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 3p = 4q and 9p = 8q - 12, find the value of pq.
12
7
-7
-12
Correct answer is A
9p = 8q - 12
9p = 2(4q) - 12
9p = 2(3q) - 12
9p = 6p - 12
3p = -12
p = -4
\(\frac{3 \times -4}{4} = \frac{4q}{4}\)
q = 13
pq = -3 x -4
= 12
5%
10%
12%
13%
Correct answer is B
15 = (\(\frac{6,900 - C.P \times 100}{C.P}\))
15 C.P = 690000 - C.P 100
C.P = \(\frac{690000}{115}\)
C.P = N6,000
%profit = \(\frac{6,600 - 6,000}{6,000}\) x 100
= \(\frac{600}{6,000}\) x 100
= 10%
Simplify: \(\log_{10}\) 6 - 3 log\(_{10}\) 3 + \(\frac{2}{3} \log_{10} 27\)
3 \(\log_{10}^2\)
\(\log_{10}^2\)
\(\log_{10}^3\)
2 \(\log_{10}^3\)
Correct answer is B
log\(_{10}\) 6 - log\(_{10}\)3\(^3\) + log\(_{10}\) (\(\sqrt[3]{27}\))\(^2\)
= log \(_{10}\) 6 - log \(_{10}\) 27 + log\(_{10}\) 9
= log\(_{10}\) \(\frac{6 \times 9}{27}\)
= log\(_{10}\)2
Solve 4x^{2}\) - 16x + 15 = 0.
x = 1\(\frac{1}{2}\) or x = -2\(\frac{1}{2}\)
x = 1\(\frac{1}{2}\) or x = 2\(\frac{1}{2}\)
x = 1\(\frac{1}{2}\) or x = -1\(\frac{1}{2}\)
x = -1\(\frac{1}{2}\) or x -2\(\frac{1}{2}\)
Correct answer is B
4x\(^2\) - 16x + 15 = 0
(2x - 3)(2x - 5) = 0
x = 1\(\frac{1}{2}\) or x = 2\(\frac{1}{2}\)
H = \(\frac{p}{4y^2}\)
H = \(\frac{2p}{y^2}\)
H = \(\frac{p}{2y^2}\)
H = \(\frac{p}{y^2}\)
Correct answer is C
H \(\propto\) \(\frac{p}{y^2}\)
H = \(\frac{pk}{y^2}\)
1 = \(\frac{8k}{2^2}\)
k = \(\frac{4}{8}\)
= \(\frac{1}{2}\)
H = \(\frac{p}{2y^2}\)