# 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

**23.**Solved the equation \(2x^2 – x – 6\) = 0

**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\(\sqrt{3m}\)