WAEC Mathematics Past Questions & Answers - Page 229

1,141.

The ages of three men are in the ratio 3:4:5. If the difference between the ages of the oldest and youngest is 18 years, find the sum of the ages of the three men

A.

45 years

B.

72 years

C.

108 years

D.

216 years

Correct answer is C

Given that the ages are in the ratio 3: 4: 5.

Let the sum of their ages be t.

\(\therefore\) The youngest age = \(\frac{3}{12} t\)

The eldest age = \(\frac{5}{12} t\)

\(\implies \frac{5}{12} t - \frac{3}{12} t = 18\)

\(\frac{2t}{12} = 18 \implies t = \frac{18 \times 12}{2}\)

t = 108 years.

The sum of their ages = 108 years.

1,142.

If \(104_x = 68\), find the value of x

A.

5

B.

7

C.

8

D.

9

Correct answer is C

\(104_x = 68\\
1 \times x^2 + 0 \times x + 4 \times x^0 = 68\\
x^2 = 68 - 4; x^2 = 64\\
x = \sqrt{64}=8\)

1,143.

Which of the following numbers is perfect cube?

A.

350

B.

504

C.

950

D.

1728

Correct answer is D

No explanation has been provided for this answer.

1,144.

Simplify \(\left(\frac{16}{81}\right)^{-\frac{3}{4}}\times \sqrt{\frac{100}{81}}\)

A.

\(\frac{80}{243}\)

B.

\(\frac{1}{64}\)

C.

\(\frac{25}{6}\)

D.

\(\frac{15}{4}\)

Correct answer is D

\(\left(\frac{16}{81}\right)^{-\frac{3}{4}}\times \sqrt{\frac{100}{81}}\\
\frac{1}{\left(\sqrt[4]{\frac{16}{81}}\right)^3}\times \frac{10}{9}=\frac{1}{\left(\frac{2}{3}\right)^3}\times\frac{10}{9}\\
=\frac{27}{8}\times \frac{10}{9}=\frac{15}{4}\)

1,145.

What fraction must be subtracted from the sum of \(2\frac{1}{6}\) and \(2\frac{7}{12}\) to give \(3\frac{1}{4}\)?

A.

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

B.

\(\frac{1}{2}\)

C.

\(1\frac{1}{6}\)

D.

\(1\frac{1}{2}\)

Correct answer is D

\(2\frac{1}{6} + 2\frac{7}{12}\)

= \(\frac{13}{6} + \frac{31}{12}\)

= \(\frac{26 + 31}{12}\)

= \(\frac{57}{12} = \frac{19}{4}\)

\(\frac{19}{4} - 3\frac{1}{4}\)

= \(\frac{19}{4} - \frac{13}{4}\)

= \(\frac{6}{4}\)

= \(1\frac{1}{2}\)