Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 22 Dec 2011
Posts: 211

If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?
[#permalink]
Show Tags
15 Oct 2012, 04:17
Question Stats:
64% (02:15) correct 36% (02:44) wrong based on 315 sessions
HideShow timer Statistics
If x and y are positive and x^2 * y^2 = 18 – 3xy, then x^2 =? A. \(\frac{(183y)}{y^3}\) B. \(\frac{18}{y^2}\) C. \(\frac{18}{(y^2+3y)}\) D. \(\frac{9}{y^2}\) E. \(\frac{36}{y^2}\)
Official Answer and Stats are available only to registered users. Register/ Login.




Math Expert
Joined: 02 Sep 2009
Posts: 59685

If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?
[#permalink]
Show Tags
15 Oct 2012, 04:22
Jp27 wrote: If x and y are positive and x^2 * y^2 = 18 – 3xy, then x^2 =?
A. \(\frac{(183y)}{y^3}\)
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 \(xy=3\) > \(x=\frac{3}{y}\) > \(x^2=\frac{9}{y^2}\). Answer: D. Hope it's clear.
_________________




Senior Manager
Joined: 13 Aug 2012
Posts: 397
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
\(x^2y^2 + 3xy = 18\) \(xy (xy + 3) = 18\)
I thought of 2 positive numbers such as \(n\) and \(n + 3\) whose product is 18 > \(3\) and \(6\)
\(xy = 3\) \(x^2 = 9/y^2\)
Answer: D




Manager
Joined: 22 Dec 2011
Posts: 211

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. (183y)/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 \(xy=3\) > \(x=\frac{3}{y}\) > \(x^2=\frac{9}{y^2}\). Answer: D. 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: 59685

Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?
[#permalink]
Show Tags
15 Oct 2012, 04:44
Jp27 wrote: Bunuel wrote: Jp27 wrote: If x and y are positive and x^2 * y^2 = 18 – 3xy, then x^2 =?
A. (183y)/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 \(xy=3\) > \(x=\frac{3}{y}\) > \(x^2=\frac{9}{y^2}\). Answer: D. 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=29PS algebra questions: search.php?search_id=tag&tag_id=50Hope it helps.
_________________



Senior Manager
Joined: 13 Aug 2012
Posts: 397
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
\(x^2y^2+3xy18=0\) \((xy)^2+3xy18=0\) \((xy  3)(xy + 6)=0\) \(xy = 3 & xy = 6\) Since x and y are positive, we choose xy = 3.
\(xy = 3\) \(x = \frac{3}{y}\) \(x^2 = \frac{9}{y^2}\)
Answer: D



Intern
Joined: 11 Oct 2013
Posts: 18
Location: United Kingdom
Concentration: General Management, Leadership
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. (183y)/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 \(xy=3\) > \(x=\frac{3}{y}\) > \(x^2=\frac{9}{y^2}\). Answer: D. Hope it's clear. very elegant&simple...if you spot one variable in two like it



Manager
Joined: 07 Apr 2015
Posts: 152

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: 59685

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 = (183y)/y^2. We don't have (183y)/y^2 among answer choices, thus we need other approach.
_________________



Manager
Joined: 07 Apr 2015
Posts: 152

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 = (183y)/y^2. We don't have (183y)/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



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9871
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
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 = (183y)/y^2. We don't have (183y)/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
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 15704
Location: United States (CA)

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 Answer C: 18/10 = 9/5 Answer D: 9/4 Final Answer: GMAT assassins aren't born, they're made, Rich
_________________
Contact Rich at: Rich.C@empowergmat.comThe 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: 162

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. \(\frac{(183y)}{y^3}\)
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 \(xy=3\) > \(x=\frac{3}{y}\) > \(x^2=\frac{9}{y^2}\). Answer: D. 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 (xy3) (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.



Intern
Joined: 21 Jan 2015
Posts: 14
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.



Rice (Jones) School Moderator
Joined: 18 Jun 2018
Posts: 316
Location: United States (AZ)
Concentration: Finance, Healthcare
GPA: 3.36

Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?
[#permalink]
Show Tags
14 Nov 2019, 11:42
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




Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?
[#permalink]
14 Nov 2019, 11:42






