rxs0005 wrote:

If a is a positive number less than 10, is c greater than the average (arithmetic mean) of a and 10?

(1) On the number line, c is closer to 10 than it is to a.

(2) 2c – 10 is greater than a.

Given: \(0<a<10\). Question: is \(c\) greater than the average (arithmetic mean) of \(a\) and 10? --> or is \(c>\frac{a+10}{2}=average\)? --> or is \(2c>a+10\)?

(1) On the number line, c is closer to 10 than it is to a.

Number line approach:a-----average-----10----- (average of a and 10 is halfway between a and 10). So the question ask whether c is either in the BLUE or GREEN area.

As, c is closer to 10 than it (c) is to a then this statement directly tells us that c is either in the BLUE or GREEN area. Sufficient.

Algebraic approach:c is closer to 10 than it is to a, means that the distance between c and 10 is less than the distance between c and a. So, \(|10-c|<|c-a|\). Now, as c is closer to 10 than it is to a, then c>a, so \(|c-a|=c-a\) --> two cases for 10-z:

A. \(c\leq{10}\) --> \(|10-c|=10-c\) --> \(|10-c|<|c-a|\) becomes: \(10-c<c-a\) --> \(2c>10+a\). Answer to the question YES.

B. \(c>{10}\) --> in this case \(2c>20\) and as \(a<10\), then \(a+10<20\), hence \(2c>10+a\). Answer to the question YES.

(2) 2c – 10 is greater than a --> \(2c-10>a\) --> \(c>\frac{a+10}{2}=average\), again directly tells us that c is greater than the average of a and 10. Sufficient.

Answer: D.

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