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.
i and iv
ii
iii
iv
Correct answer is C
m + y + x = 180o(sum of < s on straight line)
y = n(vertically opp. angle)
m + n + x = 180o
In the diagram, MQ//RS, < TUV = 70o and < RLV = 30o. Find the value of x
150o
110o
100o
95o
Correct answer is C
L + 30o - 180o(Sum of < s on straight line)
L = 180o - 30o = 150o
L = q = 150o(opposite < s are equal)
y = b = 30o(alt. < s)
b + c = 180o(sum of < s on str. line)
30o + c 180
c = 180 - 30
c = 150o
b = a = 30o (opp < s are equal)
c = d = 150o (opp < s are equal)
a + k + 70o = 180o (sum of < s on \(\bigtriangleup\))
30o + k + 70o = 180
k + 100o = 180
k = 180 - 100
k = 80o
x + 80o = 180(sum of < s on straight line)
x = 180o - 80o
x = 100o
The diagram is a circle centre O. If < SPR = 2m and < SQR = n, express m in terms of n
m = \(\frac{n}{2}\)
m = 2n
m = n - 2
m = n + 2
Correct answer is A
If < SPR = 2m then < SQR = 2m but < AQR was n
n = 2m
m = \(\frac{n}{2}\)
In the diagram, O is a circle centre of the circle PQRS and < PSR = 86o. If < PQR = xo, find x
108o
172o
130o
50o
Correct answer is B
No explanation has been provided for this answer.
In the diagram, |QR| = 10m, |SR| = 8m
< QPS = 30o, < QRP = 90o and |PS| = x, Find x
1.32m
6.32m
9.32m
17.32
Correct answer is C
In right angled \(\bigtriangleup\)QPR
tan 30o = \(\frac{10}{x + 8}\)
(x + 8) tan 30 = 10
x + 8 = \(\frac{10}{0.5773}\)
x +8 = 17.3
x = 17.3 - 8
x = 9.32