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Given: \(5a=9b=15c\) --> least common multiple of 5, 9, and 15 is 45 hence we can write: \(5a=9b=15c=45x\), for some number \(x\) --> \(a=9x\), \(b=5x\) and \(c=3x\).

(1) 3c-a=5c-3b --> \(9x-9x=15x-15x\) --> \(0=0\) (this means that ANY appropriate values of a, b, and c satisfy the given statement). Not sufficient.

(2) 6cb=10a --> \(6*3x*5x=10*9x\) --> \(90x^2=90x\) --> \(x=0\) or \(x=1\). Not sufficient.

(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.

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

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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

*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]

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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]

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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?
_________________

(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]

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30 Aug 2014, 07:44

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

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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

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

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16 May 2015, 05:17

Here's the simplest way that I've found to solve the problem.

(1) 3c - a = 5c -3b Simplifies to 3b = 2c + a Multiply equation by 3 9b = 6c + 3a Substitute 5a for 9b 5a = 6c + 3a Simplify 2a = 6c 5a=15c, so we 2a always equals 6c 6c = 6c This expression is always true, so (1) does not help us figure out a+b+c at all.

(2) 3cb = 5a I converted this equation to look more similar to the question. 1/5 (15c) 1/9 (9b) = 5a 1/5 (5a) 1/9 (5a) = 5a 1/45 (25a^2) = 5a 1/9 (a^2) = a 1/9a^2 - a = 0 a(1/9a - 1) = 0 a = 0, 9. NOT SUFFICIENT

Answer is E. Many of the other strategies to get E are correct, but I would not think to do them on the test. (I guess I need to keep studying) This way is just basic basic algebra.

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

(1) 3c - a = 5c - 3b

(2) 6cb = 10a

Follow posting guidelines (link in my signatures) and do not disable the timer.

As for your question, you are given 5a = 9b = 15c ---> 5a = 9b = 15c = k --> a = k/5, b=k/9 and c=k/15. Thus, a+b+c = k *(some fixed number). Thus for determining the sufficiency, you need to find 1 unique value of 'k'.

Per statement 1, 3c - a = 5c - 3b --> substitute the values of a,b,c from above, you will see that you get 0=0 thus this means that this statement is NOT sufficient to find the value of 'k'.

Per statement 2, 6cb = 10a ---> substitute the values of a,b,c you get : k^2=45*k ---> k = 0 or k=45. Thus you still do not know 1 unique value of 'k'.

Combining the 2 statements, you still get k=0 or k=45, making E the correct answer.

When you modify the original condition and the question, divide 5a=9b=15c with 45. You get a/9=b/5=c/3 and suppose it k. That is, a/9=b/5=c/3=k and from a=9k, b=5k, c=3k, there are 4 variables(a,b,c,k) and 3 equations(a=9k, b=5k, c=3k), which should match with the number of equations. So you need 1 more equation. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer. In 1), when substituting a,b,c, it becomes 3(3k)-9k=5(3k)-3(5k) and 0=0, which is satisfied with all k. So it is not unique and therefore not sufficient. In 2) when substituting a,b,c, it becomes 6(3k)(5k)=10(9k) and k^2=k, k(k-1)=0, k=0,1 where the value of k is not unique and therefore not sufficient. Even in 1) & 2), 1) k=all numbers 2) k=0,1, which is not unique and not sufficient. Therefore the answer is E.

-> For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
_________________

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

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02 Jan 2016, 10:14

Bunuel wrote:

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

Given: \(5a=9b=15c\) --> least common multiple of 5, 9, and 15 is 45 hence we can write: \(5a=9b=15c=45x\), for some number \(x\) --> \(a=9x\), \(b=5x\) and \(c=3x\).

(1) 3c-a=5c-3b --> \(9x-9x=15x-15x\) --> \(0=0\) (this means that ANY appropriate values of a, b, and c satisfy the given statement). Not sufficient.

(2) 6cb=10a --> \(6*3x*5x=10*9x\) --> \(90x^2=90x\) --> \(x=0\) or \(x=1\). Not sufficient.

(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 Bunnel,

Can you please help me understand why A is not sufficient. With the solution you have, it comes out that a=b=c=0 so why is a+b+c = 0 wrong.

Given: \(5a=9b=15c\) --> least common multiple of 5, 9, and 15 is 45 hence we can write: \(5a=9b=15c=45x\), for some number \(x\) --> \(a=9x\), \(b=5x\) and \(c=3x\).

(1) 3c-a=5c-3b --> \(9x-9x=15x-15x\) --> \(0=0\) (this means that ANY appropriate values of a, b, and c satisfy the given statement). Not sufficient.

(2) 6cb=10a --> \(6*3x*5x=10*9x\) --> \(90x^2=90x\) --> \(x=0\) or \(x=1\). Not sufficient.

(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 Bunnel,

Can you please help me understand why A is not sufficient. With the solution you have, it comes out that a=b=c=0 so why is a+b+c = 0 wrong.

Thanks in advance!

Did we get that a=b=c=0? NO.

Given that 5a=9b=15c, 3c-a=5c-3b would be true for ANY a, b, and c. Try several numbers to check.
_________________

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