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

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

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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. \(\frac{36}{y^2}\)
[Reveal] Spoiler: OA
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If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ? [#permalink]

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New post 15 Oct 2012, 04:22
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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 \(xy=3\) --> \(x=\frac{3}{y}\) --> \(x^2=\frac{9}{y^2}\).

Answer: D.

Hope it's clear.
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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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New post 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 \(xy=3\) --> \(x=\frac{3}{y}\) --> \(x^2=\frac{9}{y^2}\).

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
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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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New post 15 Oct 2012, 04:44
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Expert's post
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 \(xy=3\) --> \(x=\frac{3}{y}\) --> \(x^2=\frac{9}{y^2}\).

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=29
PS algebra questions: search.php?search_id=tag&tag_id=50

Hope it helps.
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Resources:
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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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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New post 10 Dec 2012, 23:57
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\(x^2y^2 + 3xy = 18\)
\(xy (xy + 3) = 18\)

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
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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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\(x^2y^2+3xy-18=0\)
\((xy)^2+3xy-18=0\)
\((xy - 3)(xy + 6)=0\)
\(xy = 3 & xy = -6\) Since x and y are positive, we choose xy = 3.

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

Answer: D
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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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New post 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 \(xy=3\) --> \(x=\frac{3}{y}\) --> \(x^2=\frac{9}{y^2}\).

Answer: D.

Hope it's clear.


very elegant&simple...if you spot one variable in two :wink:
like it :wink:
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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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New post 11 Jun 2015, 03:21
I don't understand why you can not divide by y^2 here? Pls help
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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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New post 11 Jun 2015, 03:36
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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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New post 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
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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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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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New post 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:
[Reveal] Spoiler:
D


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Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?   [#permalink] 06 Mar 2018, 13:33
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