Paper 1  Objectives  49 Questions
WASSCE/WAEC MAY/JUNE
Year: 2014
Level: SHS
Time:
Type: Question Paper
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These are the three things to do in test preparation for higher grades and excellent performance.
There's no secret to passing WAEC or BECE apart from the merit of hard work and adequate preparation
The goal of mocks tests is to create a benchmarking tool to help students assess their performances.
#  Question  Ans 

1. 
Simplify 10\(\frac{2}{5}  6 \frac{2}{3} + 3\) A. 6\(\frac{4}{15}\) B. 6\(\frac{11}{15}\) C. 7\(\frac{4}{15}\) D. 7\(\frac{11}{15}\)
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Detailed Solution10\(\frac{2}{5}  6 \frac{2}{3} + 3\)\(\frac{52}{5}  \frac{20}{3} + \frac{3}{1}\) = \(\frac{156  100 + 45}{15}\) \(\frac{156 + 45  100}{15}\) = \(\frac{201  100}{15}\) = \(\frac{101}{15}\) = 6\(\frac{11}{15}\) 

2. 
If 23_{x} = 32_{5}, find the value of x A. 7 B. 6 C. 5 D. 4
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Detailed Solution23_{x} = 32_{5}2 \(\times x^1 + 3 \times x^0 = 3 \times 5^1 + 2 \times 5^0\) = 2x + 3 = 15 + 2 2x + 3 = 17 2x = 17  3 2x = 14 x = \(\frac{14}{2}\) x = 7 

3. 
The volume of a cube is 512cm^{3}. Find the length of its side A. 6cm B. 7cm C. 8cm D. 9cm
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Detailed Solutionvolume of cube = L x L x L512cm^{3} = L^{3} L^{3} = 512cm^{3} L = 3\(\sqrt{512}\) L (512)^{\(\frac{1}{3}\)} = (2^{9})^{\(\frac{1}{3}\)} 2^{3} = 8cm 

4. 
If one student is selected at random, find the probability that he/she scored at most 2 marks A. \(\frac{11}{18}\) B. \(\frac{11}{20}\) C. \(\frac{7}{22}\) D. \(\frac{5}{19}\)
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Detailed Solutionat most 2 marks = 5 + 2 + 4 students = 11 studentsprobability(at most 2 marks) = \(\frac{11}{20}\) 

5. 
Simplify: \(\sqrt{12} ( \sqrt{48}  \sqrt{3}\)) A. 18 B. 16 C. 14 D. 12
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Detailed Solution\(\sqrt{12} ( \sqrt{48}  \sqrt{3}\))\(\sqrt{4 \times 3} (6 \times 3  \sqrt{3}) = 2 \sqrt{3}(4 \sqrt{3}  \sqrt{3})\) = 2\(\sqrt{3} \times \sqrt{3} (4  1) 2\sqrt{9}(3) = 2 \times 3 \times 3 = 18\) 

6. 
Given that x > y and 3 < y, which of the following is/are true? i. y > 3 ii. x < 3 iii. x > y > 3 A. i B. i and ii C. i and iii D. i, ii and iii
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Detailed Solutionx > y and 3 < y; then 3 < y means that y > 3x > 3 to give the possible x > y > 3 

7. 
Three quarters of a number added to two and a half of the number gives 13. Find the number A. 4 B. 5 C. 6 D. 7
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Detailed Solutionlet the number be x2\(\frac{1}{2}x + \frac{3}{4}x = 13\) \(\frac{5}{2}x + \frac{3}{4}x = 13\) multiply through by 4 4(\(\frac{5}{2}\))x + 4(\(\frac{3}{4}\))x = 13 x 4 2(5x) + 3x = 52 10x + 3x = 52 13x = 52 x = \(\frac{52}{13}\) x = 4 

8. 
If x = {0, 2, 4, 6}, y = {1, 2, 3, 4} and z = {1, 3} are subsets of u = {x:0 \(\geq\) x \(\geq\) 6}, find x \(\cap\) (Y' \(\cup\) Z) A. {0, 2, 6} B. {1, 3} C. {0, 6) D. {9}
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Detailed Solutionx = {0, 2, 4, 6}; y = {1, 2, 3, 4}; z = {1, 3}u = {0, 1, 2, 3, 4, 5, 6} y' = {0, 5, 6} to find x \(\cap\) (Y' \(\cup\) Z) first find y' \(\cup\) z = {0, 1, 3, 5, 6} then x \(\cap\) (Y' \(\cup\) Z) = {0, 6} 

9. 
Find the truth set of the equation x^{2} = 3(2x + 9) A. {x : x = 3, x = 9} B. {x : x = 3, x = 9} C. {x : x = 3, x = 9} D. {x : x = 3, x = 9}
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Detailed Solutionx^{2} = 3(2x + 9)x^{2} = 6x + 27 x^{2}  6x  27 = 0 x^{2}  9x + 3x  27 = 0 x(x  9) + 3(x  9) = (x + 3)(x  9) = 0 x + 3 = 0 or x  9 = 0 x = 3 or x = 9 x = 3, x = 9 

10. 
The coordinates of points P and Q are (4, 3) and (2, 1) respectively. Find the shortest distance between P and Q. A. 10\(sqrt{2}\) B. 4\(sqrt{5}\) C. 5\(sqrt{2}\) D. 2\(sqrt{5}\)
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Detailed Solutionp(4, 3) Q(2  1)distance = \(\sqrt{(x_2  x_1)^2 + (Y_2  y_1)^2}\) = \(\sqrt{(2  4)^2 + (1  3)^2}\) = \(\sqrt{(2)^2 = (4)^2}\) = \(\sqrt{4 + 16}\) = \(\sqrt{20}\) = \(\sqrt{4 \times 5}\) = 2\(\sqrt{5}\) 
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