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

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New post 05 Feb 2007, 08:42
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(y+3)(y-1) - (y-2)(y-1) = r(y-1)

what is the value of y?

(1) r^2=25
(2) r=5


This question is from an old ETS test. I dont understand the OA, please provide explanations with your answer. Thanks!
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New post 05 Feb 2007, 10:05
a rather weird question....

i guess answer is E.

with some algebra, stem becomes 5(y-1)=r(y-1)

so if r can be 5 then y can be anything....
neither statements rule out the possibility that r=5 hence y can be anything. insufficient.

if a statement would say r>5 (or anything tha 5 is not a solution to it) then y must be 1.
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New post 05 Feb 2007, 14:16
From the question, we can reduce the equation to be 5(y-1) = r(y-1)

Info (1); r^2 = 25
r = -5, 5
For r=-5; 5(y-1) = -5(y-1)
y-1 = 1-y
2y = 2; y = 1
For r = 5; 5(y-1) = 5(y-1)
y could be any numbers. INSUFF

Info (2); r = 5
this is one of the solution from info (1). y could be any numbers. INSUFF

Info (1) + (2); r = 5; same solution y could be any numbers. INSUFF

E. is the answer.
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New post 05 Feb 2007, 14:30
(E) as well... Nothing to add to the given explanations :)
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New post 05 Feb 2007, 14:32
Agreed the answer will be E indeed as Y can be anything, I also get the equation as:

5(y-1) = r(y-1)
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New post 05 Feb 2007, 15:25
ahh, I see. I've never come across a question like this. It didnt occur to me that Y could be anything. Sometimes the simple things elude me in GMAT.

OA is E

Thanks.
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Re: (y+3)(y-1) - (y-2)(y-1) = r(y-1) what is the value of y? (1) [#permalink]

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Re: (y+3)(y-1) - (y-2)(y-1) = r(y-1) what is the value of y? (1)   [#permalink] 22 Feb 2016, 11:14
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