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MGMAT Adv Quant - WS#5, Question 50 -- (x-y)² vs (y-x)²

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by rishimaharaj » Thu Oct 13, 2011 6:56 pm
From MGMAT Advanced Quant Supplement, Chapter 9, WorkoutSet #5, page 241, #50:
If 8xy³ + 8x³y = (2x²y²)/2^-³, what is the value of xy?
  • y > x
  • x < 0
My Solution:
8xy³ + 8x³y = (2x²y²) / 2^-³
2³xy³ + 2³x³y = 2^4 * x² * y²
2³(xy³ + x³y) = 2^4 * x² * y² ; div both sides by 2³
xy³ + x³y = 2x²y²
xy³ + x³y - 2x²y² = 0
xy(y² + x² - 2xy) = 0
xy(x - y)(x - y) = 0
xy(x - y)² = 0

Statement 1: y > x means that x - y < 0, which means that x - y cannot be 0, thus xy = 0.
Sufficient.

Statement 2: x < 0. No info about y.
Insufficient.

I chose answer choice A, which is the correct answer.


My question though, is about the factoring that I did. In the book, the explanation factors it like this:
(xy)(y - x)² = 0.
How do we know which should be be subtracted from which? In this particular case, it works out to be the same, because the result is squared.
Has anyone seen an official question where the order of the x and y matters?
Will it ever matter?

Thanks,
--Rishi


Note: Mistyped RHS of problem. Fixed from (2x²y²)/2^-³ to be (2x²y²)/2^-².
Last edited by rishimaharaj on Fri Oct 14, 2011 5:29 pm, edited 2 times in total.
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by sl750 » Fri Oct 14, 2011 3:48 am
Is the exponent on the R.H.S (2)^-2 or (2)^-3 ?
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by gmatclubmember » Fri Oct 14, 2011 5:17 am
Since its a square and its always going to be positive so it wont really matter weather u write (x-y)^2 or (y-x)^2.
a lil' Thank note goes a long way :)!!
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by rishimaharaj » Fri Oct 14, 2011 5:38 am
sl750 wrote:Is the exponent on the R.H.S (2)^-2 or (2)^-3 ?
On the right, the denominator is 2^(-2).
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by sl750 » Fri Oct 14, 2011 9:34 am
@rishimaharaj

In the second line of your calculation you've arrived at the below equation

2³xy³ + 2³x³y = 2^4 * x² * y² . Shouldn't the R.H.S be 2^3*x^2*y^2?
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by rishimaharaj » Fri Oct 14, 2011 3:03 pm
sl750 wrote:In the second line of your calculation you've arrived at the below equation
2³xy³ + 2³x³y = 2^4 * x² * y² . Shouldn't the R.H.S be 2^3*x^2*y^2?
Hello SL750,
I think that it should be 2^4:

Right Hand Side
(2^1 * x² * y²) / (2^-3)
=(2^1 * x² * y²) / (1 / 2^3)
=(2^1 * x² * y²) * 2^3
=2^1*2^3 * x² * y²
=2^(1+3) * x² * y²
=2^4 * x² * y²

--Rishi
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by n@resh » Fri Oct 14, 2011 4:32 pm
rishimaharaj wrote:
sl750 wrote:In the second line of your calculation you've arrived at the below equation
2³xy³ + 2³x³y = 2^4 * x² * y² . Shouldn't the R.H.S be 2^3*x^2*y^2?
Hello SL750,
I think that it should be 2^4:

Right Hand Side
(2^1 * x² * y²) / (2^-3)
=(2^1 * x² * y²) / (1 / 2^3)
=(2^1 * x² * y²) * 2^3
=2^1*2^3 * x² * y²
=2^(1+3) * x² * y²
=2^4 * x² * y²

--Rishi
But in the above problem it's 2^-2 i.e. 8xy³ + 8x³y = (2x²y²)/2^-²
Hence the whole expression becomes 8xy(x^2 + y^2) = 2^3 (xy)^2
=> xy( x^2 + y^2 -xy) = 0
=> xy= 0 or (x-y)^2 +xy = 0.
BUT, if it's 2^-3, i.e. 8xy³ + 8x³y = (2x²y²)/2^-3
i.e. xy(x^2 + Y^2) = 2(xy)^2
=> xy(x^2 + y^2 - 2xy) = 0
=> xy(x - y)^2 = 0
so, xy = 0 or (x-y)^2 = 0 , here one of the statment must be true.

from stat 1: y>x, (x-y)^2 can't be zero because it will turn x = y!. so,must be xy = 0.
Hence Answer A!
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by rishimaharaj » Fri Oct 14, 2011 5:33 pm
n@resh wrote: But in the above problem it's 2^-2 i.e. 8xy³ + 8x³y = (2x²y²)/2^-²
Ah! Fixed that!

But still, though, any comments on the question:
(xy)(y - x)² = 0.
How do we know which should be be subtracted from which? In this particular case, it works out to be the same, because the result is squared.
Has anyone seen an official question where the order of the x and y matters?
Will it ever matter?


Thanks
--Rishi
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by n@resh » Fri Oct 14, 2011 6:46 pm
rishimaharaj wrote:
n@resh wrote: But in the above problem it's 2^-2 i.e. 8xy³ + 8x³y = (2x²y²)/2^-²
Ah! Fixed that!

But still, though, any comments on the question:
(xy)(y - x)² = 0.
How do we know which should be be subtracted from which? In this particular case, it works out to be the same, because the result is squared.
Has anyone seen an official question where the order of the x and y matters?
Will it ever matter?


Thanks
--Rishi
of course,subtraction does matter: x - y is different from y - x. BUT if you are performing subtraction in the square form then it won't matter!
(x-y)^2 = (y-x)^2 => ( -(x-y))^2 => (x-y)^2
Hope you got me!
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by rohit_gmat » Sat Oct 15, 2011 5:57 am
rishimaharaj wrote:
n@resh wrote: But in the above problem it's 2^-2 i.e. 8xy³ + 8x³y = (2x²y²)/2^-²
Ah! Fixed that!

But still, though, any comments on the question:
(xy)(y - x)² = 0.
How do we know which should be be subtracted from which? In this particular case, it works out to be the same, because the result is squared.
Has anyone seen an official question where the order of the x and y matters?
Will it ever matter?


Thanks
--Rishi

hi rishi.. the order shud never matter...
firstly mathematically, they are the same..
second, they also lead to the same thing (which in this case is) ==> x = y (OR y = x)
but be careful when you dun have zero on the other side (so u get + or -) which then boils down to the same thing e.g.
(x - y)²=4 ... x -y = + 2 or x - y = - 2 ..... x = y + 2 or y = 2 + x
(y -x )² = 4 ... y -x = +2 or y - x = -2 ..... y = x + 2 or y + 2 = x

hope this helps
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