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Mathematics (Core)
Paper 1 | Objectives | 50 Questions
WASSCE/WAEC MAY/JUNE
Year: 2018
Level: SHS
Time:
Type: Question Paper
Answers provided
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Paper has been added to Cart!
Paper 1 | Objectives | 50 Questions
WASSCE/WAEC MAY/JUNE
Year: 2018
Level: SHS
Time:
Type: Question Paper
Answers provided
No description provided
This paper is yet to be rated
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| # | Question | Ans |
|---|---|---|
| 1. |
Simplify: \(\sqrt{108} + \sqrt{125} - \sqrt{75}\) A. \(\sqrt{3} + 5\sqrt{5}\) B. \(6 \sqrt{3} - 5 \sqrt{5}\) C. \(6 \sqrt{3} + \sqrt{2}\) D. \(6\sqrt{3} - \sqrt{2}\)
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Detailed Solution\(\sqrt{108} + \sqrt{125} - \sqrt{75}\)= \(\sqrt{3 \times 36} + \sqrt{5 \times 25} - \sqrt{3 \times 25}\) = \(6 \sqrt{3} + 5 \sqrt{5} - 5 \sqrt{3}\) = \(\sqrt{3} + 5\sqrt{5}\) |
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| 2. |
Evaluate: \((64^{\frac{1}{2}} + 125^{\frac{1}{3}})^2\) A. 121 B. 144 C. 169 D. 196
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Detailed Solution\([64^{\frac{1}{2}} + 125^{\frac{1}{3}}]^2\) = \([\sqrt{64} + \sqrt[3] {125}]^2\)\([8 + 5]^2\) = \([13]^2\) = 169 |
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| 3. |
Given that y varies inversely as the square of x. If x = 3 when y = 100, find the equation connecting x and y. A. \(yx^2 = 300\) B. \(yx^2 = 900\) C. y = \(\frac{100x}{9}\) D. \(y = 900x^2\)
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Detailed SolutionY \(\alpha \frac{1}{x^2} \rightarrow y = \frac{k}{x^2}\)If x = 3 and y = 100, then, \(\frac{100}{1} = \frac{k}{3^2}\) \(\frac{100}{1} = \frac{k}{9}\) k = 100 x 9 = 900 Substitute 900 for k in y = \(\frac{k}{x^2}\); y = \(\frac{900}{x^2}\) = \(yx^2 = 900\) |
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| 4. |
Find the value of x for which \(32_{four} = 22_x\) A. three B. five C. six D. seven
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Detailed Solution\(32_4 = 22_x\)\(3 \times 4^1 + 2 \times 4^o\) = \(2 \times x^1 + 2 \times x^o\) 12 + 2 x 1 = 2x + 2 x 1 14 = 2x + 2 14 - 2 = 2x 12 = 2x x = \(\frac{12}{2}\) x = 6 |
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| 5. |
Simplify; 2\(\frac{1}{4} \times 3\frac{1}{2} \div 4 \frac{3}{8}\) A. \(\frac{5}{9}\) B. 1\(\frac{1}{5}\) C. 1\(\frac{1}{4}\) D. 1\(\frac{4}{5}\) |
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| 6. |
There are 250 boys and 150 girls in a school, if 60% of the boys and 40% of the girls play football, what percentage of the school play football? A. 40.0% B. 42.2% C. 50.0% D. 52.5%
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Detailed SolutionPopulation of school = 250 + 150 = 40060% of 250 = \(\frac{\text{60%}}{\text{100%}}\) x 250 = 150 40% of 150 = \(\frac{\text{40%}}{\text{100%}}\) x 150 = 60 Total number of students who plays football; 150 + 60 = 210 Percentage of school that play football; \(\frac{210}{400}\) x 100% = 52.5% |
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| 7. |
If \(\log_{10}\)(6x - 4) - \(\log_{10}\)2 = 1, solve for x. A. 2 B. 3 C. 4 D. 5
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Detailed Solution\(\log_{10}\)(6x - 4) - \(\log_{10}\)2 = 1\(\log_{10}\)(6x - 4) - \(\log_{10}\)2 = \(\log_{10}\)10 \(\log_{10}\)\(\frac{6x - 4}{2}\) - \(\log_{10}\)10 \(\frac{6x - 4}{2}\) = 10 6x - 4 = 2 x 10 = 20 6x = 20 + 4 6x = 20 x = \(\frac{24}{6}\) x = 4 |
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| 8. |
If F = \(\frac{9}{5}\)C + 32, find C when F = 98.6 A. 30 B. 37 C. 39 D. 41
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Detailed SolutionF = \(\frac{9}{5}\)C + 32When F = 98.6 98.6 = \(\frac{9}{5}\)C + 32 98.6 - 32 = \(\frac{9}{5}\)C 66.6= \(\frac{9}{5}\)C 66.6 x 5 = 9C C = \(\frac{66.6 \times 5}{9}\) = 37 |
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| 9. |
If y + 2x = 4 and y - 3x = -1, find the value of (x + y) A. 3 B. 2 C. 1 D. -1
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Detailed Solutiony + 2x = 4 .....(1)y - 3x = -1 ......(2) Subtract (2) from (1) 2x - (-3x) = 4 - (-1) 2x + 3x = 4 + 1 5x = 5 X = \(\frac{5}{5}\) = 1 Substitute 1 for x in (1); y + 2(1) = 4 y + 2 = 4 y = 4 - 2 = 2 Hence, (x + y) = (1 + 2) = 3 |
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| 10. |
If x : y : z = 3 : 3 : 4, evaluate \(\frac{9x + 3y}{6z - 2y}\) A. 1\(\frac{1}{2}\) B. 2 C. 2\(\frac{1}{2}\) D. 3
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Detailed SolutionIf x : y : z = 3 : 3 : 4, evaluate \(\frac{9x + 3y}{6x - 2y}\)\(\frac{x}{y}\) = \(\frac{2}{3}\) and \(\frac{y}{z}\) = \(\frac{3}{4}\) Thus; x = \(\frac{2}{3}T_1\) and z = \(\frac{3}{5}T_1\) y = \(\frac{3}{7}T_2\) and z = \(\frac{4}{7}T_2\) Using y = y \(\frac{3}{5}T_1\) = \(\frac{3}{7}T_2\); \(\frac{T_1}{T_2}\) = \(\frac{3}{7}\) x \(\frac{5}{3}\) \(\frac{T_1}{T_2}\) = \(\frac{15}{21}\) \(T_1\) = 15 and \(T_2\) = 21 Therefore; x = \(\frac{2}{5}\) x 15 = 6 y = \(\frac{3}{5}\) x 15 = 9 y = \(\frac{3}{7}\) x 21 = 9 (again) z = \(\frac{4}{7}\) x 21 = 12 Hence; \(\frac{9x + 3y}{6z - 2y}\) = \(\frac{9(6) + 3(9)}{6(12) - |
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