2012 - WAEC Mathematics Past Questions and 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
Users' Answers & Comments2
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\(\sqrt{5}\)
Users' Answers & Comments= \(\frac{\sqrt{5} \times 2\sqrt{6}}{\sqrt{2} \times \sqrt{3}}\)
= \(\frac{\sqrt{5} \times 2 \sqrt{6}}{\sqrt{6}}\)
= 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
Users' Answers & Commentsand y = \(\frac{1190}{70}\) = 17
ratio of enrollment of x to y = 15:17
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
Users' Answers & Comments= 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
Users' Answers & Comments= \(\frac{200}{8}\)
= 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%
Users' Answers & CommentsA = 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
Users' Answers & Comments= \(\frac{1}{2}\)
\(\frac{5}{3x} = \frac{1}{3}\)
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)
Users' Answers & Comments= \(\frac{6(3k + 1) - (3k - 1)}{3k + 1}\)
= 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 = \(\frac{2qr - rs}{6}\)
Users' Answers & Commentsq = \(\frac{6p + rs}{2r}\)
6p + rs = 2qr
6p = 2qr - rs
p = \(\frac{2qr - rs}{6}\)