# gre q

#### sfspitfire23

If 3/5 = x/y, then
Col A: x - 3
Col B: y – 5

Which col is larger/are they the same/not enough information.

Now, I can solve each and get x-3=(3y/5)-3 and y-5=(5x/3)-5. In this case, the answer is not enough info because it depends on the values of x and y.

But then, You could also do this:

3y=5x. Now by subtracting 15 on both sides
3y-15=5x-15 => 3(y-5)=5(x-3)

=>(y-5)/(x-3)=5/3 so (y-5) is greater than (x-3) so answer is 'B'

Is the second method here the correct way to do this problem?

#### undefined

MHF Hall of Honor
If 3/5 = x/y, then
Col A: x - 3
Col B: y – 5

Which col is larger/are they the same/not enough information.

Now, I can solve each and get x-3=(3y/5)-3 and y-5=(5x/3)-5. In this case, the answer is not enough info because it depends on the values of x and y.

But then, You could also do this:

3y=5x. Now by subtracting 15 on both sides
3y-15=5x-15 => 3(y-5)=5(x-3)

=>(y-5)/(x-3)=5/3 so (y-5) is greater than (x-3) so answer is 'B'

Is the second method here the correct way to do this problem?
The easiest way in my opinion is to look at some small examples, for example (x,y) = (3,5) where clearly those two quantities are equal, then (x,y) = (6,10) where clearly those two quantities are not equal, therefore there is not enough information.

#### Also sprach Zarathustra

3/5=x/y=(x-3)/(y-5)<1