Year :
2008
Title :
Mathematics (Core)
Exam :
WASSCE/WAEC MAY/JUNE
Paper 1 | Objectives
1 - 10 of 45 Questions
| # | Question | Ans |
|---|---|---|
| 1. |
If x% of 240 equals 12, find x A. x = 1 B. x = 3 C. x = 5 D. x = 7 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 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\(\sqrt{2}\) 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 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 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 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 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 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 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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| 10. |
Factorize 5y2 + 2ay - 3a2 A. (a - y)(5y - 3a) B. (y - a)(5y - 3a) C. (y - a)(5y + 3a) D. (y + a)(5y - 3a) Detailed Solution5y2 + 2ay - 3a2 = 5y2 + 5ay - 3a2= 5y(y + a) - 3a(y + a) = (y + a)(5y - 3a) |
| 1. |
If x% of 240 equals 12, find x A. x = 1 B. x = 3 C. x = 5 D. x = 7 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 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\(\sqrt{2}\) 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 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 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.) |
| 6. |
If p-2g + 1 = g + 3p and p - 2 = 0, find g A. -2 B. -1 C. 1 D. 2 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 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 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 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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| 10. |
Factorize 5y2 + 2ay - 3a2 A. (a - y)(5y - 3a) B. (y - a)(5y - 3a) C. (y - a)(5y + 3a) D. (y + a)(5y - 3a) Detailed Solution5y2 + 2ay - 3a2 = 5y2 + 5ay - 3a2= 5y(y + a) - 3a(y + a) = (y + a)(5y - 3a) |