Mathematics (Core)
Paper 1 | Objectives | 49 Questions
WASSCE/WAEC MAY/JUNE
Year: 2017
Level: SHS
Time:
Type: Question Paper
Answers provided
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Paper 1 | Objectives | 49 Questions
WASSCE/WAEC MAY/JUNE
Year: 2017
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. |
Express 0.0000407, correct to 2 significant figures A. 0.0 B. 0.00004 C. 0.000041 D. 0.0000407
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Detailed Solution0.0000407 to 2 s.f0.000041 (2 s.f) |
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| 2. |
If x varies inversely as y and y varies directly as z, what is the relationship between x and z? A. x \(\alpha\) z B. x \(\alpha\) \(\frac{1}{z}\) C. a \(\alpha\) z\(^2\) D. x \(\alpha\) \(\frac{1}{z^2}\)
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Detailed Solution\(x \propto \frac{1}{y}\), y \(\propto\) zx = \(\frac{k}{y}\) y = mz Since y = mz, x = \(\frac{k}{mz}\), where k and m are constants. Hence, x \(\propto\) \(\frac{1}{z}\) |
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| 3. |
Evaluate \(\frac{3\frac{1}{4} \times 1\frac{3}{5}}{11\frac{1}{3} - 5 \frac{1}{3}}\) A. \(\frac{14}{15}\) B. \(\frac{13}{15}\) C. \(\frac{4}{5}\) D. \(\frac{11}{15}\)
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Detailed Solution\(\frac{3\frac{1}{4} \times 1\frac{3}{5}}{11\frac{1}{3} - 5 \frac{1}{3}}\) = \(\frac{\frac{26}{5}}{\frac{18}{3}}\) = \(\frac{26}{5} \div \frac{18}{3}\)= \(\frac{13}{15}\) |
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| 4. |
The ages of Tunde and Ola are in the ratio 1:2. If the ratio of Ola's age to Musa's age is 4:5, what is the ratio of Tunde's age to Musa's age? A. 1 : 4 B. 1 : 5 C. 2 : 5 D. 5 : 2
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Detailed SolutionTunde: Ola \(\to\) 1 : 2 ; Ola; Musa \(\to\) 4 : 5\(\frac{1}{2}\) x \(\frac{4}{5}\) = \(\frac{2}{5}\) |
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| 5. |
If M = {x : 3 \(\leq\) x < 8} and N = {x : 8 < x \(\leq\) 12}, which of the following is true?i. 8 \(\in\) M \(\cap\) Nii. 8 \(\in\) M \(\cup\) Niii. M \(\cap\) N = \(\varnothing\) A. iii only B. i and ii C. ii and iii only D. i, ii and iii
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Detailed SolutionM = {3, 4, 5, 6, 7,}, N = {9, 10, 11, 12} |
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| 6. |
Given that a = log 7 and b = \(\log\) 2, express log 35 in terms of a and b. A. a + b + 1 B. ab - 1 C. a - b + 1 D. b - a + 1
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Detailed Solution\(\frac{\log 7 \times \log 10}{\log 2}\)log 7 x log 10 \(\div\) log 2 a + 1 - b a - b + 1 |
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| 7. |
If x = \(\frac{2}{3}\) and y = - 6, evaluate xy - \(\frac{y}{x}\) A. 0 B. 5 C. 8 D. 9
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Detailed Solutionx = \(\frac{2}{3}\) and y = - 6xy - \(\frac{y}{x}\) \(\frac{2}{3} - (6)^2 - (-6) \div \frac{2}{3}\) = -4 - (6) x \(\frac{3}{2}\) = -4 - (-6) x \(\frac{3}{2}\) = -4 - (-9) = -4 + 9 = 5 |
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| 8. |
Solve the equation: \(\frac{1}{5x} + \frac{1}{x}\)= 3 A. \(\frac{1}{5}\) B. \(\frac{2}{5}\) C. \(\frac{3}{5}\) D. \(\frac{4}{5}\)
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Detailed Solution\(\frac{1}{5x} + \frac{1}{x}\)= 3\(\frac{1 + 5}{5x}\) = 3 6 = 15x x = \(\frac{6}{15}\) = \(\frac{2}{5}\) |
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| 9. |
A sum of N18,100 was shared among 5 boys and 4 girls with each boy taking N20.00 more than each girl. Find a boy's share. A. N1,820.00 B. 2,000.00 C. N2,020.00 D. N2,040.00
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Detailed SolutionLet a girl's share = x + 204x + 5(x + 20) = 18,100 4x + 5x + 100 = 18,100 9x + 100 = 18,100 9x = 18,000 x = \(\frac{18,000}{9}\) x = 2,000 \(\therefore\) Each boy gets N(2,000 + 20) = N2,020. |
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| 10. |
One factor of \(7x^2 + 33x - 10\) is A. 7x + 5 B. x - 2 C. 7x - 2 D. x - 5
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Detailed Solution\(7x^2 + 33x - 10\)\(7x^2 + 35x - 2x - 10\) 7x (x + 5) - 2 (x + 5) (7x - 2) (x + 5) |
Preview displays only 10 out of the 49 Questions