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DS: Difficult question- help needs.

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Intern
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DS: Difficult question- help needs. [#permalink] New post 08 Aug 2004, 22:52
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
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Hi,

Any idea for the following question ? What's correct answer and explain ?

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

[/b]
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 [#permalink] New post 08 Aug 2004, 23:07
I think it is C.

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

(1) t>0 -->
with (r+t)>0 & (r-t)>0 <--> r> - t & r> t --> r>t

OR (r+t)<0 & (r-t)<0 <--> r<-t & r<t ---> r< t

So (1) is insufficient.

(2) r>0 -->
with (r+t)>0 & (r-t) > 0 <--> r > - t & r> t

or (r+t)<0 & (r-t)<0 <--> r < -t & r<t

So (2) is insufficient

(1) & (2) --> with (r+t)>0 & (r-t)> 0 --> r>t
or (r+t)<0 & (r-t)<0 --> impossible cos r+t always > 0

---> sufficient

So the ans is C
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 [#permalink] New post 09 Aug 2004, 00:42
b?

Given: r+t/r-t>0
Question: r>t?

For r+t/r-t to be positive,
-> Option 1: r is postive and > t
OR
-> Option 2: r is negative and < t

The fact that t is greater than zero (St: 1) doesn't do much help, since it is subtracted in denominator.
The fact that r is greater than zero (St: 2) removes option 2 and assures r>t

took more than 2 minutes though :cry:
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 [#permalink] New post 09 Aug 2004, 02:57
1) if both t and r are positive, then r>t
but if t>0 and r is a big negative then r<t
NOT SUFFICIENT

2) if r>0 then it is impossible that r<t because this would make (r+t)/(r-t) negative. Hence, it must be r>t. SUFFICIENT

Therefore, B
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 [#permalink] New post 09 Aug 2004, 07:37
very good question..thanks for posting.
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 [#permalink] New post 09 Aug 2004, 08:05
Yes, good question. I got B also by plugging in numbers. Took about 2 min.
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Best Regards,

Paul

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 [#permalink] New post 09 Aug 2004, 16:25
nice question. I was tempted to go on with C. B is my choice too
  [#permalink] 09 Aug 2004, 16:25
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DS: Difficult question- help needs.

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