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.
\(\frac{1}{2}\)
\(\frac{2}{3}\)
\(\frac{3}{8}\)
\(\frac{3}{11}\)
Correct answer is C
40 = 20 - x + x + 35 - x
40 = 55 - x
x = 55 - 40
= 15
∴ P(both) \(\frac{15}{40}\)
= \(\frac{3}{8}\)
\(\frac{3}{2}\)
\(\frac{9}{4}\)
\(\frac{5}{2}\)
3
Correct answer is B
\(\begin{array}{c|c} x & f & fx & \bar{x} - x & (\bar{x} - x)^2 & f(\bar{x} - x)^2 \\ \hline 1 & 2 & 2 & -2 & 4 & 8\\ 2 & 1 & 2 & -1 & 1 & 1\\ 3 & 2 & 6 & 0 & 0 & 0\\ 4 & 1 & 4 & 1 & 1 & 1\\ 2 & 2 & 10 & 2 & 4 & 8\\ \hline & 8 & 24 & & & 18 \end{array}\)
x = \(\frac{\sum fx}{\sum f}\)
= \(\frac{24}{8}\)
= 3
Variance (62) = \(\frac{\sum f(\bar{x} - x)^2}{\sum f}\)
= \(\frac{18}{8}\)
= \(\frac{9}{4}\)
10\(\frac{1}{2}\)
11\(\frac{1}{2}\)
12
13
Correct answer is C
Median = L + [\(\frac{\frac{N}{2} - f}{fm}\)]h
N = Sum of frequencies
L = lower class boundary of median class
f = sum of all frequencies below L
fm = frequency of modal class and
h = class width of median class
Median = 11 + [\(\frac{\frac{50}{2} - 21}{20}\)]5
= 11 + (\(\frac{25 - 21}{20}\))5
= 11 + (\(\frac{(4)}{20}\))
11 + 1 = 12
Find the mean deviation of the set of numbers 4, 5, 9
zero
2
5
6
Correct answer is B
x = \(\frac{\sum x}{N}\)
= \(\frac{18}{3}\)
= 6
\(\begin{array}{c|c} x & x - x & x - x \\ \hline 4 & -2 & 2\\ 5 & 1 & 1\\ 9 & 3 & 3\\ \hline & & 6\end{array}\)
M.D = \(\frac{|x - x|}{N}\)
= \(\frac{6}{3}\)
= 2
1
2
3
4
Correct answer is B
| x | 1 | 2 | 3 | 4 | 5 | Total |
| f | y + 2 | y - 1 | 2y - 3 | y + 4 | 3y - 4 | 8y - 2 |
| fx | y + 2 | 2y - 2 | 6y - 9 | 4y + 16 | 15y - 20 | 28y - 13 |
Mean = \(\frac{\sum fx}{\sum f}\)
\(\therefore \frac{28y - 13}{8y - 2} = \frac{43}{14}\)
\(\implies 14(28y - 13) = 43(8y - 2)\)
\(392y - 182 = 344y - 86\)
\(392y - 344y = -86 + 182 \implies 48y = 96\)
\(y = 2\)