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The average (arithmetic mean) of the multiples of 6

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The average (arithmetic mean) of the multiples of 6 [#permalink] New post 28 May 2010, 12:57
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A
B
C
D
E

Difficulty:

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Question Stats:

72% (02:16) correct 28% (01:21) wrong based on 71 sessions
The average (arithmetic mean) of the multiples of 6 that are greater than 0 and less than 1,000 is

A. 499
B. 500
C. 501
D. 502
E. 503
[Reveal] Spoiler: OA
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Re: The average (arithmetic mean) of the multiples of 6 [#permalink] New post 28 May 2010, 13:16
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perseverant wrote:
The average (arithmetic mean) of the multiples of 6 that are greater than 0 and less than 1,000 is

499
500
501
502
503


Could someone explain what the quickest way to solve this is?
Thanks!


Multiples of 6 represent arithmetic progression (aka evenly spaced set). In any evenly spaced set the arithmetic mean (average) is equal to the median and can be calculated by the formula mean=median=\frac{a_1+a_n}{2}, where a_1 is the first term and a_n is the last term.

First term is 6 and last term is 996 (the last even multiple of 3 below 1000). So mean=\frac{6+996}{2}=501.

Answer: C.
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Re: The average (arithmetic mean) of the multiples of 6 [#permalink] New post 28 May 2010, 13:56
Thank you! this is very helpful!

First term is 6 and last term is 996 (the last even multiple of 3 below 1000). So mean=\frac{6+996}{2}=501.
I assume you meant multiple of 6 instead of 3.
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Re: The average (arithmetic mean) of the multiples of 6 [#permalink] New post 28 May 2010, 14:12
Expert's post
perseverant wrote:
Thank you! this is very helpful!

First term is 6 and last term is 996 (the last even multiple of 3 below 1000). So mean=\frac{6+996}{2}=501.
I assume you meant multiple of 6 instead of 3.


No. This is how I found the last multiple of 6 below 1000: it would be the last EVEN multiple of 3 (thus multiple of 6) below 1000 .
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Re: The average (arithmetic mean) of the multiples of 6 [#permalink] New post 09 Nov 2014, 02:58
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Re: The average (arithmetic mean) of the multiples of 6   [#permalink] 09 Nov 2014, 02:58
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