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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?
110
174
154
165
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 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
By: Amit Kumar ProfileResourcesReport error
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