Year :
2010
Title :
Mathematics (Core)
Exam :
WASSCE/WAEC MAY/JUNE
Paper 1 | Objectives
21 - 30 of 49 Questions
| # | Question | Ans |
|---|---|---|
| 21. |
The sum of the exterior of an n-sided convex polygon is half the sum of its interior angle. find n A. 6 B. 8 C. 9 D. 12 Detailed Solutionsum of exterior angles = 360oSum of interior angle = (n - 2) x 180 360 = \(\frac{1}{2}\) x(n - 2) x 180(90o) 360 = \(\frac{1}{2}\) x(n - 2) x 90o \(\frac{360}{90}\) = a - 2 4 = n - 2 n = 4 + 2 = 6 |
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| 22. |
If y = \(\frac{y(2\sqrt{x^2 + m})}{3N}\), make x the subject of the formular A. \(\frac{\sqrt{9y^2 N^2 - 2m}}{3}\) B. \(\frac{\sqrt{9y^2 N^2 - 4m}}{2}\) C. \(\frac{\sqrt{9y^2 N^2 - 3m}}{2}\) D. \(\frac{\sqrt{9y^2 N - 3m}}{2}\) Detailed Solutiony = \(\frac{y(2\sqrt{x^2 + m})}{3N}\)3yN = 2(\(\sqrt{x^2 + m})\) \(\frac{3yN}{2} = \sqrt{x^2 + m}\) (\(\frac{3yN}{2})^2 = ( \sqrt{x^2 + m})\) \(\sqrt{\frac{9y^2N^2}{4} - \frac{m}{1}}\) x = \(\frac{\sqrt{9Y^2N^2 - 4m}}{4}\) x = \(\frac{\sqrt{9y^2N^2 - 4m}}{2}\) |
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| 23. |
The nth term of the sequence -2, 4, -8, 16.... is given by A. Tn = 2n B. Tn = (-2)n C. Tn = (-2n) D. Tn = n Detailed Solutionsequence: -2, 4, -8, 16........{GP}a = -2; r = \(\frac{4}{-2}\) = -2 nth term Tn = arn-1 Tn = (-2)(-2)^n-1 Tn = (-2)1 + n - 1 Tn = (-2)n |
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| 24. |
How many times, correct to the nearest whole number, will a man run round circular track of diameter 100m to cover a distance of 1000m? A. 3 B. 4 C. 5 D. 6 Detailed SolutionNo. of times = \(\frac{\text{Total distance}}{\text{Circumference of circle}}\)= \(\frac{\text{Total distance}}{\pi d}\) = \(\frac{1000m}{\frac{22}{7} \times 100m}\) = \(\frac{1000 \times 7}{2200} = 3.187\) = 3(approx.) nearest whole no. |
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| 25. |
Bola sold an article for N6,900.00 and made a profit of 15%. If he sold it for N6,600.00 he would make a A. profit of 13% B. loss of 12% C. loss of 10% D. loss of 5% Detailed Solutions.p = N6900%profit = 15% %profit = \(\frac{s.p - c.p}{c.p}\) x 100% 15% = \(\frac{6900 - c.p}{c.p}\) x 100% \(\frac{15}{100}\)c.p = N6900 - c.p 0.15 c.p = N6900 - c.p 1.15c.p + c.p = N6900 c.p = \(\frac{6900}{1.15}\) = 6000.00 Now new S.P = N6600 profit = s.p - c.p = 6000 - 6600 = 600 %profit = \(\frac{600}{6600}\) x 100% = 10% |
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| 26. |
The mean age of R men in a club is 50 years, Two men aged 55 and 63, left the club and the mean age reduced by 1 year. Find the value of R A. 18 B. 20 C. 22 D. 28 Detailed Solutionmean age = \(\frac{\text{sum of ages}}{\text{no. of men}}\)50 = \9\frac{sum}{R}\) sum = 50R.....(1) Sum of ages of the men that left = 55 + 63 = 188 remaining sum = 50R - 118 remaining no. of men = R - 2 now mean age = 50 - 1 = 49 years 49 = \(\frac{50R - 118}{R - 2}\) 49(R - 2) = 50R - 118 49R - 50R = -188 - 98 -R = -20 R = 20 |
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| 27. |
\(\begin{array}{c|c} x & 0 & 2 & 4 & 6\\ \hline y & 1 & 2 & 3 & 4\end{array}\). A. y = 2x B. y = x + 1 C. y = x D. y = \(\frac{1}{2}x + 1\) Detailed Solutiony = mx + c; when x = 0; y = 11 = m(0) + c; 1 = 0 + c; c = 1 when x = 2; y = 2 2 = m(2) + c; 2 = 2m + c; but c = 1 2 = 2m + 1 2 - 1 = 2m 2m = 1 m = \(\frac{1}{2}\) y = \(\frac{1}{2}\)x + 1 |
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| 28. |
What is the value of x when y = 5?y = \(\frac{1}{2}\) x + 1 A. 8 B. 9 C. 10 D. 11 Detailed Solutionwhen y = 5; x = ?; y = \(\frac{1}{2}\)x + 15 = \(\frac{1}{2}\)x + 1 5 - 1 = \(\frac{1}{2}\)x 4 = \(\frac{1}{2}\)x x = 4 x 2 x = 8 |
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| 29. |
The subtraction below is in base seven. Find the missing number. A. 2 B. 3 C. 4 D. 5 Detailed Solution5 1 6 2seven-2 6 4 4seven -------- 2 2 1 5 -------- the missing number is 2 |
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| 30. |
If the sum of the roots of the equation (x - p)(2x - 1) - 0 is 1, find the value of x A. 1\(\frac{1}{2}\) B. \(\frac{1}{2}\) C. -\(\frac{3}{2}\) D. -1\(\frac{1}{2}\) Detailed Solution(x - p)(2x + 1) = 02x2 + x - 2px - p = 0 2x2 + x (1 - 2p) - p = 0 2x2 - (2p - 1)x - p = 0 divide through by 2 x2 - \(\frac{(2p - 1)}{2}\)x - \(\frac{p}{2}\) = 0 compare to x2 - (sum of roots)x + product of roots = 0 sum of roots = \(\frac{2p - 1}{2}\) But sum of roots = 1 Given; \(\frac{2p - 1}{2}\) = 1 2p - 1 = 2 x 1 2p - 1 = 2 2p = 2 + 1 = 3 p = \(\fr |
| 21. |
The sum of the exterior of an n-sided convex polygon is half the sum of its interior angle. find n A. 6 B. 8 C. 9 D. 12 Detailed Solutionsum of exterior angles = 360oSum of interior angle = (n - 2) x 180 360 = \(\frac{1}{2}\) x(n - 2) x 180(90o) 360 = \(\frac{1}{2}\) x(n - 2) x 90o \(\frac{360}{90}\) = a - 2 4 = n - 2 n = 4 + 2 = 6 |
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| 22. |
If y = \(\frac{y(2\sqrt{x^2 + m})}{3N}\), make x the subject of the formular A. \(\frac{\sqrt{9y^2 N^2 - 2m}}{3}\) B. \(\frac{\sqrt{9y^2 N^2 - 4m}}{2}\) C. \(\frac{\sqrt{9y^2 N^2 - 3m}}{2}\) D. \(\frac{\sqrt{9y^2 N - 3m}}{2}\) Detailed Solutiony = \(\frac{y(2\sqrt{x^2 + m})}{3N}\)3yN = 2(\(\sqrt{x^2 + m})\) \(\frac{3yN}{2} = \sqrt{x^2 + m}\) (\(\frac{3yN}{2})^2 = ( \sqrt{x^2 + m})\) \(\sqrt{\frac{9y^2N^2}{4} - \frac{m}{1}}\) x = \(\frac{\sqrt{9Y^2N^2 - 4m}}{4}\) x = \(\frac{\sqrt{9y^2N^2 - 4m}}{2}\) |
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| 23. |
The nth term of the sequence -2, 4, -8, 16.... is given by A. Tn = 2n B. Tn = (-2)n C. Tn = (-2n) D. Tn = n Detailed Solutionsequence: -2, 4, -8, 16........{GP}a = -2; r = \(\frac{4}{-2}\) = -2 nth term Tn = arn-1 Tn = (-2)(-2)^n-1 Tn = (-2)1 + n - 1 Tn = (-2)n |
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| 24. |
How many times, correct to the nearest whole number, will a man run round circular track of diameter 100m to cover a distance of 1000m? A. 3 B. 4 C. 5 D. 6 Detailed SolutionNo. of times = \(\frac{\text{Total distance}}{\text{Circumference of circle}}\)= \(\frac{\text{Total distance}}{\pi d}\) = \(\frac{1000m}{\frac{22}{7} \times 100m}\) = \(\frac{1000 \times 7}{2200} = 3.187\) = 3(approx.) nearest whole no. |
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| 25. |
Bola sold an article for N6,900.00 and made a profit of 15%. If he sold it for N6,600.00 he would make a A. profit of 13% B. loss of 12% C. loss of 10% D. loss of 5% Detailed Solutions.p = N6900%profit = 15% %profit = \(\frac{s.p - c.p}{c.p}\) x 100% 15% = \(\frac{6900 - c.p}{c.p}\) x 100% \(\frac{15}{100}\)c.p = N6900 - c.p 0.15 c.p = N6900 - c.p 1.15c.p + c.p = N6900 c.p = \(\frac{6900}{1.15}\) = 6000.00 Now new S.P = N6600 profit = s.p - c.p = 6000 - 6600 = 600 %profit = \(\frac{600}{6600}\) x 100% = 10% |
| 26. |
The mean age of R men in a club is 50 years, Two men aged 55 and 63, left the club and the mean age reduced by 1 year. Find the value of R A. 18 B. 20 C. 22 D. 28 Detailed Solutionmean age = \(\frac{\text{sum of ages}}{\text{no. of men}}\)50 = \9\frac{sum}{R}\) sum = 50R.....(1) Sum of ages of the men that left = 55 + 63 = 188 remaining sum = 50R - 118 remaining no. of men = R - 2 now mean age = 50 - 1 = 49 years 49 = \(\frac{50R - 118}{R - 2}\) 49(R - 2) = 50R - 118 49R - 50R = -188 - 98 -R = -20 R = 20 |
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| 27. |
\(\begin{array}{c|c} x & 0 & 2 & 4 & 6\\ \hline y & 1 & 2 & 3 & 4\end{array}\). A. y = 2x B. y = x + 1 C. y = x D. y = \(\frac{1}{2}x + 1\) Detailed Solutiony = mx + c; when x = 0; y = 11 = m(0) + c; 1 = 0 + c; c = 1 when x = 2; y = 2 2 = m(2) + c; 2 = 2m + c; but c = 1 2 = 2m + 1 2 - 1 = 2m 2m = 1 m = \(\frac{1}{2}\) y = \(\frac{1}{2}\)x + 1 |
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| 28. |
What is the value of x when y = 5?y = \(\frac{1}{2}\) x + 1 A. 8 B. 9 C. 10 D. 11 Detailed Solutionwhen y = 5; x = ?; y = \(\frac{1}{2}\)x + 15 = \(\frac{1}{2}\)x + 1 5 - 1 = \(\frac{1}{2}\)x 4 = \(\frac{1}{2}\)x x = 4 x 2 x = 8 |
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| 29. |
The subtraction below is in base seven. Find the missing number. A. 2 B. 3 C. 4 D. 5 Detailed Solution5 1 6 2seven-2 6 4 4seven -------- 2 2 1 5 -------- the missing number is 2 |
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| 30. |
If the sum of the roots of the equation (x - p)(2x - 1) - 0 is 1, find the value of x A. 1\(\frac{1}{2}\) B. \(\frac{1}{2}\) C. -\(\frac{3}{2}\) D. -1\(\frac{1}{2}\) Detailed Solution(x - p)(2x + 1) = 02x2 + x - 2px - p = 0 2x2 + x (1 - 2p) - p = 0 2x2 - (2p - 1)x - p = 0 divide through by 2 x2 - \(\frac{(2p - 1)}{2}\)x - \(\frac{p}{2}\) = 0 compare to x2 - (sum of roots)x + product of roots = 0 sum of roots = \(\frac{2p - 1}{2}\) But sum of roots = 1 Given; \(\frac{2p - 1}{2}\) = 1 2p - 1 = 2 x 1 2p - 1 = 2 2p = 2 + 1 = 3 p = \(\fr |