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