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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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(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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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

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

gmatclubot

Re: If a > b > c > 0, is c < 3 ?
[#permalink]
29 Mar 2016, 00:27

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