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805+ (Hard)|   Algebra|      
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dvinoth86
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If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x? Updated on: 21 May 2018, 00:30
Originally posted by dvinoth86 on 23 Feb 2012, 20:27.
Last edited by Bunuel on 21 May 2018, 00:30, edited 5 times in total.
Edited the question and added the OA
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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43% (02:14) correct 57% (02:37) wrong based on 686 sessions

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If \(8xy^3 + 8x^3y=\frac{2x^2y^2}{2^{-3}}\), what is the value of xy?

(1) y > x
(2) x < 0
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Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
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If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x? 24 Feb 2012, 00:18
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\(8x*y^3 + 8x^3*y = \frac{2x^2*y^2}{2^{(-3)}}\), What is xy?


\(8xy^3+8x^3y=\frac{2x^2*y^2}{2^{-3}}\);

\(8xy^3+8x^3y=2x^2*y^2*8\);

Reduce by 8: \(xy^3+x^3y=2x^2*y^2\);

Rearrange: \(xy^3+x^3y-2x^2*y^2=0\);

Factor out \(xy\): \(xy(y^2+x^2-2xy)=0\);

\(xy(y-x)^2=0\);

Either \(xy=0\) or \(y-x=0\) (\(x=y\)).


(1) y > x --> \(y\neq{x}\), which means that \(xy=0\). Sufficient.

(2) x < 0. Clearly not sufficient.

Answer: A.

Hope it's clear.
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If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x? 16 Mar 2012, 00:04
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if the question stem says y=x and option A says y>x.. How can er say xy=0 . am unable to understand

Bunuel
8x*y^3 + 8x^3*y = 2x^2*y^2 / 2^(-3), What is xy?

\(8xy^3+8x^3y=\frac{2x^2*y^2}{2^{-3}}\) --> \(8xy^3+8x^3y=2x^2*y^2*8\) --> reduce by 8: \(xy^3+x^3y=2x^2*y^2\) --> rearrange: \(xy^3+x^3y-2x^2*y^2=0\) --> factor out \(xy\): \(xy(y^2+x^2-2xy)=0\) --> \(xy(y-x)^2=0\) --> either \(xy=0\) or \(y-x=0\) (\(x=y\)).

(1) y > x --> \(y\neq{x}\), which means that \(xy=0\). Sufficient.
(2) x < 0. Clearly not sufficient.

Answer: A.

Hope it's clear.

P.S. dvinoth86 please check the questions before posting and format them correctly. Thank you.
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If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x? 16 Mar 2012, 02:18
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if the question stem says y=x and option A says y>x.. How can er say xy=0 . am unable to understand

Bunuel
8x*y^3 + 8x^3*y = 2x^2*y^2 / 2^(-3), What is xy?

\(8xy^3+8x^3y=\frac{2x^2*y^2}{2^{-3}}\) --> \(8xy^3+8x^3y=2x^2*y^2*8\) --> reduce by 8: \(xy^3+x^3y=2x^2*y^2\) --> rearrange: \(xy^3+x^3y-2x^2*y^2=0\) --> factor out \(xy\): \(xy(y^2+x^2-2xy)=0\) --> \(xy(y-x)^2=0\) --> either \(xy=0\) or \(y-x=0\) (\(x=y\)).

(1) y > x --> \(y\neq{x}\), which means that \(xy=0\). Sufficient.
(2) x < 0. Clearly not sufficient.

Answer: A.

Hope it's clear.

P.S. dvinoth86 please check the questions before posting and format them correctly. Thank you.
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The stem DOES NOT say that \(x=y\), it says: EITHER \(xy=0\) OR \(x=y\).

(1) says \(y > x\), which means that \(x=y\) is not possible, hence \(xy=0\).

Hope it's clear.
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If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x? 24 Dec 2012, 06:32
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7. Algebra



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If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x? 25 Dec 2012, 13:18
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Bunuel
If \(8xy^3 + 8x^3y=\frac{2x^2y^2}{2^{-3}}\), what is the value of xy?

\(8xy^3 + 8x^3y=\frac{2x^2y^2}{2^{-3}}\) --> \(8xy^3 + 8x^3y=8*2x^2y^2=0\) --> reduce by 8 and re-arrange: \(xy^3+x^3y-2x^2y^2\) --> factor out xy: \(xy(y^2+x^2-2xy)=0\) --> \(xy(y-x)^2=0\) --> \(xy=0\) or \(y-x=0\).

(1) y > x. Since \(y>x\), then \(y-x\neq{0}\), thus \(xy=0\). Sufficient.

(2) x < 0. Not sufficient.

Answer: A.
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This solution does not answer to the question. Question asks about value xy.
if y>x then xy can't be equal 0. It can't be discussed.
Really I don't understand your solution please explain
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If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x? 26 Dec 2012, 03:43
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Bunuel
If \(8xy^3 + 8x^3y=\frac{2x^2y^2}{2^{-3}}\), what is the value of xy?

\(8xy^3 + 8x^3y=\frac{2x^2y^2}{2^{-3}}\) --> \(8xy^3 + 8x^3y=8*2x^2y^2=0\) --> reduce by 8 and re-arrange: \(xy^3+x^3y-2x^2y^2\) --> factor out xy: \(xy(y^2+x^2-2xy)=0\) --> \(xy(y-x)^2=0\) --> \(xy=0\) or \(y-x=0\).

(1) y > x. Since \(y>x\), then \(y-x\neq{0}\), thus \(xy=0\). Sufficient.

(2) x < 0. Not sufficient.

Answer: A.

This solution does not answer to the question. Question asks about value xy.
if y>x then xy can't be equal 0. It can't be discussed.
Really I don't understand your solution please explain
Show more

You should read a solution carefully.

From the stem we have that either \(xy=0\) or \(y-x=0\).

(1) says that y > x, so y - x > 0, which means that \(y-x\neq{0}\), thus \(xy=0\).
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If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x? 14 Nov 2014, 06:23
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Bunuel
8x*y^3 + 8x^3*y = 2x^2*y^2 / 2^(-3), What is xy?

\(8xy^3+8x^3y=\frac{2x^2*y^2}{2^{-3}}\) --> \(8xy^3+8x^3y=2x^2*y^2*8\) --> reduce by 8: \(xy^3+x^3y=2x^2*y^2\) --> rearrange: \(xy^3+x^3y-2x^2*y^2=0\) --> factor out \(xy\): \(xy(y^2+x^2-2xy)=0\) --> \(xy(y-x)^2=0\) --> either \(xy=0\) or \(y-x=0\) (\(x=y\)).

(1) y > x --> \(y\neq{x}\), which means that \(xy=0\). Sufficient.
(2) x < 0. Clearly not sufficient.

Answer: A.

Hope it's clear.

P.S. dvinoth86 please check the questions before posting and format them correctly. Thank you.
Show more

Hi Bunuel,

I did a blunder while reducing 8x*y^3 + 8x^3*y = 2x^2*y^2 / 2^(-3) but not sure the reason for that.
i started by cancelling out the xy from both sides to get (x-y)^2 = 0 at the end.
Why cant we cancel the x and y ? Please help.
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If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x? 14 Nov 2014, 06:28
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Bunuel
8x*y^3 + 8x^3*y = 2x^2*y^2 / 2^(-3), What is xy?

\(8xy^3+8x^3y=\frac{2x^2*y^2}{2^{-3}}\) --> \(8xy^3+8x^3y=2x^2*y^2*8\) --> reduce by 8: \(xy^3+x^3y=2x^2*y^2\) --> rearrange: \(xy^3+x^3y-2x^2*y^2=0\) --> factor out \(xy\): \(xy(y^2+x^2-2xy)=0\) --> \(xy(y-x)^2=0\) --> either \(xy=0\) or \(y-x=0\) (\(x=y\)).

(1) y > x --> \(y\neq{x}\), which means that \(xy=0\). Sufficient.
(2) x < 0. Clearly not sufficient.

Answer: A.

Hope it's clear.

P.S. dvinoth86 please check the questions before posting and format them correctly. Thank you.

Hi Bunuel,

I did a blunder while reducing 8x*y^3 + 8x^3*y = 2x^2*y^2 / 2^(-3) but not sure the reason for that.
i started by cancelling out the xy from both sides to get (x-y)^2 = 0 at the end.
Why cant we cancel the x and y ? Please help.
Show more

If you divide (reduce) 8x*y^3 + 8x^3*y = 2x^2*y^2/2^(-3), by xy you assume, with no ground for it, that xy does not equal to zero thus exclude a possible solution (notice that xy=0 satisfies 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.

Check more tips on Algebra here: algebra-tips-and-hints-175003.html

Hope it helps.
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If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x? 13 Aug 2015, 18:19
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I factored out the xy from both sides and that screwed up my equation. Note to self: dont factor out things when it may lead to division by zero!
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If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x? 22 Nov 2017, 09:24
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St2: if x < 0, and y not equal to 0 , then LHS is negative but RHS is positive, which is not possible so y has to be 0.
and hence xy = 0. this sums to Answer D.

Where am I getting it wrong.

Bunuel
\(8x*y^3 + 8x^3*y = \frac{2x^2*y^2}{2^{(-3)}}\), What is xy?

\(8xy^3+8x^3y=\frac{2x^2*y^2}{2^{-3}}\);

\(8xy^3+8x^3y=2x^2*y^2*8\);

Reduce by 8: \(xy^3+x^3y=2x^2*y^2\);

Rearrange: \(xy^3+x^3y-2x^2*y^2=0\);

Factor out \(xy\): \(xy(y^2+x^2-2xy)=0\);

\(xy(y-x)^2=0\);

Either \(xy=0\) or \(y-x=0\) (\(x=y\)).


(1) y > x --> \(y\neq{x}\), which means that \(xy=0\). Sufficient.

(2) x < 0. Clearly not sufficient.

Answer: A.

Hope it's clear.
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If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x? 22 Nov 2017, 09:34
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St2: if x < 0, and y not equal to 0 , then LHS is negative but RHS is positive, which is not possible so y has to be 0.
and hence xy = 0. this sums to Answer D.

Where am I getting it wrong.

Bunuel
\(8x*y^3 + 8x^3*y = \frac{2x^2*y^2}{2^{(-3)}}\), What is xy?

\(8xy^3+8x^3y=\frac{2x^2*y^2}{2^{-3}}\);

\(8xy^3+8x^3y=2x^2*y^2*8\);

Reduce by 8: \(xy^3+x^3y=2x^2*y^2\);

Rearrange: \(xy^3+x^3y-2x^2*y^2=0\);

Factor out \(xy\): \(xy(y^2+x^2-2xy)=0\);

\(xy(y-x)^2=0\);

Either \(xy=0\) or \(y-x=0\) (\(x=y\)).


(1) y > x --> \(y\neq{x}\), which means that \(xy=0\). Sufficient.

(2) x < 0. Clearly not sufficient.

Answer: A.

Hope it's clear.
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You can always check your theoretical deduction by simple number plugging.

Fro (2) plug x = y = -1 or x = y = -2.
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If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x? 30 Nov 2017, 06:00
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Thanks

I understood now...


Bunuel
Barui
St2: if x < 0, and y not equal to 0 , then LHS is negative but RHS is positive, which is not possible so y has to be 0.
and hence xy = 0. this sums to Answer D.

Where am I getting it wrong.

Bunuel
\(8x*y^3 + 8x^3*y = \frac{2x^2*y^2}{2^{(-3)}}\), What is xy?

\(8xy^3+8x^3y=\frac{2x^2*y^2}{2^{-3}}\);

\(8xy^3+8x^3y=2x^2*y^2*8\);

Reduce by 8: \(xy^3+x^3y=2x^2*y^2\);

Rearrange: \(xy^3+x^3y-2x^2*y^2=0\);

Factor out \(xy\): \(xy(y^2+x^2-2xy)=0\);

\(xy(y-x)^2=0\);

Either \(xy=0\) or \(y-x=0\) (\(x=y\)).


(1) y > x --> \(y\neq{x}\), which means that \(xy=0\). Sufficient.

(2) x < 0. Clearly not sufficient.

Answer: A.

Hope it's clear.

You can always check your theoretical deduction by simple number plugging.

Fro (2) plug x = y = -1 or x = y = -2.
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If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x? 26 Mar 2018, 00:28
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This is an excellent question.
The trick is in simplyfing the question. Once you paraphrase the question into simple terms, it becomes,
xy*(x-y)^2=0.


8x∗y3+8x3∗y=2x2∗y22(−3)8x∗y3+8x3∗y=2x2∗y22(−3), What is xy?


(1) y > x ==>. Sufficient. Because, if y is greater than 1)x then x can be equal to Y.
so only one equation left. xy=0.

(2) x < 0. Clearly not sufficient.

Answer: A.
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If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x? 13 Apr 2019, 04:55
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Bunuel
\(8x*y^3 + 8x^3*y = \frac{2x^2*y^2}{2^{(-3)}}\), What is xy?


\(8xy^3+8x^3y=\frac{2x^2*y^2}{2^{-3}}\);

\(8xy^3+8x^3y=2x^2*y^2*8\);

Reduce by 8: \(xy^3+x^3y=2x^2*y^2\);

Rearrange: \(xy^3+x^3y-2x^2*y^2=0\);

Factor out \(xy\): \(xy(y^2+x^2-2xy)=0\);

\(xy(y-x)^2=0\);

Either \(xy=0\) or \(y-x=0\) (\(x=y\)).


(1) y > x --> \(y\neq{x}\), which means that \(xy=0\). Sufficient.

(2) x < 0. Clearly not sufficient.

Answer: A.

Hope it's clear.
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This may be me being extremely dim but can someone explain why \(xy\): \(xy(y^2+x^2-2xy)=0\); factors to \(xy(y-x)^2=0\) and not \(xy(x-y)^2=0\) I've written it out a lot and it seems to have the same product?
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If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x? 28 Dec 2019, 21:01
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Bunuel
\(8x*y^3 + 8x^3*y = \frac{2x^2*y^2}{2^{(-3)}}\), What is xy?


\(8xy^3+8x^3y=\frac{2x^2*y^2}{2^{-3}}\);

\(8xy^3+8x^3y=2x^2*y^2*8\);

Reduce by 8: \(xy^3+x^3y=2x^2*y^2\);

Rearrange: \(xy^3+x^3y-2x^2*y^2=0\);

Factor out \(xy\): \(xy(y^2+x^2-2xy)=0\);

\(xy(y-x)^2=0\);

Either \(xy=0\) or \(y-x=0\) (\(x=y\)).


(1) y > x --> \(y\neq{x}\), which means that \(xy=0\). Sufficient.

(2) x < 0. Clearly not sufficient.

Answer: A.

Hope it's clear.

This may be me being extremely dim but can someone explain why \(xy\): \(xy(y^2+x^2-2xy)=0\); factors to \(xy(y-x)^2=0\) and not \(xy(x-y)^2=0\) I've written it out a lot and it seems to have the same product?
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I have the same question
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If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x? 03 Mar 2020, 07:43
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strawberryjelly
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\(8x*y^3 + 8x^3*y = \frac{2x^2*y^2}{2^{(-3)}}\), What is xy?


\(8xy^3+8x^3y=\frac{2x^2*y^2}{2^{-3}}\);

\(8xy^3+8x^3y=2x^2*y^2*8\);

Reduce by 8: \(xy^3+x^3y=2x^2*y^2\);

Rearrange: \(xy^3+x^3y-2x^2*y^2=0\);

Factor out \(xy\): \(xy(y^2+x^2-2xy)=0\);

\(xy(y-x)^2=0\);

Either \(xy=0\) or \(y-x=0\) (\(x=y\)).


(1) y > x --> \(y\neq{x}\), which means that \(xy=0\). Sufficient.

(2) x < 0. Clearly not sufficient.

Answer: A.

Hope it's clear.

This may be me being extremely dim but can someone explain why \(xy\): \(xy(y^2+x^2-2xy)=0\); factors to \(xy(y-x)^2=0\) and not \(xy(x-y)^2=0\) I've written it out a lot and it seems to have the same product?
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The bigger question is would it really matter in this case?

If it is xy(x-y)^2 =0 or it is xy(y-x)^2 =0 ???

Nope it will not matter, as eventually both the expressions will boil down to xy=0 or x=y. :) ;)

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If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x? 07 Feb 2023, 12:14
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Bunuel
\(8x*y^3 + 8x^3*y = \frac{2x^2*y^2}{2^{(-3)}}\), What is xy?


\(8xy^3+8x^3y=\frac{2x^2*y^2}{2^{-3}}\);

\(8xy^3+8x^3y=2x^2*y^2*8\);

Reduce by 8: \(xy^3+x^3y=2x^2*y^2\);

Rearrange: \(xy^3+x^3y-2x^2*y^2=0\);

Factor out \(xy\): \(xy(y^2+x^2-2xy)=0\);

\(xy(y-x)^2=0\);

Either \(xy=0\) or \(y-x=0\) (\(x=y\)).


(1) y > x --> \(y\neq{x}\), which means that \(xy=0\). Sufficient.

(2) x < 0. Clearly not sufficient.

Answer: A.

Hope it's clear.
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Hi Bunuel please advise the way we factor out 8 on both side of equation why are we not factoring out xy from both sides of equation

xy^3+x^3y=2x^2*y^2
Xy(Y^2+x^2)=2xy * xy
Y^2+x^2= 2xy
(Y-x)^2=0

I can see that we don't anything to related with xy from this but. I still curious to find out we cannot factor out xy

Bunuel

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If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x? 07 Feb 2023, 12:27
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puneetfitness
Bunuel
\(8x*y^3 + 8x^3*y = \frac{2x^2*y^2}{2^{(-3)}}\), What is xy?


\(8xy^3+8x^3y=\frac{2x^2*y^2}{2^{-3}}\);

\(8xy^3+8x^3y=2x^2*y^2*8\);

Reduce by 8: \(xy^3+x^3y=2x^2*y^2\);

Rearrange: \(xy^3+x^3y-2x^2*y^2=0\);

Factor out \(xy\): \(xy(y^2+x^2-2xy)=0\);

\(xy(y-x)^2=0\);

Either \(xy=0\) or \(y-x=0\) (\(x=y\)).


(1) y > x --> \(y\neq{x}\), which means that \(xy=0\). Sufficient.

(2) x < 0. Clearly not sufficient.

Answer: A.

Hope it's clear.

Hi Bunuel please advise the way we factor out 8 on both side of equation why are we not factoring out xy from both sides of equation

xy^3+x^3y=2x^2*y^2
Xy(Y^2+x^2)=2xy * xy
Y^2+x^2= 2xy
(Y-x)^2=0

I can see that we don't anything to related with xy from this but. I still curious to find out we cannot factor out xy

Bunuel

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You can factor out xy but you cannot reduce by xy and as to why you cannot do that is explained here: https://gmatclub.com/forum/if-8xy-3-8x- ... l#p1442635

Hope it helps.
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If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x? 19 Mar 2024, 23:59
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