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 figure, if XZ is 10cm, calculate RY in cm
10
10(1 - \(\frac{1}{\sqrt{3}}\))
10(1 - \(\frac{1}{\sqrt{2}}\))
10[1 - \(\sqrt{3}\)]
Correct answer is B
RY = YZ - RZ
= 10 tan 45 - 10 tan 30
= 10(1 - \(\frac{1}{\sqrt{3}}\))
The diagram is a circle with centre O. Find the area of the shaded portion
9(\(\pi - 2) cm^2\)
18\(\pi cm^2\)
9\(\pi cm^2\)
36\(\pi cm^2\)
Correct answer is A
Area of the quadrant = \(\frac{1}{4} \pi r^2 = \frac{1}{4} \pi (6)^2\)
= \(\frac{36 \pi}{4} = 9 \pi \)
Area of the triangle = \(\frac{1}{2} \times 6 = 18 \times \sin 90^o = 18\)
Area of shaded portion =(9 \(\pi \) - 18)cm^2\) =
9(\(\pi - 2)cm^2\).
QS = PS\(\sqrt{SR}\)
QS = \(\sqrt{(PS SR)}\)
QS = 2\(\sqrt{(PS SR)}\)
QS = \(\frac{1}{\sqrt{2}}\)\(\sqrt{(PS SR)}\)
Correct answer is B
No explanation has been provided for this answer.
In the figure, TSP = 100° and PRQ = 20°. Find PQR.
140o
120o
75o
30o
Correct answer is A
PSR = 80°
\(\therefore\) SRP = 20°
\(\implies\) RPQ = 20°
x = 140°
The sketch is the curve of y = ax2 + bx + c. Find a, b and c respectively
1, 0, -4
-2, 2, -4
0, 1, -4
2, -2, -4
2, -2, -4
Correct answer is A
Given the graph and the curve y = ax2 + bx + c the roots are x - 2 and 2 while its equation (x + 2)(x - 2) = y
y = x2 - 4 i.e. y = x2 + 0x - 4
a = 1, b = 0 and c = -4