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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number-line-problem-22709.htmlx-is-a-positive-number-less-than-86563.html600-level-question-95138.html?hilit=halfway%20between%20greater%20closerHope it helps.

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