The ratio of the radii of two circles is 1 : 3. The ratio of their areas is [#1703]
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Q1. The ratio of the radii of two circles is 1 : 3. The ratio of their areas is
Q1. The ratio of the radii of two circles is 1 : 3. The ratio of their areas is
(A) 1:6
(A) 1:6
(A) 1:6
(B) 2:9
(B) 2:9
(B) 2:9
(C) 1:9
(C) 1:9
(C) 1:9
(D) 6:9
(D) 6:9
(D) 6:9
Answer: (C) 1:9
Answer: (C) 1:9
Answer: (C) 1:9
Area = πr2
π12 : π32
= π1 : π9
= 1:9
Area = πr2 π12 : π32 = π1 : π9 = 1:9
Area = πr2 π12 : π32 = π1 : π9 = 1:9
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Related MCQ Quizzes
Q1. 5 : 27 :: 9 : ________
Q1. 5 : 27 :: 9 : ________
Fill the blank.
Fill the blank.
Fill the blank.
(A) 83
(A) 83
(A) 83
(B) 81
(B) 81
(B) 81
(C) 36
(C) 36
(C) 36
(D) 18
(D) 18
(D) 18
Answer: (A) 83
Answer: (A) 83
Answer: (A) 83
5 : (52 + 2) = 5 : 27
Hence
9 : (92 + 2) = 9 : 83
5 : (52 + 2) = 5 : 27 Hence 9 : (92 + 2) = 9 : 83
5 : (52 + 2) = 5 : 27 Hence 9 : (92 + 2) = 9 : 83
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Q2. When a number is divided by 893 the remainder is 193. If the same number is divided by 47, the remainder will be
Q2. When a number is divided by 893 the remainder is 193. If the same number is divided by 47, the remainder will be
(A) 3
(A) 3
(A) 3
(B) 25
(B) 25
(B) 25
(C) 5
(C) 5
(C) 5
(D) 33
(D) 33
(D) 33
Answer: (C) 5
Answer: (C) 5
Answer: (C) 5
5
5
5
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Q3. A number is as much greater than 31 as is less 55. Then the number is
Q3. A number is as much greater than 31 as is less 55. Then the number is
(A) 39
(A) 39
(A) 39
(B) 32
(B) 32
(B) 32
(C) 43
(C) 43
(C) 43
(D) 47
(D) 47
(D) 47
Answer: (C) 43
Answer: (C) 43
Answer: (C) 43
43
43
43
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Q4. Two numbers are in the ratio of 2 : 3 and the product of their LCM and HCF is 96. The sum of the numbers is
Q4. Two numbers are in the ratio of 2 : 3 and the product of their LCM and HCF is 96. The sum of the numbers is
(A) 8
(A) 8
(A) 8
(B) 12
(B) 12
(B) 12
(C) 20
(C) 20
(C) 20
(D) 36
(D) 36
(D) 36
Answer: (C) 20
Answer: (C) 20
Answer: (C) 20
=> 2x * 3x = 96
=> 6x2 = 96
=> x2 = 96/6
=> x2 = 16
=> x = 4
2x + 3x = 5x = 5 * 4 = 20
=> 2x * 3x = 96 => 6x2 = 96 => x2 = 96/6 => x2 = 16 => x = 4 2x + 3x = 5x = 5 * 4 = 20
=> 2x * 3x = 96 => 6x2 = 96 => x2 = 96/6 => x2 = 16 => x = 4 2x + 3x = 5x = 5 * 4 = 20
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Q5. The largest of 26, 35, 44 and 53 is
Q5. The largest of 26, 35, 44 and 53 is
(A) 26
(A) 26
(A) 26
(B) 35
(B) 35
(B) 35
(C) 44
(C) 44
(C) 44
(D) 53
(D) 53
(D) 53
Answer: (C) 44
Answer: (C) 44
Answer: (C) 44
26 = 64
35 = 243
44 = 256
53 = 125
26 = 64 35 = 243 44 = 256 53 = 125
26 = 64 35 = 243 44 = 256 53 = 125
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Q6. ‘A’ starts his journey at 1:00 p.m. from a location P with a speed of 1 m/sec. ‘B’ starts his journey from the same location P and along the same direction at 1:10 p.m. with a speed of 2 m/sec. If ‘B’ meets ‘A’ at the location Q, then the distance PQ is :
Q6. ‘A’ starts his journey at 1:00 p.m. from a location P with a speed of 1 m/sec. ‘B’ starts his journey from the same location P and along the same direction at 1:10 p.m. with a speed of 2 m/sec. If ‘B’ meets ‘A’ at the location Q, then the distance PQ is :
(A) 1.5 km
(A) 1.5 km
(A) 1.5 km
(B) 1.75 km
(B) 1.75 km
(B) 1.75 km
(C) 1.2 km
(C) 1.2 km
(C) 1.2 km
(D) 1.25 km
(D) 1.25 km
(D) 1.25 km
Answer: (C) 1.2 km
Answer: (C) 1.2 km
Answer: (C) 1.2 km
A starts journey before B at a speed of 1 m/s. Hence A will be ahead of B
(10*60)s * 1m/s = 600m
Let A covers a distance of X after starting of B,
Then X + 600m = 2X
=> 2X - X = 600m
=> X = 600m
Hence B will cover a distance of 2X = 2 * 600m = 1200m = 1.2 km
A starts journey before B at a speed of 1 m/s. Hence A will be ahead of B (10*60)s * 1m/s = 600m Let A covers a distance of X after starting of B, Then X + 600m = 2X => 2X - X = 600m => X = 600m Hence B will cover a distance of 2X = 2 * 600m = 1200m = 1.2 km
A starts journey before B at a speed of 1 m/s. Hence A will be ahead of B (10*60)s * 1m/s = 600m Let A covers a distance of X after starting of B, Then X + 600m = 2X => 2X - X = 600m => X = 600m Hence B will cover a distance of 2X = 2 * 600m = 1200m = 1.2 km
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Q7. If the sum of five consecutive numbers is 190, then the lowest number amongst them is
Q7. If the sum of five consecutive numbers is 190, then the lowest number amongst them is
(A) 40
(A) 40
(A) 40
(B) 19
(B) 19
(B) 19
(C) 38
(C) 38
(C) 38
(D) 36
(D) 36
(D) 36
Answer: (D) 36
Answer: (D) 36
Answer: (D) 36
36
36
36
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Q8. The product of two consecutive odd numbers is 19043. Which is the smaller number?
Q8. The product of two consecutive odd numbers is 19043. Which is the smaller number?
(A) 133
(A) 133
(A) 133
(B) 131
(B) 131
(B) 131
(C) 137
(C) 137
(C) 137
(D) 129
(D) 129
(D) 129
Answer: (C) 137
Answer: (C) 137
Answer: (C) 137
137
137
137
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Q9. What is the term for a angle greater than 90 degrees but less than 180 degrees?
Q9. What is the term for a angle greater than 90 degrees but less than 180 degrees?
(A) Acute angle
(A) Acute angle
(A) Acute angle
(B) Right angle
(B) Right angle
(B) Right angle
(C) Obtuse angle
(C) Obtuse angle
(C) Obtuse angle
(D) Straight angle
(D) Straight angle
(D) Straight angle
Answer: (C) Obtuse angle
Answer: (C) Obtuse angle
Answer: (C) Obtuse angle
An obtuse angle is an angle greater than 90 degrees but less than 180 degrees, like the angle formed by two walls that meet at a corner.
An obtuse angle is an angle greater than 90 degrees but less than 180 degrees, like the angle formed by two walls that meet at a corner.
An obtuse angle is an angle greater than 90 degrees but less than 180 degrees, like the angle formed by two walls that meet at a corner.
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Q10. The smallest among is
Q10. The smallest among is
(A)
(A)
(A)
(B)
(B)
(B)
(C)
(C)
(C)
(D)
(D)
(D)
Answer: (C)
Answer: (C)
Answer: (C)
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Related Questions
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