Paper 1 | Objectives | 50 Questions
WASSCE/WAEC MAY/JUNE
Year: 2019
Level: SHS
Time:
Type: Question Paper
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1. |
Express, correct to three significant figures, 0.003597. A. 0.359 B. 0.004 C. 0.00360 D. 0.00359
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Detailed Solution0,00 3597 = 0.00360 to 3 s.f |
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2. |
Evaluate: (0.064) - \(\frac{1}{3}\) A. \(\frac{5}{2}\) B. \(\frac{2}{5}\) C. -\(\frac{2}{5}\) D. -\(\frac{5}{2}\)
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Detailed Solution(0.064)\(^{- \frac{1}{3}}\)= (\(\frac{64}{1000}\))\(^{-\frac{1}{3}}\) = 3\(\sqrt{\frac{1000}{64}}\) = \(\frac{10}{4}\) = \(\frac{5}{2}\) |
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3. |
Solve: \(\frac{y + 1}{2} - \frac{2y - 1}{3}\) = 4 A. y = 19 B. y = -19 C. y = -29 D. y = 29
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Detailed Solution\(\frac{y + 1}{2} - \frac{2y - 1}{3}\) = \(\frac{4}{1}\)- \(\frac{3(y + 1) - 2(2y - 1)}{6} = \frac{4}{1}\) 3y + 3 - 4y + 2 = 24 - y + 5 = 24 - y = 24 - 5 = 19 y = - 19 |
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4. |
Simplify, correct to three significant figures, (27.63)\(^2\) - (12.37)\(^2\) A. 614 B. 612 C. 611 D. 610
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Detailed Solution(27.63)\(^2\) - (12.37)\(^2\)= (27.63 + 12.37)(27.63 - 12.37) = 40 x 15.26 = 610 |
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5. |
If 7 + y = 4 (mod 8), find the least value of y, 10 \(\leq y \leq 30\) A. 11 B. 13 C. 19 D. 21
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Detailed Solution7 + y = 4 (mod 8)y = 4 - 7 (mod 8) y = -3 + 8 (mod 8) y = 5 + 8 (mod 8) y = 13 |
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6. |
If T = {prime numbers} and M = {odd numbers} are subsets of \(\mu\) = {x : 0 < x < 10} and x is an integer, find (T\(^{\prime}\) \(\mu\) M\(^{\prime}\)). A. {4, 6, 8, 10} B. {1 C. {1, 2, 4, 6, 8, 10} D. {1, 2, 3, 5, 7, 8, 9}
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Detailed SolutionT = {2, 3, 5, 7}M = {1, 3, 5, 7, 9} \(\mu\) = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} T\(^{\prime}\) = = {1, 4, 6, 8, 9, 10} M\(^{\prime}\) = {2, 4, 6, 8, 10} (T\(^{\prime}\) \(\cap\) M\(^{\prime}\)) = {4, 6, 8, 10} |
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7. |
Evaluate; \(\frac{\log_3 9 - \log_2 8}{\log_3 9}\) A. -\(\frac{1}{3}\) B. \(\frac{1}{2}\) C. \(\frac{1}{3}\) D. -\(\frac{1}{2}\)
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Detailed Solution\(\frac{\log_3 9 - \log_2 8}{\log_3 9}\)= \(\frac{\log_3 3^2 - \log_2 2^3}{\log_3 3^2}\) = \(\frac{2 -3}{2}\) = \(\frac{-1}{2}\) |
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8. |
If 23\(_y\) = 1111\(_{\text{two}}\), find the value of y A. 4 B. 5 C. 6 D. 7
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Detailed Solution23\(_y\) = 1111\(_{\text{two}}\),2 x y\(^1\) + 3 x y\(^0\) = 1 x 2\(^3\) + 1 x 2\(^1\) + 1 x 2\(^o\) 2y + 3 = 8 + 4 + 2 + 1 2y + 3 = 15 \(\frac{2y}{2}\) \(\frac{12}{2}\) y = 6 |
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9. |
If 6, P, and 14 are consecutive terms in an Arithmetic Progression (AP), find the value of P. A. 9 B. 10 C. 6 D. 8
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Detailed Solution6, p, 1414 - p = p - 6 14 + 6 = p - 6 14 + 6 = p + p \(\frac{2p}{2}\) = \(\frac{20}{2}\) p = 10 |
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10. |
Evaluate: 2\(\sqrt{28} - 3\sqrt{50} + \sqrt{72}\) A. 4\(\sqrt{7} - 21 \sqrt{2}\) B. 4\(\sqrt{7} - 11 \sqrt{2}\) C. 4\(\sqrt{7} - 9 \sqrt{2}\) D. 4\(\sqrt{7} + \sqrt{2}\)
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Detailed Solution2\(\sqrt{28} - 3\sqrt{50} + \sqrt{22}\)4\(\sqrt{7} - 15\sqrt{2} + 6\sqrt{2}\) 6\(\sqrt{7} - 9\sqrt{2}\) |
Preview displays only 10 out of the 50 Questions