + 27
Correct answer is C
y = x3 - 27, y = -27 \(\to\) (0, -27)
when y = 0, x = 3 (3, 0)
The bar chart shows the distribution of marks in a class test. How many students took the test?
15
20
25
30
Correct answer is B
Number f students that took the test = \(\sum f\). Where f is the frequencies
= 2 + 5 + 0 + 3 + 4 + 0 + 0 + 2 + 3 + 1 + 2 = 20
From the figure, calculate TH in centimeters
\(\frac{5}{\sqrt{3} + 1}\)
\(\frac{5}{\sqrt{3} - 1}\)
\(\frac{5}{\sqrt{3}}\)
\(\frac{\sqrt{3}}{5}\)
Correct answer is B
TH = tan 45o, TH = QH
\(\frac{TH}{5 + QH}\) = tan 30o
TH = (b + QH) tan 30o
QH = 56 (5 + QH) \(\frac{1}{\sqrt{3}}\)
QH(1 - \(\frac{1}{\sqrt{3}}\)) = \(\frac{5}{\sqrt{3}}\)
QH = \(\frac{5\sqrt{3}}{\sqrt{3}} - \frac{1}{\sqrt{3}}\)
= \(\frac{5}{\sqrt{3} - 1}\)