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factoring and canceling exponents (I think)

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by AndyWRX » Wed Nov 05, 2008 10:33 am
DS question

If xy does not equal zero, what is the value of

x^4y^2 - (xy)^2
--------------------
x^3 y^2


1) x=2

2) y = 8

This question has to do with canceling out the variable y but I do not know how to manipulate this equation to eliminate y.

Can someone please help me manipulate this equation, particularly how to change (xy)^2

Thanks, I will post the answer later.

Andy
Its been 20 years since I last did this stuff and I had forgotten how much fun it is
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Source: — Data Sufficiency |
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by dhanda.arun » Wed Nov 05, 2008 10:39 am
x^4y^2 - (xy)^2
--------------------
x^3 y^2

This reduces to x^2y^2(x^2 -1)
-------------------
x^3y^2

cancel x^2y^2 from num and denominator
we have
[x^2-1]/x
this is equal to x-1/x
and hence can be solved when x is known
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by AndyWRX » Wed Nov 05, 2008 10:48 am
Thanks for the reply but I still not sure I understand the process of reducing the equation, would you mind explaining how you reduced the top line of the equation.

Just for clarity, perhaps I should have written it with spaces

x^4 y^2 - (xy)^2
-------------------
x^3 y^2

Thanks

Andy
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by dhanda.arun » Wed Nov 05, 2008 11:53 am
x^4y^2 - (xy)^2

take the second term (xy)^2 = x^2 y^2
and
x^4 in first term can be written as X^2 * X^2

so the equation can be written as

X^2 * X^2 * y^2 - x^2 y^2

we can take out x^2 y^2
so the equn becomes x^2 y^2(X^2 - 1)

if u have difficulty in factoring out,
consider x^2 y^2 = a
so the equn is a.X^2 -a
take a common..........!!!
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by cramya » Wed Nov 05, 2008 2:29 pm
I would go with A) sicne the y's cancel out but x remains. We need to know x to determine the value of the expression.

Hope I am not missing something.


Please post OA's using spoiler function when posting the question if possible
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by cramya » Wed Nov 05, 2008 2:32 pm
x^4 y ^ 2 - x^2 y ^2 / x ^ 3 y ^ 2 ( (xy) ^ 2 = x^2 y ^ 2 using Exponent rule: (ab)^n = a^n b ^ n)

Take y ^ 2 i.e common from the numerator x^4 y ^ 2 - x^2 y ^2

= y ^ 2 ( x^4 - x^2) / x ^ 3 y ^ 2
y^2 in the Numerator and Denominator cancels leaving us with

( x^4 - x^2) / x ^ 3

Stmt I is sufficient since it gives value of x and therefore we can calculate the value of the expression.

Hope this helps!
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Re: factoring and canceling exponents (I think)

by aj5105 » Thu May 07, 2009 10:53 pm
Simplifying, we get:

(x^4y^2 - x^2y^2) / x^3y^2

Taking x^2y^2 common, [x^2y^2 (x^2 - 1)]/x^3y^2

(x^2 - x) / x

x - 1

Hence (A) is sufficient.



AndyWRX wrote:DS question

If xy does not equal zero, what is the value of

x^4y^2 - (xy)^2
--------------------
x^3 y^2


1) x=2

2) y = 8

This question has to do with canceling out the variable y but I do not know how to manipulate this equation to eliminate y.

Can someone please help me manipulate this equation, particularly how to change (xy)^2

Thanks, I will post the answer later.

Andy
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by cooo3 » Fri Mar 19, 2010 3:47 am
isnt the answer supposed to be x-1/x ?
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