31 - 40 of 50 Questions
# | Question | Ans |
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31. |
If y = x sinx, find dy/dx when x = π/2. A. -π/2 B. -1 C. 1 D. π/2 Detailed Solutiony = x sin xdy/dx = 1 x sinx + x cosx = sinx + x cosx At x = π/2, = sin (π/2) + (π/2) cos (π/2) = 1 + (π/2) x (0) = 1 |
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32. |
Find the dimensions of a rectangle of greatest area which has a fixed perimeter p. A. square of sides p B. square of sides 2p C. square of sides (p/2) D. square of sides (p/4) Detailed SolutionLet the rectangle be a square of sides p/4.So that perimeter of square = 4p 4 x (p/4) = p. |
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33. |
Given the scores: 4, 7, 8, 11, 13, 8 with corresponding frequencies: 3, 5, 2, 7, 2, 1 respectively. Find the square of the mode. A. 49 B. 121 C. 25 D. 64 Detailed SolutionMode = score with highest frequency = 11.Square of 11 = 121 |
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34. |
Given the scores: 4, 7, 8, 11, 13, 8 with corresponding frequencies: 3, 5, 2, 7, 2, 1 respectively. The mean score is A. 7.0 B. 8.7 C. 9.5 D. 11.0 Detailed SolutionMean = ∑fx/∑fMean = (4x3) + (7x5) + ... + (8x1) all divided by 20 Mean = 174/20 = 8.7 |
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35. |
Teams P and Q are involved in a game of football. What is the probability that the game ends in a draw? A. 2/3 B. 1/2 C. 1/3 D. 1/4 Detailed SolutionP (games end in draw)=> Team P wins and Q wins P(P wins) = 1/2 P(Q wins) = 1/2 Therefore P (games ends in draw) = 1/2 x 1/2 = 1/4 |
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36. |
If 6Pr = 6, find the value of 6Pr+1 A. 30 B. 33 C. 35 D. 15 Detailed Solution6Pr = 6 Thus r = 1N.B 6P1 = \(\frac{\text{6!}}{\text{(6 - 1)!}}\) = \(\frac{\text{6!}}{\text{5!}}\) = \(\frac{6 \times \text{5!}}{\text{5!}}\) = 6 6Pr + 1 = 6P2 = \(\frac{\text{6!}}{(6 - 2)\text{!}} = \frac{6\text{!}}{\text{4!}}\) = \(\frac{6 \times 5 \times \text{4!}}{\text{4!}}\) = 30 |
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37. |
Find the variance of 2, 6, 8, 6, 2 and 6 A. 6 B. 5 C. √6 D. √5 Detailed Solutionmean of X = x = (2+6+8+6+2+6)/6 = 30/6 = 5X → 2,6,8,6,2,6 X - x = -3, 1, 3, 1, 3, 1 (X - x)2 = 9, 1, 9, 1, 9, 1 ∑(X-x) = 9+1+9+1+9+1 = 30 Variance = ∑(X-x)2/n = 30/6 = 5 |
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38. |
Find the number of ways of selecting 8 subjects from 12 subjects for an examination A. 490 B. 495 C. 496 D. 498 Detailed SolutionCombination is the number (n) of ways of selecting a number of (m) of objects from n\(^{12}C_{8}\frac{12!}{8!(12-8)!}\\=\frac{12!}{8!4!}\\\frac{(12\times 11\times 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1)}{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}\\After\hspace{1mm} cancelling \hspace{1mm}out \hspace{1mm}we\hspace{1mm} have\\11\times 5\times 9 = 495 \) |
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39. |
Find the range of 1/6, 1/3, 3/2, 2/3, 8/9 and 4/3 A. 3/4 B. 5/6 C. 7/6 D. 4/3 Detailed Solution1/6, 1/3, 3/2, 2/3, 8/9 and 4/3L.C.M of the denominators = 36 The fraction now becomes 6/36, 12/36, 54/36, 24/36, 32/36 and 48/36 Highest fraction = 54/36 Lowest fraction = 6/36 Range = Highest - Lowest (54/36) - (6/36) = 48/36 = 4/3 |
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40. |
Triangle SPT is the solution of the linear inequalities A. 2y - x - 2 ≤ 0, y + 2x + 2 ≥ 0, -2 ≤ x ≤ - 1 B. -2 ≤ x ≤ 2, y ≥ 0, y + 2x + 2 ≤ 0, x ≤ 0 C. 2y - x - 2 ≤ 0, y + 2x + 2 ≤ 0, y ≥ 0, x ≤ 0 D. 2y - x - 2 ≥ 0, y + 2x + 2 ≤ 0, x ≤ 0 Detailed Solution2y - x - 2 ≤ 0 2y ≤ 2 + x y ≤ 2/2 + x/2 y ≤ 1 + 1/2x y + 2x + 2 ≥ 0, y ≥ -2 -2x ∴ the solution of the inequalities 2y - x - 2 ≤ 0, y + 2x + 2 ≥ 0 = -2 ≤ x ≤ -1 |
31. |
If y = x sinx, find dy/dx when x = π/2. A. -π/2 B. -1 C. 1 D. π/2 Detailed Solutiony = x sin xdy/dx = 1 x sinx + x cosx = sinx + x cosx At x = π/2, = sin (π/2) + (π/2) cos (π/2) = 1 + (π/2) x (0) = 1 |
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32. |
Find the dimensions of a rectangle of greatest area which has a fixed perimeter p. A. square of sides p B. square of sides 2p C. square of sides (p/2) D. square of sides (p/4) Detailed SolutionLet the rectangle be a square of sides p/4.So that perimeter of square = 4p 4 x (p/4) = p. |
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33. |
Given the scores: 4, 7, 8, 11, 13, 8 with corresponding frequencies: 3, 5, 2, 7, 2, 1 respectively. Find the square of the mode. A. 49 B. 121 C. 25 D. 64 Detailed SolutionMode = score with highest frequency = 11.Square of 11 = 121 |
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34. |
Given the scores: 4, 7, 8, 11, 13, 8 with corresponding frequencies: 3, 5, 2, 7, 2, 1 respectively. The mean score is A. 7.0 B. 8.7 C. 9.5 D. 11.0 Detailed SolutionMean = ∑fx/∑fMean = (4x3) + (7x5) + ... + (8x1) all divided by 20 Mean = 174/20 = 8.7 |
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35. |
Teams P and Q are involved in a game of football. What is the probability that the game ends in a draw? A. 2/3 B. 1/2 C. 1/3 D. 1/4 Detailed SolutionP (games end in draw)=> Team P wins and Q wins P(P wins) = 1/2 P(Q wins) = 1/2 Therefore P (games ends in draw) = 1/2 x 1/2 = 1/4 |
36. |
If 6Pr = 6, find the value of 6Pr+1 A. 30 B. 33 C. 35 D. 15 Detailed Solution6Pr = 6 Thus r = 1N.B 6P1 = \(\frac{\text{6!}}{\text{(6 - 1)!}}\) = \(\frac{\text{6!}}{\text{5!}}\) = \(\frac{6 \times \text{5!}}{\text{5!}}\) = 6 6Pr + 1 = 6P2 = \(\frac{\text{6!}}{(6 - 2)\text{!}} = \frac{6\text{!}}{\text{4!}}\) = \(\frac{6 \times 5 \times \text{4!}}{\text{4!}}\) = 30 |
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37. |
Find the variance of 2, 6, 8, 6, 2 and 6 A. 6 B. 5 C. √6 D. √5 Detailed Solutionmean of X = x = (2+6+8+6+2+6)/6 = 30/6 = 5X → 2,6,8,6,2,6 X - x = -3, 1, 3, 1, 3, 1 (X - x)2 = 9, 1, 9, 1, 9, 1 ∑(X-x) = 9+1+9+1+9+1 = 30 Variance = ∑(X-x)2/n = 30/6 = 5 |
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38. |
Find the number of ways of selecting 8 subjects from 12 subjects for an examination A. 490 B. 495 C. 496 D. 498 Detailed SolutionCombination is the number (n) of ways of selecting a number of (m) of objects from n\(^{12}C_{8}\frac{12!}{8!(12-8)!}\\=\frac{12!}{8!4!}\\\frac{(12\times 11\times 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1)}{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}\\After\hspace{1mm} cancelling \hspace{1mm}out \hspace{1mm}we\hspace{1mm} have\\11\times 5\times 9 = 495 \) |
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39. |
Find the range of 1/6, 1/3, 3/2, 2/3, 8/9 and 4/3 A. 3/4 B. 5/6 C. 7/6 D. 4/3 Detailed Solution1/6, 1/3, 3/2, 2/3, 8/9 and 4/3L.C.M of the denominators = 36 The fraction now becomes 6/36, 12/36, 54/36, 24/36, 32/36 and 48/36 Highest fraction = 54/36 Lowest fraction = 6/36 Range = Highest - Lowest (54/36) - (6/36) = 48/36 = 4/3 |
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40. |
Triangle SPT is the solution of the linear inequalities A. 2y - x - 2 ≤ 0, y + 2x + 2 ≥ 0, -2 ≤ x ≤ - 1 B. -2 ≤ x ≤ 2, y ≥ 0, y + 2x + 2 ≤ 0, x ≤ 0 C. 2y - x - 2 ≤ 0, y + 2x + 2 ≤ 0, y ≥ 0, x ≤ 0 D. 2y - x - 2 ≥ 0, y + 2x + 2 ≤ 0, x ≤ 0 Detailed Solution2y - x - 2 ≤ 0 2y ≤ 2 + x y ≤ 2/2 + x/2 y ≤ 1 + 1/2x y + 2x + 2 ≥ 0, y ≥ -2 -2x ∴ the solution of the inequalities 2y - x - 2 ≤ 0, y + 2x + 2 ≥ 0 = -2 ≤ x ≤ -1 |