GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 23 Sep 2019, 13:38

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

### Hide Tags

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

### Show Tags

15 Oct 2012, 04:17
2
21
00:00

Difficulty:

65% (hard)

Question Stats:

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

### HideShow timer Statistics

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: 58116
If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?  [#permalink]

### Show Tags

15 Oct 2012, 04:22
10
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.
_________________
Senior Manager
Joined: 13 Aug 2012
Posts: 405
Concentration: Marketing, Finance
GPA: 3.23
Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?  [#permalink]

### Show Tags

10 Dec 2012, 23:57
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 $$6$$

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

_________________
Impossible is nothing to God.
##### General Discussion
Veritas Prep GMAT Instructor
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]

### Show Tags

11 Jun 2015, 21:45
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.
_________________
Karishma
Veritas Prep GMAT Instructor

Math Expert
Joined: 02 Sep 2009
Posts: 58116
Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?  [#permalink]

### Show Tags

15 Oct 2012, 04:44
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.
_________________
Senior Manager
Joined: 13 Aug 2012
Posts: 405
Concentration: Marketing, Finance
GPA: 3.23
Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?  [#permalink]

### Show Tags

18 Dec 2012, 21:45
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.

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

_________________
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]

### Show Tags

15 Oct 2012, 04:38
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
Intern
Joined: 11 Oct 2013
Posts: 18
Location: United Kingdom
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]

### Show Tags

16 Dec 2013, 19:21
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...
Manager
Joined: 07 Apr 2015
Posts: 155
Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?  [#permalink]

### Show Tags

11 Jun 2015, 03:21
I don't understand why you can not divide by y^2 here? Pls help
Math Expert
Joined: 02 Sep 2009
Posts: 58116
Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?  [#permalink]

### Show Tags

11 Jun 2015, 03:36
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.
_________________
Manager
Joined: 07 Apr 2015
Posts: 155
Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?  [#permalink]

### Show Tags

11 Jun 2015, 04:34
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
EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 15052
Location: United States (CA)
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]

### Show Tags

06 Mar 2018, 13:33
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
_________________
Contact Rich at: Rich.C@empowergmat.com

The Course Used By GMAT Club Moderators To Earn 750+

souvik101990 Score: 760 Q50 V42 ★★★★★
ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★
Manager
Joined: 23 Apr 2018
Posts: 121
If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?  [#permalink]

### Show Tags

07 Apr 2019, 10:15
@
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
Posts: 125
Location: Italy
If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?  [#permalink]

### Show Tags

08 Apr 2019, 07:29
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.
_________________
If you like my post, Kudos are appreciated! Thank you.

MEMENTO AUDERE SEMPER
Intern
Joined: 21 Jan 2015
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]

### Show Tags

11 Apr 2019, 04:03
"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.
If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?   [#permalink] 11 Apr 2019, 04:03
Display posts from previous: Sort by