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.
Solve the simultaneous equations for x in x2 + y - 8 = 0, y + 5x - 2 = 0
-28, 7
6, -28
6, -1
-1, 7
3, 2
Correct answer is C
x2 + y - 8 = 0, y + 5x - 2 = 0
Rearranging, x2 + y = 8.....(i)
5x + y = 2.......(ii)
Subtract eqn(ii) from eqn(i)
x2 - 5x - 6 = 0
(x - 6)(x + 1) = 0
x = 6, -1
u = \(\frac{8}{v^3}\)
v = \(\frac{8}{u^2v^3}\)
u = 8v3
v = 8u2
Correct answer is A
W \(\alpha\) \(\frac{1}{v}\)u \(\alpha\) w3
w = \(\frac{k1}{v}\)
u = k2w3
u = k2(\(\frac{k1}{v}\))3
= \(\frac{k_2k_1^2}{v^3}\)
k = k2k1k2
u = \(\frac{k}{v^3}\)
k = uv3
= (1)(2)3
= 8
u = \(\frac{8}{v^3}\)
Simplify log10 a\(\frac{1}{3}\) + \(\frac{1}{4}\)log10 a - \(\frac{1}{12}\)log10a7
1
\(\frac{7}{6}\)log10 a
zero
10
Correct answer is C
log10 a\(\frac{1}{3}\) + \(\frac{1}{4}\)log10 a - \(\frac{1}{12}\)log10a7 = log10 a\(\frac{1}{3}\) + log10\(\frac{1}{4}\) - log10 a\(\frac{7}{12}\)
= log10 a\(\frac{7}{12}\) - log10 a\(\frac{7}{12}\)
= log10 1 = 0
Find x if (x\(_4\))\(^2\) = 100100\(_2\)
6
12
100
210
110
Correct answer is B
x\(_4\) = x \(\times\) 4\(^0\) = x in base 10.
100100\(_2\) = 1 x 2\(^5\) + 1 x 2\(^2\)
= 32 + 4
= 36 in base 10
\(\implies\) x\(^2\) = 36
x = 6 in base 10.
Convert 6 to a number in base 4.
| 4 | 6 |
| 4 | 1 r 2 |
| 0 r 1 |
= 12\(_4\)
47
50
48\(\frac{1}{2}\)
48
49
Correct answer is C
By re-arranging 21, 23, 29, 40, 41, 43| 47, 50| 50, 55, 56, 70, 70, 73
The median = \(\frac{47 + 50}{2}\)
\(\frac{97}{2}\)
= 48\(\frac{1}{2}\)