Paper 1 | Objectives | 47 Questions
WASSCE/WAEC MAY/JUNE
Year: 1990
Level: SHS
Time:
Type: Question Paper
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Past questions are effective for revisions for all tests including WAEC, BECE, SAT, TOEFL, GCSE, IELTS
# | Question | Ans |
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1. |
Simplify 125\(^{\frac{-1}{3}}\) x 49\(^{\frac{-1}{2}}\) x 10\(^0\) A. 350 B. 35 C. 1/35 D. 1/350
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Detailed Solution125\(^{\frac{-1}{3}}\) x 49\(^{\frac{-1}{2}}\) x 10\(^0\)= 5\(^{-1}\) x 7\(^{-1}\) x 1 = \(\frac{1}{35}\) |
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2. |
If 3\(^{2x}\) = 27, what is x? A. 1 B. 1.5 C. 4.5 D. 18 E. 40.5
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Detailed Solution3\(^{2x}\) = 273\(^{2x}\) = 3\(^3\) 2x = 3 x = 1.5 |
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3. |
Express 0.00562 in standard form A. 5.62 x 10-3 B. 5.62 x 10-2 C. 562 x 10-2 D. 5.62 x 102 E. 5.62 x 103
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Detailed Solution0.00562 = 5.62 x 10\(^{-3}\) |
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4. |
Given that 1/3log10 P = 1, find the value of P A. 1/10 B. 3 C. 10 D. 100 E. 1000
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Detailed Solution1/3log10P = 1log10P1/3 = log1010 P1/3 = 10 P = 1000 |
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5. |
Simplify \(\frac{\log \sqrt{8}}{\log 8}\) A. 1/3 B. 1/2 C. 1/3log√2 D. 1/3log√8 E. 1/2log√2
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Detailed Solution\(\frac{\log \sqrt{8}}{\log 8}\)= \(\frac{\log 8^{\frac{1}{2}}}{log 8}\) = \(\frac{\frac{1}{2} \log 8}{\log 8}\) = \(\frac{1}{2}\) |
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6. |
Evaluate using the logarithm table, log(0.65)2 A. 1.6258 B. 0.6272 C. 0.6258 D. 3.6258 E. 1.6272
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Detailed Solutionlog(0.65)2 = 2log(0.65) but log0.65 = 1.8129∴2 x 1.8129 = 3.6258 |
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7. |
If log x = \(\bar{2}.3675\) and log y = 0.9750, what is the value of x + y? Correct to three significant figures A. 1.18 B. 1.31 C. 9.03 D. 9.44 E. 9.46
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Detailed Solutionlog x = \(\bar{2}.3675\) ; log y = 0.9750\(x = 10^{\bar{2}.3675} = 0.02331 \) \(y = 10^{0.9750} = 9.441 \) \(x + y = 9.4641 \approxeq 9.46\) |
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8. |
While doing his physics practical, Idowu recorded a reading as 1.12cm instead of 1.21cm. Calculate his percentage error A. 1.17% B. 6.38% C. 7.44% D. 8.035% E. 9.00%
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Detailed Solution%error = \(\frac{1.21 - 1.12}{1.21} \times 100%\)= \(\frac{9}{121} \times 100%\) = 7.44% |
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9. |
Find the 4th term of an A.P, whose first term is 2 and the common difference is 0.5 A. 0.5 B. 2.5 C. 3.5 D. 4 E. 4.5
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Detailed Solution\(U_{n} = a + (n - 1)d\)\(U_{4} = 2 + (4 - 1) \times 0.5\) = \(2 + 1.5\) = 3.5 |
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10. |
From the graph determine the roots of the equation y = 2x2 + x - 6 A. -3, 4 B. -2, -6 C. -2, 1.5 D. -1, 1 E. 2, 1.5 |
C |
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