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# If the average (arithmetic mean) of 5 consecutive integers i

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If the average (arithmetic mean) of 5 consecutive integers i [#permalink]  19 Mar 2018, 11:18
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Question Stats:

72% (00:43) correct 28% (00:31) wrong based on 25 sessions
If the average (arithmetic mean) of 5 consecutive integers is 12, what is the sum of the least and greatest of the 5 integers?

(A) 24
(B) 14
(C) 12
(D) 11
(E) 10
[Reveal] Spoiler: OA

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Re: If the average (arithmetic mean) of 5 consecutive integers i [#permalink]  02 Apr 2018, 19:31
1
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METHOD ONE:

The slowest way to solve this is to make some kind of equation as follows. The 5 terms are consecutive integers, so if we call the first n, the second would be n+1, the third n+2, etc. So if the average is 12, we have a big unwieldy equation like this:

(n + n+1 + n+2 + n+3 + n+4)/5 = 12

Eventually, we'll find that the smallest is 10, so the largest is 14, so they total 24. Yuck, though.

METHOD TWO:

An easier way would be to write down the 5 consecutive integers. If you know that the average of any evenly spaced set is the same as its middle term, you can immediately just write down 10, 11, 12, 13, and 14. Add the first and last to get 24.

METHOD THREE:

That wasn't bad, but the fastest and easiest way is this: if you know that the average of any evenly spaced set is the same as the average of the first and last terms, then we know that the average of the first and last terms is 12. Since there are 2, we can just multiply the average by 2 to get 24, so the answer is A.
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Re: If the average (arithmetic mean) of 5 consecutive integers i [#permalink]  11 Jan 2019, 10:27
SherpaPrep wrote:
METHOD ONE:

The slowest way to solve this is to make some kind of equation as follows. The 5 terms are consecutive integers, so if we call the first n, the second would be n+1, the third n+2, etc. So if the average is 12, we have a big unwieldy equation like this:

(n + n+1 + n+2 + n+3 + n+4)/5 = 12

Eventually, we'll find that the smallest is 10, so the largest is 14, so they total 24. Yuck, though.

METHOD TWO:

An easier way would be to write down the 5 consecutive integers. If you know that the average of any evenly spaced set is the same as its middle term, you can immediately just write down 10, 11, 12, 13, and 14. Add the first and last to get 24.

METHOD THREE:

That wasn't bad, but the fastest and easiest way is this: if you know that the average of any evenly spaced set is the same as the average of the first and last terms, then we know that the average of the first and last terms is 12. Since there are 2, we can just multiply the average by 2 to get 24, so the answer is A.

3rd one is great.
Re: If the average (arithmetic mean) of 5 consecutive integers i   [#permalink] 11 Jan 2019, 10:27
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