31 - 40 of 47 Questions
# | Question | Ans |
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31. |
The area of a circular plate is one-sixteenth the surface area of a ball of a ball, If the area of the plate is given as P cm2, then the radius of the ball is A. \(\frac{2P}{\pi}\) B. \(\frac{P}{\sqrt{\pi}}\) C. \(\frac{P}{\sqrt{2\pi}}\) D. 2\(\frac{P}{\pi}\) Detailed SolutionSurface area of a sphere = 4\(\pi\)r2\(\frac{1}{16}\) of 4\(\pi\)r2 = \(\frac{\pi r^2}{4}\) P = \(\frac{\pi r^2}{4}\) r = 2\(\frac{P}{\pi}\) |
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32. |
Show that \(\frac{\sin 2x}{1 + \cos x}\) + \(\frac{sin2 x}{1 - cos x}\) is A. sin x B. cos2x C. 2 D. 3 Detailed Solution\(\frac{\sin^{2} x}{1 + \cos x} + \frac{\sin^{2} x}{1 - \cos x}\)\(\frac{\sin^{2} x (1 - \cos x) + \sin^{2} x (1 + \cos x)}{1 - \cos^{2} x}\) = \(\frac{\sin^{2} x - \cos x \sin^{2} x + \sin^{2} x + \sin^{2} x \cos x}{\sin^{2} x}\) (Note: \(\sin^{2} x + \cos^{2} x = 1\)). = \(\frac{2 \sin^{2} x}{\sin^{2} x}\) = 2. |
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33. |
The first term of an Arithmetic progression is 3 and the fifth term is 9. Find the number of terms in the progression if the sum is 81 A. 12 B. 27 C. 9 D. 4 E. 36 Detailed Solution1st term a = 3, 5th term = 9, sum of n = 81nth term = a + (n - 1)d, 5th term a + (5 - 1)d = 9 3 + 4d = 9 4d = 9 - 3 d = \(\frac{6}{4}\) = \(\frac{3}{2}\) = 6 Sn = \(\frac{n}{2}\)(6 + \(\frac{3}{4}\)n - \(\frac{3}{2}\)) 81 = \(\frac{12n + 3n^2}{4}\) - 3n = \(\frac{3n^2 + 9n}{4}\) 3n2 + 9n = 324 3n2 + 9n - 324 = 0 By almighty formula positive no. n = 9 = 3 |
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34. |
Simplify T = \(\frac{4R_2}{R_1^{-1} + R_2^{-1} + 4R_3^{-1}}\) A. \(\frac{4R_1 \times R_2 R_3}{R_2R_3 + R_1R_3 + 4R_1 R_2}\) B. \(\frac{R_1 R_2 R_3}{R_2R_3 + R_1R_2 + 4R_1 R_2}\) C. \(\frac{16R_1 R_2 R_3}{R_2R_3 + R_1R_2 + R_1 R_2}\) D. \(\frac{4R_1 R_2 R_3}{4R_2R_3 + R_1R_2 + 4R_1 R_2}\) Detailed SolutionT = \(\frac{4R_2}{R_1^{-1} + R_2^{-1} + 4R_3^{-1}}\) = \(\frac{4R_2}{\frac{1}{R_1} + \frac{1}{R_2} + \frac{4}{R_3}}\)= \(\frac{4R_2}{\frac{R_2R_3 + R_1R_3 + 4R_1R_2}{R_1R_2R_3}}\) = \(\frac{4R_2 \times R_1 R_2 R_3}{R_2R_3 + R_1R_3 + 4R_1 R_2}\) = \(\frac{4R_1 \times R_2 R_3}{R_2R_3 + R_1R_3 + 4R_1R_2}\) T = \(\frac{4R_1 \times R_2 R_3}{R_2R_3 + R_1R_3 + 4R_1 R_2}\) |
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35. |
If b = a + cp and r = ab + \(\frac{1}{2}\)cp2, express b2 in terms of a, c, r. A. b2 = aV + 2cr B. b2 = ar + 2c2r C. b2 = a2 = \(\frac{1}{2}\) cr2 D. b2 = \(\frac{1}{2}\)ar2 + c E. b2 = 2cr - a2 Detailed Solutionb = a + cp....(i)r = ab + \(\frac{1}{2}\)cp2.....(ii) expressing b2 in terms of a, c, r, we shall first eliminate p which should not appear in our answer from eqn, (i) b - a = cp = \(\frac{b - a}{c}\) sub. for p in eqn.(ii) r = ab + \(\frac{1}{2}\)c\(\frac{(b - a)^2}{\frac{ab + b^2 - 2ab + a^2}{2c}}\) 2cr = 2ab + b2 - 2ab + a2 b2 = 2cr - a2 |
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36. |
Multiply x2 + x + 1 by x2 - x + 1 A. x4 - x + x2 B. x4 - x2 + x2 C. x4 + x2 + 1 D. x4 + x2 Detailed Solution(x2 + x + 1)( x2 - x + 1)= x2(x2 + x + 1) - x(x2 + x + 1) + (x3 + x + 1) = x4 - x3 + x2 + x3 - x2 - x + 1 = x4 + x2 + 1 |
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37. |
A baking recipe calls for 2.5kg of sugar and 4.5kg of flour. With this recipe some cakes were baked using 24.5kg of a mixture of sugar and flour. How much sugar was used? A. 12.25kg B. 6.75kg C. 8.75kg D. 15.75kg E. 8.25kg Detailed SolutionSugar : flour = 2.5 : 4.5Total = 7 sugar used = \(\frac{2.5}{7}\) x 24.5 = \(\frac{61.25}{7}\) = 8.75 |
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38. |
The difference between 4\(\frac{5}{7}\) and 2\(\frac{1}{4}\) greater by A. \(\frac{23}{28}\) B. \(\frac{24}{28}\) C. \(\frac{50}{56}\) D. \(\frac{27}{28}\) E. \(\frac{48}{56}\) Detailed SolutionDifference between 4\(\frac{5}{7}\) and 2\(\frac{1}{4}\)\(\frac{33}{7}\) - \(\frac{9}{4}\) = \(\frac{69}{24}\) The sum of \(\frac{1}{14}\) and 1\(\frac{1}{14}\) + \(\frac{3}{2}\) = \(\frac{11}{7}\) \(\frac{69}{28}\) - \(\frac{11}{7}\) = \(\frac{25}{28}\) = \(\frac{50}{56}\) |
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39. |
What is the size of an exterior angle of a regular pentagon? A. 36o B. 60o C. 72o D. 120o E. 360o Detailed SolutionPentagon is a polygon with 5 sides. Sum of interior angles of polygon (regular) = (2n - 4) x 90o= 5 = (10 - 4) x 90o = 540o each interior angle = \(\frac{540}{5}\) = 108o = 180o let x represent ext. angle x + 108o = 180v x = 180o - 108o = 72o |
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40. |
The sum of the root of a quadratic equation is \(\frac{5}{2}\) and the product of its root is 4. The quadratic equation is A. x + 8x2 - 5 = 0 B. 2x2 - 5x + 8 = 0 C. 2x2 - 8x + 5 = 0 D. x2 + 8x - 5 = 0 Detailed Solutionx2 - (sum of roots) x + (product of roots) = 0x2 - \(\frac{5x}{2}\) + 4 = 0 2x2 - 5x + 8 = 0 |
31. |
The area of a circular plate is one-sixteenth the surface area of a ball of a ball, If the area of the plate is given as P cm2, then the radius of the ball is A. \(\frac{2P}{\pi}\) B. \(\frac{P}{\sqrt{\pi}}\) C. \(\frac{P}{\sqrt{2\pi}}\) D. 2\(\frac{P}{\pi}\) Detailed SolutionSurface area of a sphere = 4\(\pi\)r2\(\frac{1}{16}\) of 4\(\pi\)r2 = \(\frac{\pi r^2}{4}\) P = \(\frac{\pi r^2}{4}\) r = 2\(\frac{P}{\pi}\) |
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32. |
Show that \(\frac{\sin 2x}{1 + \cos x}\) + \(\frac{sin2 x}{1 - cos x}\) is A. sin x B. cos2x C. 2 D. 3 Detailed Solution\(\frac{\sin^{2} x}{1 + \cos x} + \frac{\sin^{2} x}{1 - \cos x}\)\(\frac{\sin^{2} x (1 - \cos x) + \sin^{2} x (1 + \cos x)}{1 - \cos^{2} x}\) = \(\frac{\sin^{2} x - \cos x \sin^{2} x + \sin^{2} x + \sin^{2} x \cos x}{\sin^{2} x}\) (Note: \(\sin^{2} x + \cos^{2} x = 1\)). = \(\frac{2 \sin^{2} x}{\sin^{2} x}\) = 2. |
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33. |
The first term of an Arithmetic progression is 3 and the fifth term is 9. Find the number of terms in the progression if the sum is 81 A. 12 B. 27 C. 9 D. 4 E. 36 Detailed Solution1st term a = 3, 5th term = 9, sum of n = 81nth term = a + (n - 1)d, 5th term a + (5 - 1)d = 9 3 + 4d = 9 4d = 9 - 3 d = \(\frac{6}{4}\) = \(\frac{3}{2}\) = 6 Sn = \(\frac{n}{2}\)(6 + \(\frac{3}{4}\)n - \(\frac{3}{2}\)) 81 = \(\frac{12n + 3n^2}{4}\) - 3n = \(\frac{3n^2 + 9n}{4}\) 3n2 + 9n = 324 3n2 + 9n - 324 = 0 By almighty formula positive no. n = 9 = 3 |
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34. |
Simplify T = \(\frac{4R_2}{R_1^{-1} + R_2^{-1} + 4R_3^{-1}}\) A. \(\frac{4R_1 \times R_2 R_3}{R_2R_3 + R_1R_3 + 4R_1 R_2}\) B. \(\frac{R_1 R_2 R_3}{R_2R_3 + R_1R_2 + 4R_1 R_2}\) C. \(\frac{16R_1 R_2 R_3}{R_2R_3 + R_1R_2 + R_1 R_2}\) D. \(\frac{4R_1 R_2 R_3}{4R_2R_3 + R_1R_2 + 4R_1 R_2}\) Detailed SolutionT = \(\frac{4R_2}{R_1^{-1} + R_2^{-1} + 4R_3^{-1}}\) = \(\frac{4R_2}{\frac{1}{R_1} + \frac{1}{R_2} + \frac{4}{R_3}}\)= \(\frac{4R_2}{\frac{R_2R_3 + R_1R_3 + 4R_1R_2}{R_1R_2R_3}}\) = \(\frac{4R_2 \times R_1 R_2 R_3}{R_2R_3 + R_1R_3 + 4R_1 R_2}\) = \(\frac{4R_1 \times R_2 R_3}{R_2R_3 + R_1R_3 + 4R_1R_2}\) T = \(\frac{4R_1 \times R_2 R_3}{R_2R_3 + R_1R_3 + 4R_1 R_2}\) |
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35. |
If b = a + cp and r = ab + \(\frac{1}{2}\)cp2, express b2 in terms of a, c, r. A. b2 = aV + 2cr B. b2 = ar + 2c2r C. b2 = a2 = \(\frac{1}{2}\) cr2 D. b2 = \(\frac{1}{2}\)ar2 + c E. b2 = 2cr - a2 Detailed Solutionb = a + cp....(i)r = ab + \(\frac{1}{2}\)cp2.....(ii) expressing b2 in terms of a, c, r, we shall first eliminate p which should not appear in our answer from eqn, (i) b - a = cp = \(\frac{b - a}{c}\) sub. for p in eqn.(ii) r = ab + \(\frac{1}{2}\)c\(\frac{(b - a)^2}{\frac{ab + b^2 - 2ab + a^2}{2c}}\) 2cr = 2ab + b2 - 2ab + a2 b2 = 2cr - a2 |
36. |
Multiply x2 + x + 1 by x2 - x + 1 A. x4 - x + x2 B. x4 - x2 + x2 C. x4 + x2 + 1 D. x4 + x2 Detailed Solution(x2 + x + 1)( x2 - x + 1)= x2(x2 + x + 1) - x(x2 + x + 1) + (x3 + x + 1) = x4 - x3 + x2 + x3 - x2 - x + 1 = x4 + x2 + 1 |
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37. |
A baking recipe calls for 2.5kg of sugar and 4.5kg of flour. With this recipe some cakes were baked using 24.5kg of a mixture of sugar and flour. How much sugar was used? A. 12.25kg B. 6.75kg C. 8.75kg D. 15.75kg E. 8.25kg Detailed SolutionSugar : flour = 2.5 : 4.5Total = 7 sugar used = \(\frac{2.5}{7}\) x 24.5 = \(\frac{61.25}{7}\) = 8.75 |
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38. |
The difference between 4\(\frac{5}{7}\) and 2\(\frac{1}{4}\) greater by A. \(\frac{23}{28}\) B. \(\frac{24}{28}\) C. \(\frac{50}{56}\) D. \(\frac{27}{28}\) E. \(\frac{48}{56}\) Detailed SolutionDifference between 4\(\frac{5}{7}\) and 2\(\frac{1}{4}\)\(\frac{33}{7}\) - \(\frac{9}{4}\) = \(\frac{69}{24}\) The sum of \(\frac{1}{14}\) and 1\(\frac{1}{14}\) + \(\frac{3}{2}\) = \(\frac{11}{7}\) \(\frac{69}{28}\) - \(\frac{11}{7}\) = \(\frac{25}{28}\) = \(\frac{50}{56}\) |
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39. |
What is the size of an exterior angle of a regular pentagon? A. 36o B. 60o C. 72o D. 120o E. 360o Detailed SolutionPentagon is a polygon with 5 sides. Sum of interior angles of polygon (regular) = (2n - 4) x 90o= 5 = (10 - 4) x 90o = 540o each interior angle = \(\frac{540}{5}\) = 108o = 180o let x represent ext. angle x + 108o = 180v x = 180o - 108o = 72o |
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40. |
The sum of the root of a quadratic equation is \(\frac{5}{2}\) and the product of its root is 4. The quadratic equation is A. x + 8x2 - 5 = 0 B. 2x2 - 5x + 8 = 0 C. 2x2 - 8x + 5 = 0 D. x2 + 8x - 5 = 0 Detailed Solutionx2 - (sum of roots) x + (product of roots) = 0x2 - \(\frac{5x}{2}\) + 4 = 0 2x2 - 5x + 8 = 0 |