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

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

by shargaur » Tue Aug 04, 2009 11:20 am
if (y+3)(y-1) -(y-2)(y-1)=r(y-1) what is value of y?
1) r^2=25
2) r=5

The[spoiler] OA=E[/spoiler]. I got to OA also. but i have different approach of solving

let given eqn be solved to (y-1)(y+3 -y+2 -r)=0
=> (y-1)(5-r)=0
it says y=1..

so there is no need for any of the statement. Answer should be D
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by srisanj4 » Sun Aug 09, 2009 5:02 pm
y could be any value to satisfy r = 5

hence, the answer is E
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Re: (y+3)(y-1) -(y-2)(y-1)=r(y-1)

by tohellandback » Sun Aug 09, 2009 10:07 pm
shargaur wrote:if (y+3)(y-1) -(y-2)(y-1)=r(y-1) what is value of y?
1) r^2=25
2) r=5

The[spoiler] OA=E[/spoiler]. I got to OA also. but i have different approach of solving

let given eqn be solved to (y-1)(y+3 -y+2 -r)=0
=> (y-1)(5-r)=0
it says y=1..

so there is no need for any of the statement. Answer should be D
it does not say y=1

it says:
either, Y=1 OR r=5 OR both

1)r=5 or -5
if r=5, Y=any value is possible
NOT SUFF

2)same as 1
combined same as above

so E
The powers of two are bloody impolite!!
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by csabiga » Mon Aug 10, 2009 7:02 am
Only y=1 can satisfy r^2=25
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