# 2014 - WAEC Mathematics Past Questions & Answers - page 1

1
Simplify 10$$\frac{2}{5} - 6 \frac{2}{3} + 3$$
A
6$$\frac{4}{15}$$
B
6$$\frac{11}{15}$$
C
7$$\frac{4}{15}$$
D
7$$\frac{11}{15}$$
CORRECT OPTION: b
10$$\frac{2}{5} - 6 \frac{2}{3} + 3$$

$$\frac{52}{5} - \frac{20}{3} + \frac{3}{1}$$

= $$\frac{156 - 100 + 45}{15}$$

$$\frac{156 + 45 - 100}{15}$$

= $$\frac{201 - 100}{15}$$

= $$\frac{101}{15}$$

= 6$$\frac{11}{15}$$
2
If 23x = 325, find the value of x
A
7
B
6
C
5
D
4
CORRECT OPTION: a
23x = 325

2 $$\times x^1 + 3 \times x^0 = 3 \times 5^1 + 2 \times 5^0$$

= 2x + 3 = 15 + 2

2x + 3 = 17

2x = 17 - 3

2x = 14

x = $$\frac{14}{2}$$

x = 7
3
The volume of a cube is 512cm3. Find the length of its side
A
6cm
B
7cm
C
8cm
D
9cm
CORRECT OPTION: c
volume of cube = L x L x L

512cm3 = L3

L3 = 512cm3

L = 3$$\sqrt{512}$$

L (512)$$\frac{1}{3}$$

= (29)$$\frac{1}{3}$$

23 = 8cm
4
If one student is selected at random, find the probability that he/she scored at most 2 marks
A
$$\frac{11}{18}$$
B
$$\frac{11}{20}$$
C
$$\frac{7}{22}$$
D
$$\frac{5}{19}$$
CORRECT OPTION: b
at most 2 marks = 5 + 2 + 4 students = 11 students

probability(at most 2 marks) = $$\frac{11}{20}$$
5
Simplify: $$\sqrt{12} ( \sqrt{48} - \sqrt{3}$$)
A
18
B
16
C
14
D
12
CORRECT OPTION: a
$$\sqrt{12} ( \sqrt{48} - \sqrt{3}$$)

$$\sqrt{4 \times 3} (6 \times 3 - \sqrt{3}) = 2 \sqrt{3}(4 \sqrt{3} - \sqrt{3})$$

= 2$$\sqrt{3} \times \sqrt{3} (4 - 1) 2\sqrt{9}(3) = 2 \times 3 \times 3 = 18$$
6
Given that x > y and 3 < y, which of the following is/are true? i. y > 3 ii. x < 3 iii. x > y > 3
A
i
B
i and ii
C
i and iii
D
i, ii and iii
CORRECT OPTION: c
x > y and 3 < y; then 3 < y means that y > 3

x > 3 to give the possible

x > y > 3
7
Three quarters of a number added to two and a half of the number gives 13. Find the number
A
4
B
5
C
6
D
7
CORRECT OPTION: a
let the number be x

2$$\frac{1}{2}x + \frac{3}{4}x = 13$$

$$\frac{5}{2}x + \frac{3}{4}x = 13$$

multiply through by 4

4($$\frac{5}{2}$$)x + 4($$\frac{3}{4}$$)x = 13 x 4

2(5x) + 3x = 52

10x + 3x = 52

13x = 52

x = $$\frac{52}{13}$$

x = 4
8
If x = {0, 2, 4, 6}, y = {1, 2, 3, 4} and z = {1, 3} are subsets of u = {x:0 $$\geq$$ x $$\geq$$ 6}, find x $$\cap$$ (Y' $$\cup$$ Z)
A
{0, 2, 6}
B
{1, 3}
C
{0, 6)
D
{9}
CORRECT OPTION: c
x = {0, 2, 4, 6}; y = {1, 2, 3, 4}; z = {1, 3}

u = {0, 1, 2, 3, 4, 5, 6}

y' = {0, 5, 6}

to find x $$\cap$$ (Y' $$\cup$$ Z)

first find y' $$\cup$$ z = {0, 1, 3, 5, 6}

then x $$\cap$$ (Y' $$\cup$$ Z) = {0, 6}
9
Find the truth set of the equation x2 = 3(2x + 9)
A
{x : x = 3, x = 9}
B
{x : x = -3, x = -9}
C
{x : x = 3, x = -9}
D
{x : x = -3, x = 9}
CORRECT OPTION: d
x2 = 3(2x + 9)

x2 = 6x + 27

x2 - 6x - 27 = 0

x2 - 9x + 3x - 27 = 0

x(x - 9) + 3(x - 9) = (x + 3)(x - 9) = 0

x + 3 = 0 or x - 9 = 0

x = -3 or x = 9

x = -3, x = 9
10
The coordinates of points P and Q are (4, 3) and (2, -1) respectively. Find the shortest distance between P and Q.
A
10$$sqrt{2}$$
B
4$$sqrt{5}$$
C
5$$sqrt{2}$$
D
2$$sqrt{5}$$
CORRECT OPTION: d
p(4, 3) Q(2 - 1)

distance = $$\sqrt{(x_2 - x_1)^2 + (Y_2 - y_1)^2}$$

= $$\sqrt{(2 - 4)^2 + (-1 - 3)^2}$$

= $$\sqrt{(-2)^2 = (-4)^2}$$

= $$\sqrt{4 + 16}$$

= $$\sqrt{20}$$

= $$\sqrt{4 \times 5}$$

= 2$$\sqrt{5}$$
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