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# If a > b > c > 0, is c < 3 ?

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If a > b > c > 0, is c < 3 ?  [#permalink]

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Updated on: 12 Aug 2014, 03:52
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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

Originally posted by yezz on 16 Aug 2009, 00:45.
Last edited by Bunuel on 12 Aug 2014, 03:52, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: If a > b > c > 0, is c < 3 ?  [#permalink]

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12 Aug 2014, 04:04
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If a > b > c > 0, is c < 3 ?

(1) 1/a > 1/3 --> cross-multiply (notice that we can safely do this since we know that a > 0): a < 3. As c < a, then c < a < 3. Sufficient.

(2) 1/a + 1/b + 1/c = 1. If c is more than or equal to 3, then 1/c is less than or equal to 1/3 (for example, if c is 3, then 1/c is 1/3 and if c is 4, then 1/c is 1/4). In this case both 1/b and 1/a would be less than 1/3. So, 1/a + 1/b + 1/c = (less than 1/3) + (less than 1/3) + (less than or equal to 1/3) = (less than 1), which contradicts the given statement. Therefore our assumption that c could be more than or equal to 3 was wrong, which implies that c < 3. Sufficient.

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

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16 Aug 2009, 08:30
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1
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: If a > b > c > 0, is c < 3 ?  [#permalink]

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16 Aug 2009, 08:43
scarish wrote:
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.

Thats D.
good job for good question.
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Re: If a > b > c > 0, is c < 3 ?  [#permalink]

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

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

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04 Aug 2014, 18:36
Can some one explain Stm 2 better than the above explanations
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Re: If a > b > c > 0, is c < 3 ?  [#permalink]

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05 Aug 2014, 05:04
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ashutoshbarawkar wrote:
Can some one explain Stm 2 better than the above explanations

Statement 2 : 1/a + 1/b + 1/c = 1

given a>b>c and all are positive thus == >1/c > 1/b > 1/a --(i)

let me assume values

1/a 1/b 1/c C
0.1 0.2 0.7 10/7
0.2 0.3 0.5 10/5
0.3 0.4 0.4 ---> This violates the condition--(i)

Thus C value will always be <3 ..

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

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10 Nov 2015, 16:56
2
ashutoshbarawkar wrote:
Can some one explain Stm 2 better than the above explanations

definitely you can get c<3 from the condition:

a>b>c>0

so 1/a<1/b<1/c

because 1/a+1/b+1/c=1

if you replace 1/a with 1/c, and 1/b with 1/c, you will get a sum bigger than 1:

1/c+1/c+1/c>1/a+1/b+1/c=1

so 3/c>1

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

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10 Nov 2015, 17:19
3
yezz wrote:
If a > b > c > 0, is c < 3 ?

(1) 1/a > 1/3
(2) 1/a + 1/b + 1/c = 1

Here's a slightly different approach . . .

Target question: Is c < 3?

Given: 0 < c < b < a

Statement 1: 1/a > 1/3
Since we can be certain that a is positive, it's safe to take the inequality 1/a > 1/3 and multiply both sides by a to get: 1 > a/3
Likewise, we can take 1 > a/3 and multiply both sides by 3 to get: 3 > a
If 3 > a and c < a, then we can conclude that c < 3
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: (1/a) + (1/b) + (1/c) = 1
IMPORTANT: If 0 < c < a, we can conclude that 1/a < 1/c
Likewise, since 0 < c < b, we can conclude that 1/b < 1/c

In other words, 1/c is BIGGER than both 1/a and 1/b
So, if we take the equation (1/a) + (1/b) + (1/c) = 1 and replace both 1/a and 1/b with 1/c, the resulting sum will be BIGGER than 1
That is, (1/c) + (1/c) + (1/c) > 1
Simplify to get: 3/c > 1
Since we know that c is positive, it's safe to take the inequality and multiply both sides by c to get: 3 > c
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

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

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29 Mar 2016, 01:27
a>b>c>0 is c>3?

1) 1/a >1/3
since a is positive we can multiply both sides by a . therefore a <3
3<a<c , and hence c >3 ...sufficient

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

we have to prove c>3 ... also means 1/c >1/3 ...also means 1/c>33.33% ( since 1/3 is 33.33% for getting rid of fraction, i have a phobia of fractions )
also 1/c>1/b>1/a ( rephrasing given data, it says 1/c should be greatest)
now back to st. 2 we can write this as 1/a + 1/b +1/c =100%
as we have to prove 1/c >33.33% take a case where 1/c is 30% .. then 35+35+30 =100 ( cannot happen since 1/c should be greatest)
1/c has to greater than equal to 40% in this case (21+39+40) so 1/c >33.33 %
sufficient.

ans is D
i hope i am right
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Re: If a > b > c > 0, is c < 3 ?  [#permalink]

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16 Mar 2018, 01:49
Hi,

is it possible to do the following for the 2nd statement:

[a + b + c][/abc] = 1

a+b+c = abc

this forces us to use numbers for a, b, and c, that will both add and multiply to give the same answer. Is this a viable approach?
Re: If a > b > c > 0, is c < 3 ?   [#permalink] 16 Mar 2018, 01:49
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