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# M30-22

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Math Expert
Joined: 02 Sep 2009
Posts: 47037

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13 Oct 2014, 05:56
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Difficulty:

75% (hard)

Question Stats:

51% (01:17) correct 49% (01:43) wrong based on 47 sessions

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What is the value of $$xy$$?

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

(2) $$xy>-9.5$$

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Math Expert
Joined: 02 Sep 2009
Posts: 47037

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13 Oct 2014, 07:47
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.

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08 Jul 2015, 11:14
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?

Math Expert
Joined: 02 Sep 2009
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08 Jul 2015, 11:21
1
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 ($$x^2y^2+2xy\pi-3\pi^2 = 0$$).
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Joined: 11 Nov 2014
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GMAT 1: 660 Q48 V31
GMAT 2: 720 Q50 V37
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08 Jul 2015, 14:12
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 ?
Math Expert
Joined: 02 Sep 2009
Posts: 47037

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09 Jul 2015, 01:56
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.
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09 Jul 2015, 04:58
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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30 Nov 2016, 09:45
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.
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23 Aug 2017, 07:58
1
I think this is a high-quality question and I agree with explanation.
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Joined: 13 Oct 2017
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15 Feb 2018, 02:51
Hi Bunuel,

I'm completely lost as to how to factor out statement 1.

I did the following:

(xy)^2 + 2xy - 3pie^2 = 0
xy(xy + 2) - 3pie^2 = 0

I was then confused as to where to go next...I ended up guessing and picking C, but from this working, it appears that xy could be 3pie^2 or (-2) and combined with statement 2...would lead me to the correct answer of E.

However, according to your explanation the way I factored statement 1 was incorrect.

I've tried numerous times to factor out statement 1 the way you have but have had no luck thus far. How did you end up getting
(xy + 3pie) (xy - pie)

Thanks again.

Tosin
Math Expert
Joined: 02 Sep 2009
Posts: 47037

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15 Feb 2018, 03:14
ttaiwo wrote:
Hi Bunuel,

I'm completely lost as to how to factor out statement 1.

I did the following:

(xy)^2 + 2xy - 3pie^2 = 0
xy(xy + 2) - 3pie^2 = 0

I was then confused as to where to go next...I ended up guessing and picking C, but from this working, it appears that xy could be 3pie^2 or (-2) and combined with statement 2...would lead me to the correct answer of E.

However, according to your explanation the way I factored statement 1 was incorrect.

I've tried numerous times to factor out statement 1 the way you have but have had no luck thus far. How did you end up getting
(xy + 3pie) (xy - pie)

Thanks again.

Tosin

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06 May 2018, 09:03
1
Bunuel wrote:
What is the value of $$xy$$?

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

(2) $$xy>-9.5$$

Responding to a pm:

Put xy = z to make it easier to understand. It becomes just another quadratic

$$z^2 + 2z\pi - 3\pi^2 = 0$$
$$z^2 + 3z\pi - z\pi - 3\pi^2 = 0$$
$$z ( z + 3\pi) - \pi(z + 3\pi) = 0$$
$$(z + 3\pi)*(z - \pi) = 0$$
$$z = \pi, -3\pi$$

Two values for xy. Not sufficient.

(2) $$xy>-9.5$$
Not sufficient alone

Using both, z can still be $$\pi$$ or$$-3\pi$$ ( which is -9.4 something).
Not sufficient.

P. S. - Will respond to all PMs in the coming days (was travelling so was unable to get to the requests).
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Re: M30-22   [#permalink] 06 May 2018, 09:03
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# M30-22

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