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.
10(sin 35o - sin 55o)
10 cos35o - cos 55o
10(tan35o - tan55o)
10(cot35o - cot55o)
Correct answer is D
x = 10 cot35o - 10 cot55o
= 10(cot35o - cot55o)
If x : y = 5 : 12 and z = 52cm, find the perimeter of the triangle.
68cm
84cm
100cm
120cm
Correct answer is D
13 = 52
1 = \(\frac{52}{13}\)
= 4
5 + 12 + 13 = 30
Total perimeter = 30 x 4
= 120cm
In XYZ, y = z = 30° and XZ = 3cm. Find YZ
\(\frac{\sqrt{3}}{2}\) cm
3\(\frac{\sqrt{3}}{2}\) cm
3\(\sqrt{3}\) cm
2\(\sqrt{3}\) cm
Correct answer is C
Y \(\to\) 30°
X \(\to\) 120°
Z \(\to\) 30°
\(\frac{3}{\text{sin 30}}\) = \(\frac{YZ}{\text{sin 120}}\)
YZ = \(\frac{3\times \sin 120}{\sin 30}\)
\(3 \times \frac{\sqrt{3}}{2} \times 2 = 3\sqrt{3}\)
PQRS is a rhombus. If PR\(^2\) + QS\(^2\) = kPQ\(^2\), determine k.
1
2
3
4
Correct answer is D
PR\(^2\) + QS\(^2\) = kPQ\(^2\)
SQ\(^2\) = SR\(^2\) + RQ\(^2\)
PR\(^2\) + SQ\(^2\) = PQ\(^2\) + SR\(^2\) + 2RQ\(^2\)
= 2PQ\(^2\) + 2RQ\(^2\)
= 4PQ\(^2\)
∴ K = 4
A regular polygon of (2k + 1) sides has 140° as the size of each interior angle. Find k
4
4\(\frac{1}{2}\)
8
8\(\frac{1}{2}\)
Correct answer is A
A regular has all sides and all angles equal. If each interior angle is 140° each exterior angle must be
180° - 140° = 40°
The number of sides must be \(\frac{360^o}{40^o}\) = 9 sides
hence 2k + 1 = 9
2k = 9 - 1
8 = 2k
k = \(\frac{8}{2}\)
= 4