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.
21.The total surface area of a hemispher is 75\(\pi cm^2\). Find the radius.
- A. 5.0 cm
- B. 7.0 cm
- C. 8.5 cm
- D. 12.0 cm
Correct Option: Answer is A
Total surface area of hemisphere is
3\(\pi r^2\) = 75\(\pi cm^2\)
\(r^2\) = \(\frac{75 \pi}{3 \pi}\)
\(r^2\) = 25
r = \(\sqrt{25}\)
r = 5cm
22. Find the value of x for which \(\frac{x – 5}{x(x – 1)}\) is defined
- A. 0 or 5
- B. -5 or 5
- C. -11 or 5
- D. 0 or 1
Correct Option: Answer is A
The expression \(\frac{x – 5}{x(x – 1)}\) is defined whten
x(x – 5) = 0.38
either x = 0 or x – 5 = 0
Hence, x = 0 or x = 5
- A. x = \(\frac{-3}{2}\) or 2
- B. x = -2 or \(\frac{3}{2}\)
- C. x = -3 or 2
- D. x = 3 or -2
Correct Option: Answer is A
\(2x^2 – x – 6\) = 0
\(2x^2 – 4x + 3x – 6\) = 0
2x(x – 2) + 3(x – 2) = 0
(2x + 3) (x – 2) = 0
Either; 2x + 3 = 0 or x – 2 = 0
x = \(\frac{-3}{2}\) or x = 2
24. Factorise completely the expression
\((x + 2)^2\) – \((2x + 1)^2\)
- A. (3x + 2)(1 – x)
- B. (3x + 2)(2x + 1)
- C. 3\((x + 2)^2\)
- D. 3(x + 1)(1 – x)
Correct Option: Answer is D
\((x + 2)^2\) – \((2x + 1)^2\)
= \((x^2 + 4x + 4) – (4x^2 + 4x + 1)\)
= \(x^2 \) + 4x + 4 – 4 \(x^2 \) – 4x – 1
= -3 \(x^2 \) + 3
= 3 – 3 \(x^2 \)
= 3(1 – \(x^2 \))
= 3(1 + x)(1 – x)
25. Find the \(n^{th}\) term of the sequence 2 x 3, 4 x 6, 8 x 9, 16 x 12…
- A. 2\(^n\) x 3(n + 1)
- B. 2\(^n\) x 3n
- C. 2\(^n\) x 3\(^n\)
- D. 2\(^n\) x 3\(^n – 1\)
Correct Option: Answer is B
2 x 3, 4 x 6, 8 x 9, 16 x 12,…
2\(^1\) x 3 x 1, 2\(^2\) x 3 x 2, 2\(^3\) x 3 x 3, 2\(^4\) x 3 x 4,…. 2\(^n\) x 3n
26. If 3x\(^o\) 4(mod 5), find the least value of x
- A. 1
- B. 2
- C. 3
- D. 4
Correct Option: Answer is C
3x ≡ 4(mod 5)
In modulo 5, multiples of 5 that give solution to the given equation are 5, 20, 35, 50,… but 5 will yield the least value of x.
Thus; 3x = 4 + 5 = 9
x = 9/3
x = 3
27. Find the inter-quartile range of 1, 3, 4, 5, 8, 9, 10, 11, 12, 14, 16
- A. 6
- B. 7
- C. 8
- D. 9
Correct Option: Answer is B
28. If x : y = \(\frac{1}{4} : \frac{3}{8}\) and y : z = \(\frac{1}{3} : \frac{4}{9}\), find x : z
- A. 2:3
- B. 3:4
- C. 3:8
- D. 1:2
Correct Option: Answer is D
\(\frac{x}{y}\) = \(\frac{1}{4} \div \frac{3}{8}\) = \(\frac{1}{4} \times \frac{8}{3}\) = \(\frac{2}{3}\)
\(\frac{y}{z}\) = \(\frac{1}{3} \div \frac{4}{9}\) = \(\frac{1}{3} \times \frac{9}{4}\) = \(\frac{3}{4}\)
But,
x = \(\frac{2}{5}T_1\), y = \(\frac{3}{5}T_1\)
y = \(\frac{3}{7}T_2\), z = \(\frac{4}{7}T_2\)
Using y = y
\(\frac{3}{5}T_1\) = x = \(\frac{3}{7}T_2\)
\(\frac{T_1}{T_2}\) = \(\frac{3}{7}\) x \(\frac{5}{3}\) = \(\frac{15}{21}\)
\(T_1 = 15\) and \(T_2 = 21\)
Thus , x = \(\frac{2}{5}\) x 15 = 6
y = \(\frac{3}{5}\) x 15 = 9
y = \(\frac{3}{7}\) x 21 = 9
z = \(\frac{4}{7}\) x 21 = 12
Hence; x : z = 6 : 12
= 1 : 2
29. Expression 0.612 in the form \(\frac{x}{y}\), where x and y are integers and y \(\neq\) 0
- A. \(\frac{153}{250}\)
- B. \(\frac{68}{111}\)
- C. \(\frac{61}{100}\)
- D. \(\frac{21}{33}\)
Correct Option: Answer is A
0.612 = \(\frac{0.612}{1}\) x \(\frac{1000}{1000}\)
= \(\frac{612}{1000}\)
= \(\frac{153}{250}\)
30. The angle of elevation of the top of a tree from a point 27m away and on the same horizontal ground as the foot of the tree is 30\(^o\). Find the height of the tree.
- A. 27m
- B. 13.5 \(\sqrt{3m}\)
- C. 13.5 \(\sqrt{2m}\)
- D. 9\(\sqrt{3m}\)
Correct Option: Answer is D
From the diagram above,
tan 30\(^o\) = \(\frac{h}{27}\)
h = 27 tan 30\(^o\)
= 27 x \(\frac{1}{\sqrt{3}}\)
= \(\frac{27}{\sqrt{3}}\) x \(\frac{\sqrt{3}}{\sqrt{3}}\)
= \(\frac{27 \sqrt{3}}{3}\)
= 9\(\sqrt{3m}\)