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Three children A, B and C have candies such that the ratio of candies 15 Jul 2019, 06:22
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68% (02:10) correct 32% (02:32) wrong based on 247 sessions

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Three children A, B and C have candies such that the ratio of candies with A to the total candies with other two is 2 : 5, and the ratio of candies with C to the total candies with other two is 4 : 9. The minimum possible number of total candies with A, B and C is

(A) 64
(B) 91
(C) 72
(D) 56
(E) 45
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Three children A, B and C have candies such that the ratio of candies 15 Jul 2019, 07:08
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The ratio of A to the rest is 2 to 5, so A has 2/7 of all the candies. The ratio of C to the rest is 4 to 9, so C has 4/13 of all the candies. So the total number of candies is divisible by both 7 and 13, and the smallest possible total is 91.
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Three children A, B and C have candies such that the ratio of candies 15 Jul 2019, 09:00
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Since A /All = 2/7
and C / All = 4/13

Total no = should be divisible by both 7 and 13.

Min total no of candles = 91

Answer should be B
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Three children A, B and C have candies such that the ratio of candies 15 Jul 2019, 12:14
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since A/B+C=2/5, then A/All=2/7, similarly C/all=4/13,

now the minimum possible candies will be LCM of 7 & 13, i.e 91

hence answer is 91
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Three children A, B and C have candies such that the ratio of candies 22 Oct 2024, 14:33
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Three children A, B and C have candies such that the ratio of candies with A to the total candies with other two is 2 : 5, and the ratio of candies with C to the total candies with other two is 4 : 9. The minimum possible number of total candies with A, B and C is

When we're told that the numbers of two things a and b are in a ratio a:b, we know that the number of a things is a multiple a, the number of b things is a multiple of b, and the total number of things is a multiple of a + b.

Thus, since the ratio of A's candies to B's and C's candies is 2:5, we know that the total number of candies is a multiple of 2 + 5 = 7.

Similarly, since the ratio of C's candies to A's and B's is 4:9, we know that the total number of candies is a multiple of 4 + 9 = 13.

Thus, the minimum total number of candies must be the least common multiple of 7 and 13, which is 91.

(A) 64
(B) 91
(C) 72
(D) 56
(E) 45

Correct answer: B
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Three children A, B and C have candies such that the ratio of candies 18 Nov 2025, 16:09
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