Mathematics questions and answers

Mathematics Questions and Answers

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.

2,956.

A man is four times as old as his son. The difference between their ages is 36 years Find the sum of their ages

A.

45 years

B.

48 years

C.

60 years

D.

74 years

Correct answer is C

Let the sons age be x. The father is 4x ∴ 4x - x = 36; 3x = 36; x = 12 The son is 12 years and the father is 12 x 4 = 48. The sum of their ages (12 + 48) years = 60years

2,957.

Evaluate \(\frac{1}{2}+\frac{3}{4}of\frac{2}{5}\div 1\frac{3}{5}\)

A.

\(\frac{15}{16}\)

B.

\(\frac{11}{16}\)

C.

\(\frac{49}{50}\)

D.

\(3\frac{1}{5}\)

Correct answer is B

\(\frac{1}{2} + (\frac{3}{4} \text{ of } \frac{2}{5}) \div 1\frac{3}{5}\)

= \(\frac{1}{2} + (\frac{3}{4} \times \frac{2}{5}) \div \frac{8}{5}\)

= \(\frac{1}{2} + \frac{3}{10} \div \frac{8}{5}\)

= \(\frac{1}{2} + (\frac{3}{10} \times \frac{5}{8})\)

= \(\frac{1}{2} + \frac{3}{16}\)

= \(\frac{11}{16}\)

2,958.

The nth term of a sequence is \(2^{2n-1}\). Which term of the sequence is \(2^9?\)

A.

3rd

B.

4th

C.

5th

D.

6th

Correct answer is C

\(T_{n} = 2^{2n - 1}\)

\(2^{2n - 1} = 2^9\)

\(2n - 1 = 9 \implies 2n = 9 + 1\)

\(2n = 10 \implies n = 5\)

The 5th term = 2\(^9\)

2,959.

Evaluate \(5\frac{2}{5}\times \left(\frac{2}{3}\right)^2\div\left(1\frac{1}{2}\right)^{-1}\)

A.

\(\frac{8}{25}\)

B.

\(\frac{12}{25}\)

C.

\(3\frac{3}{5}\)

D.

\(4\frac{1}{8}\)

Correct answer is C

\(5\frac{2}{5}\times \left(\frac{2}{3}\right)^2 ÷  \left(1\frac{1}{2}\right)^{-1}\\
\frac{27}{5}\times \frac{4}{9} \times \frac{3}{2}=3\frac{3}{5}
\)

2,960.

A boy estimated his transport fare for a journey as N190 instead of N200. Find the percentage error in his estimate

A.

95%

B.

47.5%

C.

5.26%

D.

5%

Correct answer is D

The percentage error is \(\frac{error}{actual}\times \frac{100}{1}\%\\
=\frac{200-190}{200}\times \frac{100}{1}\\
=\frac{10}{200}\times \frac{100}{1}\% = 5\%\)