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If (r+t)/(r-t)>0, is r>t? (1) t>0 (2) r>0

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gmatblast
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If (r+t)/(r-t)>0, is r>t? (1) t>0 (2) r>0 [#permalink] New post   24 Aug 2004, 12:42
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E

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If (r+t)/(r-t)>0, is r>t?

(1) t>0
(2) r>0



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Re: If (r+t)/(r-t)>0, is r>t? (1) t>0 (2) r>0 [#permalink] New post   24 Aug 2004, 14:16
A is insuff. R can be positive and large then T, or negative and smaller than T.

B IS sufficient.

I choose B. ( if T is Bigger than R, than numerator is positive, denom is negative) By contradiction.
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Re: If (r+t)/(r-t)>0, is r>t? (1) t>0 (2) r>0 [#permalink] New post   25 Aug 2004, 11:52
Yes..it is B I believe. First I was tempted to go for E.
Agree with the reasons given by others..

gmatblast wrote:
If (r+t)/(r-t)>0, is r>t?

(1) t>0
(2) r>0
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Re: If (r+t)/(r-t)>0, is r>t? (1) t>0 (2) r>0 [#permalink] New post   25 Aug 2004, 18:23
1.5 min
B
plugged in 2 for t and got my answer by some minor plumbing for r
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Re: If (r+t)/(r-t)>0, is r>t? (1) t>0 (2) r>0 [#permalink] New post   26 Aug 2004, 19:41
(r+t)/(r-t) > 0
(r+t) > 0
r > -t

From (1), if t > 0, then t will have values 1,2,3... and -t will have values -1,-2,-3...
But we do not have a range for r. r could be having values 0,1,2,3,4... in which case it satisfies the equation
Conversely, r could be -3 when t is -2 and so smaller than -t
Therefore (1) is not sufficient.

From (2), we are told r > 0, so we know for sure it is bigger than any negative value and satisfies the equation.

And when we satisfy (r+t)/(r-t)>0, we will know r > t because if this isn't so, (r+t)/(r-t) may be < 0.



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Re: If (r+t)/(r-t)>0, is r>t? (1) t>0 (2) r>0 [#permalink]
26 Aug 2004, 19:41
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