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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?"

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...

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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

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

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:

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?"

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.

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

Show Tags

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:

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?"

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.

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...
_________________

[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

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

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11 Dec 2013, 03:14

1

This post was BOOKMARKED

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:

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?"

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.

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.

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:

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?"

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.

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.

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.

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

Show Tags

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:

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?"

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.

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

Show Tags

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