# Mathematics 2018 Past Questions | WAEC

Study the following Mathematics past questions and answers for JAMB, WAEC NECO and Post-JAMB. Get prepared with official past questions and answers for upcoming examinations.

- Simplify: \(\sqrt{108} + \sqrt{125} – \sqrt{75}\)

**A.**\(\sqrt{3} + 5\sqrt{5}\)**B.**\(6 \sqrt{3} – 5 \sqrt{5}\)**C.**\(6 \sqrt{3} + \sqrt{2}\)**D.**\(6\sqrt{3} – \sqrt{2}\)

**Correct Option: Answer is A**

\(\sqrt{108} + \sqrt{125} – \sqrt{75}\)

= \(\sqrt{3 \times 36} + \sqrt{5 \times 25} – \sqrt{3 \times 25}\)

= \(6 \sqrt{3} + 5 \sqrt{5} – 5 \sqrt{3}\)

= \(\sqrt{3} + 5\sqrt{5}\)

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

**A.**121**B.**144**C.**169**D.**196

**Correct Option: Answer is C**

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

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

= 169

**3. **Given that y varies inversely as the square of x. If x = 3 when y = 100, find the equation connecting x and y.

**A.**\(yx^2 = 300\)**B.**\(yx^2 = 900\)**C.**y = \(\frac{100x}{9}\)**D.**\(y = 900x^2\)

**Correct Option: 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 = \(\frac{900}{x^2}\)

= \(yx^2 = 900\)

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

**A.**three**B.**five**C.**six**D.**seven

**Correct Option: 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

**5. **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 Option: 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 \(\frac{4}{5}\)

**6. **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 Option: Answer is D**

Population of school = 250 + 150 = 400

60% of 250 = 60%/100% x 250 = 150

40% of 150 = 40%/100% x 150 = 60

Total number of students who plays football;

150 + 60 = 210

Percentage of school that play football;

210/400 x 100% = 52.5%

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

**A.**2**B.**3**C.**4**D.**5

**Correct Option: 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

**8. **If F = \(\frac{9}{5}\)C + 32, find C when F = 98.6

A. 30

B. 37

C. 39

D. 41

**Correct Option: Answer is B**

F = \(\frac{9}{5}\)C + 32

When F = 98.6

98.6 = \(\frac{9}{5}\)C + 32

98.6 – 32 = \(\frac{9}{5}\)C

66.6= \(\frac{9}{5}\)C

66.6 x 5 = 9C

C = \(\frac{66.6 \times 5}{9}\)

= 37

**9. **If y + 2x = 4 and y – 3x = -1, find the value of (x + y)

**A.**3**B.**2**C.**1**D.**-1

**Correct Option: Answer is A**

y + 2x = 4 …..(1)

9 – 3x = -1 ……(2)

Substract (2) from (1)

2x – (-3x) = 4 – (-1)

2x + 3x = 4 + 1

5x = 5

X = \(\frac{5}{5}\)

= 1

Substitute 1 for x in (1);

y + 2(1) = 4

y + 2 = 4

y = 4 – 2 = 2

Hence, (x + y) = (1 + 2)

= 3

**10. **If x : y : z = 3 : 3 : 4, evaluate \(\frac{9x + 3y}{6x – 2y}\)

**A.**1\(\frac{1}{2}\)**B.**2**C.**2\(\frac{1}{2}\)**D.**3

**Correct Option: Answer is A**

If x : y : z = 3 : 3 : 4, evaluate \(\frac{9x + 3y}{6x – 2y}\)

\(\frac{x}{y}\) = \(\frac{2}{3}\) and \(\frac{y}{z}\) = \(\frac{3}{4}\)

Thus; x = \(\frac{2}{3}T_1\) and z = \(\frac{3}{5}T_1\)

y = \(\frac{3}{7}T_2\) and z = \(\frac{4}{7}T_2\)

Using y = y

\(\frac{3}{5}T_1\) = \(\frac{3}{7}T_2\); \(\frac{T_1}{T_2}\) = \(\frac{3}{7}\) x \(\frac{5}{3}\)

\(\frac{T_1}{T_2}\) = \(\frac{15}{21}\)

\(T_1\) = 15 and \(T_2\) = 21

Therefore;

x = \(\frac{2}{5}\) x 15 = 6

y = \(\frac{3}{5}\) x 15 = 9

y = \(\frac{3}{7}\) x 21 = 9 (again)

z = \(\frac{4}{7}\) x 21 = 12

Hence;

\(\frac{9x + 3y}{6z – 2y}\) = \(\frac{9(6) + 3(9)}{6(12) – 2(9)}\)

\(\frac{54 + 27}{72 – 18}\) = \(\frac{81}{54}\) = \(\frac{3}{2}\)

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