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.
\(\frac{7}{10}\)
\(\frac{3}{5}\)
\(\frac{4}{5}\)
\(\frac{3}{10}\)
Correct answer is B
Simple Space: (1, 2, 3, 4, 5, 6, 7, 8, 9, 10 = 10)
Prime: (2, 3, 5, 7)
multiples of 3: (3, 6, 9)
Prime or multiples of 3: (2, 3, 5, 6, 7, 9 = 6)
Probability = \(\frac{6}{10}\)
= \(\frac{3}{5}\)
Without using table, calculate the value of 1 + sec2 30o
2\(\frac{1}{3}\)
\(\frac{2}{15}\)
\(\frac{5}{3}\)
3\(\frac{1}{2}\)
Correct answer is A
1 + sec2 30o = sec 30o
= \(\frac{2}{\sqrt{3}}\)
\(\frac{(2)^2}{3}\)
= \(\frac{4}{3}\)
1 + sec2 30o = sec 30o
= 1 + \(\frac{4}{3}\)
= 2\(\frac{1}{3}\)
S 18o W
S 72o W
S 72o E
S 27o E
S 27o W
Correct answer is B
B = Bird ; H = Hunter.
Bearing of the hunter from the bird = S 72° W.
234.00 cm3
526.50 cm3
166.00 cm3
687cm3
Correct answer is A
Let x represent total vol. 2 : 3 = 2 + 3 = 5
\(\frac{3}{5}\)x = 351
x = \(\frac{351 \times 5}{3}\)
= 585
Volume of smaller block = \(\frac{2}{5}\) x 585
= 234.00cm\(^3\)
If two fair coins are tossed, what is the probability of getting at least one head?
\(\frac{1}{4}\)
\(\frac{1}{2}\)
1
\(\frac{43}{78}\)
\(\frac{3}{4}\)
Correct answer is E
Prob. of getting at least one head
Prob. of getting one head + prob. of getting 2 heads
= \(\frac{1}{4}\) + \(\frac{2}{4}\)
= \(\frac{3}{4}\)