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ds - equation algebra

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Master | Next Rank: 500 Posts
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by ccassel » Thu Apr 07, 2011 8:22 am
Hi,

How would you answer this question?

The values of x and y vary with the value of z so that each addictive increase of 2 in the value of z corresponds to the value of x increasing by a factor of 2 and the value of y increasing by a factor of 3. If x and y are positive for each z>0, what is the value of x/(x+y) when z=12?

(1) when z=6, x=5y
(2) when z=0, x=y+1

Cheers,
Last edited by ccassel on Thu Apr 07, 2011 12:16 pm, edited 1 time in total.
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Source: — Data Sufficiency |
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by manpsingh87 » Thu Apr 07, 2011 10:31 am
ccassel wrote:Hi,

How would you answer this question?

The values of x and y vary with the value of z so that each addictive increase of 2 in the value of z corresponds to the value of x increasing by a factor of 2 and the value of y increasing by a factor of 3. If x and y are positive for each z>0, what is the value of x/(x+y) when z=12?

(1) when z=6, x=5y
(2) when x=0, x=y+1

Cheers,
wow.. a tough one really, but i will try to answer it..!!
as each addictive increase of 2 in the value of z corresponds to the value of x increasing by a factor of 2 therefore algebraically it can be presented as z+2n=2^n*x; here n represents the no. of times addition takes place.
similarly for y we have z+2n=3^n*y;

now consider statement 1) when z=6, x=5y; (assume y=k at z=6) also as per the question we have to find the value of the expression x/x+y at z=12, therefore in the expression z+2n=2^n*x; n=3, now substituting n=3 we have
6+6=2^3*5k; also 6+6=3^3*k
i.e. at z=12 we have x=40k, and y=27k
now substituting these values in x/x+y we have 40/67; hence 1) alone is sufficient.

now consider 2)x=0, x=y+1; now as we must now the value of z to calculate the value of x and y at z=12, therefore 2) is not sufficient to answer the question henceA

anyways what's the source of this question..!!!
O Excellence... my search for you is on... you can be far.. but not beyond my reach!
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by clock60 » Thu Apr 07, 2011 12:08 pm
i also got A for this, but i think there is a typo in 2 st without info about z st 2 is obviously insuff, any way
(1) z=6, x=5y
z=6+2, 2x=2*5*y,and 3y
z=6+2+2, 2*2x=2*2*5y and 3*3y
z=6+2+2+2, 2*2*2x=2*2*2*5y, and 3*3*3y
and now simply insert
(2^3*5y)/(2^3*5y+3^3y) from here we can cancel y and left with fixed value for fraction

p.s i guess that 2 st must be written when z=0, x=y+1,
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by ccassel » Thu Apr 07, 2011 12:13 pm
I thought the group might find that one interesting. I am not sure of the original source of the question. It was in a pack used as a teaching aid.

The correct answer is A.
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by ccassel » Thu Apr 07, 2011 12:16 pm
Yes you are correct. Statement 2 should read.... when z=0, x=y+1. Careless typo. Sorry.
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by clock60 » Thu Apr 07, 2011 12:29 pm
ok let us come to the end
(2) when z=0, x=y+1
using the same logic
z=0+2, 2x=2(y+1),and 3y
z=0+2*6. 2^6(y+1) and (3^6)*y
x/(x+y)=2^6*(y+1)/(2*6*(y+1)+3^6*y)-there is no way to cancel unknow y so insuff
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by ccassel » Thu Apr 07, 2011 1:21 pm
Does anyone have a plug-in way to answer this question?
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