Paper 1  Objectives  48 Questions
WASSCE/WAEC MAY/JUNE
Year: 2013
Level: SHS
Time:
Type: Question Paper
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#  Question  Ans 

1. 
Multiply 2.7 x 10^{4} by 6.3 x 10^{6} and leave your answers in standard form A. 1.7 x 10^{3} B. 1.70 x 10^{3} C. 1.701 x 10^{3} D. 17.01 x 10^{3}
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Detailed Solution2.7 x 10^{4} x 6.3 x 10^{6}= 2.7 x 6.3 x 10^{4} x 10^{6} = 17.01 x 10^{4 + 6} = 17.01 x 10^{2} = 1.701 x 10^{1} x 10^{2} = 1.701 x 10^{1 + 2} = 1.701 x 10^{3} 

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) = 33^{2(2  x)} = 3 2(2  x) = 1 4  2x = 1 2x = 1  4 2x = 3 x = \(\frac{3}{2}\) x = \(\frac{3}{2}\) 

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 \(6\frac{3}{4} \div \frac{3}{4}\) = \(\frac{27}{4} \div \frac{3}{4}\) = \(\frac{27}{4} \times \frac{4}{3}\) = \(\frac{9 \times 1}{1 \times 1} = \frac{9}{1}\) \(\frac{9}{2} \div \frac{9}{1} = \frac{9}{2} \times \frac{1}{9}\) = \(\frac{1}{2}\) 

5. 
If Un = n(n^{2} + 1), evaluate U5  U4 A. 18 B. 56 C. 62 D. 80
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Detailed SolutionUn = n(n^{2} + 1)U^{5} = 5(^{2} + 1) = 5(25 + 1) = 5(26) = 130 U^{4} = 4(4^{2} + 1) = 4(16 + 1) = 4(17) = 68 U5  U4 = 130  68 = 62 

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 < 

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\(\frac{5}{9}\)% 

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) 

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\(\frac{x^2  y^2}{(x + y)^2} + \frac{(x  y)^2}{(3x + 3y)}\)\(\frac{(x + y)(x  y)}{(x + y)(x + y)} + \frac{(x  y)(x  y)}{3(x + y)}\) = \(\frac{3}{x  y}\) 

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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