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Manager  Joined: 22 Dec 2011
Posts: 216
If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?  [#permalink]

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21 00:00

Difficulty:   65% (hard)

Question Stats: 65% (02:16) correct 35% (02:46) wrong based on 281 sessions

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If x and y are positive and x^2 * y^2 = 18 – 3xy, then x^2 =?

A. $$\frac{(18-3y)}{y^3}$$

B. $$\frac{18}{y^2}$$

C. $$\frac{18}{(y^2+3y)}$$

D. $$\frac{9}{y^2}$$

E. $$\frac{36}{y^2}$$
Math Expert V
Joined: 02 Sep 2009
Posts: 58116
If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?  [#permalink]

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1
Jp27 wrote:
If x and y are positive and x^2 * y^2 = 18 – 3xy, then x^2 =?

A. $$\frac{(18-3y)}{y^3}$$

B. $$\frac{18}{y^2}$$

C. $$\frac{18}{(y^2+3y)}$$

D. $$\frac{9}{y^2}$$

E. $$\frac{36}{y^2}$$

$$(xy)^2+3(xy)-18=0$$ --> solving for $$xy$$ --> $$xy=-6$$ (not a valid solution: $$xy$$ must be positive as both unknowns are positive) or $$xy=3$$ --> so $$xy=3$$ --> $$x=\frac{3}{y}$$ --> $$x^2=\frac{9}{y^2}$$.

Hope it's clear.
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Senior Manager  Joined: 13 Aug 2012
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Concentration: Marketing, Finance
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Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?  [#permalink]

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5
$$x^2y^2 + 3xy = 18$$
$$xy (xy + 3) = 18$$

I thought of 2 positive numbers such as $$n$$ and $$n + 3$$ whose product is 18 --> $$3$$ and $$6$$

$$xy = 3$$
$$x^2 = 9/y^2$$

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##### General Discussion
Veritas Prep GMAT Instructor D
Joined: 16 Oct 2010
Posts: 9647
Location: Pune, India
Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?  [#permalink]

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2
noTh1ng wrote:
Bunuel wrote:
noTh1ng wrote:
I don't understand why you can not divide by y^2 here? Pls help

If you divide x^2 * y^2 = 18 – 3xy by y^2 you get x^2 = 18/y^2 - 3/y = (18-3y)/y^2. We don't have (18-3y)/y^2 among answer choices, thus we need other approach.

I sometimes don't get the GMAC..

just because it's not among the answer choices doesnt make it wrong imho

If an expression is not in the options, it doesn't make the expression wrong. You might just need to manipulate it further.

Here, dividing by y^2 doesn't work: $$x^2 * y^2 = 18 – 3xy$$
When you divide by y^2, you get $$x^2 = 18/y^2 – 3x/y$$
How do you separate the x and y since you need to write x^2 in terms of y only?
You will need to divide by xy and then solve for it.

By the way, GMAC is absolutely reasoning based. If something confuses you, especially in Quant, it means you have missed a point.
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Math Expert V
Joined: 02 Sep 2009
Posts: 58116
Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?  [#permalink]

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1
Jp27 wrote:
Bunuel wrote:
Jp27 wrote:
If x and y are positive and x^2 * y^2 = 18 – 3xy, then x^2 =?

A. (18-3y)/y^3
B. 18/y^2
C. 18/(y^2+3y)
D. 9/y^2
E. 36/y^2

$$(xy)^2+3(xy)-18=0$$ --> solving for $$xy$$ --> $$xy=-6$$ (not a valid solution: $$xy$$ must be positive as both unknowns are positive) or $$xy=3$$ --> so $$xy=3$$ --> $$x=\frac{3}{y}$$ --> $$x^2=\frac{9}{y^2}$$.

Hope it's clear.

Hi Bunuel - thanks for your swift response.

Where I can get similar question like these for practice?

I have done the roots and equation set in your signature, are there any other sets that I can use for practice? Sincere thanks!

cheers

Check our question banks (viewforumtags.php) for more questions.

DS algebra questions: search.php?search_id=tag&tag_id=29
PS algebra questions: search.php?search_id=tag&tag_id=50

Hope it helps.
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Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?  [#permalink]

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1
1
$$x^2y^2+3xy-18=0$$
$$(xy)^2+3xy-18=0$$
$$(xy - 3)(xy + 6)=0$$
$$xy = 3 & xy = -6$$ Since x and y are positive, we choose xy = 3.

$$xy = 3$$
$$x = \frac{3}{y}$$
$$x^2 = \frac{9}{y^2}$$

_________________
Impossible is nothing to God.
Manager  Joined: 22 Dec 2011
Posts: 216
Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?  [#permalink]

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Bunuel wrote:
Jp27 wrote:
If x and y are positive and x^2 * y^2 = 18 – 3xy, then x^2 =?

A. (18-3y)/y^3
B. 18/y^2
C. 18/(y^2+3y)
D. 9/y^2
E. 36/y^2

$$(xy)^2+3(xy)-18=0$$ --> solving for $$xy$$ --> $$xy=-6$$ (not a valid solution: $$xy$$ must be positive as both unknowns are positive) or $$xy=3$$ --> so $$xy=3$$ --> $$x=\frac{3}{y}$$ --> $$x^2=\frac{9}{y^2}$$.

Hope it's clear.

Hi Bunuel - thanks for your swift response.

Where I can get similar question like these for practice?

I have done the roots and equation set in your signature, are there any other sets that I can use for practice? Sincere thanks!

cheers
Intern  Joined: 11 Oct 2013
Posts: 18
Location: United Kingdom
Concentration: General Management, Leadership
GMAT 1: 490 Q32 V25 GPA: 3.9
WE: Other (Other)
Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?  [#permalink]

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Bunuel wrote:
Jp27 wrote:
If x and y are positive and x^2 * y^2 = 18 – 3xy, then x^2 =?

A. (18-3y)/y^3
B. 18/y^2
C. 18/(y^2+3y)
D. 9/y^2
E. 36/y^2

$$(xy)^2+3(xy)-18=0$$ --> solving for $$xy$$ --> $$xy=-6$$ (not a valid solution: $$xy$$ must be positive as both unknowns are positive) or $$xy=3$$ --> so $$xy=3$$ --> $$x=\frac{3}{y}$$ --> $$x^2=\frac{9}{y^2}$$.

Hope it's clear.

very elegant&simple...if you spot one variable in two like it _________________
Good things come to those who wait… greater things come to those who get off their ass and do anything to make it happen...
Manager  Joined: 07 Apr 2015
Posts: 155
Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?  [#permalink]

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I don't understand why you can not divide by y^2 here? Pls help
Math Expert V
Joined: 02 Sep 2009
Posts: 58116
Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?  [#permalink]

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noTh1ng wrote:
I don't understand why you can not divide by y^2 here? Pls help

If you divide x^2 * y^2 = 18 – 3xy by y^2 you get x^2 = 18/y^2 - 3/y = (18-3y)/y^2. We don't have (18-3y)/y^2 among answer choices, thus we need other approach.
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Manager  Joined: 07 Apr 2015
Posts: 155
Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?  [#permalink]

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Bunuel wrote:
noTh1ng wrote:
I don't understand why you can not divide by y^2 here? Pls help

If you divide x^2 * y^2 = 18 – 3xy by y^2 you get x^2 = 18/y^2 - 3/y = (18-3y)/y^2. We don't have (18-3y)/y^2 among answer choices, thus we need other approach.

I sometimes don't get the GMAC..

just because it's not among the answer choices doesnt make it wrong imho
EMPOWERgmat Instructor V
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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?  [#permalink]

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Hi All,

This question has a "quirky" design element to it - it takes a bit longer than normal to solve AND you might have to TEST VALUES TWICE to get to the correct answer.

You could TEST Y = 3, then X = 1. Then, the answer to the question is also 1 (1² = 1). Plugging Y = 3 into the answer choices gets us 2 answers that match:

Answer C = 18/(3² + 9) = 18/18 = 1
Answer D = 9/3² = 9/9 = 1

None of the other answers matches, so we're down to 2 choices. At this point, you can either guess one of the 2 or go back and TEST a new value.

If we use Y = 2 in the original equation, then we have
4X² = 18 - 6X
4X² +6X - 18 = 0
2X² +3X - 9 = 0

Factoring this down might seem a little strange, but it DOES lead to a solution…
(X + 3)(2X - 3) = 0
X = -3 or +3/2

Since we're told that X and Y are both POSITIVE, X = 3/2

So now we plug Y = 2 into the answers and we're looking for (3/2)² = 9/4

Answer C: 18/10 = 9/5

GMAT assassins aren't born, they're made,
Rich
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Manager  S
Joined: 23 Apr 2018
Posts: 121
If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?  [#permalink]

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@
Bunuel wrote:
Jp27 wrote:
If x and y are positive and x^2 * y^2 = 18 – 3xy, then x^2 =?

A. $$\frac{(18-3y)}{y^3}$$

B. $$\frac{18}{y^2}$$

C. $$\frac{18}{(y^2+3y)}$$

D. $$\frac{9}{y^2}$$

E. $$\frac{36}{y^2}$$

$$(xy)^2+3(xy)-18=0$$ --> solving for $$xy$$ --> $$xy=-6$$ (not a valid solution: $$xy$$ must be positive as both unknowns are positive) or $$xy=3$$ --> so $$xy=3$$ --> $$x=\frac{3}{y}$$ --> $$x^2=\frac{9}{y^2}$$.

Hope it's clear.

IT's a really basic thing I want to ask you, but I must before my exam, I feel...
How do you equate the 2 equations here..
like can you tell me, how is (xy-3) (xy+6) derived ..

I understand that once we open the brackets, we get the above equation, also that 1 entity must be negative and 1 be positive as we have +3 and -18.

But, I ask, how 6 and 3.. is there any other way I am missing ? or is this the right way only..

I welcome any sort of help that will make this crystal clear for me
Manager  G
Joined: 21 Feb 2019
Posts: 125
Location: Italy
If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?  [#permalink]

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Call $$xy = t$$ and solve the equation. You'll get:

$$t^2 +3t -18 = 0$$
$$(t + 6) (t - 3) = 0$$. $$6 - 3 = 3$$; $$6 * -3 = -18$$

$$xy = -6$$
$$xy = 3$$

Since you consider only positive values, $$x = \frac{3}{y}$$, so $$x^2 = \frac{9}{y^2}$$.

Hope it's clear.
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Intern  B
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Posts: 10
Location: India
Concentration: Finance, Technology
WE: Engineering (Computer Software)
If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?  [#permalink]

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"Plug in the value" is a good approach for this one. Take x = 3 and y = 1. Plug in the values and we'll get 9 as the answer. Moreover, from the options only one option D satisfies that.
On the more regular approach, I really like the method of taking xy as one variable T and then soving the quadratic equation thus formed.

P.S. :- This is my first post. Please give kudos if this helps you and help me grow here. Thanks in advance. If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?   [#permalink] 11 Apr 2019, 04:03
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