Paper 1 | Objectives | 50 Questions
WASSCE/WAEC MAY/JUNE
Year: 2020
Level: SHS
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Type: Question Paper
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1. |
Evaluate and correct to two decimal places, 75.0785 - 34.624 + 8.83 A. 30.60 B. 50.29 C. 50.28 D. 30.62
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Detailed Solution75.0785- 34.624 -------------- 40.4545 + 9.83 -------------- 50.28 to 2d.p |
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2. |
If X = {x : x < 7} and Y = {y:y is a factor of 24} are subsets of \(\mu\) = {1, 2, 3...10} find X \(\cap\) Y. A. {2, 3, 4, 6} B. {1, 2, 3, 4, 6} C. {2, 3, 4, 6, 8} D. {1, 2, 3, 4, 6, 8}
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Detailed Solution\(\mu\) = {1, 2, 3, 4..., 10}X = {1, 2, 3, 4, 5, 6} Y = {1, 2, 3, 4, 6, 8} Therefore; X \(\cap\) Y = {1, 2, 3, 4, 6} |
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3. |
Simplify; [(\(\frac{16}{9}\))\(^{\frac{-3}{2}}\) x 16\(^{\frac{-3}{2}}\)]\(^{\frac{1}{3}}\) A. \(\frac{3}{4}\) B. \(\frac{9}{16}\) C. \(\frac{3}{8}\) D. \(\frac{1}{4}\)
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Detailed Solution[(\(\frac{16}{9}\))\(^{\frac{-3}{2}}\) x 16\(^{\frac{-3}{2}}\)]\(^{\frac{1}{3}}\)= [(\(\frac{9}{16}\))]\(^{\frac{3}{2}}\) x [(\(\frac{1}{16}\))\(^{\frac{3}{4}}\)]\(^{\frac{1}{3}}\) = [(\(\sqrt{\frac{9}{10}}\))\(^3\) x (4\(\sqrt{\frac{1}{16}})^3\)]\(^{\frac{1}{3}}\) = [(\(\frac{3}{4})^3 \times (\frac{1}{2})^3\)]\(^\frac{1}{3}\) (\(\frac{27}{64} \times \frac{1}{8}\))\(^\frac{1}{3}\) = \({3}\sqrt{\frac{27}{64} \times \frac{1}{8}}\) = \(\frac{3}{4} \times \frac{1}{2}\) = \(\frac{3}{8}\) |
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4. |
Express 1 + 2 log10\(^3\) in the form log10\(^9\) A. log10\(^{90}\) B. log10\(^{19}\) C. log10\(^{9}\) D. log10\(^{6}\)
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Detailed Solution1 + 2log\(_{10}^3\)= log\(_{10}^{10} + log_{10}^{3^2}\) = log\(_{10}^{10} + log_{10}^{9}\) = log\(_{10}^{10 \times 90}\) = log\(_{10}^{90}\) |
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5. |
If 101\(_{\text{two}}\) + 12y = 3.3\(_{\text{five}}\). Find the value of y A. 8 B. 7 C. 6 D. 5
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Detailed Solution012 + 01 = 01101\(_2\) + 12\(_y\) = 2.3\(_5\) 1 x 2\(^o\) + 0 x 2\(^o\) + 1 x2\(^2\) + 1 x y\(^o\) + 2 x y\(^1\) = 3 x 5\(^o\) + 3 x 5\(^1\) 1 + 4 + 1 + 2y = 3 + 15 6 + 2y = 18 2y = 18 - 6 \(\frac{2y}{2} = \frac{12}{2}\) y = 6 |
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6. |
An amount of N550,000.00 was realized when a principal, x was saved at 2% simple interest for 5 years. Find the value of x A. N470,000.00 B. N480,000.00 C. N490,000.00 D. N500,000.00
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Detailed SolutionS.I = \(\frac{x \times 2 \times 5}{100}\) = 0.1xA = P + S.I 550,000 = x + 0.1x \(\frac{550,000}{1.1} = \frac{1.1x}{1.1}\) x = N500,000 |
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7. |
Given that \(\frac{\sqrt{3} + \sqrt{5}}{\sqrt{5}}\) = x + y\(\sqrt{15}\), find the value of (x + y) A. 1\(\frac{3}{5}\) B. 1\(\frac{2}{5}\) C. 1\(\frac{1}{5}\) D. \(\frac{1}{5}\)
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Detailed Solution\(\frac{\sqrt{3} + \sqrt{5}}{\sqrt{5}}\) = x + y\(\sqrt{15}\)cross multiply to have: \(\sqrt{3}\) + \(\sqrt{5}\) = x\(\sqrt{5}\) + 5y\(\sqrt{3}\) Collect like roots : x\(\sqrt{5}\) = \(\sqrt{5}\) → x = 1 5y\(\sqrt{3}\) = \(\sqrt{3}\) → y = \(\frac{1}{5}\) ∴ ( x + y ) = 1 + \(\frac{1}{5}\) = 1\(\frac{1}{5}\) |
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8. |
If x = 3 and y = -1, evaluate 2(x\(^2\) - y\(^2\)) A. 24 B. 22 C. 20 D. 16
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Detailed Solution2(\(x^2 - y^2\))= 2(x + y)(x - y) = 2(3 + (-1))(3 - (-1)) = 2(2)(4) = 16 |
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9. |
Solve 3x - 2y = 10 and x + 3y = 7 simultaneously A. x = -4 and y = 1 B. x = -1 and y = -4 C. x = 1 and y = 4 D. x = 4 and y = 1
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Detailed Solution3x - 2y = 10 - - x 3x + 3y = 7 ---x 2 ------------------------ 9x - 6y = 30 2x + 6y = 14 ------------------------- \(\frac{11x}{11} \frac{44}{11}\) x = 4 From x + 3y = 7 3y = 7 - 4 \(\frac{3y}{3}\) = \(\frac{3}{3}\) y = 1 |
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10. |
The implication x \(\to\) y is equivalent to A. ~ y \(\to\) ~ x B. y \(\to\) ~ x C. ~ x \(\to\) ~ y D. y \(\to\) x |
A |
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