Year :
2005
Title :
Mathematics (Core)
Exam :
WASSCE/WAEC MAY/JUNE
Paper 1 | Objectives
21 - 30 of 47 Questions
| # | Question | Ans |
|---|---|---|
| 21. |
Solve the equation 2x - 3y = 22; 3x + 2y = 7 A. -5 B. -4 C. 4 D. 5 Detailed Solution2x - 3y = 22 ---- eqn I3x + 2y = 7 ---- eqn II multiply eqn I by 2 4x - 6y = 44 ---- eqn III multiply eqn II by 3 9x + 6y = 21 ---- eqn IV Adding eqn III and IV => 13x = 65 => x = 5 Substituting 5 for x in eqn II 3x5 + 2y = 7 => 2y = -8 y = -4 |
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| 22. |
Solve the equation \(\frac{2y-1}{3} - \frac{3y-1}{4} = 1\) A. -8 B. -13 C. 13 D. 19 Detailed Solution\(\frac{4(2y-1)-3(3y-1)}{12}=12 \Rightarrow 12 8y - 4 - 9y + 3 = 12 \\\Rightarrow -y-1=12\Rightarrow -y=13\Rightarrow y=-13\) |
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| 23. |
If \(log_9x= 1.5\),find x A. 36 B. 27 C. 24.5 D. 13.5 Detailed Solution\(log_9x= 1.5\Rightarrow x = 9^{1.5} = 9^{\frac{3}{2}}=(3^2)^{\frac{3}{2}}=3^3=27\) |
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| 24. |
A sequence is given by \(2\frac{1}{2}, 5, 7\frac{1}{2}, .....\) if the nth term is 25, find n A. 9 B. 10 C. 12 D. 15 Detailed Solution\(a = 2\frac{1}{2}, nth = a + (n-1)d \Rightarrow 25 = 2\frac{1}{2} + (n-1)2\frac{1}{2}\\25 = \frac{5}{2}+(n-1)\frac{5}{2} \Rightarrow 22\frac{1}{2} = \frac{5n-5}{2}\Rightarrow \frac{45}{2} = \frac{5n-5}{2}\\ 45 = 5n - 5 \Rightarrow 5n = 50 \Rightarrow n = 10\) |
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| 25. |
Given that the root of an the equation \(2x^2 + (k+2)x+k=0\) is 2, find the value of k A. -4 B. -2 C. -1 D. \(-\frac{1}{4}\) Detailed SolutionSubstituting for x in the equation\(2(2)^2 + (k+2)2+k = 0 \Rightarrow 8 +2k + 4 + k =0 \Rightarrow 3k =-12; k=-4\) |
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| 26. |
Find the mean of the numbers 1, 3, 4, 8, 8, 4 and 7 A. 4 B. 5 C. 6 D. 7 Detailed Solutionmean \(=\frac{1+3+4+8+8+4+7}{7}=\frac{35}{7}=5\) |
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| 27. |
What is the total surface area of a closed cylinder of height 10cm and diameter 7cm? [Take \(\pi = \frac{22}{7}\)] A. 77cm2 B. 227cm2 C. 297cm2 D. 374cm2 Detailed SolutionT.S.A of s closed cylinder = \(2\pi r(r+h)\\=\frac{2}{1}\times \frac{22}{7} \times \frac{7}{2}\left(\frac{7}{2}+\frac{10}{1}\right)=\frac{22}{1}\left(\frac{27}{2}\right)=27\time 11=297cm^2\) |
|
| 28. |
An arc of a circle of radius 14cm subtends angle 300o at the center. Find the perimeter of the sector formed by the arc A. 14.67cm B. 42.67cm C. 101.33cm D. 543.33cm Detailed SolutionThe perimeter of the sector \(=2r+\frac{\theta}{360}\times 2\pi r \\\Rightarrow 28 + \frac{300}{360} \times \frac{2}{1} \times \frac{22}{7}\times \frac{14}{1} = \frac{220}{3}+28\\ 73.133+28=101.33cm\) |
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| 29. |
Which of the following statements describes the locus of a point R which moves in a plane such that its equidistant from two intersecting lines? A. the bisector of the angle formed by the lines B. the point of intersection of the two lines C. A cone, with two intersecting lines as slant height D. A circle, with the point of intersection of the two lines as the center |
A |
| 30. |
P = {3, 9, 11, 13} and Q = {3, 7, 9, 15} are subset of the universal set ξ = {1, 3, 7, 9, 11, 13, 15} find PI ∩ QI A. {3, 9} B. {5, 7, 9} C. {1} D. {1, 11} Detailed SolutionPI = {1, 7, 15} ; QI = {1, 13, 11}PI ∩ QI = {1} |
| 21. |
Solve the equation 2x - 3y = 22; 3x + 2y = 7 A. -5 B. -4 C. 4 D. 5 Detailed Solution2x - 3y = 22 ---- eqn I3x + 2y = 7 ---- eqn II multiply eqn I by 2 4x - 6y = 44 ---- eqn III multiply eqn II by 3 9x + 6y = 21 ---- eqn IV Adding eqn III and IV => 13x = 65 => x = 5 Substituting 5 for x in eqn II 3x5 + 2y = 7 => 2y = -8 y = -4 |
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| 22. |
Solve the equation \(\frac{2y-1}{3} - \frac{3y-1}{4} = 1\) A. -8 B. -13 C. 13 D. 19 Detailed Solution\(\frac{4(2y-1)-3(3y-1)}{12}=12 \Rightarrow 12 8y - 4 - 9y + 3 = 12 \\\Rightarrow -y-1=12\Rightarrow -y=13\Rightarrow y=-13\) |
|
| 23. |
If \(log_9x= 1.5\),find x A. 36 B. 27 C. 24.5 D. 13.5 Detailed Solution\(log_9x= 1.5\Rightarrow x = 9^{1.5} = 9^{\frac{3}{2}}=(3^2)^{\frac{3}{2}}=3^3=27\) |
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| 24. |
A sequence is given by \(2\frac{1}{2}, 5, 7\frac{1}{2}, .....\) if the nth term is 25, find n A. 9 B. 10 C. 12 D. 15 Detailed Solution\(a = 2\frac{1}{2}, nth = a + (n-1)d \Rightarrow 25 = 2\frac{1}{2} + (n-1)2\frac{1}{2}\\25 = \frac{5}{2}+(n-1)\frac{5}{2} \Rightarrow 22\frac{1}{2} = \frac{5n-5}{2}\Rightarrow \frac{45}{2} = \frac{5n-5}{2}\\ 45 = 5n - 5 \Rightarrow 5n = 50 \Rightarrow n = 10\) |
|
| 25. |
Given that the root of an the equation \(2x^2 + (k+2)x+k=0\) is 2, find the value of k A. -4 B. -2 C. -1 D. \(-\frac{1}{4}\) Detailed SolutionSubstituting for x in the equation\(2(2)^2 + (k+2)2+k = 0 \Rightarrow 8 +2k + 4 + k =0 \Rightarrow 3k =-12; k=-4\) |
| 26. |
Find the mean of the numbers 1, 3, 4, 8, 8, 4 and 7 A. 4 B. 5 C. 6 D. 7 Detailed Solutionmean \(=\frac{1+3+4+8+8+4+7}{7}=\frac{35}{7}=5\) |
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| 27. |
What is the total surface area of a closed cylinder of height 10cm and diameter 7cm? [Take \(\pi = \frac{22}{7}\)] A. 77cm2 B. 227cm2 C. 297cm2 D. 374cm2 Detailed SolutionT.S.A of s closed cylinder = \(2\pi r(r+h)\\=\frac{2}{1}\times \frac{22}{7} \times \frac{7}{2}\left(\frac{7}{2}+\frac{10}{1}\right)=\frac{22}{1}\left(\frac{27}{2}\right)=27\time 11=297cm^2\) |
|
| 28. |
An arc of a circle of radius 14cm subtends angle 300o at the center. Find the perimeter of the sector formed by the arc A. 14.67cm B. 42.67cm C. 101.33cm D. 543.33cm Detailed SolutionThe perimeter of the sector \(=2r+\frac{\theta}{360}\times 2\pi r \\\Rightarrow 28 + \frac{300}{360} \times \frac{2}{1} \times \frac{22}{7}\times \frac{14}{1} = \frac{220}{3}+28\\ 73.133+28=101.33cm\) |
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| 29. |
Which of the following statements describes the locus of a point R which moves in a plane such that its equidistant from two intersecting lines? A. the bisector of the angle formed by the lines B. the point of intersection of the two lines C. A cone, with two intersecting lines as slant height D. A circle, with the point of intersection of the two lines as the center |
A |
| 30. |
P = {3, 9, 11, 13} and Q = {3, 7, 9, 15} are subset of the universal set ξ = {1, 3, 7, 9, 11, 13, 15} find PI ∩ QI A. {3, 9} B. {5, 7, 9} C. {1} D. {1, 11} Detailed SolutionPI = {1, 7, 15} ; QI = {1, 13, 11}PI ∩ QI = {1} |