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

Thanks Bunnuel.

My answer was B.

Using the original question, I solved for a:

5a = 9b
a=9b/5

Then I substituted a into statement 2:

6cb = 10a
6cb = 10(9b/5)
6c = 90/5
30c = 90
c = 3

Once I have c , I calculated for a and b:

5a = 15(3)
a = 45/5
a = 9

9b = 15(3)
b = 5

a+b+c = 17

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.

I made the same mistake! Thanks Brunel for pointing it out

My take is E.

5a=9b=15c
LCM(5,9,15) = 45.
=> a = 9x, b = 5y, and c = 3z

1) 3c-a=5c-3b
=> 6z = 15y-9x
or 2z = 5y - 3x

y and x must be odd for LHS to be even. so multiple solutions exist.

2) 6cb = 10a
6*5y*3z = 10*9x
x = yz

x=y=z=0
or x=y=z=1
not sufficient.

1)+2)
x=y=z=0
or x=y=z=1
not sufficient.

Hence, E.

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

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.

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.

Hope this helps.

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.

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.

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.