Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

13 Jul 2013, 23:30

8

This post received KUDOS

Expert's post

5

This post was BOOKMARKED

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.

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

Show Tags

22 Aug 2013, 10:38

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]

Show Tags

22 Aug 2013, 10:41

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]

Show Tags

24 Aug 2013, 05:57

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]

Show Tags

24 Aug 2013, 07:01

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]

Show Tags

25 Aug 2013, 07:29

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]

Show Tags

30 Aug 2014, 08:44

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]

Show Tags

04 Sep 2014, 18:45

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]

Show Tags

16 May 2015, 06:17

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.

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

Show Tags

19 Dec 2015, 05:56

Expert's post

sara86 wrote:

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.

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

Show Tags

21 Dec 2015, 22:05

Expert's post

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

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

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]

Show Tags

02 Jan 2016, 11:14

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.

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

Show Tags

03 Jan 2016, 10:47

Expert's post

neeraj609 wrote:

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.

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. _________________

Post your Blog on GMATClub We would like to invite all applicants who are applying to BSchools this year and are documenting their application experiences on their blogs to...

Since the value of the NZ Dollar is much lower than the Pound, foreign currency exchange rates and how to pay MBA tuition fees are obviously of much concern...