is xy > 0 ?

[joehurundas]

Is xy > 0 ?
(1) x - y > -2
(2) x - 2y < -6

[purnendu.shukla]

I think answer is "C"
combining A&B y>4
which -> x>2 so xy>0 suff

[vivekcall81]

C cannot be the answer if you x=3 and y=5
yes XY>0 but out it in to the equation
x-y=3-5=-2 which is equal but not .-2
hence not satisfied and E should be the answer.

[adiagr]

C cannot be the answer if you x=3 and y=5
yes XY>0 but out it in to the equation
x-y=3-5=-2 which is equal but not .-2
hence not satisfied and E should be the answer.

Vivek,

Given (1) and (2), we have to see whether xy>0.

Now x=3, y=5

x-y=-2, which means that (1) itself is not satisfied. so you cannot take this pair at all.

From (1) and (2) together

y>4 and x>2........so xy>0.

It appears Ans is C.

[RonPurewal]

Vivek,

Given (1) and (2), we have to see whether xy>0.

Now x=3, y=5

x-y=-2, which means that (1) itself is not satisfied. so you cannot take this pair at all.

yes.

@ vivek, at this point i would recommend a careful study of the first principles of data sufficiency -- it appears that you are still a bit hazy on how the data sufficiency problems work. without that sort of baseline understanding, you shouldn't advance to solving problems like this one yet.

--

for any other readers reading this thread, the key to solving the statements together is the usual key to combining inequalities: IF THE INEQUALITIES ARE BOTH ">" OR BOTH "<", YOU CAN ADD THEM TOGETHER.

the original versions of statements 1 and 2 have opposite inequality signs, so multiply the second one by -1 in order to achieve ">" on both:

(1) x - y > -2
(2) 2y - x > 6

then add these together, to give y > 4. the rest of the solution is as posted above by other users.

[joehurundas]

thanks all for your contributions to solving the problem; the trick, as explained by RonPurewal, lies in reversing the sign before addition.
OA is C

[mschwrtz]

glad Ron (et alia) could help

[jigar24]

Hi Ron, but could you be more clear on how to eliminate 'B' first?.. I am able to eliminate 'A' (and so also 'D') easily but getting stuck on eliminating 'B' (only statement 2 alone is sufficient).
Thanks

My approach:

In order for xy>0 both x and y either have to positive or both negative.. So, I tired to come up with numbers 1) both with same signs 2) Both with different signs .. if both these type work in a statement then it implies given statement is NOT sufficient..

In the process, I could very easily prove statement 1 insufficient but getting stuck (and taking an eternity) on statement 2 ... cant come up with examples fast enough..

Please help

[RonPurewal]

Hi Ron, but could you be more clear on how to eliminate 'B' first?.. I am able to eliminate 'A' (and so also 'D') easily but getting stuck on eliminating 'B' (only statement 2 alone is sufficient).
Thanks

My approach:

In order for xy>0 both x and y either have to positive or both negative.. So, I tired to come up with  numbers 1) both with same signs  2) Both with different signs .. if both these type work in a statement then it implies given statement is NOT sufficient.. 

In the process, I could very easily prove statement 1 insufficient but  getting stuck (and taking an eternity) on statement 2 ... cant come up with examples fast enough..

Please help

this task is a lot easier if you solve the inequality for one of the variables, i.e., x < -6 + 2y.  now, if you plug in something like 10 for y, you get x < 14, for which all three signs (positive, negative, and zero) are possible.  so, insufficient.

alternatively, if you don't think of solving the original inequality for x, just find a solution in which both numbers are positive (say, 10 and 10), and then realize that you can just make one of the numbers 0.  that's good enough, since 0 > 0 is false.

TAKEAWAY:
there are three signs, not just two; positive, negative, and zero.
don't forget that zero is a sign!

[jigar24]

Ron, I understood the first part that you mentioned.. That will surely make life much easier.. One has to try to get +ve value on the right hand side of '<' sign and a -ve value on the right hand side of '>' sign, in order to include all three signs and prove the statement insufficient.. Right??

However, I am still a bit unclear about your second explanation:

"alternatively, if you don't think of solving the original inequality for x, just find a solution in which both numbers are positive (say, 10 and 10), and then realize that you can just make one of the numbers 0. that's good enough, since 0 > 0 is false."

Could you please explain this once more, with an example??

[jigar24]

Thanks

[RonPurewal]

Ron, I understood the first part that you mentioned.. That will surely make life much easier.. One has to try to get +ve value on the right hand side of '<' sign and a -ve value on the right hand side of '>' sign, in order to include all three signs and prove the statement insufficient.. Right??

However, I am still a bit unclear about your second explanation:

"alternatively, if you don't think of solving the original inequality for x, just find a solution in which both numbers are positive (say, 10 and 10), and then realize that you can just make one of the numbers 0. that's good enough, since 0 > 0 is false."

Could you please explain this once more, with an example??

there's already an example in there.

if x = 10 and y = 10, then statement 2 is satisfied, and the answer to the prompt question is "YES; xy > 0".

now we need a NO (this is how number-plugging on data sufficiency works; you want to try to prove "insufficient", since further YES's won't help you).
one easy way to get a NO will be to set one of x or y to 0, since 0 is not greater than 0.
so, say, x = 0, and y = 10. then statement 2 is true again, but, NO, xy is not > 0 this time.

therefore, insufficient.

--

GRAPHICAL SOLUTION

if you know what the graphs of these inequalities look like -- they're shaded on one side of a straight line -- then you'll know that any such inequality will have to shade in at least two neighboring quadrants.
however, in any 2 neighboring quadrants, xy will have opposite signs; so we actually know that this statement is insufficient for ANY inequality of the form Ax + By < C or Ax + By > C.

cool stuff.