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.
7.3, 5.9
7.7, 12.5
12.5, 7.7
5.9, 7.3
Correct answer is C
\(\frac{PQ}{PN} = \frac{PM}{PR} = \frac{QM}{NR}\)
\(\frac{4.8}{12} = \frac{5}{PR}\)
PR = \(\frac{5 \times 12}{4.8} = \frac{50}{4}\)
= 12.5
\(\frac{PQ}{PN} = \frac{PM}{PT} = \frac{TM}{NT}\)
\(\frac{PT}{12} = \frac{5}{PR}\)
PT2 = 60
PT = \(\sqrt{60}\)
= 7.746
= 7.7
In the figure above, Find the value of x
130o
110o
100o
90o
Correct answer is A
No explanation has been provided for this answer.
Find the curved surface area of the frustrum in the figure
16\(\pi \sqrt{10}\)cm2
20\(\pi \sqrt{10}\)cm2
24\(\pi \sqrt{10}\)cm2
36\(\pi \sqrt{10}\)cm2
Correct answer is B
\(\frac{x}{4} = \frac{6 + x}{6}\)
6x = 4(6 + x) = 24 + 4x
x = 12cm
CSA = \(\pi RL - \pi rl\)
= \(\pi (6) \sqrt{{18^2} + 6^2} - \pi \times 4 \times \sqrt{{12^2} + 4^2}\)
= 6\(\pi \sqrt{360} - 4 \pi \sqrt{160}\)
= 36\(\pi \sqrt{10} - 16 \pi \sqrt{10}\)
= 20\(\pi \sqrt{10}\)cm2
In the figure, PQR is a semicircle. Calculate the area of the shaded region
125\(\frac{2}{7}\)2
149\(\frac{2}{7}\)cm2
234\(\frac{1}{7}\)cm2
267\(\frac{1}{2}\)cm2
Correct answer is A
No explanation has been provided for this answer.
In the figure, YXZ = 30o, XYZ = 105o and XY = 8cm. Calculate YZ
16\(\sqrt{2}\)cm
8\(\sqrt{2}\)cm
4\(\sqrt{2}\)cm
22cm
Correct answer is C
yzx + 105o + 30o = 180o
yzx = 180 - 155 = 45o
\(\frac{yz}{sin 30^o} = \frac{8}{sin 45^o}\)
yz = \(\frac{8 \sin 30}{sin 45}\)
= 8(\(\frac{1}{2}) = \frac{8}{1} \times \frac{1}{2} \times \frac{\sqrt{2}}{1}\)
= 4 \(\div\) \(\frac{1}{\sqrt{2}}\)
= 4\(\sqrt{2}\)cm