Bunuel wrote:
Spaniard wrote:
Bunuel wrote:
12. If 6a=3b=7c, what is the value of a+b+c?
Given: 6a=3b=7c --> least common multiple of 6, 3, and 7 is 42 hence we ca write: 6a=3b=7c=42x, for some number x --> a=7x, b=14x and c=6x.
(1) ac=6b --> 7x*6x=6*14x --> x^2=2x --> x=0 or x=2. Not sufficient.
(2) 5b=8a+4c --> 5*14x=8*7x+4*14x --> 70x=80x --> 10x=0 --> x=0 --> a=b=c=0 --> a+b+c=0. Sufficient.
Answer: B.
Given: 6a=3b=7c --> least common multiple of 6, 3, and 7 is 42 hence we ca write: 6a=3b=7c=42x, for some number x --> a=7x, b=14x and c=6x.
(1) ac=6b --> 7x*6x=6*14x --> x^2=2x --> x=0 or x=2. Not sufficient.
(2) 5b=8a+4c --> 5*14x=8*7x+4*14x --> 70x=80x --> 10x=0 --> x=0 --> a=b=c=0 --> a+b+c=0. Sufficient.
Answer: B.
Hi Bunuel,
In the 1st condition if we take ac = 6b and hence multiply the the given expression 6a=3b=7c by 2 we get 12a=6b=14c.
Now substituting 12a=ac=14c
I get 12a=ac --> c = 12 and similarly a = 14 and hence b= 28.
Can you please point where am I going wrong?
Regards
ac=12a (here you can not reduce by a and write c=12 as you exclude possibility of a=0) --> a(c-12)=0 --> either a=0 OR c=12. So, we get either a=b=c=0 or a=14, b=28 and c=12.
Does this make sense?
I had missed that possibility. Thanks a lot.