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MGMAT Test question wrong?

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by Rastis » Mon Sep 15, 2014 10:54 am
If x and y are positive and x2y2 = 18 - 3xy, then x2 =

A) (18 - 3y)/y^3

B) 18/y^2

C) 18/(y^2 + 3y)

D) 9/y^2

E) 36/y^2

They say the answer is D but I believe the answer is B. If you plug in 3 for "y" and solve you get x=2 (because x,y are positive). If you plug 3 into B you get an answer of 2. Am I wrong?
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by gmatcracker0123 » Mon Sep 15, 2014 11:59 am
Rastis wrote:If x and y are positive and x2y2 = 18 - 3xy, then x2 =

A) (18 - 3y)/y^3

B) 18/y^2

C) 18/(y^2 + 3y)

D) 9/y^2

E) 36/y^2

They say the answer is D but I believe the answer is B. If you plug in 3 for "y" and solve you get x=2 (because x,y are positive). If you plug 3 into B you get an answer of 2. Am I wrong?
Is the question x^2 * y^2 = 18 - 3xy then x^2 = ?
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by Rastis » Mon Sep 15, 2014 12:06 pm
Yes. Sorry for the confusion.
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by gmatcracker0123 » Mon Sep 15, 2014 12:33 pm
Following your approach
When y=3 --> x=2

As you have been asked the value of x^2, the answers should after plugging in y=3 should yield 4 and not 2.
Ans E satisfies this value as 36/(y^2) = 36/(3^2) = 36/9 = 4 = 2^2 = x^2
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by Brent@GMATPrepNow » Mon Sep 15, 2014 12:36 pm
Rastis wrote:If x and y are positive and x²y² = 18 - 3xy, then x² =

A) (18 - 3y)/y³

B) 18/y²

C) 18/(y² + 3y)

D) 9/y²

E) 36/y²
x²y² = 18 - 3xy = (xy)² = 18 - 3xy
To make it easier to solve this equation, let k = xy
We get: k² = 18 - 3k
Rewrite as: k² + 3k - 18 = 0
Factor: (k + 6)(k - 3)
So, EITHER k = -6 OR k = 3
This means that EITHER xy = -6 OR xy = 3

case a: xy = -6
Since x and y are both POSITIVE, xy cannot equal -6.
So, we can rule out case a.

case b: xy = 3
Since x and y are both POSITIVE, it's possible for xy to equal 3.

If xy = 3, then ( xy)² = 3²
Simplify to get: x²y² = 9
Divide both sides by y² to get: x² = [spoiler]9/y²[/spoiler]

Answer: D

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by Brent@GMATPrepNow » Mon Sep 15, 2014 12:42 pm
Rastis wrote:If x and y are positive and x²y² = 18 - 3xy, then x² =

A) (18 - 3y)/y³

B) 18/y²

C) 18/(y² + 3y)

D) 9/y²

E) 36/y²

They say the answer is D but I believe the answer is B. If you plug in 3 for "y" and solve you get x=2 (because x,y are positive). If you plug 3 into B you get an answer of 2. Am I wrong?
The statement in green is not true.
If we let y = 3, we get: x²(3)² = 18 - 3x(3)
Simplify: 9x² = 18 - 9x
Rearrange: 9x² + 9x - 18 = 0
Divide both sides by 9 to get: x² + x - 2 = 0
Factor to get: (x + 2)(x - 1) = 0
So, x = -2 or x = 1
Since x is POSITIVE, it must be the case that x = 1 (when y = 3)

Cheers,
Brent
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by [email protected] » Mon Sep 15, 2014 1:06 pm
Hi Rastis,

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.

If you TEST Y = 3, then X = 1 (as Brent pointed out). Then, the answer to the question is also 1 (1² = 1). Plugging Y = 3 into the answer choices gets us 2 answers that match:

C = 18/(3² + 9) = 18/18 = 1
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: D

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by j_shreyans » Tue Sep 16, 2014 10:07 pm
Hi Guys ,

We can also solve this question through middle term :

x^2y^2=18-3xy

x^2y^2+3xy-18=0

x^2y^2+6xy-3xy-18=0
xy(xy+6)-3(xy+6)=0

(xy-3)(xy+6)=0

xy=3 and xy=-6

as given x and y are positive so we will take xy=3


now for xy=3

x must be 1 or 3

put both the value in options you will get the answer .

option D satisfy
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