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M27-21

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Bunuel
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M27-21 [#permalink] New post  16 Sep 2014, 01:27
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23% (01:55) correct 77% (02:25) wrong based on 185 sessions
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If \(6a=3b=7c\), what is the value of \(a+b+c\)?


(1) \(ac=6b\)

(2) \(5b=8a+4c\)
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Bunuel
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Re M27-21 [#permalink] New post  16 Sep 2014, 01:28
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If \(6a=3b=7c\), what is the value of \(a+b+c\)?

Given: \(6a = 3b = 7c\). The least common multiple of 6, 3, and 7 is 42, so we can write: \(6a = 3b = 7c = 42x\), for some number \(x\). Thus, \(a = 7x\), \(b = 14x\), and \(c = 6x\).

(1) \(ac=6b\)

\(7x*6x=6*14x\);

\(x^2=2x\);

\(x=0\) or \(x=2\).

Not sufficient.

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

\(5*14x=8*7x+4*6x\);

\(70x=80x\);

\(10x=0\);

\(x=0\)

As \(x = 0\), we find that \(a = b = c = 0\) and \(a + b + c = 0\).

Sufficient.


Answer: B
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Bunuel
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Re: M27-21 [#permalink] New post  14 Nov 2014, 02:33
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Alaukik wrote:
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.


You cannot reduce ac = 12a 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. So, the first statement is not sufficient.

Hope it's clear.
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Re: M27-21 [#permalink] New post  13 Nov 2014, 20:50
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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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mh72
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Re: M27-21 [#permalink] New post  24 May 2016, 07:23
Great problem!
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psaikrishna90
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Re: M27-21 [#permalink] New post  05 Oct 2018, 16:53
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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Mansoor50
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Re: M27-21 [#permalink] New post  07 Oct 2018, 05:56
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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chetan2u
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Re: M27-21 [#permalink] New post  26 Jan 2022, 00:30
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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


B
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Bunuel
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Re: M27-21 [#permalink] New post  28 Apr 2023, 13:47
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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]
28 Apr 2023, 13:47
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