If a>b>c>0, is c <3? (1) 1/a>1/3 (2) 1/a+1/b+1/c=1

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If a>b>c>0, is c <3? (1) 1/a>1/3 (2) 1/a+1/b+1/c=1 [#permalink]

16 Aug 2009, 00:45

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4)If a>b>c>0, is c <3?
(1) 1/a>1/3
(2) 1/a+1/b+1/c=1

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Re: 4)If a>b>c>0, is c <3? [#permalink]

16 Aug 2009, 08:30

IMO....D

(1) 1/a>1/3
Solution: 0 < a < 3
Since a > b > c > 0....and a < 3, then c must be < 3.
Sufficient.

2) 1/a+1/b+1/c=1

1 = 1/3 + 1/3 + 1/3
if C=3 then 1/a < 1/b < 1/3
Hence, the sum of the fraction can never reach 1....

Therefore, c has to be < 3 in order to get the sum of fractions reach 1
Sufficient.
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Re: 4)If a>b>c>0, is c <3? [#permalink]

16 Aug 2009, 08:43

scarish wrote:

Thats D.
good job for good question. 8-)

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Re: 4)If a>b>c>0, is c <3? [#permalink]

23 Oct 2009, 18:01

Stmt 2 is not crystal clear..

We have here,a+b+c=abc which is only true when a,b,c=3.However,from the stem we know that a>b>c..hmmm..getting there but missing something...
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Re: 4)If a>b>c>0, is c <3? [#permalink]

23 Oct 2009, 20:32

humm , for stmt 2 , is there any other way to solve the problem other than one mentioned by scarish

ab+bc+ab = abc , now can i solve from this ?
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Re: 4)If a>b>c>0, is c <3? [#permalink] 23 Oct 2009, 20:32

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If a>b>c>0, is c <3? (1) 1/a>1/3 (2) 1/a+1/b+1/c=1

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If a>b>c>0, is c <3? (1) 1/a>1/3 (2) 1/a+1/b+1/c=1

If a>b>c>0, is c <3? (1) 1/a>1/3 (2) 1/a+1/b+1/c=1

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