If A : B = 3, then 144A : 36B is [#1697]
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Q1. If A : B = 3, then 144A : 36B is
Q1. If A : B = 3, then 144A : 36B is
(A) 4
(A) 4
(A) 4
(B) 12
(B) 12
(B) 12
(C) 3
(C) 3
(C) 3
(D) 16
(D) 16
(D) 16
Answer: (B) 12
Answer: (B) 12
Answer: (B) 12
A : B = 3
=> = 3
=
= x 3
= 4 x 3
= 12
A : B = 3 => = 3 = = x 3 = 4 x 3 = 12
A : B = 3 => = 3 = = x 3 = 4 x 3 = 12
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Related MCQ Quizzes
Q1. When a number is divided by 893 the remainder is 193. If the same number is divided by 47, the remainder will be
Q1. When a number is divided by 893 the remainder is 193. If the same number is divided by 47, the remainder will be
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Answer: (C) 5
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5
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Q2. If the sum of five consecutive numbers is 190, then the lowest number amongst them is
Q2. If the sum of five consecutive numbers is 190, then the lowest number amongst them is
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Answer: (D) 36
36
36
36
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Q3. p, q, r are three numbers such that the LCM of p and q is q and the LCM of q and r is r. The LCM of p, q and r will be
Q3. p, q, r are three numbers such that the LCM of p and q is q and the LCM of q and r is r. The LCM of p, q and r will be
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(A) q
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Answer: (B) r
LCM will be r.
px= q and qy = r, hence r = pxy.
LCM will be r. px= q and qy = r, hence r = pxy.
LCM will be r. px= q and qy = r, hence r = pxy.
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Q4. ‘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 :
Q4. ‘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
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(B) 1.75 km
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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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Q5. Consider the four numbers given numbers. If the digits of each number are arranged in ascending order, which new number would be the smallest?
Q5. Consider the four numbers given numbers. If the digits of each number are arranged in ascending order, which new number would be the smallest?
I. 376
II. 629
III. 921
IV. 397
I. 376 II. 629 III. 921 IV. 397
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(B) II
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(C) I
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Answer: (A) III
Answer: (A) III
Answer: (A) III
376 -> 367
629 -> 269
921 -> 129 *
397 -> 379
376 -> 367 629 -> 269 921 -> 129 * 397 -> 379
376 -> 367 629 -> 269 921 -> 129 * 397 -> 379
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Q6. 3/4 of a number is 19 less than the original number. The number is
Q6. 3/4 of a number is 19 less than the original number. The number is
(A) 62
(A) 62
(A) 62
(B) 64
(B) 64
(B) 64
(C) 79
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(D) 76
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(D) 76
Answer: (D) 76
Answer: (D) 76
Answer: (D) 76
76
76
76
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Q7. The remainder when –76 is divided by 3, is
Q7. The remainder when –76 is divided by 3, is
(A) -1
(A) -1
(A) -1
(B) 1
(B) 1
(B) 1
(C) 2
(C) 2
(C) 2
(D) -2
(D) -2
(D) -2
Answer: (C) 2
Answer: (C) 2
Answer: (C) 2
2
Because, Reminder must be positive and it should be less then the divisor
2 Because, Reminder must be positive and it should be less then the divisor
2 Because, Reminder must be positive and it should be less then the divisor
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Q8. The missing term in the sequence 7, 11, 19, 31, ______, 67 is
Q8. The missing term in the sequence 7, 11, 19, 31, ______, 67 is
(A) 43
(A) 43
(A) 43
(B) 47
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(C) 51
(D) 45
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Answer: (B) 47
Answer: (B) 47
Answer: (B) 47
47
47
47
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Q9. Two numbers are in the ratio 2 : 3 and the product of their HCF and LCM is 96. The sum of the two numbers is
Q9. Two numbers are in the ratio 2 : 3 and the product of their HCF and LCM is 96. The sum of the two numbers is
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(A) 18
(A) 18
(B) 20
(B) 20
(B) 20
(C) 22
(C) 22
(C) 22
(D) 24
(D) 24
(D) 24
Answer: (B) 20
Answer: (B) 20
Answer: (B) 20
2:3 = 8:12
N1*N2 = HCF*LCM
8*12 = 96
N1+N2 = 8+12 = 20
2:3 = 8:12 N1*N2 = HCF*LCM 8*12 = 96 N1+N2 = 8+12 = 20
2:3 = 8:12 N1*N2 = HCF*LCM 8*12 = 96 N1+N2 = 8+12 = 20
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Q10. A number is as much greater than 31 as is less 55. Then the number is
Q10. A number is as much greater than 31 as is less 55. Then the number is
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(A) 39
(A) 39
(B) 32
(B) 32
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(C) 43
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Answer: (C) 43
Answer: (C) 43
43
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43
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