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

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

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

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

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New post 13 Jul 2013, 23:51
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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
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Re: If 5a=9b=15c, what is the value of a+b+c? [#permalink]

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New post 14 Jul 2013, 00:12
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jabronyo wrote:
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


The problem with your solution is that you cannot reduce 6cb = 10(9b/5) by b. Doing so you exclude b=0 case:

6cb = 10(9b/5) --> cb=3b --> b(c-3)=0 --> b=0 or c=3.

Hope it's clear.
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Re: If 5a = 9b = 15c, what is the value of a + b + c? [#permalink]

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New post 21 Aug 2013, 20:12
We have from question that a=3c, 3b=5c, 5a=9b

Statement 1

3c - a = 5c - 3b is the same as 3b-a=2c, substitute a with 3c, we get 5c=3b. Hence Statement 1 is Not Sufficient.

Statement 2

6cb = 10a is the same as 6cb = 30c,
Dividing both sides with 3c gives 2b = 10 and b = 5

With the value of b we can find that c = 3 and a = 9.

Hence statement 2 is Sufficient.

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

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atewari wrote:
We have from question that a=3c, 3b=5c, 5a=9b

Statement 1

3c - a = 5c - 3b is the same as 3b-a=2c, substitute a with 3c, we get 5c=3b. Hence Statement 1 is Not Sufficient.

Statement 2

6cb = 10a is the same as 6cb = 30c,
Dividing both sides with 3c gives 2b = 10 and b = 5

With the value of b we can find that c = 3 and a = 9.

Hence statement 2 is Sufficient.

Thus the answer, B.


The correct answer is E, not B. Check here: if-5a-9b-15c-what-is-the-value-of-a-b-c-155926.html#p1246081 and here: if-5a-9b-15c-what-is-the-value-of-a-b-c-155926.html#p1246110

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

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

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New post 22 Aug 2013, 09:41
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


The correct answer is E, not B. Check:
if-5a-9b-15c-what-is-the-value-of-a-b-c-155926.html#p1246081
if-5a-9b-15c-what-is-the-value-of-a-b-c-155926.html#p1246110
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Re: If 5a=9b=15c, what is the value of a+b+c? [#permalink]

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

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

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New post 25 Aug 2013, 06:29
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:
\(a=b=c=0\)
\(a=9\), \(b=5\), \(c=3\)
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Re: If 5a=9b=15c, what is the value of a+b+c? [#permalink]

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New post 25 Aug 2013, 19:11
Bunuel wrote:
jabronyo wrote:
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


The problem with your solution is that you cannot reduce 6cb = 10(9b/5) by b. Doing so you exclude b=0 case:

6cb = 10(9b/5) --> cb=3b --> b(c-3)=0 --> b=0 or c=3.

Hope it's clear.


I made the same mistake! Thanks Brunel for pointing it out
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Re: If 5a=9b=15c, what is the value of a+b+c? [#permalink]

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

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New post 31 Aug 2014, 02:27
jabronyo 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.


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

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New post 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.

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

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

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New post 19 Dec 2015, 04:56
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.

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

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New post 21 Dec 2015, 21:05
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.
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Re: If 5a=9b=15c, what is the value of a+b+c? [#permalink]

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New post 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.

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

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New post 03 Jan 2016, 09:47
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.
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Re: If 5a=9b=15c, what is the value of a+b+c?   [#permalink] 03 Jan 2016, 09:47
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