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If P<R<T, is the average of them is equal to 5/2R? 1)

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If P<R<T, is the average of them is equal to 5/2R? 1) [#permalink] New post 06 Jan 2005, 10:29
If P<R<T, is the average of them is equal to 5/2R?

1) P+R+T=5R
2) P+T=3R

explanation plz
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 [#permalink] New post 06 Jan 2005, 10:44
"D". Both r sufficient. Avg = P+R+T / 3

1) Avg = 5R/3 ...Ans is no as 5R/3 not equal to 5/2R --- So sufficient

2) Avg = 3R+R/ 3 = 4R/3 ...Ans is again no --- So sufficient


Hence "D".
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 [#permalink] New post 06 Jan 2005, 10:50
yaa, answer is D

thx
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 [#permalink] New post 06 Jan 2005, 10:53
I get D like banerjeea_98 does, only if I assume P, R, S are all non-zero.
Otherwise we can't tell as answer will vary => E
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 [#permalink] New post 06 Jan 2005, 11:24
nocilis please elobrate, I cannot see how answer would vary?
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 [#permalink] New post 06 Jan 2005, 12:11
Answer is D
and all of them cannot be 0 as one is less than another
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 [#permalink] New post 06 Jan 2005, 12:32
I think only R cannot be zero, otherwise both 1) and 2) do no make sense.

So this left either P or T can be zero, and let's look at each case:

T is zero:

T=0, R=-1, P=-4, P<R<T, and this matches 1) where P+R+T=5R. The average is -5/3 which is not equal to 5/2R. So 1) alone is sufficient to answer the question. We can eliminate B, C, & E.

P is zero:

P=0, R=1, T=3, P<R<T, and this matches 2) where P+T=3R. The average is 4/3 which is not equal to 5/2R. So 2) alone is good too. We can eliminate A and the answer is D.
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 [#permalink] New post 06 Jan 2005, 13:35
D is not correct, you assume R is not zero. but there is no conclusive definition saying R is not equal to zero.

I see no evidence to make that leap of faith, however, if R is not zero and you can prove thats that case, then D is correct!
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 [#permalink] New post 06 Jan 2005, 14:08
I don't think we sud think that R can be zero as the question wud then be asking us if the avg of these three numbers is infinity (5/2R)...doesn't make sense to think R can be zero, atleast to me.
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 [#permalink] New post 06 Jan 2005, 17:45
1)
Let us take an example
P = 1, R =2, T= 7, which satisfies P+R+T=5R
Then, average of them 'is' 10/3 => it is 'not' 5/2R

Let us take another example
P = -1, R =0, T= 1, which satisfies P+R+T=5R
Then, average of them 'is' 0 => it 'is' equal to 5/2 R =0

Answer varies

2)
P = 1, R =2, T= 5, which satisfies P+T=3R
Then, average of them 'is' 8/3 => it 'is not' 5/2R

Let us take another example
P = -1, R =0, T= 1, which satisfies P+T=3R
Then, average of them 'is' 0 => it 'is' equal to 5/2 R =0

Answer varies

1) and 2) taken together
P+R+T=5R
P+T=3R
=> 4R = 5R
=> R = 0
Therefore, P =-T and it will provide an average of 5/2 R = 5/2(0)=0 always

Answer is C (not E as I wrote before)

If the question said the numbers are positive or non-zero, D would have been right
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 [#permalink] New post 06 Jan 2005, 18:04
I have assumed that the question says 'is the average of them is equal to (5/2)* R? and not 5/(2*R). DLMD Is this correct?

(I have assumed so because if R were at the denominator, usually there is a statement stating that R<>0, and in this problem there was no such statement.)
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 [#permalink] New post 07 Jan 2005, 08:06
interesting one. at a Glance I picked D too. but nocilis showed it using 0's. well what is the OA?
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 [#permalink] New post 08 Jan 2005, 09:59
Thanks nocilis good explanation!!!
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 [#permalink] New post 10 Jan 2005, 09:38
OA is D, and agree with nocilis's excellent answer. Retract my reasoning...
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 [#permalink] New post 10 Jan 2005, 19:13
DLMD, pls tell us the OA because i am confused seeing E, C and D all over the place. Thanks
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 [#permalink] New post 11 Jan 2005, 09:40
His reply on the 3rd floor told us the OA is D...
  [#permalink] 11 Jan 2005, 09:40
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