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?	Pick.
A) 110
B) 127
C) 154
D) 165
E) 145
DETAILED SOLUTION   Given, 7A = 5127   42A = 30127   Also given 6127 = 11C   30127 = 55C   Therefore 42A = 30127 = 55C   Let 42A = 30127 = 55C = k   Rewriting, A/55 = 127/30 = C/42   The least possible integral values for A, 127, C will be A = 42; 127 = 30 and C = 55   Total = 42 + 30 + 55= 
<gadget id="calculator">42 + 30 + 55</gadget>
<output>127</output>
127   Correct Answer: 127
<result>B</result>
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?