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
Past questions are effective for revisions for all tests including WAEC, BECE, SAT, TOEFL, GCSE, IELTS
Scholarship in Norway universities are open for application in 2022 also for developing countries.
Make sure you study hard but not into the late-night hours to give your body the enough rest you need.
| # | 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