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The average of 5 consecutive integers starting with m as the first int

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The average of 5 consecutive integers starting with m as the first int  [#permalink]

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New post 14 Feb 2016, 04:37
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The average of 5 consecutive integers starting with m as the first integer is n. What is the average of 9 consecutive integers that start with (m + 2)?

A. m + 4
B. n + 6
C. n + 3
D. m + 5
E. n + 4

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Re: The average of 5 consecutive integers starting with m as the first int  [#permalink]

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New post 14 Feb 2016, 07:55
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Bunuel wrote:
The average of 5 consecutive integers starting with m as the first integer is n. What is the average of 9 consecutive integers that start with (m + 2)?

A. m + 4
B. n + 6
C. n + 3
D. m + 5
E. n + 4



Hi,
in consecutive numbers,
1) if there are odd numbers, the middle number will be the average..
2) if there are even numbers, the average of middle two numbers is the average..
3) or average of the smallest and largest number


If m is the first integer, m+2 is the middle number, which would also be the average of these 5 consecutive numbers..
so n=m+2..

now there are 9 consecutive numbers starting from m+2, which is also equal to n...
in these numbers, n+4 will be the middle number and hence the average..
E

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The average of 5 consecutive integers starting with m as the first int  [#permalink]

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New post 14 Feb 2016, 08:33
1
Bunuel wrote:
The average of 5 consecutive integers starting with m as the first integer is n. What is the average of 9 consecutive integers that start with (m + 2)?

A. m + 4
B. n + 6
C. n + 3
D. m + 5
E. n + 4


Using number properties: the average value of an even list of consecutive numbers will be the middle number.

list of numbers: m m+1 m+2 m+3 m+4
average given in question stem: n
average as per the number properties rule: m+2

Therefore m+2 = n

New list: " 9 consecutive integers that start with (m + 2)" ... since m+2 = n: n n+1 n+2 n+3 n+4 n+5 n+6 n+7 n+8
Average (using the same number properties rule): n n+1 n+2 n+3 n+4 n+5 n+6 n+7 n+8

Answer: E
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Re: The average of 5 consecutive integers starting with m as the first int  [#permalink]

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New post 04 Jun 2017, 22:56
We can also plug in the values. Assume m=1. So 5 consecutive integers starting with m are:
1, 2, 3, 4, 5 .. and their average is obviously 3. So n=3.

Now if we take 9 consecutive integers starting with m+2 (m+2=1+2=3). These 9 integers would be:
3, 4, 5, 6, 7, 8, 9, 10, 11.

Their average is 7. So if we are looking for that option where if we put m=1 OR n=3, we should get the answer as '7'.

Only option E satisfies this. So E answer
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Re: The average of 5 consecutive integers starting with m as the first int  [#permalink]

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New post 06 Jun 2017, 15:43
1
1
Bunuel wrote:
The average of 5 consecutive integers starting with m as the first integer is n. What is the average of 9 consecutive integers that start with (m + 2)?

A. m + 4
B. n + 6
C. n + 3
D. m + 5
E. n + 4


If the first of the five consecutive integers is m, then the four remaining integers are m + 1, m + 2, m + 3, and m + 4. Since the average of any number of consecutive integers is also the median, the average is m + 2. So, n = m + 2.

Likewise, if the first of the nine consecutive integers is m + 2, then the eight remaining integers are m + 3, m + 4, m + 5, m + 6, m + 7, m + 8, m + 9, and m + 10. Again, since the average of any number of consecutive integers is the median, the average is m + 6. Since m + 2 = n, m + 6 = n + 4. So, n + 4 is the average of the nine consecutive integers.

Answer: E.
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Re: The average of 5 consecutive integers starting with m as the first int  [#permalink]

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New post 27 Dec 2018, 17:04
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Re: The average of 5 consecutive integers starting with m as the first int &nbs [#permalink] 27 Dec 2018, 17:04
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