If x + y > 0, is xy < 0 ? (1) x^(2y) < 1 (2) x + 2y < 0

If \(x + y > 0\), is \(xy < 0\) ?

We are looking for an answer to - Do x and y have opposite sign?


(1) \(x^{2y} < 1\)
We can have numerous combinations.
x=2 and y=-1.....Answer is YES.
x=1/2 and y=3/2......Answer is NO.

(2) \(x + 2y < 0\)
Since x+y>0, we can write x+y>0>x+2y.
We can also add similar sides to get x+y+0>x+2y+0 or x+y>x+2y
\(x+y>x+2y\) or \(y<0.\)
But x+y>0, so surely x>0.
Thus x*y < 0.
Sufficient


B
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Question: is xy <0?

Key answer: Do x and y have opposite signs?

1. Statement 1: x^2y < 1

Since x is raised to an even power, the result will always be positive. To evaluate if x and y have different signs, we can consider two cases: y>0 and y<0

If y <0, the statement would be true for all values of x such that x > 1 or x < -1

If y >0, the statement would be true for all values of x between - 1 and 1

Hence, we cannot determine if the signs are opposite - INSUFFICIENT

Statement 2: x + 2y <0

This can be rearranged to read:

x < -2y

Now we have three possibilities

If y >0, x <0
If y <0, x >0
If y = 0, x <0 but from the question, we know that x + y >0, therefore this case cannot be considered true (This is important, I missed it in my first shot at the question!)

Hence, x and y will have opposite signs

Hence, xy<0

SUFFICIENT

Answer is B

https://gmatclub.com/forum/if-x-y-0-is-xy-0-1-x-2y-1-2-x-2y-385600.html?fl=similar

given that x+y > 0

Statement 1 ------ x^2y < 1
we can make multiple cases but lets look at the easy cases
Case 1 ---- x < 0 and y < 0 it will suffice the statement and we get xy > 0
Case 2 ---- x > 0 and y <0 - it will suffice the given statement and we get xy < 0
We do not get a conclusive answer
INSUFFICIENT

Statement 2 ----- given that x + 2y < 0 ---- from the question we are also given that x + y > 0
lets substract the inequalities with opposite signs
x + 2y - x - y < 0 ---------- y < 0
if y < 0 and x + y > 0 --- the it follows that x > 0 in all cases
hence xy < 0
SUFFICIENT

Answer - B

hello chetan2u

why cant we take equal to(XY=0) case 1 for option 2, to make it insufficient.
i am confused here, a little guidance will be helpful.

If x+y>0, is xy<0 ?
(1) x2y < 1
(2) x+2y < 0

Option (2) : x+2y < 0 , given : x+y>0 , que : XY< 0 ?

Case 1: X= 1/2 and Y=0 thus X+Y > 0 and XY=0
Case 2: X= 1/3 and Y=1/4 thus X+Y > 0 and XY>0

For xy=0, at least one of x or y should be 0.
Now, in the initial statement x+y>0, meaning if any of x or y is ZERO, the other is POSITIVE.
Also, in the statement II, x+2y<0, meaning if any of x or y is ZERO, the other is NEGATIVE .
Both the above statements contradict each other, so \(xy\neq 0\)
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