Is x + y 0? (1) xy 0 (2) x^3y^2 0

Question

Is  x + y > 0 ?

(1)  xy > 0

(2)  x^3y^2 > 0

TestPrepUnlimited · Answer

Currently, there are no conditions on either x or y.

Statement 1: 
This tells us x and y must have the same sign. We can have both x and y positive so that  x + y > 0 , or both negative so that  x + y < 0 . Insufficient.

Statement 2: 
Since  x^2  and  y^2  both must be positive, we are allowed to divide both sides by those terms. Then we can get  x > 0 . We still don't know anything about y so insufficient.

Combined: 
Since x is positive, combined with (1) y must also be positive. Then  x + y > 0  must be true. Sufficient.

AlphaCentauri4ly · Answer

The only way x+y is always greater than zero is when both x & y are positive. so we can rephrase this as "Are both x & y positive?"

From (1)

We know that xy > 0,

That means both x & y have same signs i.e. x=+ve, y=+ve or x=-ve, y=-ve. (not sufficient)

From (2)

We know that x^3y^2 > 0

In order for this to be true, x must be +ve because y^2 is always +ve regardless of sign of y and the product must be greater than zero which means positive.

Since this says nothing about sign of y, not sufficient.

From (1) & (2)

we know that both x & y are positive, since from (2) x is +ve and from (1) x & y have same signs.

So the answer is "C".

Kinshook · Answer

Asked: Is x + y > 0 ? (1) xy > 0 Either x> 0 & y>0; x + y > 0 or x < 0 & y < 0 ; x + y < 0 NOT SUFFICIENT (2) x^3y^2 > 0 x > 0 y may be > 0 or < 0 x + y may be > 0 or x + y < 0 NOT SUFFICIENT (1) + (2) (1) xy > 0 (2) x^3y^2 > 0 Since x>0 -> y>0; x + y > 0 SUFFICIENT IMO C

dimri10 · Answer

Statement 1: +*+ or -*- can bring a positive result. So not sufficient
2:
Divide by x^2*y^2 both sides to get X>0 , but nothing about Y.
Combined: X is + ( from stat. 2) , so only option +*+ is acceptable in statement 1. So both sufficient. X and y are> than 0 so X+y is > than 0

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satya2029 · Answer

From 1
Both can be negative, resulting x+y<0
Both can be postive, resulting x+y>0
not sufficient
From 2
as y^2 is always positive, for this to be true x>0
Not sufficient
From 1 and 2
x>0 and y>0 as xy>0
sufficient
C:)

Is x + y > 0? (1) xy > 0 (2) x^3y^2 > 0

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