Mathematics (Core)
Paper 1 | Objectives | 49 Questions
WASSCE/WAEC MAY/JUNE
Year: 2012
Level: SHS
Time:
Type: Question Paper
Answers provided
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Paper 1 | Objectives | 49 Questions
WASSCE/WAEC MAY/JUNE
Year: 2012
Level: SHS
Time:
Type: Question Paper
Answers provided
No description provided
This paper is yet to be rated
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| # | Question | Ans |
|---|---|---|
| 1. |
Express 302.10495 correct to five significant figures A. 302.10 B. 302.11 C. 302.105 D. 302.1049 |
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\(\sqrt{5}\)
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Detailed Solution\(\frac{3\sqrt{5} \times 4\sqrt{6}}{2 \sqrt{2} \times 3\sqrt{2}}\)= \(\frac{\sqrt{5} \times 2\sqrt{6}}{\sqrt{2} \times \sqrt{3}}\) = \(\frac{\sqrt{5} \times 2 \sqrt{6}}{\sqrt{6}}\) = 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
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Detailed SolutionDivide cash enrollment by their common factors then the answer of each division gives the required ratio. i.e. x = \(\frac{1050}{70}\) = 15and y = \(\frac{1190}{70}\) = 17 ratio of enrollment of x to y = 15:17 |
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| 4. |
Convert 3510 to number in base 2 A. 1011 B. 10011 C. 100011 D. 11001 |
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
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Detailed SolutionT4 = 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 |
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| 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
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Detailed Solution\(\frac{720 \times 1000}{8 \times 60 \times 60}\) = \(\frac{20 \times 10}{8}\)= \(\frac{200}{8}\) = 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%
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Detailed SolutionA = 3,500, P = N2,500A = 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% |
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| 8. |
Solve for x in the equation; \(\frac{1}{x} + \frac{2}{3x} = \frac{1}{3}\) A. 5 B. 4 C. 3 D. 1
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Detailed Solution\(\frac{1}{8} + \frac{2}{3x} = \frac{1}{3}\)= \(\frac{1}{2}\) \(\frac{5}{3x} = \frac{1}{3}\) 3x = 15 x = \(\frac{15}{3}\) = 5 |
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| 9. |
Simplify: \(\frac{54k^2 - 6}{3k + 1}\) A. 6(1 - 3k2) B. 6(3k2 - 1) C. 6(3k - 1) D. 6(1 - 3k)
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Detailed Solution\(\frac{54k^2 - 6}{3k + 1} = \frac{6(9k^2 - 1)}{3k + 1}\)= \(\frac{6(3k + 1) - (3k - 1)}{3k + 1}\) = 6(3k - 1) |
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| 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 = \(\frac{2qr - rs}{6}\)
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Detailed Solutionq = \(\frac{3p}{r} + \frac{s}{2}\)q = \(\frac{6p + rs}{2r}\) 6p + rs = 2qr 6p = 2qr - rs p = \(\frac{2qr - rs}{6}\) |
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