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.
\(\frac{\sqrt{3}}{2}\)
\(\frac{3}{2}\)
3
\(\sqrt{3}\)cm
Correct answer is D
Length of chord = \(2r \times \sin(\frac{\theta}{2})\)
= \( 2 \times \sqrt{3} \times \sin(\frac{60}{2})\)
= \(2 \times \sqrt{3} \times \frac{1}{2}\)
= \(\sqrt{3}\) cm.
In the diagram, QTR is a straight line and < PQT = 30o. find the sin of < PTR
\(\frac{8}{15}\)
\(\frac{2}{3}\)
\(\frac{3}{4}\)
\(\frac{15}{16}\)
Correct answer is C
\(\frac{10}{\sin 30^o} = \frac{15}{\sin x} = \frac{10}{0.5} = \frac{15}{\sin x}\)
\(\frac{15}{20} = \sin x\)
sin x = \(\frac{15}{20} = \frac{3}{4}\)
N.B x = < PRQ
TQ is tangent to circle XYTR, < YXT = 32o, RTQ = 40o. find < YTR
108o
121o
140o
148o
Correct answer is A
< TWR = < QTR = 40o (alternate segment)
< TWR = < TXR = 40o(Angles in the same segments)
< YXR = 40o + 32o = 72o
< YXR + < YTR = 180o(Supplementary)
72o + < YTR = 180o
< YTR = 180o - 72o
= 108o
35cm2
65cm2
70cm2
140cm2
Correct answer is C
A\(\bigtriangleup\) = \(\frac{1}{2}\) x 8 x h = 20
= \(\frac{1}{2}\) x 8 x h = 4h
h = \(\frac{20}{4}\)
= 5cm
A\(\bigtriangleup\)(PQTS) = L x H
A\(\bigtriangleup\)PQRT = A\(\bigtriangleup\)QSR + A\(\bigtriangleup\)PQTS
20 + 50 = 70cm\(^2\)
ALTERNATIVE METHOD
A\(\bigtriangleup\)PQRT = \(\frac{1}{2}\) x 5 x 28
= 70cm\(^2\)
In diagram, PQ || ST and < PQR = 120o, < RST = 130o, find the angle marked x
50
65
70
80
Correct answer is C
x + 60o + 50o = 180o
x = 110o = 180o
x = 180o - 110o
= 70o