Year :
2018
Title :
Mathematics (Core)
Exam :
WASSCE/WAEC MAY/JUNE
Paper 1 | Objectives
11 - 20 of 50 Questions
| # | Question | Ans |
|---|---|---|
| 11. |
Simplify; \(\frac{2 - 18m^2}{1 + 3m}\) A. \(2 (1 + 3m)\) B. \(2 (1 + 3m^2)\) C. \(2(1 - 3m)\) D. \(2(1 - 3m^2)\) Detailed Solution\(\frac{2 - 18m^2}{1 + 3m}\) = \(\frac{2(1 - 9)m^2}{1 + 3m}\)= \(\frac{2(1 + 3m)(1 - 3m)}{1 + 3m}\) = \(2(1 - 3m)\) |
|
| 12. |
A curve is such that when y = 0, x = -2 or x = 3. Find the equation of the curve. A. y = \(x^2 - 5x - 6\) B. y = \(x^2 + 5x - 6\) C. y = \(x^2 + x - 6\) D. y = \(x^2 - x - 6\) |
|
| 13. |
The volume of a cylindrical tank, 10m high is 385 m\(^2\). Find the diameter of the tank. [Take \(\pi = \frac{22}{7}\)] A. 14m B. 10m C. 7m D. 5m Detailed SolutionVolume of a cylinder = \( \pi r^2\)h385 = \(\frac{22}{7}\) x \(r^2\) x 10 385 x 7 = 22 x \(r^2\) x 10 \(r^2\) = \(\frac{385 \times 7}{22 \times 10}\) = 12.25 r = \(\sqrt{12.25}\) = 3.5m Hence, diameter of tank = 2r = 2 x 3.5 = 7m |
|
| 14. |
The surface area of a sphere is \(\frac{792}{7} cm^2\). Find, correct to the nearest whole number, its volume. [Take \(\pi = \frac{22}{7}\)] A. 113\(cm^3\) B. 131\(cm^3\) C. 311\(cm^3\) D. 414\(cm^3\) Detailed SolutionSurface area of a sphere = \(4 \pi r^2\)\(4 \pi r^2\) = \(\frac{792}{7}cm^2\) 4 x \(\frac{22}{7}\) x \(r^2\) = \(\frac{792}{7}\) \(r^2\) = \(\frac{792}{7}\) x \(\frac{7}{4 \times 22}\) = 9 r = \(\sqrt{9}\) = 3cm Hence, volume of sphere = \(\frac{4}{3} \pi r^3\) = \(\frac{4}{3} \times \frac{22}{7} \times 3 \times 3 \times 3 \) = \(\frac{4 \times 22 \times 9}{7}\) \(\approx\) = 113.143 = 113\(cm^3\) (to the nearest whole number) |
|
| 15. |
The angles of a polygon are x, 2x, 2x, (x + \(30^o\)), (x + \(20^o\)) and (x - \(10^o\)). Find the value of x A. \(45^o\) B. \(95^o\) C. \(84^o\) D. \(85^o\) Detailed Solutionx + 2x + 2x + (x + \(30^o\)) + (x + \(20^o\)) + (x - \(10^o\)) = (2n - 4) x \(90^o\)8x + 50 \(^o\) - 10\(^o\) = (2 x 6 -4) x 90\(^o\) 8x + 40\(^o\) = 8 x 90\(^o\) = 720\(^o\) 8x = 720\(^o\) - 40\(^o\) = 680\(^o\) x = \(\frac{680^o}{8}\) = 85\(^o\) |
|
| 16. |
If M and N are the points (-3, 8) and (5, -7) respectively, find |MN| A. 8 units B. 11 units C. 15 units D. 17 units Detailed Solution|MN| = \(\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\)= \(\sqrt{(-3 -5)^2 + (8 - 7)^2}\) = \(\sqrt{(-8)^2 + (8 + 7)^2}\) = \(\sqrt{64 + (15)^2}\) = \(\sqrt{64 + 225}\) = \(\sqrt{289}\) = 17 units |
|
| 17. |
The equation of the line through the points (4,2) and (-8, -2) is 3y = px + q, where p and q are constants. Find the value of p. A. 1 B. 2 C. 3 D. 9 Detailed SolutionUsing the two - point from\(\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}\) \(\frac{y - 2}{-2 - 2} = \frac{x - 4}{-8 - 4}\) \(\frac{y - 2}{-4} = \frac{x - 4}{-12}\) \(\frac{-12(y -2)}{-4}\) = x - 4 3(y -2) = x -4 3y - 6 = x - 4 3y = x - 4 + 6 3y = x + 2... By comparing the equations; 3y = px + , p = 1 |
|
| 18. |
The mean of 1, 3, 5, 7 and x is 4. Find the value of x A. 2 B. 4 C. 6 D. 8 Detailed SolutionMean = \(\frac{\sum x}{n}\)4 = \(\frac{1 + 3 + 5 + 7 + x}{5}\) 4 x 5 = 16 + x 20 - 16 = x 4 = x x = 4 |
|
| 19. |
 The table shows the distribution of goals scored by 25 teams in a football competition. Calculate the probability that a team selected at randon scored either 4 or 7 goals. A. \(\frac{9}{25}\) B. \(\frac{1}{5}\) C. \(\frac{6}{25}\) D. \(\frac{2}{5}\) Detailed SolutionProb. (team scored 4 goals) = Prob. (team scored 7 goals) = \(\frac{3}{25}\)Hence, probability that a team selected at random scored either 4 or 7 goals; = \(\frac{6}{25} + \frac{3}{25}\) = \(\frac{9}{25}\) |
|
| 20. |
 The table shows the distribution of goals scored by 25 teams in a football competition. Calculate the probability that a team selected at random scored at most 3 goals. A. \(\frac{3}{25}\) B. \(\frac{1}{5}\) C. \(\frac{6}{25}\) D. \(\frac{2}{5}\) Detailed SolutionNo. of teams that scored at most 3 goals = 3 + 1 + 6 = 10Hence, probability that a team selected at random scored at most 3 goals = \(\frac{10}{25}\) = \(\frac{2}{5}\) |
| 11. |
Simplify; \(\frac{2 - 18m^2}{1 + 3m}\) A. \(2 (1 + 3m)\) B. \(2 (1 + 3m^2)\) C. \(2(1 - 3m)\) D. \(2(1 - 3m^2)\) Detailed Solution\(\frac{2 - 18m^2}{1 + 3m}\) = \(\frac{2(1 - 9)m^2}{1 + 3m}\)= \(\frac{2(1 + 3m)(1 - 3m)}{1 + 3m}\) = \(2(1 - 3m)\) |
|
| 12. |
A curve is such that when y = 0, x = -2 or x = 3. Find the equation of the curve. A. y = \(x^2 - 5x - 6\) B. y = \(x^2 + 5x - 6\) C. y = \(x^2 + x - 6\) D. y = \(x^2 - x - 6\) |
|
| 13. |
The volume of a cylindrical tank, 10m high is 385 m\(^2\). Find the diameter of the tank. [Take \(\pi = \frac{22}{7}\)] A. 14m B. 10m C. 7m D. 5m Detailed SolutionVolume of a cylinder = \( \pi r^2\)h385 = \(\frac{22}{7}\) x \(r^2\) x 10 385 x 7 = 22 x \(r^2\) x 10 \(r^2\) = \(\frac{385 \times 7}{22 \times 10}\) = 12.25 r = \(\sqrt{12.25}\) = 3.5m Hence, diameter of tank = 2r = 2 x 3.5 = 7m |
|
| 14. |
The surface area of a sphere is \(\frac{792}{7} cm^2\). Find, correct to the nearest whole number, its volume. [Take \(\pi = \frac{22}{7}\)] A. 113\(cm^3\) B. 131\(cm^3\) C. 311\(cm^3\) D. 414\(cm^3\) Detailed SolutionSurface area of a sphere = \(4 \pi r^2\)\(4 \pi r^2\) = \(\frac{792}{7}cm^2\) 4 x \(\frac{22}{7}\) x \(r^2\) = \(\frac{792}{7}\) \(r^2\) = \(\frac{792}{7}\) x \(\frac{7}{4 \times 22}\) = 9 r = \(\sqrt{9}\) = 3cm Hence, volume of sphere = \(\frac{4}{3} \pi r^3\) = \(\frac{4}{3} \times \frac{22}{7} \times 3 \times 3 \times 3 \) = \(\frac{4 \times 22 \times 9}{7}\) \(\approx\) = 113.143 = 113\(cm^3\) (to the nearest whole number) |
|
| 15. |
The angles of a polygon are x, 2x, 2x, (x + \(30^o\)), (x + \(20^o\)) and (x - \(10^o\)). Find the value of x A. \(45^o\) B. \(95^o\) C. \(84^o\) D. \(85^o\) Detailed Solutionx + 2x + 2x + (x + \(30^o\)) + (x + \(20^o\)) + (x - \(10^o\)) = (2n - 4) x \(90^o\)8x + 50 \(^o\) - 10\(^o\) = (2 x 6 -4) x 90\(^o\) 8x + 40\(^o\) = 8 x 90\(^o\) = 720\(^o\) 8x = 720\(^o\) - 40\(^o\) = 680\(^o\) x = \(\frac{680^o}{8}\) = 85\(^o\) |
| 16. |
If M and N are the points (-3, 8) and (5, -7) respectively, find |MN| A. 8 units B. 11 units C. 15 units D. 17 units Detailed Solution|MN| = \(\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\)= \(\sqrt{(-3 -5)^2 + (8 - 7)^2}\) = \(\sqrt{(-8)^2 + (8 + 7)^2}\) = \(\sqrt{64 + (15)^2}\) = \(\sqrt{64 + 225}\) = \(\sqrt{289}\) = 17 units |
|
| 17. |
The equation of the line through the points (4,2) and (-8, -2) is 3y = px + q, where p and q are constants. Find the value of p. A. 1 B. 2 C. 3 D. 9 Detailed SolutionUsing the two - point from\(\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}\) \(\frac{y - 2}{-2 - 2} = \frac{x - 4}{-8 - 4}\) \(\frac{y - 2}{-4} = \frac{x - 4}{-12}\) \(\frac{-12(y -2)}{-4}\) = x - 4 3(y -2) = x -4 3y - 6 = x - 4 3y = x - 4 + 6 3y = x + 2... By comparing the equations; 3y = px + , p = 1 |
|
| 18. |
The mean of 1, 3, 5, 7 and x is 4. Find the value of x A. 2 B. 4 C. 6 D. 8 Detailed SolutionMean = \(\frac{\sum x}{n}\)4 = \(\frac{1 + 3 + 5 + 7 + x}{5}\) 4 x 5 = 16 + x 20 - 16 = x 4 = x x = 4 |
|
| 19. |
The table shows the distribution of goals scored by 25 teams in a football competition. Calculate the probability that a team selected at randon scored either 4 or 7 goals. A. \(\frac{9}{25}\) B. \(\frac{1}{5}\) C. \(\frac{6}{25}\) D. \(\frac{2}{5}\) Detailed SolutionProb. (team scored 4 goals) = Prob. (team scored 7 goals) = \(\frac{3}{25}\)Hence, probability that a team selected at random scored either 4 or 7 goals; = \(\frac{6}{25} + \frac{3}{25}\) = \(\frac{9}{25}\) |
|
| 20. |
The table shows the distribution of goals scored by 25 teams in a football competition. Calculate the probability that a team selected at random scored at most 3 goals. A. \(\frac{3}{25}\) B. \(\frac{1}{5}\) C. \(\frac{6}{25}\) D. \(\frac{2}{5}\) Detailed SolutionNo. of teams that scored at most 3 goals = 3 + 1 + 6 = 10Hence, probability that a team selected at random scored at most 3 goals = \(\frac{10}{25}\) = \(\frac{2}{5}\) |