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.
12
27
9
4
36
Correct answer is C
1st term a = 3, 5th term = 9, sum of n = 81
nth term = a + (n - 1)d, 5th term a + (5 - 1)d = 9
3 + 4d = 9
4d = 9 - 3
d = \(\frac{6}{4}\)
= \(\frac{3}{2}\)
= 6
Sn = \(\frac{n}{2}\)(6 + \(\frac{3}{4}\)n - \(\frac{3}{2}\))
81 = \(\frac{12n + 3n^2}{4}\) - 3n
= \(\frac{3n^2 + 9n}{4}\)
3n2 + 9n = 324
3n2 + 9n - 324 = 0
By almighty formula positive no. n = 9
= 3
Show that \(\frac{\sin 2x}{1 + \cos x}\) + \(\frac{sin2 x}{1 - cos x}\) is
sin x
cos2x
2
3
Correct answer is C
\(\frac{\sin^{2} x}{1 + \cos x} + \frac{\sin^{2} x}{1 - \cos x}\)
\(\frac{\sin^{2} x (1 - \cos x) + \sin^{2} x (1 + \cos x)}{1 - \cos^{2} x}\)
= \(\frac{\sin^{2} x - \cos x \sin^{2} x + \sin^{2} x + \sin^{2} x \cos x}{\sin^{2} x}\)
(Note: \(\sin^{2} x + \cos^{2} x = 1\)).
= \(\frac{2 \sin^{2} x}{\sin^{2} x}\)
= 2.
\(\frac{2P}{\pi}\)
\(\frac{P}{\sqrt{\pi}}\)
\(\frac{P}{\sqrt{2\pi}}\)
2\(\frac{P}{\pi}\)
Correct answer is D
Surface area of a sphere = 4\(\pi\)r2
\(\frac{1}{16}\) of 4\(\pi\)r2
= \(\frac{\pi r^2}{4}\)
P = \(\frac{\pi r^2}{4}\)
r = 2\(\frac{P}{\pi}\)
25cm
18cm
15cm
24cm
27cm
Correct answer is C
Area = L x B
L - B = 6
L = 6 + B
area = 135
B(6 + B) = B2 + 6B - 135 = 0
B = 9
L = 6 + 9
= 15
1.0622
10.6225
0.1328
0.3218
2.0132
Correct answer is C
\(4t^{2} + 7t - 1 = 0\)
Using a = 4, b = 7, c = -1.
\(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\)
= \(\frac{-7 \pm \sqrt{7^{2} - 4(4)(-1)}}{2(4)}\)
= \(\frac{-7 \pm \sqrt{49 + 16}}{8}\)
= \(\frac{-7 \pm \sqrt{65}}{8}\)
= \(\frac{-7 \pm 8.0623}{8}\)
The positive answer = \(\frac{-7 + 8.0623}{8} = \frac{1.0623}{8}\)
\(\approxeq 0.1328\) (4 decimal place)