# 2008 - WAEC Mathematics Past Questions and Answers - page 1

1
If x% of 240 equals 12, find x
A
x = 1
B
x = 3
C
x = 5
D
x = 7
correct option: c
x% of 240 = 12

$$\frac{x}{100} \times 240 = 12$$

x = $$\frac{12 \times 100}{240}$$

x = 5
2
Evaluate $$\frac{(3.2)^2 - (4.8)^2}{3.2 + 1.8}$$
A
-0.08
B
-1.60
C
-10.24
D
-12.80
correct option: b
$$\frac{(3.2)^2 - (4.8)^2}{3.2 + 1.8} = \frac{(3.2 - 4.8)(3.2 + 4.8)}{(3.2 + 4.8)}$$

= 3.2 - 4.8

= -1.60
3
Simplify $$\sqrt{50} + \frac{10}{\sqrt{2}}$$
A
10
B
10$$\sqrt{2}$$
C
20
D
20$$\sqrt{2}$$
correct option: b
$$\sqrt{50} + \frac{10}}{\sqrt{2}} = \(\frac{\sqrt{50}}{1} + \sqrt{10}{\sqrt{2}}$$

= $$\frac{\sqrt{50 \times 2} + 10}{\sqrt{2}}$$

= $$\frac{\sqrt{100} + 10}{\sqrt{2}}$$

= $$\frac{10 + 10}{\sqrt{2} = \frac{20}{\sqrt{2}}$$

= $$\frac{20}{\sqrt{2}}$$ \times \frac{\sqrt{2}}{\sqrt{2}}\)

= $$\frac{20\sqrt{2}}{2}$$

= 10$$\sqrt{2}$$
4
P naira invested for 4 years invested for 4 years at r% simple interest per annum yields 0.36 p naira interest. Find the value of r
A
1$$\frac{1}{9}$$
B
1$$\frac{4}{9}$$
C
9
D
11
correct option: c
I = $$\frac{PRT}{100}$$

where r = r% p.a; I = 0.36p

0.36p = $$\frac{P \times r \times 4}{100}$$

$$\frac{0.36 \times 100}{4}$$ = r

r = 9
5
A trader bought 100 tubers at 5 for N350.00. She sold them in sets of 4 for N290.00. Find her gain percent.
A
3.6%
B
3.5%
C
3.5%
D
2.55
correct option: a
Cost price, c.p = $$\frac{100}{5}$$ x N350 = N7000

Selling price, s.p = $$\frac{100}{5}$$ x N290 = N7250

%Gain = $$\frac{S.p - C.p}{C.p}$$ x 100%

= $$\frac{7250 - 7000}{7000}$$ x 100% = $$\frac{250 \times 100}{7000}$$

= 3.6% (approx.)
6
If p-2g + 1 = g + 3p and p - 2 = 0, find g
A
-2
B
-1
C
1
D
2
correct option: b
p - 2g + 1 = g + 3p.........(1)

p - 2 = 0 .........(2)

From (2), p = 2; put p = 2 into (1);

2 - 2g + 1 = g + 3(2)

3 - 2g = g + 6

-2g - g = 6 - 3

-3g = 3

g = $$\frac{3}{-3}$$

g = -1
7
Simplify $$\frac{\frac{1}{x} + \frac{1}{y}}{x + y}$$
A
$$\frac{1}{x + y}$$
B
$$\frac{1}{xy}$$
C
x + y
D
xy
correct option: b
$$\frac{\frac{1}{x} + \frac{1}{y}}{x + y}$$ = $$\frac{\frac{y + x}{xy}}{x + y}$$

= $$\frac{x + y}{xy}$$

= $$\frac{x + y}{xy} \times \frac{1}{x + y}$$

= $$\frac{1}{xy}$$
8
Simplify 3$$\sqrt{27x^3y^9}$$
A
9xy3
B
3xy6
C
3xy3
D
9y3
correct option: c
3$$\sqrt{27x^3y^9}$$ = 3$$\sqrt{27} \times 3\sqrt{3^3} \times 3\sqrt{y^9}$$

= 3 $$\times x \times y^3$$

= 3xy3
9
Given that x = 2 and y = -$$\frac{1}{4}$$, evaluate $$\frac{x^2y - 2xy}{5}$$
A
zero
B
$$\frac{1}{5}$$
C
1
D
2
correct option: a
Given; x = 2; y = $$\frac{-1}{4}$$

= $$\frac{x^2y - 2xy}{5}$$

= $$\frac{2^2(\frac{-1}{4}) - 2(2)(\frac{-1}{4})}{5}$$

= $$\frac{4(\frac{-1}{4}) + 4(\frac{-1}{4})}{5}$$

= $$\frac{1 + 1}{5}$$

= $$\frac{0}{5}$$

= 0
10
Factorize 5y2 + 2ay - 3a2
A
(a - y)(5y - 3a)
B
(y - a)(5y - 3a)
C
(y - a)(5y + 3a)
D
(y + a)(5y 3a)
correct option: d
5y2 + 2ay - 3a2 = 5y2 + 5ay - 3a2

= 5y(y + a) - 3a(y + a)

= (y + a)(5y - 3a)