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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) 3ca=5c3b (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.
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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) 3ca=5c3b > \(9x9x=15x15x\) > \(0=0\) (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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14 Jul 2013, 00:51
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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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Re: If 5a = 9b = 15c, what is the value of a + b + c? [#permalink]
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21 Aug 2013, 21:12
We have from question that a=3c, 3b=5c, 5a=9b
Statement 1
3c  a = 5c  3b is the same as 3ba=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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22 Aug 2013, 01:55
atewari wrote: We have from question that a=3c, 3b=5c, 5a=9b
Statement 1
3c  a = 5c  3b is the same as 3ba=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: if5a9b15cwhatisthevalueofabc155926.html#p1246081 and here: if5a9b15cwhatisthevalueofabc155926.html#p1246110Hope it helps.
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Re: If 5a=9b=15c, what is the value of a+b+c? [#permalink]
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22 Aug 2013, 10:38
If 5a=9b=15c, what is the value of a+b+c?
(1) 3ca=5c3b (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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Re: If 5a=9b=15c, what is the value of a+b+c? [#permalink]
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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.
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Re: If 5a=9b=15c, what is the value of a+b+c? [#permalink]
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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?
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Re: If 5a=9b=15c, what is the value of a+b+c? [#permalink]
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25 Aug 2013, 07:29
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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25 Aug 2013, 20:11
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(c3)=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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31 Aug 2014, 03:27
jabronyo wrote: If 5a=9b=15c, what is the value of a+b+c?
(1) 3ca=5c3b (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) 3ca=5c3b => 6z = 15y9x 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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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 : 3ca = 5c3b 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 6cb10a = 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



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



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Re: If 5a = 9b = 15c , what is the value of : [#permalink]
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19 Dec 2015, 05:56
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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21 Dec 2015, 22:05
If 5a=9b=15c, what is the value of a+b+c? (1) 3ca=5c3b (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(k1)=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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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) 3ca=5c3b > \(9x9x=15x15x\) > \(0=0\) (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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03 Jan 2016, 10:47
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) 3ca=5c3b > \(9x9x=15x15x\) > \(0=0\) (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, 3ca=5c3b would be true for ANY a, b, and c. Try several numbers to check.
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