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.
If D = \(\begin{bmatrix}2& -1&3\\4&1&2\\1&-3&1\\\end{bmatrix}\) Find |D|
16
14
-23
-37
Correct answer is C
\(\begin{bmatrix}2& -1&3\\4&1&2\\1&-3&1\\\end{bmatrix}\)
= 2[(1 x 1) - (2 x -3)] - (-1)[(4 x 1) - (2 x 1)] + 3[(4 x -3) - (1 x 1)]
= 2(1 - (-6)) + 1(4 - 2) + 3(-12 - 1)
= 2(1 + 6) + 1(2) + 3(-13)
= 2(7) + 1(2) + 3(-13)
= 14 + 2 - 39
∴ |D| = -23
60
15
66
4
Correct answer is C
If the majority are women implies that:
3 Women and 2 Men Or 4 Women and 1 Man
= \(^4C_3 \times^6C_2+^4C_4\times^6C_1\)
= \(4 \times 15 + 1 \times 6\)
= 60 + 6 = 66
The ages of students in a small primary school were recorded in the table below.
| Age | 5 - 6 | 7 - 8 | 9 -10 |
| Frequency | 29 | 40 | 38 |
Estimate the mean
7.7
7.5
7.8
7.6
Correct answer is A
| Class Interval | Class Mark | Frequency (f) | fx |
| 5 - 6 | 5.5 | 29 | 5.5 x 29 = 159.5 |
| 7 - 8 | 7.5 | 40 | 7.5 x 40 = 300 |
| 9 - 10 | 9.5 | 38 | 9.5 x 38 = 361 |
| \(\sum f = 107\) | \(\sum fx = 820.5\) |
Mean = \(\frac{\sum fx}{\sum f} = \frac {820.5}{107}\) = 7.7 (1 d.p)
Evaluate the following limit: \(lim_{x\to2} \frac {x^2 + 4x - 12}{x^2 - 2x}\)
4
8
0
2
Correct answer is A
\(lim_{x\to2} \frac {x^2 + 4x - 12}{x^2 - 2x}\) = \(lim_{x\to2} \frac {(x - 2)(x + 6)}{x(x - 2)}\)
\(lim_{x\to2} \frac {x + 6}{x}\)
\(\frac {2 + 6}{2} = \frac {8}{2}\) = 4
15 years
25 years
20 years
30 years
Correct answer is C
Amount (A) = Principal (P) + Interest (I) i.e. A = P + I
1 = \(\frac {PTR}{100}\)
A = 3P; T = 10 years (Given)
Since A = P + I
\(\implies 3P = P + \frac {P\times 10 \times R}{100}\)
\(\implies 3P = P + \frac {10PR}{100}\)
\(\implies 3P = P + \frac {PR}{10}\)
\(\implies 3P - P = \frac {PR}{10}\)
\(\implies 2P = \frac {PR}{10}\)
\(\implies \frac {2P}{1} = \frac {PR}{10}\)
\(\implies\) PR = 20P
\(\therefore\) R = 20%
Since the rate stays the same
A = 5P; R = 20%;T =?; A =P + I
\(\implies 5P = P + \frac {p \times T \times 20}{100}\)
\(\implies 5P - P = \frac {2PT}{10}\)
\(\implies 4P = \frac {2PT}{10}\)
\(\implies 2P = \frac {PT}{10}\)
\(\implies\) PT = 20P
\(\therefore\) T = 20 years