Mathematics (Core)
Paper 1 | Objectives | 50 Questions
WASSCE/WAEC MAY/JUNE
Year: 1994
Level: SHS
Time:
Type: Question Paper
Answers provided
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Paper 1 | Objectives | 50 Questions
WASSCE/WAEC MAY/JUNE
Year: 1994
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
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| # | Question | Ans |
|---|---|---|
| 1. |
If the 2nd and 5th terms of a G.P are 6 and 48 respectively, find the sum of the first for term A. -45 B. -15 C. 15 D. 33 E. 45
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Detailed SolutionT\(_{2}\) = ar = 6T\(_{5}\) = ar\(^{4}\) = 48 \(\frac{T_5}{T_2}\) = \(\frac{ar^{4}}{ar}\) = \(\frac{48}{6}\) = r\(^{3}\) = 8 ⇒ r = 2 S\(_{n}\) = \(\frac{a((r^n) - 1)}{r - 1}\) S\(_{4}\) = \(\frac{a((r^4) - 1)}{r - 1}\) S\(_{4}\) = \(\frac{3((2^4) - 1)}{2 - 1}\) = 3(16 -1) = 45 |
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| 2. |
If sin\( \theta \) = K find tan\(\theta\), 0° \(\leq\) \(\theta\) \(\leq\) 90°. A. 1-K B. \( \frac{k}{k - 1} \) C. \( \frac{k}{\sqrt{1 - k^2}} \) D. \( \frac{k}{1 - k} \) E. \( \frac{k}{\sqrt{ k^2 - 1}} \)
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Detailed Solution\(\sin \theta = \frac{k}{1}\)\(\implies 1^2 = k^2 + adj^2\) \(adj = \sqrt{1 - k^2}\) \(\therefore \tan \theta = \frac{k}{\sqrt{1 - k^2}}\) |
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| 3. |
Simplify: \(2\frac{1}{3} \div 2\frac{2}{3} \times 1\frac{1}{7}\) A. 0 B. 1 C. 2 D. 3 E. 4
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Detailed Solution\(2\frac{1}{3} \div 2\frac{2}{3} \times 1\frac{1}{7}\)= \(\frac{7}{3} \div \frac{8}{3} \times \frac{8}{7}\) = \(\frac{7}{3} \times \frac{3}{8} \times \frac{8}{7}\) = 1 |
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| 4. |
Which of the following is equal to \(\frac{72}{125}\) A. \( \frac{2^3 \times 3^2}{5^3} \) B. \( \frac{2^4 \times 3}{5^3} \) C. \( \frac{2^3 \times 3}{5^3} \) D. \( \frac{2^4 \times 3}{5^5} \) E. \( \frac{2^2 \times 3^2 \times 4^2}{5^2} \)
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Detailed Solution\(\frac{72}{125} = \frac{8 \times 9}{5 \times 5 \times 5}\)= \(\frac{2^3 \times 3^2}{5^3}\) |
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| 5. |
Express in \( \frac{8.75}{0.025} \)standard form A. 3.5 x 10-3 B. 3.5 x 10-2 C. 3.5 x 101 D. 3.5 x 102 E. 3.5 x 103
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Detailed Solution\(\frac{8.75}{0.025}\)= \(\frac{8750}{25}\) = \(350\) = \(3.5 \times 10^2\) |
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| 6. |
Evaluate \( \frac{27^{\frac{1}{3}}}{16^{-\frac{1}{4}}} \) A. 6 B. 5 C. 4 D. 3 E. 2
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Detailed Solution\(\frac{27^{\frac{1}{3}}}{16^{-\frac{1}{4}}} \)= \(\frac{(3^3)^{\frac{1}{3}}}{(2^4)^{-\frac{1}{4}}}\) = \(\frac{3}{2^{-1}}\) = 6 |
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| 7. |
Simplify: 16\(^{\frac{5}{4}}\) x 2\(^{-3}\) x 3\(^0\) A. 0 B. 2 C. 4 D. 10 E. 20
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Detailed Solution16\(^{\frac{5}{4}}\) x 2\(^{-3}\) x 3\(^0\)= \((2^4)^{\frac{5}{4}} \times 2^{-3} \times 1\) = \(2^5 \times 2^{-3} \times 1\) = \(2^2\) = 4. |
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| 8. |
Simplify; 2log\(_{3}\) 6 + log\(_{3}\) 12 - log\(_{3}\) 16 A. 2 B. 3 C. 2 - 2log32 D. 3 - log32 E. 4 - log32
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Detailed Solution2log\(_{3}\) 6 + log\(_{3}\) 12 - log\(_{3}\) 16= \(\log_3 6^2 + \log_3 12 - \log_3 16\) = \(\log_{3} (\frac{36 \times 12}{16})\) = \(\log_{3} (27)\) = \(\log_{3} 3^3\) = \(3 \log_{3} 3\) = 3 |
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| 9. |
What is the number whose logarithm to base 10 is \(\bar{3}.4771\)? A. 3.0 B. 0.3 C. 0.03 D. 0.003 E. 0.0003 |
D |
| 10. |
A house bought for N100,000 was later auctioned for N80,000. Find the loss percent. A. 20% B. 30% C. 40% D. 50% E. 60%
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Detailed SolutionLoss = N(100,000 - 80,000)= N20,000. Loss percentage = \(\frac{20,000}{100,000} \times 100%\) = 20% |
Preview displays only 10 out of the 50 Questions