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16 Sep 2014, 01:27
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If \(6a=3b=7c\), what is the value of \(a+b+c\)?
(1) \(ac=6b\)
(2) \(5b=8a+4c\)
(1) \(ac=6b\)
(2) \(5b=8a+4c\)
16 Sep 2014, 01:28
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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, 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
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
14 Nov 2014, 02:33
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Alaukik
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.
