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Re: If 5a=9b=15c, what is the value of a+b+c? [#permalink]
22 Aug 2013, 09:38

If 5a=9b=15c, what is the value of a+b+c?

(1) 3c-a=5c-3b (2) 6cb=10a

*Note - my answer is different from the official answer. I'm posting it here to confirm whether I'm correct or not.[/quote]

Statement 1: insufficient because having infinitely many solutions Statement 2: 6cb=10a means that 6cb=18b and 6cb=30c, so cb=3b and cb=5c c=3b/b=3 and b=5c/c=5 providing that c and b are not equal to 0. So I think ambiguity with 0 is resolved and a=9 so sum is 17. Sufficient

Re: If 5a=9b=15c, what is the value of a+b+c? [#permalink]
22 Aug 2013, 09:41

Expert's post

Temurkhon wrote:

If 5a=9b=15c, what is the value of a+b+c?

(1) 3c-a=5c-3b (2) 6cb=10a

*Note - my answer is different from the official answer. I'm posting it here to confirm whether I'm correct or not.

Statement 1: insufficient because having infinitely many solutions Statement 2: 6cb=10a means that 6cb=18b and 6cb=30c, so cb=3b and cb=5c c=3b/b=3 and b=5c/c=5 providing that c and b are not equal to 0. So I think ambiguity with 0 is resolved and a=9 so sum is 17. Sufficient

Re: If 5a=9b=15c, what is the value of a+b+c? [#permalink]
24 Aug 2013, 04:57

OA is E; As correctly pointed out by bunuel

You cannot simply reduce an expression ab=bc does not imply that a=c, as it is perfectly possible for b to be zero and var a might not be equal to var c.

_________________

--It's one thing to get defeated, but another to accept it.

Re: If 5a=9b=15c, what is the value of a+b+c? [#permalink]
24 Aug 2013, 06:01

Bunuel wrote:

(1)+(2) x=0 or x=1, thus a=b=c=0 (for x=0) --> a+b+c=0 OR a=9, b=5, c=3 (for x=1) --> a+b+c=17. Not sufficient.

Answer: E.

Hope it's clear.

Hi Bunuel,

When combining St 1 and St2, isn't the only acceptable solution is when x=0, since x=0 is common to both statements. How x=1 validates both the statements? Pls clarify?

Re: If 5a=9b=15c, what is the value of a+b+c? [#permalink]
25 Aug 2013, 06:29

Expert's post

imhimanshu wrote:

Bunuel wrote:

(1)+(2) x=0 or x=1, thus a=b=c=0 (for x=0) --> a+b+c=0 OR a=9, b=5, c=3 (for x=1) --> a+b+c=17. Not sufficient.

Answer: E.

Hope it's clear.

Hi Bunuel,

When combining St 1 and St2, isn't the only acceptable solution is when x=0, since x=0 is common to both statements. How x=1 validates both the statements? Pls clarify?

The part you are quoting gives two examples: a=b=c=0 a=9, b=5, c=3