Paper 1  Objectives  49 Questions
WASSCE/WAEC MAY/JUNE
Year: 2016
Level: SHS
Time:
Type: Question Paper
Source: Nigeria
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#  Question  Ans 

1. 
If 23_{x} + 101_{x} = 130_{x}, find the value of x A. 7 B. 6 C. 5 D. 4
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Detailed Solution23_{x} + 101_{x} = 130_{x}2 x X^{1} + 3 x X^{o} + 1 x X^{2} + 0 x X^{1} + 1 x X^{o} = 1 x X^{o} = 1 x X^{2} + 3 x X^{1} + 0 x X^{o} = X^{2} + 3x + 0 2x + 3 = x^{2} + 0 + 1 + x^{2} + 3x 2x  3x + x^{2}  x^{2} = 3  1  x = 4 x = 4 

2. 
Simplify: (\(\frac{3}{4}  \frac{2}{3}\)) x 1\(\frac{1}{5}\) A. \(\frac{1}{60}\) B. \(\frac{5}{72}\) C. \(\frac{1}{10}\) D. 1\(\frac{7}{10}\)
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Detailed Solution(\(\frac{3}{4}  \frac{2}{3}\)) x 1\(\frac{1}{5}\)= (\(\frac{9  8}{12} \times \frac{6}{5}\)) = \(\frac{1}{12} \times \frac{6}{5}\) = \(\frac{1}{10}\) 

3. 
Simplify:(\(\frac{10\sqrt{3}}{\sqrt{5}}  \sqrt{15}\))^{2} A. 75.00 B. 15.00 C. 8.66 D. 3.87
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Detailed SolutionNote that \(\frac{10\sqrt{3}}{\sqrt{5}} = \frac{10\sqrt{3}}{\sqrt{5}} \times  \frac{\sqrt{5}}{\sqrt{5}}\)= \(\frac{10\sqrt{15}}{\sqrt{5}} = 2\sqrt{15}\) hence, (\(\frac{10\sqrt{3}}{\sqrt{5}}  \sqrt{15}\))^{2} = (\(2\sqrt{15}  \sqrt{15}\))^{2} = (\(2\sqrt{15}  \sqrt{15}\))(\(2\sqrt{15}  \sqrt{15}\)) = 4\(\sqrt{15 \times 15}  2\sqrt{15 \times 15}  2\sqrt{15 x 15} + \sqrt{15 \times 15}\) = 4 x 15  2 x 15  2 x 15 + 15 = 60  30  30 + 15 = 15 

4. 
The distance, d, through which a stone falls from rest varies directly as the square of the time, t, taken. If the stone falls 45cm in 3 seconds, how far will it fall in 6 seconds? A. 90cm B. 135cm C. 180cm D. 225cm
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Detailed Solutiond \(\alpha\) t^{2}d = t^{2} k where k is a constant. d = 45cm, when t = 3s; thus 45 = 3^{2} x t k = \(\frac{45}{9}\) = 5 thus equation connecting d and t is d = 5t^{2} when t = 6s, d = 5 x 6^{2} = 5 x 36 = 180cm 

5. 
Which of following is a valid conclusion from the premise. "Nigeria footballers are good footballers"? A. Joseph plays football in Nigeria therefore he is a good footballer B. Joseph is a good footballer therefore he is a Nigerian footballer C. Joseph is a Nigerian footballer therefore he is a good footballer D. Joseph plays good football therefore he is a Nigerian footballer
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Detailed Solution 

6. 
On a map, 1cm represent 5km. Find the area on the map that represents 100km^{2}. A. 2cm^{2} B. 4cm^{2} C. 8cm^{2} D. 8cm^{2}
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Detailed SolutionOn a map, 1cm represents 5km. Then it follows that 1cm^{2} represents 25km^{2}. Acm^{2} represents 100km^{2}. By apparent crossmultiplication, 1cm^{2} x 100km^{2} = Acm^{2}x 25km^{2}therefore A = \(\frac{100}{25}\) = 4cm^{2} 

7. 
Simplify; \(\frac{3^{n  1} \times 27^{n + 1}}{81^{n}}\) A. 3^{2n} B. 9 C. 3^{n} D. 3^{n + 1}
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Detailed Solution\(\frac{3^{n  1} \times 27^{n + 1}}{81^{n}}\)= \(\frac{3^{n  1} \times 3^{3(n + 1)}}{3^{4n}}\) = 3\(^{n  1 + 3n + 3  4n}\) = 3\(^{4n  4n  1 + 3}\) = 3^{2} = 9 

8. 
What sum of money will amount to D10,400 in 5 years at 6% simple interest? A. D8,000.00 B. D10,000.00 C. D12,000.00 D. D16,000.00
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Detailed SolutionA = P + 1I = A  P = 10,400  P Now using I = \(\frac{P \times T \times R}{100}\) i.e. 10,400  P = \(\frac{P \times 5 \times 6}{100}\) = 100(10,400  P) = 30P 10(10,400  P) = 3P 104,000  10P = 3P 104,000  10P = 3P 104,000 = 3P + 10P = 104,000 = 13P P = \(\frac{104,000}{100}\) P = D8,000 

9. 
The roots of a quadratic equation are \(\frac{4}{3}\) and \(\frac{3}{7}\). Find the equation A. 21x^{2}  19x  12 = 0 B. 21x^{2} + 37x  12 = 0 C. 21x^{2}  x + 12 = 0 D. 21x^{2} + 7x  4 = 0
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Detailed SolutionLet x = \(\frac{4}{3}\), x = \(\frac{3}{7}\)Then 3x = 4, 7x = 3 3x  4 = 0, 7x + 3 = 0 (3x  4)(7x + 3) = 0 21x^{2} + 9x  28x  12 = 0 21x^{2}  19x  12 = 0 

10. 
Find the values of y for which the expression \(\frac{y^2  9y + 18}{y^2 + 4y  21}\) is undefined A. 6, 7 B. 3, 6 C. 3, 7 D. 3, 7
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Detailed Solution\(\frac{y^2  9y + 18}{y^2 + 4y  21}\)Factorize the denominator; Y^{2} + 7y  3y  21 = y(y + 7) 3 (y + 7) = (y  3)(y + 7) Hence the expression \(\frac{y^2  9y + 18}{y^2 + 4y  21}\) is undefined when y^{2} + 4y  21 = 0 ie. y = 3 or 7 
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