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# If (y+3)(y-1) (y-2)(y-1) = r(y-1), what is the value of y?

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Manager
Joined: 14 Jan 2006
Posts: 88
Schools: HKUST
If (y+3)(y-1) (y-2)(y-1) = r(y-1), what is the value of y?  [#permalink]

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01 Oct 2008, 01:41
If (y+3)(y-1) – (y-2)(y-1) = r(y-1), what is the value of y?
(1) r^2 = 25
(2) r = 5

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Manager
Joined: 30 Sep 2008
Posts: 111
Re: GMAT SET 16 - 1  [#permalink]

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01 Oct 2008, 02:06
D

(y+3)(y-1) – (y-2)(y-1) = r(y-1) <=> (y-1)(r-5) = 0

So no mater r is, y always = 5
Intern
Joined: 29 Sep 2008
Posts: 46
Re: GMAT SET 16 - 1  [#permalink]

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01 Oct 2008, 08:14
E.

solution of (y-1)(r-5) = 0 is
either y = 1 OR r = 5

I - r = 5 or -5. for r = 5, y could be any number. for r = -5, y = 1. Insufficient
II - r = 5. y could be any number
I and II - r = 5, y could be any number

Thus E.
VP
Joined: 17 Jun 2008
Posts: 1474
Re: GMAT SET 16 - 1  [#permalink]

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01 Oct 2008, 12:26
lylya4 wrote:
D

(y+3)(y-1) – (y-2)(y-1) = r(y-1) <=> (y-1)(r-5) = 0
So no mater r is, y always = 5

In the highlighted equation, either y = 1 or r = 5.

stmt1 says that r can be 5 or -5. Hence, insufficient.
stmt2 says that r is 5. Hence, y can have any value. Hence, insufficient.

Combining two statements, insufficient and answer should be E.
Intern
Status: In Pursuit of the Elusive
Joined: 02 Feb 2010
Posts: 9
Re: GMAT SET 16 - 1  [#permalink]

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16 May 2011, 03:18
(y+3)(y-1) - (y-2)(y-1)= r(y-1)
=> (y-1) {(y+3) -(Y-2)} = r(y-1)
=> y+3-y+2=r
=> r=5

We already know this from the question and the options are not providing any additional information. So E

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This is not a quality discussion. It has been retired.

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Re: GMAT SET 16 - 1 &nbs [#permalink] 16 May 2011, 03:18
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