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, QPS = SPR, PR = 9cm. PQ = 4cm and QS = 3cm, find SR.
6\(\frac{3}{4}\)cm
3\(\frac{3}{8}\)cm
4\(\frac{3}{8}\)cm
2\(\frac{3}{8}\)cm
Correct answer is A
Using angle bisector theorem: line PS bisects angle QPR
QS/QP = SR/PR
3/4 = SR/9
4SR = 27
SR = \(\frac{27}{4}\)
= 6\(\frac{3}{4}\)cm
In the figure, PT is tangent to the circle at U and QU/RS if TUR = 35º and SRU = 50º find x
95o
85o
50o
35o
Correct answer is A
Since QRU= Xo
RSU = Xo, But RSU = 180o - (50o + 35o)
= 180o - 85o
= 95o
x = 95o
In the diagram, QP//ST:PQR = 34o qrs = 73o and Rs = RT. Find SRT
68o
102o
107o
141o
Correct answer is B
Construction joins R to P such that SRP = straight line
R = 180o - 107o
< p = 180o - (107o - 34o)
108 - 141o = 39o
Angle < S = 39o (corr. Ang.) But in \(\bigtriangleup\)SRT
< S = < T = 39o
SRT = 180 - (39o + 39o)
= 180o - 78o
= 102o
84o
48o
45o
42o
Correct answer is B
OQ = OR = radii
< ROQ = 180 - 86 = 84o
\(\bigtriangleup\)OQR = Isosceles
R = Q
R + Q + 84 = 180(angle in a \(\bigtriangleup\))
2R = 96 since R = Q
R = 48o
ORQ = 48o
In the diagram, PQRS is a circle with O as centre and PQ/RT. If RTS = 32°. Find PSQ
32o
45o
58o
90o
Correct answer is C
< PSO = \(\frac{1}{2}\) < SOQ = \(\frac{1}{2}\)(180) = 90°
< RTS = < PQS = 32° (Alternative angle)
< PSQ = 90 - < PSQ = 90° - 32°
= 58°