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28 May 2010, 13:57
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C
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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
A. 499
B. 500
C. 501
D. 502
E. 503
28 May 2010, 14:16
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perseverant
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!
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.
General Discussion
28 May 2010, 14:56
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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.
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.
