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# Algebra - JSQ74

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Senior Manager
Joined: 25 Nov 2006
Posts: 333
Schools: St Gallen, Cambridge, HEC Montreal

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07 Feb 2009, 16:55
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Senior Manager
Joined: 30 Nov 2008
Posts: 482
Schools: Fuqua

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07 Feb 2009, 17:26
IMO E.

from stmt 1 -

$$x^2y = x^3y ==> x^2y - x^3y = 0 ==> x^2y(1-x) = 0$$

It is given that y <> 0 ==> from the above eq, either x = 0 or x = 1. Insufficient.

From Stmt 2 -

$$x^2/y = x^3y ==> x^2 = x^3y^2 ==> x^2 - x^3y^2 = 0 ==> x^2(1-xy^2) = 0$$

==> either $$x = 0$$ or $$1-xy^2 = 0 ==> x = 1/(y^2)$$Not sufficient.

Together also does not provide any informationt say whether x = 0 or not.
SVP
Joined: 07 Nov 2007
Posts: 1765
Location: New York

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07 Feb 2009, 20:22
lumone wrote:

each statment not sufficient

When combined

x^2y= x^2/y

--> x^2(y^2-1) =0
--> x=0 or x can be any number when Y^2=1

not suffcient

E
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SVP
Joined: 29 Aug 2007
Posts: 2457

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07 Feb 2009, 21:27
lumone wrote:

1: x^2y = x^3y
x^2 = x^3

x could be 0 or 1. not suff...

2: x^2/y = x^3y
x^2 = x^3y^2
y could be 1 or -1 but x could be 0 or 1.

Togather also same. E
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Senior Manager
Joined: 25 Nov 2006
Posts: 333
Schools: St Gallen, Cambridge, HEC Montreal

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08 Feb 2009, 12:08
OA is E.

Well done guys!
Senior Manager
Joined: 25 Nov 2006
Posts: 333
Schools: St Gallen, Cambridge, HEC Montreal

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08 Feb 2009, 12:21

I got:

x= 1/(y^2)

Thus x cannot equal 0.

I must be wrong somewhere, but where?
SVP
Joined: 07 Nov 2007
Posts: 1765
Location: New York

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08 Feb 2009, 13:14
lumone wrote:

I got:

x= 1/(y^2)

Thus x cannot equal 0.

I must be wrong somewhere, but where?

x^2/y = x^3y

you are striking out x^2 on both sides.. you can't do that unless you know x is not equal to zero.

x^2/y = x^3y
--> x^2( 1/y-yx)=0

so x can be zero or 1/y= yx --> x= 1/y^2

more solutions so insufficient.
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Re: Algebra - JSQ74   [#permalink] 08 Feb 2009, 13:14
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# Algebra - JSQ74

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