Data Sufficiency

from 1) x and y are either both positive and x>y or x and y have different signal and x> 0. more than 1 possible answer, insufficient.

from2) both x and y have the same signal negative or positive we don't know, insufficient

combining both, from 1) x is always positive and from 2) x and y both have the same signal, so both are positive. answer C

Can someone explain what happens after we combine both stmts? I'm still a little confused with agps's explanation. Thanks.

agps wrote:from 1) x and y are either both positive and x>y or x and y have different signal and x> 0. more than 1 possible answer, insufficient.

from2) both x and y have the same signal negative or positive we don't know, insufficient

combining both, from 1) x is always positive and from 2) x and y both have the same signal, so both are positive. answer C

Jameschan168

A suggestion when u find it diff to relate to variables plug in values.

Take the question in discussion -

Statement 1 gives X>Y

So X can be 6 and Y can be 3 or X can be -3 and Y can be -6

Statement 2 gives X/y >1

If you look at the statement alone again X can be 6 and Y can be 3 or X can be -3 and Y can be -6

But if you combine the statement you know X > y so for x/y to be > 1 they will have to be positive. IF they were negative X/y would be a fraction and so less than 1

Stmt 1 - both, x and y, can be positive or negative. NOT SUFF

Stmt 2 - both will be negative or positive. NOT SUFF

Together - from stmt 2 |x|>|y| and stmt 1 satisfies only when both are positive. So SUFF

Ans C

ans is correct e:
you guys did not consider x=-1/2 and y=-1.

rather x=-1 and y=-1/2

gmatguy16 wrote:rather x=-1 and y=-1/2

2x-2y will NOT be equal to 1

2*-1 - 2*(-1/2) = -2+1 = -1

Ans will be C

Thanks ri2007.

ri2007 wrote:Jameschan168

A suggestion when u find it diff to relate to variables plug in values.

Take the question in discussion -

Statement 1 gives X>Y

So X can be 6 and Y can be 3 or X can be -3 and Y can be -6

Statement 2 gives X/y >1

If you look at the statement alone again X can be 6 and Y can be 3 or X can be -3 and Y can be -6

But if you combine the statement you know X > y so for x/y to be > 1 they will have to be positive. IF they were negative X/y would be a fraction and so less than 1