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The average of six consecutive integers in increasing order of size is

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The average of six consecutive integers in increasing order of size is  [#permalink]

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New post Updated on: 17 May 2017, 07:17
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The average of six consecutive integers in increasing order of size is \(9 \frac{1}{2}\) . What is the average of the last three integers?

(A) 8
(B) 9 1/2
(C) 10
(D) 11
(E) 19

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Originally posted by rohan2345 on 17 May 2017, 06:00.
Last edited by Bunuel on 17 May 2017, 07:17, edited 1 time in total.
Renamed the topic and edited the question.
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Re: The average of six consecutive integers in increasing order of size is  [#permalink]

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New post 17 May 2017, 06:17
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Let numbers be n-2, n-1, n, n-1 , n+2,n+3
Then (6n+3)/6=19/2
6n+3=57
n= 9
Numbers are 7,8,9,10,11,12
Avg of last 3 =( 10+11+12)/3=11
Hence D

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The average of six consecutive integers in increasing order of size is  [#permalink]

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New post Updated on: 21 May 2017, 02:22
Let the six consecutive integers be n, n+1,n+2,n+3,n+4,n+5.

Therefore its average would be (n+n+1+n+2+n+3+n+4+n+5)/6

(6n+15)/6 = 19/2 ----> n+5/2 = 19/2-----> n = 14/2 = 7. (1)

Now lets find out the average of the last three digits,

(n+3+n+4+n+5)/3 = (10+11+12)/3 = 11 after Substituting the value of n from (1).

Therefore the answer is Option D.
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Originally posted by Kritesh on 17 May 2017, 06:54.
Last edited by Kritesh on 21 May 2017, 02:22, edited 1 time in total.
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Re: The average of six consecutive integers in increasing order of size is  [#permalink]

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New post 17 May 2017, 07:39
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rohan2345 wrote:
The average of six consecutive integers in increasing order of size is \(9 \frac{1}{2}\) . What is the average of the last three integers?

(A) 8
(B) 9 1/2
(C) 10
(D) 11
(E) 19


\(a + ( a + 1 ) + ( a + 2 ) + ( a + 3 ) + ( a + 4 ) + ( a + 5 ) = 57\)

Or, \(6a + 15 = 57\)

Or, \(6a = 42\)

So, \(a = 7\)

Average of Last 3 Integers is \(\frac{( a + 3 ) + ( a + 4 ) + ( a + 5 )}{3} = \frac{3a + 12}{3} = a + 4\)

Thus, the average of the last three integers \(= 7 + 4 = 11\)

Hence, answer must be (D) 11
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Re: The average of six consecutive integers in increasing order of size is  [#permalink]

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New post 18 May 2017, 09:06
rohan2345 wrote:
The average of six consecutive integers in increasing order of size is \(9 \frac{1}{2}\) . What is the average of the last three integers?

(A) 8
(B) 9 1/2
(C) 10
(D) 11
(E) 19

I like this question because it forced me to choose between a decent bit of algebra and logic/common sense.

List the numbers. We know 9.5 is the average, and because we have consecutive integers where mean = median, we know 9.5 is the "middle term."*

So thinking about a number line, 9.5 will be between the consecutive integers 9 and 10.

From there, you've used two of the six integers. They're consecutive. You need to use up the other four integers, and they need to "balance" so that 9.5 stays in the middle.

It's then easy to see that the numbers, which must be integers, are 7, 8, 9, 10, 11, 12

You can find the average of the last three numbers either by remembering that 10, 11, 12 is an arithmetic sequence, too, so median = mean (average), hence 11, OR

\(\frac{10 + 11 + 12}{3}\) = \(\frac{33}{3}\)= 11

*Also, with an even number of terms, the average will be the average of the two middle terms, so there will be three consecutive integer terms to the right of 9.5, and three to the left of 9.5. Even if this fact doesn't pop into your brain, once you start listing the first two numbers, that fact will become apparent.

Answer D. Hope it helps.
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Re: The average of six consecutive integers in increasing order of size is  [#permalink]

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New post 18 May 2017, 10:18
If a given set of numbers are in Arithmetic Progression (AP), then
their Arithmetic Mean (or Average) = their Median .
This is true for all AP series.

Now, we have 6 numbers in AP (with a common difference of 1).So Mean = 9.5 = Median
Median of a series having even number of terms (6 here) is found by taking average of middle 2 terms.
Since these are consecutive numbers and average of middle 2 terms is 9.5,
the middle 2 terms have to be 9 and 10

Thus, the 3rd and 4th terms of this consecutive number series are 9 and 10.
Since 4th term is 10, next two terms would be 11 and 12

For 10, 11, 12 (again an AP series), Average must be equal to median which is 11.

Hence answer is D.
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Re: The average of six consecutive integers in increasing order of size is  [#permalink]

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New post 18 May 2017, 10:32
If average is \(9 \frac{1}{2}\) or 19/2, since the average is
(sum of n numbers)/n .

Since we are given 6 consecutive integers, we need to get a base of 6. So 19/2 becomes 57/6.
If the sum of integers is 57,
5a+15 = 57 or a=7

Hence the last 3 integers are a+3,a+4 and a+5(10,11,12) and average is 11(Option D)
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Re: The average of six consecutive integers in increasing order of size is  [#permalink]

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