Mathematics (Core)
Paper 1 | Objectives | 48 Questions
WASSCE/WAEC MAY/JUNE
Year: 2013
Level: SHS
Time:
Type: Question Paper
Answers provided
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Paper 1 | Objectives | 48 Questions
WASSCE/WAEC MAY/JUNE
Year: 2013
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. |
Multiply 2.7 x 10-4 by 6.3 x 106 and leave your answers in standard form A. 1.7 x 103 B. 1.70 x 103 C. 1.701 x 103 D. 17.01 x 103
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Detailed Solution2.7 x 10-4 x 6.3 x 106= 2.7 x 6.3 x 10-4 x 106 = 17.01 x 10-4 + 6 = 17.01 x 102 = 1.701 x 101 x 102 = 1.701 x 101 + 2 = 1.701 x 103 |
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| 2. |
If 9(2 - x) = 3, find x A. 1 B. \(\frac{3}{2}\) C. 2 D. \(\frac{5}{2}\)
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Detailed Solution9(2 - x) = 332(2 - x) = 3 2(2 - x) = 1 4 - 2x = 1 -2x = 1 - 4 -2x = -3 x = \(\frac{-3}{-2}\) x = \(\frac{3}{2}\) |
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| 3. |
In what number base is the addition 465 + 24 + 225 = 1050? A. ten B. nine C. eight D. seven |
D |
| 4. |
Simplify \(\frac{1\frac{7}{8} \times 2\frac{2}{5}}{6\frac{3}{4} \div \frac{3}{4}}\) A. 9 B. 4\(\frac{1}{2}\) C. 2 D. \(\frac{1}{2}\)
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Detailed Solution\(\frac{1\frac{7}{8} \times 2\frac{2}{5}}{6\frac{3}{4} \div \frac{3}{4}}\)from numerator \(1 \frac{7}{8} \times 2 \frac{2}{5}\) = \(\frac{15}{8} \times \frac{12}{5}\) = \(\frac{3 \times 3}{2 \times 1} = \frac{9}{2}\) from denominator \(6\frac{3}{4} \div \frac{3}{4}\) = \(\frac{27}{4} \div \frac{3}{4}\) = \(\frac{27}{4} \times \frac{4}{3}\) = \(\frac{9 \times 1}{1 \times 1} = \frac{9}{1}\) \(\frac{9}{2} \div \frac{9}{1} = \frac{9}{2} \times \frac{1}{9}\) = \(\frac{1}{2}\) |
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| 5. |
If Un = n(n2 + 1), evaluate U5 - U4 A. 18 B. 56 C. 62 D. 80
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Detailed SolutionUn = n(n2 + 1)U5 = 5(2 + 1) = 5(25 + 1) = 5(26) = 130 U4 = 4(42 + 1) = 4(16 + 1) = 4(17) = 68 U5 - U4 = 130 - 68 = 62 |
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| 6. |
If \(\sqrt{50} - K\sqrt{8} = \frac{2}{\sqrt{2}}\), find K A. -2 B. -1 C. 1 D. 2
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Detailed Solution\(\sqrt{50} - K\sqrt{8} = \frac{2}{\sqrt{2}}\)\(\sqrt{50} - \frac{2}{\sqrt{2}}\) = K\(\sqrt{8}\) = \(\sqrt{2} \times 25 - \frac{2}{\sqrt{2}}\) = K \(\sqrt{4 \times 2}\) \(\frac{5\sqrt{2}}{1} - \frac{2}{\sqrt{2}}\) = 2K\(\sqrt{2}\) \(\frac{5\sqrt{4} - 2}{\sqrt{2}} = 2K\sqrt{2}\) \(\frac{10 - 2}{\sqrt{2}} = 2K \sqrt{2}\) \(\frac{8}{\sqrt{2}} = \frac{2K\sqrt{2}}{1}\) = 2k\(\sqrt{2} \times \sqrt{2}\) = 8 2k \(\sqrt{4}\) = 8 2k x 2 = 8 4k = 8 k = \(\frac{8}{4}\) k = 2 < |
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| 7. |
A sales boy gave a change of N68 instead of N72. Calculate his percentage error A. 4% B. 5\(\frac{5}{9}\)% C. 5\(\frac{15}{17}\)% D. 7%
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Detailed Solution% error = \(\frac{error}{\text{actual value}} \times 100\)error = N72 - N68 = 4 actual value = N72 %error = \(\frac{4}{72} \times 100\) = \(\frac{100}{18} = \frac{50}{9}\) = 5\(\frac{5}{9}\)% |
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| 8. |
Four oranges sell for Nx and three mangoes sell for Ny. Olu bought 24 oranges and 12 mangoes. How much did he pay in terms of x and y? A. N94x + 6y) B. N(6x + 4y) C. N(24x + 12y) D. N(12x + 24y)
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Detailed Solution4 oranges sell for Nx, 1 orange will sell for \(\frac{Nx}{4}\)24 oranges will sell for: \(\frac{Nx}{4} \times 24\) = n6x 3 mangoes sell for Ny, 1 mango will sell for \(\frac{Ny}{3}\) 12 mangoes will sell for \(\frac{Ny}{3} \times 12\) = 4Ny total money pay N6x + N4y = N(6x + 4y) |
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| 9. |
Simplify: \(\frac{x^2 - y^2}{(x + y)^2} + \frac{(x - y)^2}{(3x + 3y)}\) A. \(\frac{x - y}{3}\) B. x + y C. \(\frac{3}{x - y}\) D. x - y
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Detailed Solution\(\frac{x^2 - y^2}{(x + y)^2} + \frac{(x - y)^2}{(3x + 3y)}\)\(\frac{(x + y)(x - y)}{(x + y)(x + y)} + \frac{(x - y)(x - y)}{3(x + y)}\) = \(\frac{3}{x - y}\) |
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| 10. |
Solve the inequality: \(\frac{2x - 5}{2} < (2 - x)\) A. x > 0 B. x < \(\frac{1}{4}\) C. x > 2\(\frac{1}{2}\) D. x < 2\(\frac{1}{4}\)
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Detailed Solution\(\frac{2x - 5}{2} < \frac{(2 - x)}{1}\)2x - 5 < 4 - 2x 2x + 2x < 4 + 5 4x < 9 x < \(\frac{9}{4}\) x < 2\(\frac{1}{4}\) |
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