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# M30-22

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Math Expert
Joined: 02 Sep 2009
Posts: 42575

Kudos [?]: 135405 [0], given: 12692

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13 Oct 2014, 04:56
Expert's post
6
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Difficulty:

65% (hard)

Question Stats:

56% (01:27) correct 44% (01:49) wrong based on 43 sessions

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What is the value of $$xy$$?

(1) $$x^2y^2+2xy\pi-3\pi^2 = 0$$

(2) $$xy>-9.5$$
[Reveal] Spoiler: OA

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Kudos [?]: 135405 [0], given: 12692

Math Expert
Joined: 02 Sep 2009
Posts: 42575

Kudos [?]: 135405 [0], given: 12692

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13 Oct 2014, 06:47
Expert's post
1
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Official Solution:

What is the value of $$xy?$$

(1) $$x^2y^2+2xy\pi-3\pi^2 = 0$$. This is the same as $$(xy)^2+2xy\pi-3\pi^2 = 0$$.

Solve for $$xy$$ (you can denote $$xy$$ as $$a$$ and solve quadratics) or factor: $$(xy+3\pi)(xy-\pi)=0$$. This means that $$xy=-3\pi$$ or $$xy=\pi$$. Not sufficient.

(2) $$xy>-9.5$$. Clearly insufficient.

(1)+(2) From (1) $$xy=-3\pi\approx{-9.45}$$ or $$xy=\pi\approx{3.14}$$. Both of these values are more than -9.5 (both of them satisfy the second statement), hence both of them are valid. Not sufficient.

_________________

Kudos [?]: 135405 [0], given: 12692

Intern
Joined: 11 Nov 2014
Posts: 40

Kudos [?]: 35 [0], given: 28

Concentration: Technology, Strategy
GMAT 1: 660 Q48 V31
GMAT 2: 720 Q50 V37
GPA: 3.6
WE: Consulting (Consulting)

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08 Jul 2015, 10:14
Bunuel

Can you tell me where am I wrong in the following:
From the sttmt A, we actually may get (xy-π)^2 - 4π^2=0. Then (xy-π)^2 = 4π^2. From then take the sqrt from both sides to arrive at a different solution from yours. Where did I go wrong?

Kudos [?]: 35 [0], given: 28

Math Expert
Joined: 02 Sep 2009
Posts: 42575

Kudos [?]: 135405 [1], given: 12692

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08 Jul 2015, 10:21
1
KUDOS
Expert's post
michaelyb wrote:
Bunuel

Can you tell me where am I wrong in the following:
From the sttmt A, we actually may get (xy-π)^2 - 4π^2=0. Then (xy-π)^2 = 4π^2. From then take the sqrt from both sides to arrive at a different solution from yours. Where did I go wrong?

'
If you expand (xy-π)^2 - 4π^2=0 you'll get (xy)^2 - 2xyπ - 3π^2 = 0, which is not what we have i the firs statement ($$x^2y^2+2xy\pi-3\pi^2 = 0$$).
_________________

Kudos [?]: 135405 [1], given: 12692

Intern
Joined: 11 Nov 2014
Posts: 40

Kudos [?]: 35 [0], given: 28

Concentration: Technology, Strategy
GMAT 1: 660 Q48 V31
GMAT 2: 720 Q50 V37
GPA: 3.6
WE: Consulting (Consulting)

### Show Tags

08 Jul 2015, 13:12
Bunuel wrote:
michaelyb wrote:
Bunuel

Can you tell me where am I wrong in the following:
From the sttmt A, we actually may get (xy-π)^2 - 4π^2=0. Then (xy-π)^2 = 4π^2. From then take the sqrt from both sides to arrive at a different solution from yours. Where did I go wrong?

'
If you expand (xy-π)^2 - 4π^2=0 you'll get (xy)^2 - 2xyπ - 3π^2 = 0, which is not what we have i the firs statement ($$x^2y^2+2xy\pi-3\pi^2 = 0$$).

Ok, my mistake (pretty usual lately). How about (xy+π)^2 - 4π^2=0 ?

Kudos [?]: 35 [0], given: 28

Math Expert
Joined: 02 Sep 2009
Posts: 42575

Kudos [?]: 135405 [0], given: 12692

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09 Jul 2015, 00:56
michaelyb wrote:
Bunuel wrote:
michaelyb wrote:
Bunuel

Can you tell me where am I wrong in the following:
From the sttmt A, we actually may get (xy-π)^2 - 4π^2=0. Then (xy-π)^2 = 4π^2. From then take the sqrt from both sides to arrive at a different solution from yours. Where did I go wrong?

'
If you expand (xy-π)^2 - 4π^2=0 you'll get (xy)^2 - 2xyπ - 3π^2 = 0, which is not what we have i the firs statement ($$x^2y^2+2xy\pi-3\pi^2 = 0$$).

Ok, my mistake (pretty usual lately). How about (xy+π)^2 - 4π^2=0 ?

In this case you will get the same answer:

$$(xy+\pi)^2=4\pi^2$$.

$$xy+\pi=2\pi$$ OR $$xy+\pi=-2\pi$$.

$$xy=\pi$$ OR $$xy=-3\pi$$.

Hope it's clear.
_________________

Kudos [?]: 135405 [0], given: 12692

Intern
Joined: 11 Nov 2014
Posts: 40

Kudos [?]: 35 [0], given: 28

Concentration: Technology, Strategy
GMAT 1: 660 Q48 V31
GMAT 2: 720 Q50 V37
GPA: 3.6
WE: Consulting (Consulting)

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09 Jul 2015, 03:58
Actually, according to GMAT Club Math Book, the square root of an even number is always positive, so I thought that I cannot assume sqrt(4) = -2.
Please, see attached file "Square Root" - it´s a screenshot of the book.
>> !!!

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Kudos [?]: 35 [0], given: 28

Intern
Joined: 19 Jun 2015
Posts: 1

Kudos [?]: [0], given: 0

Location: India
Concentration: General Management, Strategy
GPA: 3.79

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30 Nov 2016, 08:45
Its not square root here..its square...and we can get square both from positive and negative value of a number....eg 25 can be obtained both from 5^2 & (-5)^2.

Kudos [?]: [0], given: 0

Intern
Joined: 02 Oct 2011
Posts: 15

Kudos [?]: 5 [0], given: 129

Location: Thailand
Concentration: Strategy, Finance
GMAT 1: 560 Q47 V20
GPA: 3.26

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23 Aug 2017, 06:58
I think this is a high-quality question and I agree with explanation.

Kudos [?]: 5 [0], given: 129

Re M30-22   [#permalink] 23 Aug 2017, 06:58
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# M30-22

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