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(r+t)/(r-t)>0. r>t? (1) t>0 (2) r>0

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(r+t)/(r-t)>0. r>t? (1) t>0 (2) r>0 [#permalink] New post 16 Jun 2004, 23:32
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(r+t)/(r-t)>0. r>t?

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

Last edited by sweetlife on 17 Jun 2004, 04:14, edited 1 time in total.
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 [#permalink] New post 16 Jun 2004, 23:56
Answer is C. Combing both we wud get to know that r>t.

If r is less that t then the denominator will result in negetive value, which in turn wud not satisfy more than 0.

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Correct me if i am wrong.
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 [#permalink] New post 17 Jun 2004, 02:11
It says the answer is B and it doesnt convince me.
I am seeking the answer myself. It looks like C.
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 [#permalink] New post 17 Jun 2004, 03:46
(r+t)/(r-t)>0. r>t?

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

Pardon me for the wrong answer given last time. The answer is indeed B.

Only in the case of R>0 would the equation (r+t)/(r-t)>0. r>t be satisfied. If r is less than 0, then the equation may result is either case. If R>0, then if t is more than r then, the original equation will not be satisfied. I had checked this with number subs .. for any value of R more than 0, the equation to be valid, t shoud be less than r.

Sorry for causing the confusion.
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 [#permalink] New post 24 Jun 2004, 09:48
sweetlife wrote:
Thanks. B it is...


This is probably a case where I just don't see the obviuos answer, but I think it's D:

1) t >0

let's say t=4, then we have

(r+4)/(r-4)>0

r has to be greater than 4, b/c otherwise the answer would yield a negative answer.
Say r is 3, then we have 7/-1=-7 which is not greater than 0

what am I missing?
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 [#permalink] New post 26 Jun 2004, 07:49
Hi Lastocka

You have made a mistake here. If r is less than t, then the original equation does not stand good. Taking your case, you have mentioned, if r=3, then the orginal equation (r+4)/(r-4)>0 will be less than 0. But we are not suppose to change the orginal equation in here. :)

I hope it is clear now.

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Re: DS: Inequality (1) [#permalink] New post 27 Jun 2004, 00:00
sweetlife wrote:
(r+t)/(r-t)>0. r>t?

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


(r+t)/(r-t) = 1 + 2t/(r-t) > 0 => t > 0, r = -1.01*t. => 1 is not sufficient.

2 is sufficient because from (r+t)/(r-t) > 0 and r > 0 => -r < t < r.
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 [#permalink] New post 27 Jun 2004, 12:17
sweetlife wrote:
(r+t)/(r-t)>0. r>t?

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

I have a different answer, I will go with A

If t>0, take as an example t=1 and r=3. in this case the result will be (3+1)/(3-1) =2>0 sufficient

But if r>0 this do not mean that t must be positive it could be any number less than r. for example, r=2 t=-4
(2-4)/(2+4)=-2/6 . In other cases the result will be poistive. hence, Insufficient.

Correct me if I am wrong.

good luck all
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 [#permalink] New post 27 Jun 2004, 21:57
dr_sabr wrote:
sweetlife wrote:
(r+t)/(r-t)>0. r>t?

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

I have a different answer, I will go with A

If t>0, take as an example t=1 and r=3. in this case the result will be (3+1)/(3-1) =2>0 sufficient

But if r>0 this do not mean that t must be positive it could be any number less than r. for example, r=2 t=-4
(2-4)/(2+4)=-2/6 . In other cases the result will be poistive. hence, Insufficient.

Correct me if I am wrong.

good luck all


No, dr_sabr, it is wrong. Look, the stem SHOULD be always true, and in your example it is not satisfied! But it should be!
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 [#permalink] New post 28 Jun 2004, 01:39
I didn't understand what you meant

I will take 1) first.
If t>0, take as an example t=1 and r=3. in this case the result will be (3+1)/(3-1) =2>0 sufficient. because whatever is t, r must be >t.

2) But if r>0 this do not mean that t must be positive it could be any number less than r. for example, r=2 t=-4 =>(2-4)/(2+4)=-2/6 . In another case if r=3 t=2 =>(3+2)/(3-2)=5>0 . hence, Insufficient

Correct me if I am mistaken :) .
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 [#permalink] New post 28 Jun 2004, 01:55
dr_sabr wrote:
2) But if r>0 this do not mean that t must be positive it could be any number less than r.


NOT any number less than r, but any number, for which the STEM is true! i.e., for which (r+t)/(r-t) > 0. But in your example this is not the case: r = 2, t = -4 and (r+t)/(r-t) = -1/3 < 0 and this contradicts the stem!!!

dr_sabr wrote:
for example, r=2 t=-4 =>(2-4)/(2+4)=-2/6 . In another case if r=3 t=2 =>(3+2)/(3-2)=5>0 . hence, Insufficient

Correct me if I am mistaken :) .
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 [#permalink] New post 28 Jun 2004, 03:25
Emmanuel wrote:
But in your example this is not the case: r = 2, t = -4 and (r+t)/(r-t) = -1/3 < 0 and this contradicts the stem!!!


:oops I think that I misunderstood this question, I thought the stem was r>t :lol: . Thank you Emmanuel for correcting me. I hope I don't make this mistake on the test.
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 [#permalink] New post 28 Jun 2004, 03:37
dr_sabr wrote:
Emmanuel wrote:
But in your example this is not the case: r = 2, t = -4 and (r+t)/(r-t) = -1/3 < 0 and this contradicts the stem!!!


:oops I think that I misunderstood this question, I thought the stem was r>t :lol: . Thank you Emmanuel for correcting me. I hope I don't make this mistake on the test.


I hope that you will do such questions easily on the test, once you have seen what problems can arise here.
  [#permalink] 28 Jun 2004, 03:37
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