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.
Make x the subject of the relation \(\frac{1 + ax}{1 - ax}\) = \(\frac{p}{q}\)
\(\frac{p + q}{a(p - q)}\)
\(\frac{p - q}{a(p + q)}\)
\(\frac{p - q}{apq}\)
\(\frac{pq}{a(p - q)}\)
Correct answer is B
\(\frac{1 + ax}{1 - ax}\) = \(\frac{p}{q}\) by cross multiplication,
q(1 + ax) = p(1 - ax)
q + qax = p - pax
qax + pax = p - q
∴ x = \(\frac{p - q}{a(p + q)}\)
If \(\sqrt{x^2 + 9}\) = x + 1, solve for x
5
4
3
2
1
Correct answer is B
\(\sqrt{x^2 + 9}\) = x + 1
x2 + 9 = (x + 1)2 + 1
0 = x2 + 2x + 1 - x2 - 9
= 2x - 8 = 0
2(x - 4) = 0
x = 4
If N225.00 yields N27.00 in x years simple interest at the rate of 4% per annum, find x
3
4
12
17
Correct answer is A
Principal = N255.00, Interest = 27.00
year = x Rate: 4%
∴ 1 = \(\frac{PRT}{100}\)
27 = \(\frac{225 \times 4 \times T}{100}\)
2700 = 900T
T = 3 years
Solve without using tables log5(62.5) - log5(\(\frac{1}{2}\))
3
4
5
8
Correct answer is A
log5(62.5) - log5(\(\frac{1}{2}\))
= log5\(\frac{(62.5)}{\frac{1}{2}}\) - log5(2 x 62.5)
= log5(125)
= log553 - 3log55
= 3
If 2log3 y + log3 x2 = 4, then y is
4 - log3
\(\frac{4}{log_3 x}\)
\(\frac{4}{x}\)
\(\pm\) \(\frac{9}{x}\)
Correct answer is D
2log3y + log3x2 = 4
log3y2 + log3x2 = 4
∴ log3 (x2y2) = log381(correct all to base 4)
x2y2 = 81
∴ xy = \(\pm\)9
∴ y = \(\pm\)\(\frac{9}{x}\)