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

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

by Vincen » Thu Feb 08, 2018 8:24 am
If a > b > c > 0, is c < 3 ?
$$(1)\ \ \frac{1}{a}>\frac{1}{3}\ .$$
$$(2)\ \ \frac{1}{a}\ +\ \frac{1}{b}+\frac{1}{c}=1.$$

The OA is the option D.

I know how to show that the statement (1) is sufficient. But I don't know how to show that the statement (2) is sufficient. Experts, can you give me some help? I'd be thankful.
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Source: — Data Sufficiency |

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by Brent@GMATPrepNow » Thu Feb 08, 2018 9:57 am
Vincen wrote:If a > b > c > 0, is c < 3 ?
$$(1)\ \ \frac{1}{a}>\frac{1}{3}\ .$$
$$(2)\ \ \frac{1}{a}\ +\ \frac{1}{b}+\frac{1}{c}=1.$$
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

Answer = D

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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