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value of y

by pappueshwar » Tue Mar 13, 2012 7:09 am
(y+3)(y-1) - (y-2)(y-1) = r (y-1), what is the value of y ?

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

OA IS E

IMO B... why is B incorrect
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by rajatvmittal » Tue Mar 13, 2012 7:27 am
You must have reached this stage by solving the equation: 5(y-1)=r(y-1)

Further,
5(y-1)-r(y-1)=0
(5-r)(y-1)=0 (A)
either(5-r)=0 or (y-1)=0

We need to calculate the value of y. If r is 5 as in option 2, then 5-r would be zero. If (5-r) is zero then irrespective of the value of y, the result of multiplication will be zero. I.e if r =5 then y could be 100 or 1 million. Not sufficient.
In option 1, r may be 5 or -5, then equation A would result in following
if r = 5, then y may be infinite numbers
if r = -5 then y must be 1.
Hence not sufficient.
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by pappueshwar » Tue Mar 13, 2012 7:44 am
thanks..

realized that after posting the question :)
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by killer1387 » Tue Mar 13, 2012 7:48 am
pappueshwar wrote:(y+3)(y-1) - (y-2)(y-1) = r (y-1), what is the value of y ?

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

OA IS E

IMO B... why is B incorrect
5(y-1)=r(y-1)
1) r=+5

for r=5 => infinite y

for r=-5
5y-5=-5y+5
10y=10
y=1

Insufficient

2) r=5
y=infinite values

from 1&2

r=5
hence y=infinite values

hence E is correct.
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by tpr-becky » Tue Mar 13, 2012 12:03 pm
Yes, the key to this question, beyond the math, is to be sure to answer what the test is asking - most people get hung up solving for what they think they should solve for (r) - even though the test asks for the value of y. It is a good practice to always re-read the final question.
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