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

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Manager
Joined: 22 Dec 2011
Posts: 200
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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Question Stats:

66% (02:18) correct 34% (02:47) wrong based on 404 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
Joined: 02 Sep 2009
Posts: 64249
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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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
Posts: 386
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
7
$$x^2y^2 + 3xy = 18$$
$$xy (xy + 3) = 18$$

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

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

##### General Discussion
Manager
Joined: 22 Dec 2011
Posts: 200
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
Math Expert
Joined: 02 Sep 2009
Posts: 64249
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
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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18 Dec 2012, 20:45
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.

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

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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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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
Manager
Joined: 07 Apr 2015
Posts: 150
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
Math Expert
Joined: 02 Sep 2009
Posts: 64249
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.
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Posts: 150
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
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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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11 Jun 2015, 20:45
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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Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?  [#permalink]

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06 Mar 2018, 12:33
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

GMAT assassins aren't born, they're made,
Rich
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Manager
Joined: 23 Apr 2018
Posts: 160
If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?  [#permalink]

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07 Apr 2019, 09:15
@
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
Joined: 21 Feb 2019
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If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?  [#permalink]

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08 Apr 2019, 06:29
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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If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?  [#permalink]

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11 Apr 2019, 03:03
"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.
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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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14 Nov 2019, 10:42
Let's assume y = 1 ==> $$x^{2}$$ + 3x - 18 = 0 ==> x = -6 or 3 ==> x = 3 since we are told x is positive ==> $$x^{2}$$ = $$3^{2}$$ = 9 ==> test the answers with y = 1, the option that equals 9 is the answer ==> answer is D
Senior Manager
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If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?  [#permalink]

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01 May 2020, 09:17
for ease of calculation let xy = a (a positive integer)
a^2=18-3a
a^2+3a-18=0
(a+6)(a-3)
a=3
a=-6 (not possible as xy is a +ve integer

putting in the equation in the stem
x^2=18-9/y^2
x^2=9/y^2
If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?   [#permalink] 01 May 2020, 09:17