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The question is from CAT Ratio, Mixtures and Averages. We need to find out the minimum number of total coins with A, B, C with the given ratios. CAT exam is known to test on basics rather than high funda ideas. A range of CAT questions can be asked from Ratios and Proportions, Mixtures, Alligations and Averages. Make sure you master the topics. 2IIMs CAT questions bank provides you with CAT questions that can help you gear for CAT Exam CAT 2019.

Question 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. 174
  3. 154
  4. 165

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Explanatory Answer

Method of solving this CAT Question from Ratio, Mixtures and Averages: How do we relate two ratios when they both have one common variable?

Given, 7A = 5B
or A/B = 5/7
or A : B :: 5 : 7
Also given 6B = 11C
B/C = 11/6
or B : C :: 11 : 6
Now, we have two ratios A : B and B : C. We can see that B is common. What value can B take such that it is the same in both ratios ?
The common value of B is the multiple of (7,11) or the LCM of (7,11) which is 77.
A : B
5 : 7
B : C
11 : 6
To make B common, we take LCM of 7 and 11 which is 77.

A : B
5 : 7 * 11
... B : C
11 : 6 * 7

A : B
55 : 77
B : C
77 : 42

A : B : C
55 : 77 : 42

A, B and C can be 55x, 77x and 42x respectively. The least possible integral values for A, B, C will be A = 55; B = 77; and C = 42
Total = 55 + 77 + 42 = 174

The question is "What is the minimum number of coins with A, B and C put together?"

Hence, the answer is "174".

Choice B is the correct answer.


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