Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st.

 It is currently 16 Jul 2019, 20:22

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x?

Author Message
TAGS:

Hide Tags

Manager
Joined: 19 Oct 2011
Posts: 97
Location: India
If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x?  [#permalink]

Show Tags

Updated on: 21 May 2018, 00:30
4
11
00:00

Difficulty:

95% (hard)

Question Stats:

42% (02:07) correct 58% (02:37) wrong based on 265 sessions

HideShow timer Statistics

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

(1) y > x
(2) x < 0

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
Math Expert
Joined: 02 Sep 2009
Posts: 56244
Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x?  [#permalink]

Show Tags

24 Feb 2012, 00:18
9
11
$$8x*y^3 + 8x^3*y = \frac{2x^2*y^2}{2^{(-3)}}$$, 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.

Hope it's clear.
_________________
General Discussion
Manager
Joined: 25 Aug 2011
Posts: 136
Location: India
GMAT 1: 730 Q49 V40
WE: Operations (Insurance)
Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x?  [#permalink]

Show Tags

16 Mar 2012, 00:04
1
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.

Hope it's clear.

P.S. dvinoth86 please check the questions before posting and format them correctly. Thank you.
Math Expert
Joined: 02 Sep 2009
Posts: 56244
Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x?  [#permalink]

Show Tags

16 Mar 2012, 02:18
1
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.

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.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 56244
If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x?  [#permalink]

Show Tags

24 Dec 2012, 06:32
2
3
Intern
Joined: 18 Dec 2012
Posts: 1
Re: If 8xy^3 + 8x^3*y=2x^2*y^2/ 2^-3,what is the value of x?  [#permalink]

Show Tags

25 Dec 2012, 13:18
Bunuel wrote:
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.

if y>x then xy can't be equal 0. It can't be discussed.
Math Expert
Joined: 02 Sep 2009
Posts: 56244
Re: If 8xy^3 + 8x^3*y=2x^2*y^2/ 2^-3,what is the value of x?  [#permalink]

Show Tags

26 Dec 2012, 03:43
akshin wrote:
Bunuel wrote:
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.

if y>x then xy can't be equal 0. It can't be discussed.

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 $$xy=0$$.
_________________
Intern
Joined: 25 May 2014
Posts: 46
Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x?  [#permalink]

Show Tags

14 Nov 2014, 06:23
1
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.

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.
_________________
Never Try Quitting, Never Quit Trying
Math Expert
Joined: 02 Sep 2009
Posts: 56244
Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x?  [#permalink]

Show Tags

14 Nov 2014, 06:28
1
2
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.

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.

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.
_________________
Manager
Joined: 09 Aug 2015
Posts: 84
GMAT 1: 770 Q51 V44
GPA: 2.3
Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x?  [#permalink]

Show Tags

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!
Intern
Joined: 30 Sep 2017
Posts: 38
Location: India
Concentration: Entrepreneurship, General Management
Schools: IIM Udaipur '17
GMAT 1: 700 Q50 V37
GPA: 3.7
WE: Engineering (Energy and Utilities)
Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x?  [#permalink]

Show Tags

22 Nov 2017, 09:24
1
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 wrote:
$$8x*y^3 + 8x^3*y = \frac{2x^2*y^2}{2^{(-3)}}$$, 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.

Hope it's clear.

_________________
If you like my post, motivate me by giving kudos...
Math Expert
Joined: 02 Sep 2009
Posts: 56244
Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x?  [#permalink]

Show Tags

22 Nov 2017, 09:34
Barui wrote:
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 wrote:
$$8x*y^3 + 8x^3*y = \frac{2x^2*y^2}{2^{(-3)}}$$, 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.

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.
_________________
Intern
Joined: 30 Sep 2017
Posts: 38
Location: India
Concentration: Entrepreneurship, General Management
Schools: IIM Udaipur '17
GMAT 1: 700 Q50 V37
GPA: 3.7
WE: Engineering (Energy and Utilities)
Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x?  [#permalink]

Show Tags

30 Nov 2017, 06:00
Thanks

I understood now...

Bunuel wrote:
Barui wrote:
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 wrote:
$$8x*y^3 + 8x^3*y = \frac{2x^2*y^2}{2^{(-3)}}$$, 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.

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.

_________________
If you like my post, motivate me by giving kudos...
Manager
Status: The journey is always more beautiful than the destination
Affiliations: Computer Science
Joined: 24 Apr 2017
Posts: 52
Location: India
Concentration: Statistics, Strategy
GMAT 1: 570 Q40 V28
GPA: 3.14
Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x?  [#permalink]

Show Tags

26 Mar 2018, 00:28
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.

_________________
Sky is the limit. 800 is the limit.
Intern
Joined: 24 Mar 2019
Posts: 4
GMAT 1: 710 Q47 V41
Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x?  [#permalink]

Show Tags

13 Apr 2019, 04:55
Bunuel wrote:
$$8x*y^3 + 8x^3*y = \frac{2x^2*y^2}{2^{(-3)}}$$, 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.

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?
Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x?   [#permalink] 13 Apr 2019, 04:55
Display posts from previous: Sort by