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Question

# 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?

CAT 2021

A
B
C
D
##### 165
Solution

Correct option is

(B)

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.

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

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