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

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8xy^3 + 8x^3y = 2x^2*y^2 / 2^(-3), What is xy? [#permalink]

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

(1) y > x
(2) x < 0
[Reveal] Spoiler: OA

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Last edited by Bunuel on 27 May 2013, 05:19, edited 3 times in total.
Edited the question and added the OA

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Re: DS: Inequalities [#permalink]

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Could you reformat the question? A lot of things are unclear.
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Re: DS: Inequalities [#permalink]

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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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Re: 8xy^3 + 8x^3y = 2x^2*y^2 / 2^(-3), What is xy? [#permalink]

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New post 24 Feb 2012, 09:47
wil follow from next time
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Re: DS: Inequalities [#permalink]

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New post 16 Mar 2012, 00:04
if the question stem says y=x and option A says y>x.. How can er say xy=0 . am unable to understand

Bunuel wrote:
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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Re: DS: Inequalities [#permalink]

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New post 16 Mar 2012, 02:18
devinawilliam83 wrote:
if the question stem says y=x and option A says y>x.. How can er say xy=0 . am unable to understand

Bunuel wrote:
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.


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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Re: 8xy^3 + 8x^3y = 2x^2*y^2 / 2^(-3), What is xy? [#permalink]

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Re: 8xy^3 + 8x^3y = 2x^2*y^2 / 2^(-3), What is xy? [#permalink]

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Bunuel wrote:
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.
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Re: 8xy^3 + 8x^3y = 2x^2*y^2 / 2^(-3), What is xy? [#permalink]

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Ankur9 wrote:
Bunuel wrote:
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.


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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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

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Re: 8xy^3 + 8x^3y = 2x^2*y^2 / 2^(-3), What is xy? [#permalink]

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New post 13 Aug 2015, 18:19
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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Re: 8xy^3 + 8x^3y = 2x^2*y^2 / 2^(-3), What is xy? [#permalink]

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New post 10 Apr 2016, 00:43
Bunuel,
In such questions,one must avoid cancelling out the variables right? I have seen myself doing that mistake in quite a few qsns but your approach helped me identify my mistake

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Re: 8xy^3 + 8x^3y = 2x^2*y^2 / 2^(-3), What is xy? [#permalink]

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Hello from the GMAT Club BumpBot!

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Re: 8xy^3 + 8x^3y = 2x^2*y^2 / 2^(-3), What is xy?   [#permalink] 03 Aug 2017, 12:09
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