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

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05 Jun 2013, 04:13

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This post received KUDOS

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This post was BOOKMARKED

The equation in question can be rephrased as follows: x^2 y – 6xy + 9y = 0

y(x^2 –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 “What is y?”

(1) INSUFFICIENT: This equation cannot be manipulated or combined with the original equation to solve directly for x or y. Instead, plug the two possible scenarios from the original equation into the equation from this statement:

If x = 3, then y = 3 + x = 3 + 3 = 6, so xy = (3)(6) = 18. If y = 0, then x = y – 3 = 0 – 3 = -3, so xy = (-3)(0) = 0. Since there are two possible answers, this statement is not sufficient.

(2) SUFFICIENT: If x3 < 0, then x < 0. Therefore, x cannot equal 3, and it follows that y = 0. Therefore, xy = 0.

Ans B

Please someone confirm on the solution, if the approach is correct
_________________

Correct me If I'm wrong !! looking for valuable inputs

The equation in question can be rephrased as follows: x^2 y – 6xy + 9y = 0

y(x^2 –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 “What is y?”

(1) INSUFFICIENT: This equation cannot be manipulated or combined with the original equation to solve directly for x or y. Instead, plug the two possible scenarios from the original equation into the equation from this statement:

If x = 3, then y = 3 + x = 3 + 3 = 6, so xy = (3)(6) = 18. If y = 0, then x = y – 3 = 0 – 3 = -3, so xy = (-3)(0) = 0. Since there are two possible answers, this statement is not sufficient.

(2) SUFFICIENT: If x3 < 0, then x < 0. Therefore, x cannot equal 3, and it follows that y = 0. Therefore, xy = 0.

Ans B

Please someone confirm on the solution, if the approach is correct

Yes, that's correct.

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

\(6xy=x^2y + 9y\) --> \(y(x^2-6x+9)=0\) --> \(y(x-3)^2=0\) --> either \(x=3\) or \(y=0\) (or both).

(1) y – x = 3. If \(y=0\) and \(x=-3\), then \(xy=0\) but if \(x=3\) and \(y=6\), then \(xy=18\). Not sufficient.

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

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

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19 Jun 2014, 02:10

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

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06 Jul 2014, 06:34

Bunuel wrote:

Manhnip wrote:

The equation in question can be rephrased as follows: x^2 y – 6xy + 9y = 0

y(x^2 –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 “What is y?”

(1) INSUFFICIENT: This equation cannot be manipulated or combined with the original equation to solve directly for x or y. Instead, plug the two possible scenarios from the original equation into the equation from this statement:

If x = 3, then y = 3 + x = 3 + 3 = 6, so xy = (3)(6) = 18. If y = 0, then x = y – 3 = 0 – 3 = -3, so xy = (-3)(0) = 0. Since there are two possible answers, this statement is not sufficient.

(2) SUFFICIENT: If x3 < 0, then x < 0. Therefore, x cannot equal 3, and it follows that y = 0. Therefore, xy = 0.

Ans B

Please someone confirm on the solution, if the approach is correct

Yes, that's correct.

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

\(6xy=x^2y + 9y\) --> \(y(x^2-6x+9)=0\) --> \(y(x-3)^2=0\) --> either \(x=3\) or \(y=0\) (or both).

(1) y – x = 3. If \(y=0\) and \(x=-3\), then \(xy=0\) but if \(x=3\) and \(y=6\), then \(xy=18\). Not sufficient.

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

Aaah this formatting.. I read the stem as x^(2y) and was totally stuck. Can this ambiguity in notation come in actual test ? I would have expected a bracket or at the least correct ordering like 6xy = yx^2

The equation in question can be rephrased as follows: x^2 y – 6xy + 9y = 0

y(x^2 –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 “What is y?”

(1) INSUFFICIENT: This equation cannot be manipulated or combined with the original equation to solve directly for x or y. Instead, plug the two possible scenarios from the original equation into the equation from this statement:

If x = 3, then y = 3 + x = 3 + 3 = 6, so xy = (3)(6) = 18. If y = 0, then x = y – 3 = 0 – 3 = -3, so xy = (-3)(0) = 0. Since there are two possible answers, this statement is not sufficient.

(2) SUFFICIENT: If x3 < 0, then x < 0. Therefore, x cannot equal 3, and it follows that y = 0. Therefore, xy = 0.

Ans B

Please someone confirm on the solution, if the approach is correct

Yes, that's correct.

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

\(6xy=x^2y + 9y\) --> \(y(x^2-6x+9)=0\) --> \(y(x-3)^2=0\) --> either \(x=3\) or \(y=0\) (or both).

(1) y – x = 3. If \(y=0\) and \(x=-3\), then \(xy=0\) but if \(x=3\) and \(y=6\), then \(xy=18\). Not sufficient.

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

Aaah this formatting.. I read the stem as x^(2y) and was totally stuck. Can this ambiguity in notation come in actual test ? I would have expected a bracket or at the least correct ordering like 6xy = yx^2

Edited the original post to avoid such confusions in future.

By the way x^2y means x^2*y. If it were \(x^{2y}\) it would be written x^(2y). But don't worry on the actual test the formatting will be clearer.

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

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06 Jul 2014, 20:23

Bunnuel,

Why is it wrong to divide the original equation by y? Since dividing the equation by y gives you x^2-6x+9=0 which gives you x=3, then statement 1 is sufficient to give you the value of y and therefore give you the value of xy as 18.

Statement 2 is not consistent with the original equation as it shows x^3<0 which is not possible (x=3, so 3^3=27) and is therefore insufficient. Therefore, answer should be A.

Or is it that only because statement 2 is inconsistent with the original equation and therefore we need to try a different approach?

Why is it wrong to divide the original equation by y? Since dividing the equation by y gives you x^2-6x+9=0 which gives you x=3, then statement 1 is sufficient to give you the value of y and therefore give you the value of xy as 18.

Statement 2 is not consistent with the original equation as it shows x^3<0 which is not possible (x=3, so 3^3=27) and is therefore insufficient. Therefore, answer should be A.

Or is it that only because statement 2 is inconsistent with the original equation and therefore we need to try a different approach?

If you divide (reduce) 6xy = x^2*y + 9y by y, you assume, with no ground for it, that y does not equal to zero thus exclude a possible solution (notice that both y = 0 AND x = 3 satisfy the equation).

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero. _________________

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

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06 May 2015, 00:22

Dear Bunuel, Thanks for the bringing out such fundamental issues.

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

I made the same error: jumped to eliminate y from LHS & RHS > came up with x=3, then solved for y as given in condition 1 to arrive at xy=18 hence right answer = A

Per your explanation: If you divide (reduce) 6xy = x^2*y + 9y by y, you assume, with no ground for it, that y does not equal to zero thus exclude a possible solution (notice that both y = 0 AND x = 3 satisfy the equation).

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.

It makes absolute sense and I feel like I am on a totally different wavelength and am solving the problems inconsistently very consistently! This is in-spite of the detailed guides and theoretical resources provided here. Is this just a matter of practice and training one's brain to approach a problem in a more ordered fashion or something fundamentally amiss?

Dear Bunuel, Thanks for the bringing out such fundamental issues.

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

I made the same error: jumped to eliminate y from LHS & RHS > came up with x=3, then solved for y as given in condition 1 to arrive at xy=18 hence right answer = A

Per your explanation: If you divide (reduce) 6xy = x^2*y + 9y by y, you assume, with no ground for it, that y does not equal to zero thus exclude a possible solution (notice that both y = 0 AND x = 3 satisfy the equation).

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.

It makes absolute sense and I feel like I am on a totally different wavelength and am solving the problems inconsistently very consistently! This is in-spite of the detailed guides and theoretical resources provided here. Is this just a matter of practice and training one's brain to approach a problem in a more ordered fashion or something fundamentally amiss?

Yes, I think practicing similar algebra/inequality questions should help.
_________________

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

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08 Nov 2015, 03:48

Bunuel wrote:

rohitd80 wrote:

Dear Bunuel, Thanks for the bringing out such fundamental issues.

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

I made the same error: jumped to eliminate y from LHS & RHS > came up with x=3, then solved for y as given in condition 1 to arrive at xy=18 hence right answer = A

Per your explanation: If you divide (reduce) 6xy = x^2*y + 9y by y, you assume, with no ground for it, that y does not equal to zero thus exclude a possible solution (notice that both y = 0 AND x = 3 satisfy the equation).

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.

It makes absolute sense and I feel like I am on a totally different wavelength and am solving the problems inconsistently very consistently! This is in-spite of the detailed guides and theoretical resources provided here. Is this just a matter of practice and training one's brain to approach a problem in a more ordered fashion or something fundamentally amiss?

Yes, I think practicing similar algebra/inequality questions should help.

Hi Bunel,

Although I am opening an old thread, I have a minor doubt about this question. I understand the logic that was applied in order to get to the correct answer (B), but what I am struggling with is that after simplifying the original expression to: y (x - 3 ) ^(2) = 0, does it not go w/o saying that irrespective of whether x is positive or negative, y will always be 0, and hence xy should always be 0? I don't see what "value add" does statement 2 actually provide - something which rarely happens in the actual GMAT questions. My question to you is this: do you think such questions COULD be considered fair game; in my opinion, they are slightly vague..... Cheers, Imran

Dear Bunuel, Thanks for the bringing out such fundamental issues.

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

I made the same error: jumped to eliminate y from LHS & RHS > came up with x=3, then solved for y as given in condition 1 to arrive at xy=18 hence right answer = A

Per your explanation: If you divide (reduce) 6xy = x^2*y + 9y by y, you assume, with no ground for it, that y does not equal to zero thus exclude a possible solution (notice that both y = 0 AND x = 3 satisfy the equation).

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.

It makes absolute sense and I feel like I am on a totally different wavelength and am solving the problems inconsistently very consistently! This is in-spite of the detailed guides and theoretical resources provided here. Is this just a matter of practice and training one's brain to approach a problem in a more ordered fashion or something fundamentally amiss?

Yes, I think practicing similar algebra/inequality questions should help.

Hi Bunel,

Although I am opening an old thread, I have a minor doubt about this question. I understand the logic that was applied in order to get to the correct answer (B), but what I am struggling with is that after simplifying the original expression to: y (x - 3 ) ^(2) = 0, does it not go w/o saying that irrespective of whether x is positive or negative, y will always be 0, and hence xy should always be 0? I don't see what "value add" does statement 2 actually provide - something which rarely happens in the actual GMAT questions. My question to you is this: do you think such questions COULD be considered fair game; in my opinion, they are slightly vague..... Cheers, Imran

If you'd read the thread and the solutions above carefully, you'd see that if x=3, then y can take any value, not necessarily 0.
_________________

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

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09 Nov 2015, 00:46

Hi Bunel,

Although I am opening an old thread, I have a minor doubt about this question. I understand the logic that was applied in order to get to the correct answer (B), but what I am struggling with is that after simplifying the original expression to: y (x - 3 ) ^(2) = 0, does it not go w/o saying that irrespective of whether x is positive or negative, y will always be 0, and hence xy should always be 0? I don't see what "value add" does statement 2 actually provide - something which rarely happens in the actual GMAT questions. My question to you is this: do you think such questions COULD be considered fair game; in my opinion, they are slightly vague..... Cheers, Imran[/quote]

If you'd read the thread and the solutions above carefully, you'd see that if x=3, then y can take any value, not necessarily 0.[/quote]

Bunuel, I did but there is some minor point that I am just not getting. For instance: if x3 < 0 i.e. x is negative, how do we know that y will be 0? I am not following how either the original equation - y(x+3)^2 =0 - or the above solutions are leading us to this conclusion. thanks a ton

Bunuel, I did but there is some minor point that I am just not getting. For instance: if x3 < 0 i.e. x is negative, how do we know that y will be 0? I am not following how either the original equation - y(x+3)^2 =0 - or the above solutions are leading us to this conclusion. thanks a ton

For y(x−3)^2 to be 0, either x must be 3 or y must be 0 (or both). x^3 < 0 implies that x is NOT 3, thus y MUST be 0 (in order y(x−3)^2 to be 0).
_________________

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

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09 Nov 2015, 02:53

Bunuel wrote:

mais1990 wrote:

Bunuel, I did but there is some minor point that I am just not getting. For instance: if x3 < 0 i.e. x is negative, how do we know that y will be 0? I am not following how either the original equation - y(x+3)^2 =0 - or the above solutions are leading us to this conclusion. thanks a ton

For y(x−3)^2 to be 0, either x must be 3 or y must be 0 (or both). x^3 < 0 implies that x is NOT 3, thus y MUST be 0 (in order y(x−3)^2 to be 0).

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

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

(1) y – x = 3

(2) x^3 < 0

If we modify the question, 0=x^2*y-6xy+9y, or 0=y(x^2-6x+9), or 0=y(x-3)^2, and we want to know whether y=0 or x=3. There are 2 variables (x,y) and 1 equation (0=x^2*y-6xy+9y). 2 more equations are given from the 2 conditions, so there is high chance (D) will be our answer. From condition 1, y=0 and x=-3 or x=3 and y=6. This is not sufficient as this is not unique. From condition 2, x^3<0--> x<0. But x=3>0, and y has to be y=0, and xy=0, so the condition is sufficient; the answer is (B).

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
_________________

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