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, determine the angle marked y
66o
110o
26o
70o
44o
Correct answer is A
From the diagram, < p = 44o, < Q = 70o
< y = 180o - (70 + 44o) = 66o
In the figure, QRS is a line, PSQ = 35o, SPR = 30o and O is the centre of the circle. Find OQP.
35o
30o
130o
25o
65o
Correct answer is D
< PSQ = 35o, < SPR = 30o
O is the centre of the circle, PRQ = 65o
< OQP = 90 - 65
= 25o
In the figure, PQRSTW is a regular hexagon. QS intersects RT at V. Calculate TVS
60o
90o
120o
30o
80o
Correct answer is A
From the diagram, PQRSTW is a regular hexagon.
Hexagon is a six sided polygon.
Sum of interior angles of polygon = (2n - 4)90o = [2 x 6 - 4] x 90 = 8 x 90 = 720o
each angle = \(\frac{720^o}{6} = 120^o\) and TVS = \(\frac{120}{2} = 60^o\)
Find the area of the shaded portion of the semicircular figure.
\(\frac{r^2}{4}(4 \pi - 3 \sqrt{3})\)
\(\frac{r^2}{4}(2 \pi - 3 \sqrt{3})\)
\(\frac{1}{2}r^2 \pi\)
\(\frac{1}{8}r^2 \sqrt{3}\)
\(\frac{r^2}{4}(4 \pi - 3 \sqrt{3})\)
Correct answer is B
Asector = \(\frac{60}{360} \times \pi r^2\)
= \(\frac{1}{6} \pi r^2\)
A\(\bigtriangleup\) = \(\frac{1}{2}r^2 \sin 60^o\)
\(\frac{1}{2} r^2 \times \frac{\sqrt{3}}{2} = \frac{r^2\sqrt{3}}{4}\)
A\(\text{shaded portion}\) = Asector -
A\(\bigtriangleup\)
= (\(\frac{1}{6} \pi r^2 - \frac{r^2\sqrt{3}}{4})^3\)
= \(\frac{\pi r^2}{2} - \frac{3r^2\sqrt{3}}{4}\)
= \(\frac{r^2}{4}(2 \pi - 3 \sqrt{3})\)
Using \(\bigtriangleup\)XYZ in the figure, find XYZ
29o
31o
31o 20'
31o 18'
Correct answer is C
\(\frac{\sin y}{3} = \frac{\sin 120^o}{5}\)
sin 120o = sin 60o
5 sin y = 3 sin 60o
sin y = \(\frac{3 \sin 60^o}{5}\)
\(\frac{3 \times 0.866}{5}\)
= \(\frac{2.598}{5}\)
y = sin-1 0.5196 = 30o 18'