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
The goal of mocks tests is to create a bench-marking tool to help students assess their performances.
Eat well during in test days for good grades. How can you plan your exam diet with good grades in mind?
Effective ways of coping with exam stress and test anxiety which is supported by Science
| # | 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