Paper 1  Objectives  45 Questions
WASSCE/WAEC MAY/JUNE
Year: 2015
Level: SHS
Time:
Type: Question Paper
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#  Question  Ans 

1. 
If {X: 2 d x d 19; X integer} and 7 + x = 4 (mod 9), find the highest value of x A. 2 B. 5 C. 15 D. 18 
C 
2. 
The sum 11011_{2}, 1111_{2} and 10_{m}10_{n}0_{2}. Find the value of m and n. A. m = 0, n = 0 B. m = 1, n = 0 C. m = 0, n = 1 D. m = 1, n = 1 
C 
3. 
A trader bought an engine for $15,000.00 outside Nigeria. If the exchange rate is $0.075 to N1.00, how much did the engine cost in Naira? A. N250,000.00 B. N200,000.00 C. N150,000.00 D. N100,000.00
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Detailed SolutionN1.00 = $0.075N X = $15,000 X = \(\frac{1.00 \times 15000}{0.075}\) = N200,000.00 

4. 
If \(\frac{27^x \times 3^{1  x}}{9^{2x}} = 1\), find the value of x. A. 1 B. \(\frac{1}{2}\) C. \(\frac{1}{2}\) D. 1
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Detailed Solution\(\frac{27^x \times 3^{1  x}}{9^{2x}} = 1\)\(\frac{3^{3x} \times 3^{1  x}}{3^{2(2  x)}} = 3^0\) \(3^{3x} \times 3^{1  x} \div 3^{4x} = 3^0\) \(3^{(3x + 1  x  4x)} = 3^0\) \(3^{(1  2x)} = 3^0\) since the bases are equal, 1  2x = 0  2x = 1 x = \(\frac{1}{2}\) 

5. 
Find the 7^{th} term of the sequence: 2, 5, 10, 17, 6,... A. 37 B. 48 C. 50 D. 63 
C 
6. 
Given that log_{x} 64 = 3, evaluate x log_{8} A. 6 B. 9 C. 12 D. 24
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Detailed SolutionIf log_{x} 64 = 3, then 64 = x^{3}4^{3} = x^{3} Since the indices are equal, x = 4 Hence, x log_{2}8 = log_{2}8^{x} = log_{2}8^{4} = log_{2}(2^{3})_{4} = log_{2}2^{12} = 1 log_{2} = 1(1) = 12 

7. 
If 2^{n} = y, Find 2\(^{(2 + \frac{n}{3})}\) A. 4y\(^\frac{1}{3}\) B. 4y\(^3\) C. 2y\(^\frac{1}{3}\) D. 2y\(^3\)
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Detailed SolutionIf 2^{n} = y,then, 2\(^{(2 + \frac{n}{3})}\) = 2^{2} x 2\(^\frac{n}{3}\) = 4 x (2^{n})\(^{\frac{1}{3}}\) But y = 2^{n}, hence 2\(^{(2 + \frac{n}{3})}\) = 4 x y\(^{\frac{1}{3}}\) = 4y\(^\frac{1}{3}\) 

8. 
Factorize completely: 6ax  12by  9ay + 8bx A. (2a  3b)(4x + 3y) B. (3a + 4b)(2x  3y) C. (3a  4b)(2x + 3y) D. (2a + 3b)(4x 3y)
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Detailed Solution6ax  12by  9ay + 8bx= 6ax  9ay + 8bx  12by = 3a(2x  3y) + 4b(2x  3y) = (3a + 4b)(2x  3y) 

9. 
Find the equation whose roots are \(\frac{3}{4}\) and 4 A. 4x^{2}  13x + 12 = 0 B. 4x^{2}  13x  12 = 0 C. 4x^{2} + 13x  12 = 0 D. 4x^{2} + 13x + 12 = 0
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Detailed SolutionLet x = \(\frac{3}{4}\) or x = 4i.e. 4x = 3 or x = 4 (4x  3)(x + 4) = 0 therefore, 4x^{2} + 13x  12 = 0 
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