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# The average (arithmetic mean) of the multiples of 6

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Intern
Joined: 24 Apr 2010
Posts: 26
The average (arithmetic mean) of the multiples of 6  [#permalink]

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28 May 2010, 13:57
1
4
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Difficulty:

25% (medium)

Question Stats:

75% (01:14) correct 25% (01:26) wrong based on 190 sessions

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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
##### Most Helpful Expert Reply
Math Expert
Joined: 02 Sep 2009
Posts: 49267
Re: The average (arithmetic mean) of the multiples of 6  [#permalink]

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28 May 2010, 14:16
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1
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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Intern
Joined: 24 Apr 2010
Posts: 26
Re: The average (arithmetic mean) of the multiples of 6  [#permalink]

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28 May 2010, 14:56
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.
Math Expert
Joined: 02 Sep 2009
Posts: 49267
Re: The average (arithmetic mean) of the multiples of 6  [#permalink]

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28 May 2010, 15:12
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]

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27 Feb 2017, 06:33
First term is 6 and last term is 996
Sum of an AP is {n( first term + last term)}/2
Average = {n( first term + last term)}/2n
= ( first term + last term)/2
= (6 + 996)/2
= 501. Option C
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Re: The average (arithmetic mean) of the multiples of 6  [#permalink]

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01 Mar 2017, 18:19
1
perseverant wrote:
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

Since we have an evenly spaced set, we can use the following average formula:

average = (first term in the set + last term in the set)/2

The first multiple of 6 in the set is 6 and the last multiple of 6 is 996. Thus:

average = (6 + 996)/2 = 1002/2 = 501

Answer: C
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Re: The average (arithmetic mean) of the multiples of 6 &nbs [#permalink] 01 Mar 2017, 18:19
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# The average (arithmetic mean) of the multiples of 6

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