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.
40
45
55
60
Correct answer is B
\(\frac{x + 50 + 70}{3} = 55\)
x + 120 = 3(55)
x + 120 = 165
= 165 - 120
x = 45%
\(\frac{2 + \sqrt{3}}{3}\)
\(\frac{1 - \sqrt{3}}{3}\)
\(\frac{1 + \sqrt{3}}{3}\)
\(\frac{2 - \sqrt{3}}{3}\)
Correct answer is D
Sin 60 = \(\frac{\sqrt{3}}{2}\); cos 60o = \(\frac{1}{2}\)
= \(\frac{1 - \sin 60}{1 + \cos 60} = \frac{1 - \frac{\sqrt{3}}{2}}{1 + \frac{1}{2}}\)
= \(\frac{2 - \sqrt{3}}{3}{2}\div \frac{2 + 1}{2}\)
= \(\frac{2 - \sqrt{3}}{2} \times \frac{2}{3}\)
= \(\frac{2 - \sqrt{3}}{3}\)
If M and N are two fixed points in a plane. Find the locus L = [P : PM = PN]
a line equal to MN
line parallel to MN
Perpendicular bisector of MN
A circle centre P, radius MN
Correct answer is C
Locus L = (P : PM = Pn}
For M and N being two fixed points
Since PM = PN, P is equidistant from Mand N, So L must be the perpendicular bisector of the line MN
80o
70o
65o
40o
Correct answer is C
(2x + 5)o + (xo + 20o) + x + (3x - 20)o + (x + 15)o = (n - 2) x 180
8n + 20 = (5 - 2) x 180
where n = 5(Pentagon)
8n + 20 = 3 x 180
8n + 20 = 540
8n = 540 - 20
8n = 520
x = 65o
672cm
1056cm
2112cm
4224cm
Correct answer is C
d = 42cm
Circumference = \(\pi d = \frac{22}{7} \times 42cm\)
= 22 x 6cm = 132cm
Total distance covered = circumference x number of revolutions
= 132cm x 16 = 2112cm