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# Is x + y < 0? (1) |x|*|y| > 0 (2) x^2y < 0

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VP
Joined: 23 Feb 2015
Posts: 1253
Is x + y < 0? (1) |x|*|y| > 0 (2) x^2y < 0  [#permalink]

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01 Mar 2019, 07:27
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25% (medium)

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83% (01:12) correct 17% (01:08) wrong based on 47 sessions

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Is $$x+y<0?$$

(1) $$|x|*|y|>0$$

(2) $$x^2y<0$$

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Senior Manager
Joined: 25 Feb 2019
Posts: 336
Re: Is x + y < 0? (1) |x|*|y| > 0 (2) x^2y < 0  [#permalink]

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01 Mar 2019, 08:06
from first .,
|x|*|y| >0 it means either x<0 and y<0
or x>0 or y>0

Insufficient

from second

x^2*y <0 , it means y <0

So this is also insufficinet , check x =5 and y=-2 and x=5 and x = -30

Even if we combined these two ,

assume x = 1 and y=-1 then we have x+y=0
x = 5 and y = -1 , we get x+y = 4
x=-5 and y =-1 we get x+y =-6

So imo E .
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Is x + y < 0? (1) |x|*|y| > 0 (2) x^2y < 0  [#permalink]

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01 Mar 2019, 15:46
Is $$x+y<0?$$

(1) $$|x|*|y|>0$$

(2) $$x^2y<0$$

(1) |X|*|y| > 0 therefore x,y cannot = 0. Let x =2, y = -3, then x+y = -1 and we get a YES. Let x =2, y =1, then x+y=3, and we get a NO NS

(2) x^2 * y <0. Then y <0, since a number squared is always positive. Let y =-1, and x =1, then x+y=0, and we get a No. Let y =-2, and x =1, then x +y = -1 and we get a Yes. NS.

(1) and (2) x,y cannot = 0 and y<0. Let y =-1, and x =1, as in (2), then x+y=0, and we get a No. Let y =-2, and x =1 as in (2), then x +y = -1 and we get a Yes. NS.

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Posts: 1810
Re: Is x + y < 0? (1) |x|*|y| > 0 (2) x^2y < 0  [#permalink]

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01 Mar 2019, 23:50
You might notice here that Statement 1 tells you almost nothing: the inequality:
$$|x| \times |y| > 0$$
is always true, unless x or y is zero. So Statement 1 is almost useless, and since Statement 2 only tells you that y is negative, then, since x can be any nonzero value at all using both statements, the answer is E.
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Re: Is x + y < 0? (1) |x|*|y| > 0 (2) x^2y < 0   [#permalink] 01 Mar 2019, 23:50
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