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
Five revision strategies that are strange but effective for memorization for high exams performance
Get full scholarship paid tuition from 5 countries with free education in 2022 for international students
Past questions are effective for revisions for all tests including WAEC, BECE, SAT, TOEFL, GCSE, IELTS
# | 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\(\frac{11}{15}\)
Show Content
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 23x = 325, find the value of x A. 7 B. 6 C. 5 D. 4
Show Content
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 |
|
3. |
The volume of a cube is 512cm3. Find the length of its side A. 6cm B. 7cm C. 8cm D. 9cm
Show Content
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 |
|
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. \(\frac{5}{19}\)
Show Content
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
Show Content
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
Show Content
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
Show Content
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}
Show Content
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 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}
Show Content
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 |
|
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\(sqrt{5}\)
Show Content
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\(\sqrt{5}\) |
Preview displays only 10 out of the 49 Questions