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Re: M27-21 [#permalink]
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Hey Bunuel,

Can u explain why this question turned out to ratio problem, as described.

On first inference it seems like a variable Question with three variables in an equated expression "6a = 3b =7c". And statement A says "ac = 6b" which solves to c = 12 (ac = 6 x 2a => c = 12 & b = 28 & a = 14) and a+b+c = 54.

statement 2 does not make sense in this scenario.

so answer to this comes as "A".

can you explain how to infer this question so that it dosent go either way.
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Re: M27-21 [#permalink]
Great problem!
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Re: M27-21 [#permalink]
This is a high-quality trap question. Many people just cancel out the variable in case (1) and end up marking D as answer. Thanks Bunuel.
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Re: M27-21 [#permalink]
like most questions from Bunuel, this tests your core understanding of the concept and the assumptions that go into those concepts. In this case, exclusion of the possibility that a-->0

On a side note, i did think of using the LCM but had no idea what will i do with it. this question has showed me how to!
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Re: M27-21 [#permalink]
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Bunuel wrote:
If \(6a=3b=7c\), what is the value of \(a+b+c\)?


(1) \(ac=6b\)

(2) \(5b=8a+4c\)



Just another way to look at it. Use relation between any two variables and solve the statements.
Here 6a=3b gives b=2a a very friendly ratio, so let us work on them.

(1) \(ac=6b\)
Now b=2a or 6b=12a
Thus ac = 12a, so
a=0, which will mean c=0 as 6a=7c
OR
c=12.
Insufficient

(2) \(5b=8a+4c\)
Again 3b=6a or b=2a, that is 4b=8a
Thus \(5b=8a+4c=4b+4c………..b=4c \), but 3b=7c.
Both give different relations between b and c, and only possibility is that both b and c are 0.
Thus a=b=c=0, and a+b+c also becomes 0.
Sufficient


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Re: M27-21 [#permalink]
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M27-21 [#permalink]
Expert Reply
Similar PS question to practice:

https://gmatclub.com/forum/m41-429460.html
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Re: M27-21 [#permalink]
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