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 \(\log_{10}\)(6x - 4) - \(\log_{10}\)2 = 1, solve for x.
2
3
4
5
Correct answer is C
\(\log_{10}\)(6x - 4) - \(\log_{10}\)2 = 1
\(\log_{10}\)(6x - 4) - \(\log_{10}\)2 = \(\log_{10}\)10
\(\log_{10}\)\(\frac{6x - 4}{2}\) - \(\log_{10}\)10
\(\frac{6x - 4}{2}\) = 10
6x - 4 = 2 x 10
= 20
6x = 20 + 4
6x = 20
x = \(\frac{24}{6}\)
x = 4
40.0%
42.2%
50.0%
52.5%
Correct answer is D
Population of school = 250 + 150 = 400
60% of 250 = \(\frac{\text{60%}}{\text{100%}}\) x 250 = 150
40% of 150 = \(\frac{\text{40%}}{\text{100%}}\) x 150 = 60
Total number of students who plays football;
150 + 60 = 210
Percentage of school that play football;
\(\frac{210}{400}\) x 100% = 52.5%
Simplify; 2\(\frac{1}{4} \times 3\frac{1}{2} \div 4 \frac{3}{8}\)
\(\frac{5}{9}\)
1\(\frac{1}{5}\)
1\(\frac{1}{4}\)
1\(\frac{4}{5}\)
Correct answer is D
2\(\frac{1}{4} \times 3\frac{1}{2} \div 4 \frac{3}{8}\)
= \(\frac{9}{4} \times \frac{7}{2} \div \frac{35}{8}\)
= \(\frac{9}{4} \times \frac{7}{2} \div \frac{8}{35}\)
= \(\frac{9}{5}\)
= 1 \(\frac{4}{5}\)
Find the value of x for which \(32_{four} = 22_x\)
three
five
six
seven
Correct answer is C
\(32_4 = 22_x\)
\(3 \times 4^1 + 2 \times 4^o\) = \(2 \times x^1 + 2 \times x^o\)
12 + 2 x 1 = 2x + 2 x 1
14 = 2x + 2
14 - 2 = 2x
12 = 2x
x = \(\frac{12}{2}\)
x = 6
\(yx^2 = 300\)
\(yx^2 = 900\)
y = \(\frac{100x}{9}\)
\(y = 900x^2\)
Correct answer is B
Y \(\alpha \frac{1}{x^2} \rightarrow y = \frac{k}{x^2}\)
If x = 3 and y = 100,
then, \(\frac{100}{1} = \frac{k}{3^2}\)
\(\frac{100}{1} = \frac{k}{9}\)
k = 100 x 9 = 900
Substitute 900 for k in
y = \(\frac{k}{x^2}\); y = \(\frac{900}{x^2}\)
= \(yx^2 = 900\)