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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. # If the average (arithmetic mean) of 5 consecutive integers i  Question banks Downloads My Bookmarks Reviews Important topics
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If the average (arithmetic mean) of 5 consecutive integers i [#permalink]
Expert's post 00:00

Question Stats: 70% (00:46) correct 29% (00:31) wrong based on 24 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

_________________ Manager  Joined: 15 Jan 2018
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Re: If the average (arithmetic mean) of 5 consecutive integers i [#permalink]
1
KUDOS
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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Director Joined: 09 Nov 2018
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Re: If the average (arithmetic mean) of 5 consecutive integers i [#permalink]
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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