Mathematics (Core)
Paper 1 | Objectives | 45 Questions
WASSCE/WAEC MAY/JUNE
Year: 2008
Level: SHS
Time:
Type: Question Paper
Answers provided
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Paper 1 | Objectives | 45 Questions
WASSCE/WAEC MAY/JUNE
Year: 2008
Level: SHS
Time:
Type: Question Paper
Answers provided
No description provided
This paper is yet to be rated
Make sure you study hard but not into the late-night hours to give your body the enough rest you need.
A guide to passing your exams without studying hard, how to pass any test and top your class
12 Most Effective Ways to Pass Any Exams Without Studying Hard even when you don't have enough time.
| # | Question | Ans |
|---|---|---|
| 1. |
If x% of 240 equals 12, find x A. x = 1 B. x = 3 C. x = 5 D. x = 7
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Detailed Solutionx% of 240 = 12\(\frac{x}{100} \times 240 = 12\) x = \(\frac{12 \times 100}{240}\) x = 5 |
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| 2. |
Evaluate \(\frac{(3.2)^2 - (4.8)^2}{3.2 + 4.8}\) A. -0.08 B. -1.60 C. -10.24 D. -12.80
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Detailed Solution\(\frac{(3.2)^2 - (4.8)^2}{3.2 + 1.8} = \frac{(3.2 - 4.8)(3.2 + 4.8)}{(3.2 + 4.8)}\)= 3.2 - 4.8 = -1.60 |
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| 3. |
Simplify \(\sqrt{50} + \frac{10}{\sqrt{2}}\) A. 10 B. 10\(\sqrt{2}\) C. 20 D. 20\(\sqrt{2}\)
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Detailed Solution\(\sqrt{50} + \frac{10}}{\sqrt{2}} = \(\frac{\sqrt{50}}{1} + \sqrt{10}{\sqrt{2}}\)= \(\frac{\sqrt{50 \times 2} + 10}{\sqrt{2}}\) = \(\frac{\sqrt{100} + 10}{\sqrt{2}}\) = \(\frac{10 + 10}{\sqrt{2} = \frac{20}{\sqrt{2}}\) = \(\frac{20}{\sqrt{2}}\) \times \frac{\sqrt{2}}{\sqrt{2}}\) = \(\frac{20\sqrt{2}}{2}\) = 10\(\sqrt{2}\) |
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| 4. |
P naira invested for 4 years invested for 4 years at r% simple interest per annum yields 0.36 p naira interest. Find the value of r A. 1\(\frac{1}{9}\) B. 1\(\frac{4}{9}\) C. 9 D. 11
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Detailed SolutionI = \(\frac{PRT}{100}\)where r = r% p.a; I = 0.36p 0.36p = \(\frac{P \times r \times 4}{100}\) \(\frac{0.36 \times 100}{4}\) = r r = 9 |
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| 5. |
A trader bought 100 tubers at 5 for N350.00. She sold them in sets of 4 for N290.00. Find her gain percent. A. 3.6% B. 3.5% C. 3.5% D. 2.55
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Detailed SolutionCost price, c.p = \(\frac{100}{5}\) x N350 = N7000Selling price, s.p = \(\frac{100}{5}\) x N290 = N7250 %Gain = \(\frac{S.p - C.p}{C.p}\) x 100% = \(\frac{7250 - 7000}{7000}\) x 100% = \(\frac{250 \times 100}{7000}\) = 3.6% (approx.) |
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| 6. |
If p-2g + 1 = g + 3p and p - 2 = 0, find g A. -2 B. -1 C. 1 D. 2
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Detailed Solutionp - 2g + 1 = g + 3p.........(1)p - 2 = 0 .........(2) From (2), p = 2; put p = 2 into (1); 2 - 2g + 1 = g + 3(2) 3 - 2g = g + 6 -2g - g = 6 - 3 -3g = 3 g = \(\frac{3}{-3}\) g = -1 |
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| 7. |
Simplify \(\frac{\frac{1}{x} + \frac{1}{y}}{x + y}\) A. \(\frac{1}{x + y}\) B. \(\frac{1}{xy}\) C. x + y D. xy
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Detailed Solution\(\frac{\frac{1}{x} + \frac{1}{y}}{x + y}\) = \(\frac{\frac{y + x}{xy}}{x + y}\)= \(\frac{x + y}{xy}\) = \(\frac{x + y}{xy} \times \frac{1}{x + y}\) = \(\frac{1}{xy}\) |
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| 8. |
Simplify 3\(\sqrt{27x^3y^9}\) A. 9xy3 B. 3xy6 C. 3xy3 D. 9y3
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Detailed Solution3\(\sqrt{27x^3y^9}\) = 3\(\sqrt{27} \times 3\sqrt{3^3} \times 3\sqrt{y^9}\)= 3 \(\times x \times y^3\) = 3xy3 |
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| 9. |
Given that x = 2 and y = -\(\frac{1}{4}\), evaluate \(\frac{x^2y - 2xy}{5}\) A. zero B. \(\frac{1}{5}\) C. 1 D. 2
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Detailed SolutionGiven; x = 2; y = \(\frac{-1}{4}\)= \(\frac{x^2y - 2xy}{5}\) = \(\frac{2^2(\frac{-1}{4}) - 2(2)(\frac{-1}{4})}{5}\) = \(\frac{4(\frac{-1}{4}) + 4(\frac{-1}{4})}{5}\) = \(\frac{1 + 1}{5}\) = \(\frac{0}{5}\) = 0 |
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