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.
A cylinder of height h and radius r is open at one end. Its surface area is
2\(\pi\)rh
\(\pi\)r2h
2\(\pi\)rh + \(\pi\)r2
2\(\pi\)rh + 2\(\pi\)r2
Correct answer is C
A cylinder of height h ans radius r is open at one end, its surface area is 2\(\pi\)rh + \(\pi\)r2
Simplify \(\frac{1 - x^2}{x - x^2}\), where x \(\neq\) 0
\(\frac{1}{x}\)
\(\frac{1 - x}{x}\)
\(\frac{1 + x}{x}\)
\(\frac{1}{x - 1}\)
\(\frac{-x - 1}{1}\)
Correct answer is C
\(\frac{1 - x^2}{x - x^2}\), where x = \(\neq\) 0
\(\frac{1^2 - x^2}{x - x^2}\)
= \(\frac{(1 + x)(1 - x)}{x(1 - x)}\)
= \(\frac{1 + x}{x}\)
For the set of numbers 2, 3, 5, 6, 7, 7, 8
The median is greater than the mode
The mean is greater than the mode
The mean is greater than the median
The median is equal to the mean
The mean is less than the median
Correct answer is E
Mean = 5, 42
Median = 6
Mode = 7
The mean is less than the median
24
18
12
9
8
Correct answer is B
12 men in 9 days,
1 day in \(12 \times 9\) men
In 6 days we have \(\frac{12 \times 9}{6}\) men
= \(\frac{108}{6}\)
= 18 men
Find the value of log\(_{10}\)\(\frac{1}{40}\), given that log10\(_4\) = 0.6021
1.3979
2.3979
-1.6021
2.6021
Correct answer is C
log\(_{10}\)\(\frac{1}{40}\) = log\(_{10}\)1 - (log\(_{10}\)4 + log\(_{10}\)10)
= 0 - 1.6021
= -1.6021