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tough gmat problem

by quantskillsgmat » Thu May 17, 2012 11:05 pm
Q The value of x and y vary with z so that each addictive increase of 2 in value of z correspondecnces to value of x increased by a factor of 2 and value of y increases by a factor of 3.If x and y are positive for each z>0, what is x/(x+y) when z=12
a)z=6 then x=5y
b)when z=0,x=y+1
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by joeyndu » Fri Jun 01, 2012 1:00 pm
Good question that is easier too intuit than to prove. I'll give it a shot.

So, we know that as z increases, x and y scale via multiplication, but at different rates, so if in know initial conditions I might be able to do something here.

1. I know how many times larger x is than y when z = 6. I also know that it takes 3 steps of adding 2 to get from z=6 to z= 12, so x will increase by a factor of 8 and y by a factor of 27. I can see that this will be sufficient. Here's why.
I need to know x/(x+y). Substation will work great if I have no additive terms and since I know x=5y, I won't have a problem.
8(5y)
------
(8(5y) + 27y)

All y terms so I can cancel the variable and I'll be left w/ numbers only

2. Very similar to statement one but with more steps to reach z= 12 so my factor increases are much larger. The key difference, however, is that I now have x = y+1 and when I substitute in, I'll maintain an additive term that won't allow me to reduce the y's out leaving only numbers. Therefore, insufficient

A.
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by Birottam Dutta » Fri Jun 01, 2012 11:55 pm
Q The value of x and y vary with z so that each addictive increase of 2 in value of z correspondecnces to value of x increased by a factor of 2 and value of y increases by a factor of 3.If x and y are positive for each z>0, what is x/(x+y) when z=12
a)z=6 then x=5y
b)when z=0,x=y+1

We have to find x/(x+y)when z=12

a) z=6, then x=5y.

When z increases from 6 to 12, we see than z has increased by 2 thrice.
So, X will also increase by a factor of 2 thrice and y will increase by a factor of 3 thrice.
Therefore, x becomes 2*2*2x= 8x, and y becomes 3*3*3y = 27Y
Also, we know x=5Y.
Then, x/(x+y) = 8X/(8X+27Y). We can replace x=5Y and get 8(5Y)/{8(5Y) + 27Y}. This equation is totally in Y, so Y will cancel in the numerator and denominator and we will get a ratio.

Hence, a is sufficient.

b) when z=0, x=y+1.

In this case, we see that z will increase six times be 2 to become 12. So,x will increase 6 times by a factor of 2 and y will increase six times by a factor of 3.

However, the equation will not reduce to a firm ratio because the equation relating x and y in this case contains an additive term 1, so it will not allow cancelling of Y from the numerator and denominator.

Hence, b is insufficient.
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by GMATGuruNY » Sat Jun 02, 2012 3:08 am
quantskillsgmat wrote:Q The value of x and y vary with z so that each addictive increase of 2 in value of z correspondecnces to value of x increased by a factor of 2 and value of y increases by a factor of 3.If x and y are positive for each z>0, what is x/(x+y) when z=12
a)z=6 then x=5y
b)when z=0,x=y+1
To determine x/(x+y) when z=12, we need to know x/y when z=12.
To illustrate:
If x/y = 1/2, then x/(x+y) = 1/(1+2) = 1/3.
If x/y = 2/5, then x/(x+y) = 2/(2+5) = 2/7.

Statement 1: When z=6, x=5y.
Thus, when z=6, x/y = 5.
Every time z increases by 2, x will increase by a factor of 2 and y will increase by a factor of 3.
Thus, we can determine x/y when z=12.
SUFFICIENT.

Statement 2: When z=0, x=y+1.
No way to determine x/y.
INSUFFICIENT.

The correct answer is A.
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