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.
6cm
7cm
8cm
9cm
Correct answer is C
Since area of square PQRS = 100cm2
each lenght = 10cm
Also TUYS : XQVU = 1 : 16
lengths are in ratio 1 : 4, hence TU : UV = 1: 4
Let TU = x
UV = 1: 4
hence TV = x + 4x = 5x = 10cm
x = 2cm
TU = 2cm
UV = 8cm
But TU = SY and UV = YR
YR = 8cm
7 \(\sqrt{3}\)cm
12.9cm
\(\sqrt{87}\)cm
7cm
Correct answer is B
SQ2 + OS2 = OQ2 + 52 = 142
SQ2 = 142 - 52
196 - 25 = 171
ST2 + TQ2 = SQ2
ST2 + 22 = 171
ST2 = 171 - 4
= 167
ST = \(\sqrt{167}\)
= 12.92 = 12.9cm
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