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# Is X + Y > 0

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Senior Manager
Joined: 25 Sep 2018
Posts: 423
Location: United States (CA)
Concentration: Finance, Strategy
GMAT 1: 640 Q47 V30
GPA: 3.97
WE: Investment Banking (Investment Banking)
Is X + Y > 0  [#permalink]

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29 Sep 2018, 09:07
5
00:00

Difficulty:

55% (hard)

Question Stats:

62% (01:46) correct 38% (01:43) wrong based on 41 sessions

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Is X + Y > 0

1. $$3^{(x+y)^2} >9^{xy}$$
2.$$y=3$$

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Why do we fall?...So we can learn to pick ourselves up again
Director
Joined: 19 Oct 2013
Posts: 519
Location: Kuwait
GPA: 3.2
WE: Engineering (Real Estate)
Re: Is X + Y > 0  [#permalink]

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29 Sep 2018, 09:17
1
Statement 1)

(x+y)^2 > 2xy

x^2 + 2xy + y^2 > 2xy

x^2 + y^2 > 0 x and y could be positive or negative. So x+y > 0 or x+y < 0

Insufficient

Try 2)

Even if I know y = 3 I have no information on x so insufficient.

Combine x^2 + 9 > 0

x^2 > -9 x could be -4 or 4

-4 + 3 < 0
4 + 3 > 0

Insufficient

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Is X + Y > 0  [#permalink]

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29 Sep 2018, 11:54
Abhi077 wrote:
Is X + Y > 0

1. $$3^{(x+y)^2} >9^{xy}$$
2.$$y=3$$

2) Know nothing about x -----> INSUFFICIENT
- eliminate B & D

1) 3^(x+y)^2 > 9^xy
- break bases up into their prime factors: 3^(x+y)^2 > (3^2)^xy ----> 3^(x+y)^2 > 3^2xy
- since bases are the same all we need to look at are the exponents: (x + y)^2 > 2xy
- distribute the left side: x^2 + 2xy + y^2 > 2xy
- x^2 + y^2 > 0 -----> since any number squared will always be positive or 0, all this tells us is that x & y each aren't equal to zero.

Since they can be any combo of positive and negative numbers this is insufficient ----> eliminate A

1 & 2) Since x can still be anything but zero this does not help us much.

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Is X + Y > 0   [#permalink] 29 Sep 2018, 11:54
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