This is a tricky one...let me try though!!!
Let 5a=9b=15c=k
so a=k/5 ; b=k/9 ; c=k/15
Statement 1:
3c-a=5c-3b
==> 2c = (k/3) - (k/5)
==> 2c = 2* k/15
==> c = k/15 this wat we assumed...dont arrive at anything
so i think A insufficient!!! frnds please help me to solve st.1 in a better way
Statement 2:
6cb=10a
==> 3cb = 5a
==> 3cb = k (from the assumption made in the question stem)
==> 3*(k/9)*(k/15) = k
==> k^2/45 =k
==>k^2 = 45 k (we cant cancel k since we dont know whether k[u]> [/u] 0 or k < 0)....Please correct me if i am wrong on this!!!)
==> k(k-45) = 0
so k = 0 or k = 45
not sufficient!!!
HENCE IMO E
FRNDS PLS COMMENT ON MY REASONING...IF AT ANY POINT I AM WRONG KINDLY HIGHLIGHT THAT!!
Thanks in advance,
Apoorva
If 5a = 9b = 15c, what is the value of a + b + c?
This topic has expert replies
Source: Beat The GMAT — Data Sufficiency |
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apoorva.srivastva
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Answer: E.
The important thing to note about S1 is that all it does is shuffle the numbers around. It you run through the substitutions, you get 0 = 0 in the end.
S2 has already been explained. The statements combined will yield nothing new as S1 gives no new information.
The important thing to note about S1 is that all it does is shuffle the numbers around. It you run through the substitutions, you get 0 = 0 in the end.
S2 has already been explained. The statements combined will yield nothing new as S1 gives no new information.
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rajshree.misra
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Hi,
My answer is B. i.e. statement 2 is sufficient.
The main question states that 5a=9b=15c
The second statement mentions that 6cb=10a
Since 10a = 2(5a), 10a = 18b = 30c
therefore, 6cb = 18b = 30c
If you solve 6cb = 18b, you get c = 3 and if you solve 6cb = 30c, you get b = 5. From these two numbers, you get that a = 9.
Therefore a+b+c = 17.
My answer is B. i.e. statement 2 is sufficient.
The main question states that 5a=9b=15c
The second statement mentions that 6cb=10a
Since 10a = 2(5a), 10a = 18b = 30c
therefore, 6cb = 18b = 30c
If you solve 6cb = 18b, you get c = 3 and if you solve 6cb = 30c, you get b = 5. From these two numbers, you get that a = 9.
Therefore a+b+c = 17.
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abhinav85
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IMO D.
From the given equation we get,
5a=9b=15c.
So that means 5x9=9x4=15x3.......because the LCM of 5,9 and 15 is 45.
So now look at the statement that they satisfies this!!!
From 1 we get
(1) 3c – a = 5c – 3b
that means 3x3 - 9=5x3 - 3x5 that is 0.Sufficeint.
From 2 we get,
(2) 6cb = 10a
same thing 6x3x5 = 10x9.........i.e 90 = 90. Sufficeint.
BTW what is the OA!!!
hmm I didnt consider that we cant cancel in case any of them is equal to zero.
Well done, OA is E
From the given equation we get,
5a=9b=15c.
So that means 5x9=9x4=15x3.......because the LCM of 5,9 and 15 is 45.
So now look at the statement that they satisfies this!!!
From 1 we get
(1) 3c – a = 5c – 3b
that means 3x3 - 9=5x3 - 3x5 that is 0.Sufficeint.
From 2 we get,
(2) 6cb = 10a
same thing 6x3x5 = 10x9.........i.e 90 = 90. Sufficeint.
BTW what is the OA!!!
hmm I didnt consider that we cant cancel in case any of them is equal to zero.
Well done, OA is E
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Domnu
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The answer is E. Here's a way using matrix algebra:
1) Consider the matrix
5 -9 0
0 9 -15
1 -3 2
This has determinant 0, so the information contained in part 1 is useless and is essentially already provided to us.
2) From this, we get either b = 0 or c = 3. This tells us nothing about A.
So, the answer is E.
1) Consider the matrix
5 -9 0
0 9 -15
1 -3 2
This has determinant 0, so the information contained in part 1 is useless and is essentially already provided to us.
2) From this, we get either b = 0 or c = 3. This tells us nothing about A.
So, the answer is E.
Have you wondered how you could have found such a treasure? -T
