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

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Senior Manager
Joined: 22 Dec 2011
Posts: 294

Kudos [?]: 319 [2], given: 32

If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ? [#permalink]

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15 Oct 2012, 03:17
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65% (hard)

Question Stats:

67% (01:43) correct 33% (02:15) wrong based on 249 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. $$\frac{36}{y^2}$$
[Reveal] Spoiler: OA

Kudos [?]: 319 [2], given: 32

Math Expert
Joined: 02 Sep 2009
Posts: 43334

Kudos [?]: 139626 [6], given: 12794

If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ? [#permalink]

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15 Oct 2012, 03:22
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Expert's post
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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.
_________________

Kudos [?]: 139626 [6], given: 12794

Senior Manager
Joined: 22 Dec 2011
Posts: 294

Kudos [?]: 319 [0], given: 32

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

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15 Oct 2012, 03:38
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

Kudos [?]: 319 [0], given: 32

Math Expert
Joined: 02 Sep 2009
Posts: 43334

Kudos [?]: 139626 [1], given: 12794

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

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15 Oct 2012, 03:44
1
KUDOS
Expert's post
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.
_________________

Kudos [?]: 139626 [1], given: 12794

Senior Manager
Joined: 13 Aug 2012
Posts: 456

Kudos [?]: 589 [3], given: 11

Concentration: Marketing, Finance
GPA: 3.23
Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ? [#permalink]

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10 Dec 2012, 22:57
3
KUDOS
$$x^2y^2 + 3xy = 18$$
$$xy (xy + 3) = 18$$

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

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

_________________

Impossible is nothing to God.

Kudos [?]: 589 [3], given: 11

Senior Manager
Joined: 13 Aug 2012
Posts: 456

Kudos [?]: 589 [0], given: 11

Concentration: Marketing, Finance
GPA: 3.23
Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ? [#permalink]

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18 Dec 2012, 20:45
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$$x^2y^2+3xy-18=0$$
$$(xy)^2+3xy-18=0$$
$$(xy - 3)(xy + 6)=0$$
$$xy = 3 & xy = -6$$ 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.

Kudos [?]: 589 [0], given: 11

Intern
Joined: 11 Oct 2013
Posts: 18

Kudos [?]: 33 [0], given: 34

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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16 Dec 2013, 18:21
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...

Kudos [?]: 33 [0], given: 34

Manager
Joined: 07 Apr 2015
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Kudos [?]: 74 [0], given: 185

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

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11 Jun 2015, 02:21
I don't understand why you can not divide by y^2 here? Pls help

Kudos [?]: 74 [0], given: 185

Math Expert
Joined: 02 Sep 2009
Posts: 43334

Kudos [?]: 139626 [0], given: 12794

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

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11 Jun 2015, 02:36
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.
_________________

Kudos [?]: 139626 [0], given: 12794

Manager
Joined: 07 Apr 2015
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Kudos [?]: 74 [0], given: 185

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

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11 Jun 2015, 03:34
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

Kudos [?]: 74 [0], given: 185

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7868

Kudos [?]: 18494 [1], given: 237

Location: Pune, India
Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ? [#permalink]

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11 Jun 2015, 20:45
1
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Expert's post
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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Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ? [#permalink]

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Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?   [#permalink] 01 Nov 2017, 09:31
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# If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?

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