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
Solution
Correct option is
(B)
Explanatory Answer :
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
