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by dumb.doofus » Mon Apr 27, 2009 8:39 pm
Got it.. my bad!! A it is..
Last edited by dumb.doofus on Tue Apr 28, 2009 11:04 pm, edited 1 time in total.
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by keizer Soze » Tue Apr 28, 2009 7:19 am
I just solved it this way:

1° R2-R1 = -1 | So, R1=R2+1 so, R1>R2

2° R3-R2 = 1/2 | So, R3=R2 + 1/2 so, R3>R2

3° R1=R2 + 1 and R3=R2 + 1/2 R1 is 1/2 more than R3 | So R1>R3

So I guess that it´s A the answer...
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by dendude » Tue Apr 28, 2009 9:16 am
I think I am missing something here. Do you know what is the source of this q and if there is any element missing from the q?

They way I see it is that all boils down to the value of r1
On simplifying,
r2 = (1/2) + r1
r3 = (1/6) + r1

If r1 is -ve say -(1/2)
r2 = 0 and r3 = -(1/3)
So you have r3<r1<r2

But if r1 is +ve say 1/2
r2 = 1 and r3 = 2/3
Now, r1<r3<r2

I can only conclude that r2 would be the greatest and can eliminate choices A, B and E.
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by keizer Soze » Tue Apr 28, 2009 9:43 am
Thanx dendude,

I think that somenthing is wrong with your numbers:

r2 - r1 = -1^n/n
r2= -1^1/1 + r1
r2= -1 + r1 and not r2= (1/2) + r1
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by dendude » Tue Apr 28, 2009 9:59 am
keizer Soze wrote:Thanx dendude,

I think that somenthing is wrong with your numbers:

r2 - r1 = -1^n/n
r2= -1^1/1 + r1
r2= -1 + r1 and not r2= (1/2) + r1
You're right thanks for pointing out my blunder :oops:
I assumed r2 = (-1)^2 etc....
when in fact it was r(n+1) = (-1)^n and not to the power n+1

With that in mind,
r2 = -1 + r1
r3 = (-1/2) + r1

If r1 is -ve, say -1/2
r2 = -3/2 and r3 = -1
So, r2<r3<r1

If r1 is +ve, say 1/2
r2 = -1/2 and r3 = 0
Also, r2<r3<r1

Clearly, Choice A
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by keizer Soze » Tue Apr 28, 2009 12:48 pm
Great, we are Ok then!
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