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Official Solution: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, hence we can write: \(6a=3b=7c=42x\), for some number \(x\). So, \(a=7x\), \(b=14x\) and \(c=6x\). (1) \(ac=6b\) \(7x*6x=6*14x\); \(x^2=2x\); \(x=0\) or \(x=2\). Not sufficient. (2) \(5b=8a+4c\) \(5*14x=8*7x+4*6x\); \(70x=80x\); \(10x=0\); \(x=0\), hence \(a=b=c=0\) and \(a+b+c=0\). Sufficient. Answer: B
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Re: M2721
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13 Nov 2014, 20:50
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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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(c12)=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: M2721
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14 Nov 2014, 03:27
Bunuel wrote: 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 can not reduce ac = 12a by a and write c = 12 as you exclude possibility of a=0 > a(c12)=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. Oh yes. I discounted the other possibility.. This leaves BCE as answer. can you help me understand if we can use the Statement 2 to arrive at the answer in the method i was following? So in statement 2, on sperating and adding the original equation "6a=3b=7c" , we can get "6a+3b=14c" and using equation from satement 2 we can solve both as 6a = 58c => that only possibility is zero? as "6a = 7c" already? or we go to "C" both together and substitute value c = 0 & 12 to prove that only zero makes sense and with numerical values LHS != RHS? just asking mathematically. I know answer has to come B only.



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Re: M2721
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14 Nov 2014, 05:43
Alaukik wrote: Bunuel wrote: 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 can not reduce ac = 12a by a and write c = 12 as you exclude possibility of a=0 > a(c12)=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. Oh yes. I discounted the other possibility.. This leaves BCE as answer. can you help me understand if we can use the Statement 2 to arrive at the answer in the method i was following? So in statement 2, on sperating and adding the original equation "6a=3b=7c" , we can get "6a+3b=14c" and using equation from satement 2 we can solve both as 6a = 58c => that only possibility is zero? as "6a = 7c" already? or we go to "C" both together and substitute value c = 0 & 12 to prove that only zero makes sense and with numerical values LHS != RHS? just asking mathematically. I know answer has to come B only. Not quite following what you are trying to say but from (2) the only possible case is a=b=c=0.
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Re: M2721
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14 Nov 2014, 06:32
Well I was trying to say that the equation and condition in question stem were contradicting in Statement 2. so yes, only a=b=c=0 comes as solution for "B".
Thanks.



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Re: M2721
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07 May 2016, 08:36
Bunuel wrote: Official Solution:
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, hence we can write: \(6a=3b=7c=42x\), for some number \(x\). So, \(a=7x\), \(b=14x\) and \(c=6x\). (1) \(ac=6b\) \(7x*6x=6*14x\); \(x^2=2x\); \(x=0\) or \(x=2\). Not sufficient. (2) \(5b=8a+4c\) \(5*14x=8*7x+4*6x\); \(70x=80x\); \(10x=0\); \(x=0\), hence \(a=b=c=0\) and \(a+b+c=0\). Sufficient.
Answer: B hi, Just a basic doubt.. while solving for statement 1  why can't we divide x from both sides. There by we will have a solution; x = 2. Do we always to take the possibility of a variable to be zero, if not mentioned explicitly. Thanks in advance



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Re: M2721
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07 May 2016, 09:07
arunmandapaka wrote: Bunuel wrote: Official Solution:
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, hence we can write: \(6a=3b=7c=42x\), for some number \(x\). So, \(a=7x\), \(b=14x\) and \(c=6x\). (1) \(ac=6b\) \(7x*6x=6*14x\); \(x^2=2x\); \(x=0\) or \(x=2\). Not sufficient. (2) \(5b=8a+4c\) \(5*14x=8*7x+4*6x\); \(70x=80x\); \(10x=0\); \(x=0\), hence \(a=b=c=0\) and \(a+b+c=0\). Sufficient.
Answer: B hi, Just a basic doubt.. while solving for statement 1  why can't we divide x from both sides. There by we will have a solution; x = 2. Do we always to take the possibility of a variable to be zero, if not mentioned explicitly. Thanks in advance yes, always get the variables on one side and solve for the variable if nothing is specified
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Great problem!



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Re M2721
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30 May 2017, 18:08
I think this is a poorquality question. My answer was A. Please explain what is wrong with my understanding below.  We have 6a= 3b = 7c hence , 6a = 3b & 6a = 7c → b= 2a (1) and c = 6/7a (2). From this (1) and (2), we have b/c = 2a/ 6/7a = 14/6.  now a + b + c= a + 2a + 6/7a = 27/7 a. So we need to find the value of a.
From condition (1) ac=6b → a= 6b/c = 6 * 14/6 =14. Substitute this a = 14 to a+ b +c = 27/7 a we have a + +c = 54.
From condition (2), we can not find a, Hence only A is correct answer.
Please help. Thank you.



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Re: M2721
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30 May 2017, 21:15
Labmalo wrote: I think this is a poorquality question. My answer was A. Please explain what is wrong with my understanding below.  We have 6a= 3b = 7c hence , 6a = 3b & 6a = 7c → b= 2a (1) and c = 6/7a (2). From this (1) and (2), we have b/c = 2a/ 6/7a = 14/6.  now a + b + c= a + 2a + 6/7a = 27/7 a. So we need to find the value of a.
From condition (1) ac=6b → a= 6b/c = 6 * 14/6 =14. Substitute this a = 14 to a+ b +c = 27/7 a we have a + +c = 54.
From condition (2), we can not find a, Hence only A is correct answer.
Please help. Thank you. The question is flawless. The fact that you did no understood it does not mean that it's of poor quality. Your doubt is addressed in a post above as well as in the solution itself. (1) gives TWO sets a = b = c = 0 or a = 14, b = 28, and c = 12.
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Re: M2721
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05 Oct 2018, 16:53
This is a highquality 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: M2721
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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!










