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 diagram above, TR = TS and ∠TRS = 60o. Find the value of x
30
45
60
120
150
Correct answer is C
∠TRS = ∠RST = 60o
∠PSR + ∠PQR = 180o, 120o + xo = 180o
x = 60o
In the diagram above, O is the center of the circle and ∠POR = 126°. Find ∠PQR
234o
117o
72o
63o
54o
Correct answer is B
Reflex < POR = 360° - 126° = 234°
\(\therefore\) < PQR = \(\frac{1}{2} \times 234°\)
= 117°
48o
45o
37o
32o
30o
Correct answer is C
∠RQT = 180° - (95° + 48°) = 73°
∠OQR = 90° - 37° = 53°
∠QOR = 180° - (53° + 53°) = 74°
QSR = 74°/2 = 37°
5/12cm2
7/12cm2
11/6cm2
2 1/3cm2
6 1/6cm2
Correct answer is B
Area of \(\Delta OXY\) = \(12\frac{1}{4} cm^2\)
Area of sector OXY = \(\frac{30}{360} \times \frac{22}{7} \times 7 \times 7\)
= \(\frac{77}{6} = 12\frac{5}{6} cm^2\)
Area of the shaded portion = \(12\frac{5}{6} - 12\frac{1}{4}\)
= \(\frac{7}{12} cm^2\)
The area of a parallelogram is 513cm\(^2\) and the height is 19cm. Calculate the base.
13.5cm
25cm
27cm
54cm
108cm
Correct answer is C
Area of parallelogram = base \(\times\) height.
\(513 = base \times 19 \implies base = \frac{513}{19}\)
= 27 cm