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.
Using the diagram, find the bearing of X from Y
300o
240o
120o
60o
Correct answer is B
The bearing of X from Y
= 360 - 120 = 240
In the diagram, \SQ\ = 4cm, \PT\ = 7cm. /TR/ = 5cm and ST//OR. If /SP/ = xcm, find the value of x
5.6
6.5
6.6
6.8
Correct answer is A
\(\frac{/PS/}{/PQ/} = \frac{/PT/}{/PR/}\)
\(\frac{x}{4 + x} = \frac{7}{7 + 5}\)
\(\frac{x}{4 + x} = \frac{7}{12}\)
(2x = 7)(4 + x); 12x = 28 + 7x
12x - 7x = 28
5x = 28
x = \(\frac{28}{5}\)
x = 5.6
In the diagram, IG is parallel to JE, JEF = 120o and FHG = 130o, fins the angle marked t
40o
70o
80o
100o
Correct answer is B
Now, t + 60o + 50o = 180o
t = 180o - 110o
t = 70o
In the diagram, /TP/ = 12cm and it is 6cm from O, the centre of the circle, Calculate < TOP
120o
90o
60o
45o
Correct answer is B
Tan \(\theta = \frac{6}{6} = 1\)
\(\theta\) = tab - 1(1) = 45o
< top = 20
= 2 x 45o = 90o
27cm2
30cm2
36cm2
37cm2
Correct answer is A
Area = \(\frac{1}{2}(a + b)\)h
where a = QR = 6cm
b = PA = 10cm
Area = \(\frac{1}{2}(6 + 10)\) x 3.35cm2
= \(\frac{1}{2}(16)\) x 3.35cm2
= 26.8cm2
= 27cm2 (approx.)