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# If x + y > 0, is xy < 0 ?

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Director
Joined: 31 Oct 2013
Posts: 856
Concentration: Accounting, Finance
GPA: 3.68
WE: Analyst (Accounting)
If x + y > 0, is xy < 0 ?  [#permalink]

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08 Jun 2018, 06:15
00:00

Difficulty:

55% (hard)

Question Stats:

67% (02:17) correct 33% (02:15) wrong based on 51 sessions

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If x + y > 0, is xy < 0 ?

(1) $$x^{2y} < 1$$

(2) x + 2y < 0
Intern
Joined: 01 Jan 2018
Posts: 44
Re: If x + y > 0, is xy < 0 ?  [#permalink]

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08 Jun 2018, 12:34
for xy< 0 any one of x or y should be negative.

given is x+y > 0.

Statement 1:

x^2y < 1.

x = 0.1 and y = 1 would satisfy the condition and xy is not less than 0.
x = 2 and y = -1 would also satisfy the condition and xy is less than 0.
Hence A is not sufficient.

Statement 2:

x + 2y < 0

We already know that x + y > 0. Doubling Y makes the equations -ve. Therefore Y should be negative and X is positive.
xy is < 0.

Hence 'B'.
Director
Status: Learning stage
Joined: 01 Oct 2017
Posts: 931
WE: Supply Chain Management (Energy and Utilities)
If x + y > 0, is xy < 0 ?  [#permalink]

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08 Jun 2018, 21:10
selim wrote:
If x + y > 0, is xy < 0 ?

(1) $$x^{2y} < 1$$

(2) x + 2y < 0

Given, x+y > 0

The above inequality is valid when ,
(a) x>0, $$y\leq{0}$$, |x|>|y|
(b)$$x\leq{0}$$, y>0,|x|<|y|
(c) x>0, y>0

St1, $$x^{2y} < 1$$
Or, $$(x^2)^y <1$$
so this could be valid when y<0 and x can take both the signs. so xy can be positive or negative. (e.g, when x=2, y=-1; xy<0 & when x=-2,y=-1; xy>0)
hence st1 insufficient. (As we have got Y & N answers to the question stem, no need to check other cases)

St2, x+2y < 0
Or,( x+y)+y < 0
This is statement is valid only when x>0, y<0 and |x|>|y| and this maintains the given condition (x+y)>0.
Therefore, xy<0

Hence st2 is sufficient.

Ans. B
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PKN

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Senior Manager
Joined: 29 Dec 2017
Posts: 385
Location: United States
Concentration: Marketing, Technology
GMAT 1: 630 Q44 V33
GMAT 2: 690 Q47 V37
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GPA: 3.25
WE: Marketing (Telecommunications)
If x + y > 0, is xy < 0 ?  [#permalink]

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09 Jun 2018, 03:26
If x + y > 0, is xy < 0 ?

(1) $$x^{2y} < 1$$, $$(x^{y})^{2} < 1$$, $$- 1 <x^{y} < 1$$

If x=0, y=1, then xy=0 - NO
If x=-1/2, y=2, then xy<0 - YES

Insufficient

(2) x + 2y < 0 and x + y > 0 (- x - y < 0),
Let's combine both statements
x + 2y < 0
- x - y < 0
--------------
y<0, hence from x + y > 0, x > - y (where -y>0), so it must be that x>0. xy<0

Sufficient

If x + y > 0, is xy < 0 ? &nbs [#permalink] 09 Jun 2018, 03:26
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