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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 _________________

Re: If 5a=9b=15c, what is the value of a+b+c? [#permalink]
30 Aug 2014, 07:44

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Re: If 5a=9b=15c, what is the value of a+b+c? [#permalink]
04 Sep 2014, 17:45

Hi all,

My answer is E.

Given: 5a = 9b = 15c to find: a+b+c

Let us suppose 5a = 9b = 15c = K (some constant) Therefore, a= K/5, b = K/9, and c = K/15

a+b+c = K/5 + K/9 + K/15 = 17K/45 Now we have to find K

statement 1 : 3c-a = 5c-3b 3*(K/15)-(K/5) = 5*(K/15)-3*(K/9) This reduces to 0=0 hence, statement 1 alone is not sufficient

statement 2: 6cb = 10a I'll write this as 6cb-10a = 0

6*(K/15)*(K/9)-10(K/5) = 0 => 2*(K^2)/45 - 2*K = => 2K(K/45 - 1) = either K = 0 or K = 45 so a+b+c = 17K/45 = 0 or 17 hence, statement 2 alone is not sufficient

statement 1 and statement 2 together will not suffice because statement 1 reduces to 0.

Hence, the correct answer is E.

-------------------------------------------------------------------------------- Kudos if this helps

gmatclubot

Re: If 5a=9b=15c, what is the value of a+b+c?
[#permalink]
04 Sep 2014, 17:45

Great to know you are joining Kellogg. A lot was being talked about your last minute interview on Pagalguy (all good though). It was kinda surprise that you got the...