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Mathematics (Core)
Paper 1 | Objectives | 48 Questions
WASSCE/WAEC MAY/JUNE
Year: 1993
Level: SHS
Time:
Type: Question Paper
Answers provided
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Paper 1 | Objectives | 48 Questions
WASSCE/WAEC MAY/JUNE
Year: 1993
Level: SHS
Time:
Type: Question Paper
Answers provided
No description provided
This paper is yet to be rated
Five revision strategies that are strange but effective for memorization for high exams performance
There's no secret to passing WAEC or BECE apart from the merit of hard work and adequate preparation
Revision tips on test preparation for Grade A students to. Grade A students tips for any test.
| # | Question | Ans |
|---|---|---|
| 1. |
S = {1, 2, 3, 4, 5, 6}, T = {2,4,5,7} and R = {1,4, 5}, and (S∩T) ∪ R A. {1, 4, 5} B. {2, 4, 5} C. {1, 2, 4, 5} D. {2, 3, 4, 5} E. {1, 2, 3, 4, 5}
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Detailed SolutionS = {1, 2, 3, 4, 5, 6}; T = {2, 4, 5, 7}; R = {1, 4, 5}(S∩T) ∪ R = {2, 4, 5} ∪ {1, 4, 5} = {1, 2, 4, 5} |
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| 2. |
Simplify: \(\frac{3}{4} \div 1\frac{1}{4} \times (1\frac{1}{2} - \frac{2}{3})\) A. 7/30 B. 7/24 C. 9/25 D. 1/2 E. 18/25
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Detailed Solution\(\frac{3}{4} \div 1\frac{1}{4} \times (1\frac{1}{2} - \frac{2}{3})\)\(\frac{3}{4} \div \frac{5}{4} \times (\frac{9 - 4}{6})\) = \(\frac{3}{4} \times \frac{4}{5} \times \frac{5}{6}\) = \(\frac{1}{2}\) |
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| 3. |
Solve the inequality: 3m + 3 > 9 A. m > 2 B. m > 3 C. m>4 D. m>6 E. m>12
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Detailed Solution3m + 3 > 93m > 9 - 3 3m > 6 m > 2 |
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| 4. |
Convert 89\(_{10}\) to a number in base two. A. 1101001 B. 1011001 C. 1001101 D. 101101 E. 11001
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Detailed Solution\(89_{10}\)\(89_{10} = 1011001_{2}\) |
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| 5. |
A stick of length 1.75m was measured by a boy as 1.80m. Find the percentage error in his measurement A. 27/9% B. 26/7% C. 5% D. 277/9% E. 284/7%.
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Detailed Solution% Error = Error /Actual measurement x 100/1 =
0.05/1.80
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| 6. |
The nth term of a sequence is given by (-1)\(^{n-2}\) x 2\(^{n-1}\). Find the sum of the second and third terms. A. -2 B. 1 C. 2 D. 6 E. 12
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Detailed Solutionwhen n = 2(-1)\(^{n-2}\) 2\(^{n+1}\) = 2 When n = 3 (-1)\(^{n-2}\) 2\(^{n+1}\) = -4 Sum = 2 - 4 = -2 |
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| 7. |
Simplify: \(\frac{4^{-\frac{1}{2}} \times 16^{\frac{3}{4}}}{4^{\frac{1}{2}}}\) A. 1/4 B. 0 C. 1 D. 2 E. 4
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Detailed Solution\(\frac{4^{-\frac{1}{2}} \times 16^{\frac{3}{4}}}{4^{\frac{1}{2}}}\)= \(\frac{16^{\frac{3}{4}}}{4^{\frac{1}{2}} \times 4^{\frac{1}{2}}}\) = \(\frac{(2^4)^{\frac{3}{4}}}{4^{\frac{1}{2}} \times 4^{\frac{1}{2}}}\) = \(\frac{2^3}{4}\) = 2 |
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| 8. |
Simplify: \(\frac{\log \sqrt{27}}{\log \sqrt{81}}\) A. 1/6 B. 3/8 C. 1/2 D. 3/4 E. 6
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Detailed Solution\(\frac{\log \sqrt{27}}{\log 81}\)= \(\frac{\log \sqrt{3^3}}{\log 3^4}\) = \(\frac{\log 3^{\frac{3}{2}}}{\log 3^4}\) = \(\frac{\frac{3}{2} \log 3}{4 \log 3}\) = \(\frac{\frac{3}{2}}{4}\) = \(\frac{3}{8}\) |
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| 9. |
Factorize the expression 2s\(^2\) - 3st - 2t\(^2\). A. (2s - t)(s + 2t) B. (2s + t)(s - 2t) C. (s + t)(2s - 1) D. (2s + t)(s -t) E. (2s + t)(s + 2t)
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Detailed Solution2s\(^2\) - 3st - 2t\(^2\)= 2s\(^2\) - 4st + st - 2t\(^2\) = 2s(s - 2t) + t(s - 2t) = (2s + t)(s - 2t) |
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| 10. |
Solve the equation x\(^2\) - 2x - 3 = 0 A. (-3, 1) B. (-1, -3) C. (3,1) D. (43, 0) E. (-1, 3).
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Detailed Solutionx\(^2\) - 2x - 3 = 0x\(^2\) - 3x + x - 3 = 0 x(x - 3) + 1(x - 3) = 0 (x + 1)(x - 3) = 0 x = (-1, 3) |
Preview displays only 10 out of the 48 Questions