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

really tricky

a+b+c=17 (c)/3

now (1) yields a=3b-2c which is the same equation as 5a=9b=15c

so INSUFF


(2) yields b =0 or 3; c=0 or 5--INSUFF


IMO E

hmm I didnt consider that we cant cancel in case any of them is equal to zero.
Well done, OA is E

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.

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.

you can get c only when b Not equal to Zero

rajshree.misra wrote:If you solve 6cb = 18b, you get c = 3Therefore a+b+c = 17.

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

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.

The equations can satisfied by (9,5,3) & (0,0,0). Not sure.

(E)