# CAT Practice : Averages, Ratios, Mixtures

## Number of coins

Q.24: A, B and C have a few coins with them. 7 times the number of coins that A has is equal to 5 times the number of coins B has while 6 times the number of coins B has is equal to 11 times the number of coins C has. What is the minimum number of coins with A, B and C put together?
1. 110
2. 127
3. 154
4. 165

Choice B. 127

## Detailed Solution

Given, 7A = 5B
42A = 30B
Also given 6B = 11C
30B = 55C
Therefore 42A = 30B = 55C

Let 42A = 30B = 55C = k
Rewriting, A/55 = B/30 = C/42

The least possible integral values for A, B, C will be A = 42; B = 30 and C = 55
Total = 42 + 30 + 55 = 127

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