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Value problem

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Source: — Data Sufficiency |
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Re: Value problem

by parallel_chase » Sat Sep 06, 2008 1:20 am
arorag wrote: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


IMO A
y^2-y+3y-3 - [ y^2 - y -2y +2 ] = r(y-1)

y^2 +2y -3 - y^2 + 3y -2 =r(y-1)

5y - 5 = r(y-1)

5(y-1) = r(y-1)

r=5

Statement I

r^2 = + or -5

if r=-5

5(y-1) = -5(y-1)

5y-5 = -5y +5

y=1

or

if r=5

5(y-1) = 5y-5

y could be anything 1,2,3, anything

Insufficient.

Statement II

r=5

y could be anything 1,2,3 anything

Insufficient.


Combining I & II

r=5

y could be anything, 1 ,2,3,4, 0.5

Insufficient.

Hence the answer should be E.


Whats the OA?
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by 4meonly » Sat Sep 06, 2008 6:00 am
What is the source of this Q? I think it is little bit crooked

From the main statement we have
r=5 - and this should be not changable

(1)
gives us r = +/- 5
but we have r = 5
5(y-1) = 5y-5 - nothing definite
Thus, INSUFF

(2)
r=5
Yeh, good news! We knew it!!!
So what? 5(y-1) = 5y-5 - nothing definite once again!
INSUFF

Answer E


By the way, I think that should be read in such way:

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

with answer B

arorag, could you please check?
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