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Director  Joined: 01 May 2007
Posts: 735
If x and y are positive and x^2 * y^2 = 18 – 3xy, then x^2 =  [#permalink]

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6
10 00:00

Difficulty:   65% (hard)

Question Stats: 59% (01:09) correct 41% (01:24) wrong based on 342 sessions

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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

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-x-and-y-are-positive-and-x-2y-2-18-3xy-then-x-140728.html

Originally posted by jimmyjamesdonkey on 23 Jun 2008, 18:44.
Last edited by Bunuel on 20 Nov 2013, 07:27, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
Manager  Joined: 21 Mar 2008
Posts: 184
Re: MGMAT Question 4  [#permalink]

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1
jimmyjamesdonkey wrote:
If x and y are positive and (x^2)(y^2) = 18 - 3xy, then x^2 =

(18-3y)/y^3

18/y^2

18/(y^2+3y)

9/y^2

Lost the E answer.. the equation can be solved as (xy-3)(xy+6) = 0
Now, xy ne -6 as both are positive
xy = 3 is the only solution
x^2 = 9/y^2

D

Originally posted by Sunny143 on 23 Jun 2008, 19:06.
Last edited by Sunny143 on 23 Jun 2008, 19:13, edited 1 time in total.
Director  Joined: 01 May 2007
Posts: 735
Re: MGMAT Question 4  [#permalink]

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1
D is correct, but can anyone tell me why we can't divide by Y^2, since we know y is positive?
VP  Joined: 28 Dec 2005
Posts: 1354
Re: MGMAT Question 4  [#permalink]

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how do you get started on this one ??
Current Student Joined: 28 Dec 2004
Posts: 2991
Location: New York City
Schools: Wharton'11 HBS'12
Re: MGMAT Question 4  [#permalink]

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fresinha12 wrote:
looks like D

i just tried the ans choices..

if x^2=9/y^2

x^2*y^2=9 = 18-3*3/y * y=9..works
VP  Joined: 17 Jun 2008
Posts: 1295
Re: MGMAT Question 4  [#permalink]

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Just dividing by y^2 will not give the expression for x^2 purely in terms of y.
Director  Joined: 23 Sep 2007
Posts: 693
Re: MGMAT Question 4  [#permalink]

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jimmyjamesdonkey wrote:
If x and y are positive and (x^2)(y^2) = 18 - 3xy, then x^2 =

(18-3y)/y^3

18/y^2

18/(y^2+3y)

9/y^2

Lost the E answer.. (x^2)(y^2) = 18 - 3xy ---> (x^2)(y^2) + 3xy -18 = 0 ----> (xy)^2 + 3xy -18 = 0

(xy - 3)(xy + 6) = 0
xy = 3 or - 6, but since x and y are positive, xy must be 3

substitute xy=3 back into the original equation

(x^2)(y^2) = 18 - 9 = 9 ---> x^2 = 9/(y^2)
Manager  Joined: 03 Dec 2012
Posts: 192
Re: If x and y are positive and (x^2)(y^2) = 18 - 3xy, then x^2  [#permalink]

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1
The equation x2y2 = 18 – 3xy is really a quadratic, with the xy as the variable.
x2y2 + 3xy – 18 = 0
(xy + 6)(xy – 3) = 0
xy = 3 or -6
However, we are told that x and y are positive so xy must equal 3.
Therefore, x = 3/y and x2 = 9/y2.
Alternatively, this is a VIC (variable in choice) and can be solved by plugging numbers. If we plug a
value for y and find the corresponding value of x, we can check the answers to see which one
matches the value of x.
Looking at the values 3 and 18 in the equation, a y value of 3 makes sense.
x2(3)2 = 18 – 3(x)(3)
9x2 = 18 – 9x
9x2 – 9x + 18 = 0
x2 – x + 2 = 0
(x+ 2)(x – 1) = 0
x = 1, -2
But since x cannot be negative, x = 1
If we plug y = 3 into each of the answer choices, (C) and (D) both give an x value of 1.

We must now plug another value of y to decide between (C) and (D). Ultimately, only (D) represents
the correct value each time.
The correct answer is D
Math Expert V
Joined: 02 Sep 2009
Posts: 57244
Re: If x and y are positive and x^2 * y^2 = 18 – 3xy, then x^2 =  [#permalink]

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3
1
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 $$xy=3$$ --> $$x=\frac{3}{y}$$ --> $$x^2=\frac{9}{y^2}$$.

Answer: D.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-x-and-y-are-positive-and-x-2y-2-18-3xy-then-x-140728.html
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Manager  B
Status: Student
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Location: France
Concentration: Finance, General Management
Schools: EMLYON FT'16
GMAT 1: 650 Q47 V32 GPA: 3.44
Re: If x and y are positive and x^2 * y^2 = 18 – 3xy, then x^2 =  [#permalink]

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1
My solution:

Say $$X^2Y^2 = (XY)^2 = M^2$$

Than $$M^2=18-3M$$

$$M=3=XY$$

So $$Y^2X^2=18-9 ==> X^2=9/Y^2$$

hope it helps
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Non-Human User Joined: 09 Sep 2013
Posts: 12068
Re: If x and y are positive and x^2 * y^2 = 18 – 3xy, then x^2 =  [#permalink]

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