Mathematics (Core)
Paper 1 | Objectives | 49 Questions
WASSCE/WAEC MAY/JUNE
Year: 2014
Level: SHS
Time:
Type: Question Paper
Answers provided
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Paper 1 | Objectives | 49 Questions
WASSCE/WAEC MAY/JUNE
Year: 2014
Level: SHS
Time:
Type: Question Paper
Answers provided
No description provided
This paper is yet to be rated
Past questions are effective for revisions for all tests including WAEC, BECE, SAT, TOEFL, GCSE, IELTS
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| # | 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\(\frac{11}{15}\) |
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| 2. |
If 23x = 325, find the value of x A. 7 B. 6 C. 5 D. 4
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Detailed Solution23x = 3252 \(\times x^1 + 3 \times x^0 = 3 \times 5^1 + 2 \times 5^0\) = 2x + 3 = 15 + 2 2x + 3 = 17 2x = 17 - 3 2x = 14 x = \(\frac{14}{2}\) x = 7 |
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| 3. |
The volume of a cube is 512cm3. 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 L512cm3 = L3 L3 = 512cm3 L = 3\(\sqrt{512}\) L (512)\(\frac{1}{3}\) = (29)\(\frac{1}{3}\) 23 = 8cm |
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| 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) = \(\frac{11}{20}\) |
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| 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\(\sqrt{3} \times \sqrt{3} (4 - 1) 2\sqrt{9}(3) = 2 \times 3 \times 3 = 18\) |
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| 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 |
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| 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 |
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| 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} |
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| 9. |
Find the truth set of the equation x2 = 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 Solutionx2 = 3(2x + 9)x2 = 6x + 27 x2 - 6x - 27 = 0 x2 - 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 |
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| 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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