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If 6xy = x^2y + 9y, what is the value of xy?

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If 6xy = x^2y + 9y, what is the value of xy? [#permalink]

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If 6xy = \(x^2\)y + 9y, what is the value of xy?

(1) x = –2
(2) x < 0

[Reveal] Spoiler:
The answer is D, with the following explanation:
The equation in question can be rephrased as follows:

x2y – 6xy + 9y = 0
y(x2 – 6x + 9) = 0
y(x – 3)2 = 0

Therefore, one or both of the following must be true:
y = 0 or
x = 3

It follows that the product xy must equal either 0 or 3y. This question can therefore be rephrased several different ways: "What is y?" or "Does x = 3?"

(1) SUFFICIENT: If x = -2, then:
(–2)2y – 6(–2)y + 9y = 0
4y + 12y + 9y = 0
25y = 0

Therefore y = xy = 0. Tn fact, for any value of x other than 3, y must equal zero.

(2) SUFFICIENT: x is negative. Therefore, x cannot equal 3, and it follows that y = 0. Therefore, xy = 0.

The correct answer is D.

What I did, i simply applied the info from statement 1, and got that -12Y=13Y which can only be true if Y is 0 and therefore, XY is 0.
In the second statement, i did the same and the result came to be the same. Is this faulty reasoning? Also, this is said to be 700-800 level question and my way seems just too easy to solve such a high level problem...
[Reveal] Spoiler: OA

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Last edited by reto on 16 Jul 2015, 00:55, edited 1 time in total.
proper format

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Re: If 6xy = x^2y + 9y, what is the value of xy? [#permalink]

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New post 19 Dec 2010, 16:22
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MisterEko wrote:
Guys, I just have a question if my way of solving it is reasonable and can often be applied here. I got the right answer on this one, but am not sure it would work usually:

If 6xy = x2y + 9y, what is the value of xy?

(1) x = –2

(2) x < 0

The answer is D, with the following explanation:
[Reveal] Spoiler:
The equation in question can be rephrased as follows:

x2y – 6xy + 9y = 0
y(x2 – 6x + 9) = 0
y(x – 3)2 = 0

Therefore, one or both of the following must be true:
y = 0 or
x = 3

It follows that the product xy must equal either 0 or 3y. This question can therefore be rephrased several different ways: "What is y?" or "Does x = 3?"

(1) SUFFICIENT: If x = -2, then:
(–2)2y – 6(–2)y + 9y = 0
4y + 12y + 9y = 0
25y = 0

Therefore y = xy = 0. Tn fact, for any value of x other than 3, y must equal zero.

(2) SUFFICIENT: x is negative. Therefore, x cannot equal 3, and it follows that y = 0. Therefore, xy = 0.

The correct answer is D.


What I did, i simply applied the info from statement 1, and got that -12Y=13Y which can only be true if Y is 0 and therefore, XY is 0.
In the second statement, i did the same and the result came to be the same. Is this faulty reasoning? Also, this is said to be 700-800 level question and my way seems just too easy to solve such a high level problem...


For statement (1) your reasoning looks OK. As for statement (2) what do you mean by "did the same"?

One can also do this way: \(6xy=x^2y + 9y\) --> \(y(x^2-6x+9)=0\) --> \(y(x-3)^2=0\) --> either \(x=3\) or \(y=0\) (or both).

(1) x = –2 --> so \(x\neq{3}\), then \(y=0\) and \(xy=0\). Sufficient.

(2) x < 0 --> again \(x\neq{3}\), then \(y=0\) and \(xy=0\). Sufficient.

Answer: D.

Check question #1 here: inequality-and-absolute-value-questions-from-my-collection-86939.html for harder version of the same problem.

Hope it helps.
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Re: If 6xy = x^2y + 9y, what is the value of xy? [#permalink]

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New post 19 Dec 2010, 17:01
Bunuel wrote:
MisterEko wrote:
Guys, I just have a question if my way of solving it is reasonable and can often be applied here. I got the right answer on this one, but am not sure it would work usually:

If 6xy = x2y + 9y, what is the value of xy?

(1) x = –2

(2) x < 0

The answer is D, with the following explanation:
[Reveal] Spoiler:
The equation in question can be rephrased as follows:

x2y – 6xy + 9y = 0
y(x2 – 6x + 9) = 0
y(x – 3)2 = 0

Therefore, one or both of the following must be true:
y = 0 or
x = 3

It follows that the product xy must equal either 0 or 3y. This question can therefore be rephrased several different ways: "What is y?" or "Does x = 3?"

(1) SUFFICIENT: If x = -2, then:
(–2)2y – 6(–2)y + 9y = 0
4y + 12y + 9y = 0
25y = 0

Therefore y = xy = 0. Tn fact, for any value of x other than 3, y must equal zero.

(2) SUFFICIENT: x is negative. Therefore, x cannot equal 3, and it follows that y = 0. Therefore, xy = 0.

The correct answer is D.


What I did, i simply applied the info from statement 1, and got that -12Y=13Y which can only be true if Y is 0 and therefore, XY is 0.
In the second statement, i did the same and the result came to be the same. Is this faulty reasoning? Also, this is said to be 700-800 level question and my way seems just too easy to solve such a high level problem...


For statement (1) your reasoning looks OK. As for statement (2) what do you mean by "did the same"?

One can also do this way: \(6xy=x^2y + 9y\) --> \(y(x^2-6x+9)=0\) --> \(y(x-3)^2=0\) --> either \(x=3\) or \(y=0\) (or both).

(1) x = –2 --> so \(x\neq{3}\), then \(y=0\) and \(xy=0\). Sufficient.

(2) x < 0 --> again \(x\neq{3}\), then \(y=0\) and \(xy=0\). Sufficient.

Answer: D.

Check question #1 here: inequality-and-absolute-value-questions-from-my-collection-86939.html for harder version of the same problem.

Hope it helps.


Thanks a lot. What I meant by "I did the same" is that I substituted x for -1 and tested, and result came out to be the same...
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[highlight]Monster collection of Verbal questions (RC, CR, and SC)[/highlight]
http://gmatclub.com/forum/massive-collection-of-verbal-questions-sc-rc-and-cr-106195.html#p832142

[highlight]Massive collection of thousands of Data Sufficiency and Problem Solving questions and answers:[/highlight]
http://gmatclub.com/forum/1001-ds-questions-file-106193.html#p832133

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Re: If 6xy = x^2y + 9y, what is the value of xy? [#permalink]

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Re: If 6xy = x^2y + 9y, what is the value of xy? [#permalink]

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New post 11 Dec 2013, 03:14
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Bunuel wrote:
MisterEko wrote:
Guys, I just have a question if my way of solving it is reasonable and can often be applied here. I got the right answer on this one, but am not sure it would work usually:

If 6xy = x2y + 9y, what is the value of xy?

(1) x = –2

(2) x < 0

The answer is D, with the following explanation:
[Reveal] Spoiler:
The equation in question can be rephrased as follows:

x2y – 6xy + 9y = 0
y(x2 – 6x + 9) = 0
y(x – 3)2 = 0
Therefore, one or both of the following must be true:
y = 0 or
x = 3

It follows that the product xy must equal either 0 or 3y. This question can therefore be rephrased several different ways: "What is y?" or "Does x = 3?"

(1) SUFFICIENT: If x = -2, then:
(–2)2y – 6(–2)y + 9y = 0
4y + 12y + 9y = 0
25y = 0

Therefore y = xy = 0. Tn fact, for any value of x other than 3, y must equal zero.

(2) SUFFICIENT: x is negative. Therefore, x cannot equal 3, and it follows that y = 0. Therefore, xy = 0.

The correct answer is D.


What I did, i simply applied the info from statement 1, and got that -12Y=13Y which can only be true if Y is 0 and therefore, XY is 0.
In the second statement, i did the same and the result came to be the same. Is this faulty reasoning? Also, this is said to be 700-800 level question and my way seems just too easy to solve such a high level problem...


For statement (1) your reasoning looks OK. As for statement (2) what do you mean by "did the same"?

One can also do this way: \(6xy=x^2y + 9y\) --> \(y(x^2-6x+9)=0\) --> \(y(x-3)^2=0\) --> either \(x=3\) or \(y=0\) (or both).

(1) x = –2 --> so \(x\neq{3}\), then \(y=0\) and \(xy=0\). Sufficient.

(2) x < 0 --> again \(x\neq{3}\), then \(y=0\) and \(xy=0\). Sufficient.

Answer: D.

Check question #1 here: inequality-and-absolute-value-questions-from-my-collection-86939.html for harder version of the same problem.

Hope it helps.





Hi Bunuel,

I am still not clear why do we need statements. Once we know that y=0 then no matter what the value of x is xy will always be =0. Do we need to even consider the statements. Am I thinking in the right direction or is there a disconnect somewhere.

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Re: If 6xy = x^2y + 9y, what is the value of xy? [#permalink]

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New post 11 Dec 2013, 03:31
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davidfrank wrote:
Bunuel wrote:
MisterEko wrote:
Guys, I just have a question if my way of solving it is reasonable and can often be applied here. I got the right answer on this one, but am not sure it would work usually:

If 6xy = x2y + 9y, what is the value of xy?

(1) x = –2

(2) x < 0

The answer is D, with the following explanation:
[Reveal] Spoiler:
The equation in question can be rephrased as follows:

x2y – 6xy + 9y = 0
y(x2 – 6x + 9) = 0
y(x – 3)2 = 0
Therefore, one or both of the following must be true:
y = 0 or
x = 3

It follows that the product xy must equal either 0 or 3y. This question can therefore be rephrased several different ways: "What is y?" or "Does x = 3?"

(1) SUFFICIENT: If x = -2, then:
(–2)2y – 6(–2)y + 9y = 0
4y + 12y + 9y = 0
25y = 0

Therefore y = xy = 0. Tn fact, for any value of x other than 3, y must equal zero.

(2) SUFFICIENT: x is negative. Therefore, x cannot equal 3, and it follows that y = 0. Therefore, xy = 0.

The correct answer is D.


What I did, i simply applied the info from statement 1, and got that -12Y=13Y which can only be true if Y is 0 and therefore, XY is 0.
In the second statement, i did the same and the result came to be the same. Is this faulty reasoning? Also, this is said to be 700-800 level question and my way seems just too easy to solve such a high level problem...


For statement (1) your reasoning looks OK. As for statement (2) what do you mean by "did the same"?

One can also do this way: \(6xy=x^2y + 9y\) --> \(y(x^2-6x+9)=0\) --> \(y(x-3)^2=0\) --> either \(x=3\) or \(y=0\) (or both).

(1) x = –2 --> so \(x\neq{3}\), then \(y=0\) and \(xy=0\). Sufficient.

(2) x < 0 --> again \(x\neq{3}\), then \(y=0\) and \(xy=0\). Sufficient.

Answer: D.

Check question #1 here: inequality-and-absolute-value-questions-from-my-collection-86939.html for harder version of the same problem.

Hope it helps.





Hi Bunuel,

I am still not clear why do we need statements. Once we know that y=0 then no matter what the value of x is xy will always be =0. Do we need to even consider the statements. Am I thinking in the right direction or is there a disconnect somewhere.


See the red part is not true. We don't know whether \(y=0\). From \(y(x-3)^2=0\) we have that \(x=3\) OR \(y=0\). Now, if \(x=3\), then y can be any number, thus xy is not necessarily 0.

Hope it's clear.
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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.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

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Re: If 6xy = x^2y + 9y, what is the value of xy? [#permalink]

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New post 11 Dec 2013, 05:21
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MisterEko wrote:
If 6xy = x^2y + 9y, what is the value of xy?

(1) x = –2
(2) x < 0



Responding to a pm:

Why do we need the statements?

Given: \(6xy = x^2y + 9y\)
Given: \(y(x-3)^2 = 0\)

What we know is this: EITHER y is 0 OR x is 3. It is possible that both y is 0 and x is 3 but at least one of them MUST be true. This is all we know. We don't know whether y is 0 or whether x is 3 or both.

Question: what is xy?
I cannot say yet. All I know is that either y is 0 or x is 3.
If y is 0, xy will be 0.
If x is 3, then I don't know y which could be anything so xy could be anything.
So I cannot say what xy is.

(1) x = –2

This tells me that x is not 3. If x is not 3, y MUST be 0 because one of them has to be true. If y is 0, we know for sure that xy is 0. Sufficient.

(2) x < 0

This again tells me that x is not 3. If x is not 3, y MUST be 0 because one of them has to be true. If y is 0, we know for sure that xy is 0. Sufficient.

Answer (D)
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Re: If 6xy = x^2y + 9y, what is the value of xy? [#permalink]

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New post 17 Sep 2015, 07:53
Bunuel wrote:
MisterEko wrote:
Guys, I just have a question if my way of solving it is reasonable and can often be applied here. I got the right answer on this one, but am not sure it would work usually:

If 6xy = x2y + 9y, what is the value of xy?

(1) x = –2

(2) x < 0

The answer is D, with the following explanation:
[Reveal] Spoiler:
The equation in question can be rephrased as follows:

x2y – 6xy + 9y = 0
y(x2 – 6x + 9) = 0
y(x – 3)2 = 0

Therefore, one or both of the following must be true:
y = 0 or
x = 3

It follows that the product xy must equal either 0 or 3y. This question can therefore be rephrased several different ways: "What is y?" or "Does x = 3?"

(1) SUFFICIENT: If x = -2, then:
(–2)2y – 6(–2)y + 9y = 0
4y + 12y + 9y = 0
25y = 0

Therefore y = xy = 0. Tn fact, for any value of x other than 3, y must equal zero.

(2) SUFFICIENT: x is negative. Therefore, x cannot equal 3, and it follows that y = 0. Therefore, xy = 0.

The correct answer is D.


What I did, i simply applied the info from statement 1, and got that -12Y=13Y which can only be true if Y is 0 and therefore, XY is 0.
In the second statement, i did the same and the result came to be the same. Is this faulty reasoning? Also, this is said to be 700-800 level question and my way seems just too easy to solve such a high level problem...


For statement (1) your reasoning looks OK. As for statement (2) what do you mean by "did the same"?

One can also do this way: \(6xy=x^2y + 9y\) --> \(y(x^2-6x+9)=0\) --> \(y(x-3)^2=0\) --> either \(x=3\) or \(y=0\) (or both).

(1) x = –2 --> so \(x\neq{3}\), then \(y=0\) and \(xy=0\). Sufficient.

(2) x < 0 --> again \(x\neq{3}\), then \(y=0\) and \(xy=0\). Sufficient.

Answer: D.

Check question #1 here: inequality-and-absolute-value-questions-from-my-collection-86939.html for harder version of the same problem.

Hope it helps.


why we can not cancel out Y from 6xy=x^2y + 9y ? this is not inequality..what is mistake in my thinking

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Re: If 6xy = x^2y + 9y, what is the value of xy? [#permalink]

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New post 17 Sep 2015, 19:24
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anupamadw wrote:

why we can not cancel out Y from 6xy=x^2y + 9y ? this is not inequality..what is mistake in my thinking


Usually, it is not a good idea to cancel out a variable. You lose a solution if you do.

If you do not cancel out the y, you get that this inequality: 6xy=x^2y + 9y holds when either x = 3 or y = 0.

If you do cancel out the y, you get that this inequality: 6xy=x^2y + 9y holds when x = 3.

You lost the y = 0 value. Hence, it's always better to take out y common and keep it on the side, not cancel it.
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If 6xy = x^2y + 9y, what is the value of xy? [#permalink]

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New post 31 Dec 2017, 05:42
[quote="MisterEko"]If 6xy = \(x^2\)y + 9y, what is the value of xy?

(1) x = –2
(2) x < 0

[spoiler=]The answer is D, with the following explanation:
The equation in question can be rephrased as follows:

Slightly different analysis. Bunuel / Experts pl. validate

Given \(6xy=x^2y+9y\)

Find xy= ?

Now \(6xy=x^2y+9y\) ------ (1)
=> \(6xy=y(x^2+9)\)
=> \(6x=x^2+9\)
=> \(x^2-6x+9=0\)
=> \(x^2-3x-3x+9=0\)
=> \((x-3)^2=0\)
=> Therefore x=3
=> Substituting x=3 in (1) we have
=> 0=0

So we know the value of 'x'. We need to find y=?

Statement 1 x=-2
=> Substituting in (1) we have
=> -12y=13y
=> OR 25y=0
=> OR y=0
=> Therefore xy=0 SUFFICIENT

Statement 2 x<0 i.e 'x' is -ve
=> Substituting value of 'x' in equ (1)
=> Now irrespective of the magnitude ' -x ' LHS of equ (1) will be -ve
=> i.e 6xy= -ky ---- for some constant 'k'
=> Again irrespective of the magnitude ' -x ' RHS of equ (1) will be +ve
=> i.e \(x^2y+9y=my\) ---- for some constant 'm'
=> -ky=my
=> OR my+ky=0
=> OR ny=0 ----- for some constant 'n'
=> Irrespective of the SIGN of n, y=0
=> Therefore xy=0. SUFFICIENT

Therefore "D"

Thanks
Dinesh

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If 6xy = x^2y + 9y, what is the value of xy?   [#permalink] 31 Dec 2017, 05:42
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