WAEC Mathematics Past Questions & Answers - Page 65

321.

If \(\log_{10}\)(6x - 4) - \(\log_{10}\)2 = 1, solve for x.

A.

2

B.

3

C.

4

D.

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

322.

There are 250 boys and 150 girls in a school, if 60% of the boys and 40% of the girls play football, what percentage of the school play football?

A.

40.0%

B.

42.2%

C.

50.0%

D.

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%

323.

Simplify; 2\(\frac{1}{4} \times 3\frac{1}{2} \div  4 \frac{3}{8}\)

A.

\(\frac{5}{9}\)

B.

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

C.

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

D.

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 \(\frac{4}{5}\)

324.

Find the value of x for which \(32_{four} = 22_x\)

A.

three

B.

five

C.

six

D.

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

325.

Evaluate: \((64^{\frac{1}{2}} + 125^{\frac{1}{3}})^2\)

A.

121

B.

144

C.

169

D.

196

Correct answer is C

\([64^{\frac{1}{2}} + 125^{\frac{1}{3}}]^2\) = \([\sqrt{64} + \sqrt[3] {125}]^2\) 

\([8 + 5]^2\) = \([13]^2\)

= 169