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If a is a positive number less than 10, is c greater than the average

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If a is a positive number less than 10, is c greater than the average [#permalink]

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New post 22 Jan 2018, 22:28
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If a is a positive number less than 10, is c greater than the average (arithmetic mean) of a and 10?

(1) c = 3a
(2) On the number line, c is closer to a than it is to 10.
[Reveal] Spoiler: OA

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Re: If a is a positive number less than 10, is c greater than the average [#permalink]

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New post 22 Jan 2018, 23:01
B.

From question stem we can say there are nine possible values of 'a'. the greatest is 9 and the smallest is 1. average of 1 and 10 is 5.5 and average of 9 and 10 is 9.5.

From statement 1 we can say, for smallest possible value of 'a', 'c' is 3 and for the greatest possible value of 'a', 'c' is 27.

when a=1 and c=3
average of 'a' and 10 is 5.5>c

when a=9 and c=27
average of 'a' and 10 is 9.5<c

Hence insufficient.

From statement 2, we can say c <= 5 when a = 1 and c <= 9 when a = 9.

in both the cases c < average of 'a' and 10.

Hence sufficient.
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Re: If a is a positive number less than 10, is c greater than the average [#permalink]

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New post 23 Jan 2018, 22:31
Bunuel wrote:
If a is a positive number less than 10, is c greater than the average (arithmetic mean) of a and 10?

(1) c = 3a
(2) On the number line, c is closer to a than it is to 10.


Is c > (a+10)/2 Or Is 2c > (a+10)?

(1) c = 3a, then 2c = 6a. So now the question becomes:
Is 6a > a+10 Or Is 5a > 10 Or Is a > 2?
Since comparison of a with 2 is not given, we cannot answer the question. So Insufficient.

(2) The nature of average of two numbers is such that it will lie exactly in the middle of the two numbers. So the average of a and 10 will lie exactly in the middle of a and 10 on the number line. But we are given that c is closer to a than to 10, so this means c is lesser than the average of a and 10. Thus this gives us NO as an answer to the question asked. So this is sufficient.

Hence B answer
Re: If a is a positive number less than 10, is c greater than the average   [#permalink] 23 Jan 2018, 22:31
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