# 2012 - WAEC Mathematics Past Questions & Answers - page 1

1
Express 302.10495 correct to five significant figures
A
302.10
B
302.11
C
302.105
D
302.1049
CORRECT OPTION: a
2
Simplify; $$\frac{3\sqrt{5} \times 4\sqrt{6}}{2 \sqrt{2} \times 3\sqrt{2}}$$
A
$$\sqrt{2}$$
B
$$\sqrt{5}$$
C
2$$\sqrt{2}$$
D
2$$\sqrt{5}$$
CORRECT OPTION: d
$$\frac{3\sqrt{5} \times 4\sqrt{6}}{2 \sqrt{2} \times 3\sqrt{2}}$$

= $$\frac{\sqrt{5} \times 2\sqrt{6}}{\sqrt{2} \times \sqrt{3}}$$

= $$\frac{\sqrt{5} \times 2 \sqrt{6}}{\sqrt{6}}$$

= 2$$\sqrt{5}$$
3
In 1995, the enrollment of two schools X and Y were 1,050 and 1,190 respectively. Find the ration of the enrollments of X and Y
A
50:11
B
15:17
C
13:55
D
12:11
CORRECT OPTION: b
Divide cash enrollment by their common factors then the answer of each division gives the required ratio. i.e. x = $$\frac{1050}{70}$$ = 15

and y = $$\frac{1190}{70}$$ = 17

ratio of enrollment of x to y = 15:17
4
Convert 3510 to number in base 2
A
1011
B
10011
C
100011
D
11001
CORRECT OPTION: c
5
The nth term of a sequence is Tn = 5 + (n - 1)2. Evaluate T4 - T6
A
30
B
16
C
-16
D
-30
CORRECT OPTION: c
T4 = 5 + (4 - 1)2; where n = 4

= 5 + (3)2 = 5 + 9

= 14

T = 5 + (6 - 1)2

where n = 6

= 5 + (5)2

= 5 + 25 = 30

T4 + T6 = 14 - 30

= -16
6
Mr. Manu travelled from Accra to Pamfokromb a distance of 720km in 8 hours. What will be his speed in m/s?
A
25m/s
B
150m/s
C
250m/s
D
500m/s
CORRECT OPTION: a
$$\frac{720 \times 1000}{8 \times 60 \times 60}$$ = $$\frac{20 \times 10}{8}$$

= $$\frac{200}{8}$$

= 25 m/s
7
If N2,500.00 amounted to N3,50.00 in 4 years at simple interest, find the rate at which the interest was charged
A
5%
B
7$$\frac{1}{2}$$%
C
8%
D
10%
CORRECT OPTION: d
A = 3,500, P = N2,500

A = P + I

But I = N3,500 - N2,500

I = N1,000

I = $$\frac{PRT}{100}$$

N1,000 = $$\frac{2.500 \times R \times 4}{100}$$

1000 = 100R

R = $$\frac{1000}{100}$$

= 10%
8
Solve for x in the equation; $$\frac{1}{x} + \frac{2}{3x} = \frac{1}{3}$$
A
5
B
4
C
3
D
1
CORRECT OPTION: a
$$\frac{1}{8} + \frac{2}{3x} = \frac{1}{3}$$

= $$\frac{1}{2}$$

$$\frac{5}{3x} = \frac{1}{3}$$

3x = 15

x = $$\frac{15}{3}$$

= 5
9
Simplify: $$\frac{54k^2 - 6}{3k + 1}$$
A
6(1 - 3k2)
B
6(3k2 - 1)
C
6(3k - 1)
D
6(1 - 3k)
CORRECT OPTION: c
$$\frac{54k^2 - 6}{3k + 1} = \frac{6(9k^2 - 1)}{3k + 1}$$

= $$\frac{6(3k + 1) - (3k - 1)}{3k + 1}$$

= 6(3k - 1)
10
Make p the subject of the relation: q = $$\frac{3p}{r} + \frac{s}{2}$$
A
p = $$\frac{2q - rs}{6}$$
B
p = 2qr - sr - 3
C
p = $$\frac{2qr - s}{6}$$
D
p = $$\frac{2qr - rs}{6}$$
CORRECT OPTION: d
q = $$\frac{3p}{r} + \frac{s}{2}$$

q = $$\frac{6p + rs}{2r}$$

6p + rs = 2qr

6p = 2qr - rs

p = $$\frac{2qr - rs}{6}$$
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