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The average of 5 consecutive integers starting with m as the first int 14 Feb 2016, 05:37
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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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E
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The average of 5 consecutive integers starting with m as the first int 14 Feb 2016, 08:55
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Bunuel
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
Show more


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 14 Feb 2016, 09:33
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Bunuel
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
Show more

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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The average of 5 consecutive integers starting with m as the first int 04 Jun 2017, 23:56
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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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The average of 5 consecutive integers starting with m as the first int 06 Jun 2017, 16:43
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Bunuel
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
Show more

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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The average of 5 consecutive integers starting with m as the first int 18 Oct 2021, 00:19
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Here's a beginner friendly explanation!

This is a easy level question and the ideal time to be taken in the GMAT is 30-45 seconds.

The average of any odd number of consecutive integers is the middle number. If it is 1, 2, 3, 4, 5; the average is 3(the middle number). If it is 12673, 12674, 12675, 12676, 12677; the average is 12675(the middle number; trust me, I calculated it :p)

Approach: jot down how can the 5 consecutive integers that starts with m can be. In most questions, jotting it down will give you an idea on how the question is set.

The 5 numbers can be m, m + 1, m + 2, m + 3, m + 4. As explained earlier, for the consecutive odd integers, the average is the middle term i.e., m + 2.
It is given that the average is n. So, m + 2 = n.

The 9 consecutive numbers that start with m + 2 will be m + 2, m + 3, m + 4, m + 5, m + 6, m + 7, m + 8, m + 9, m + 10. The average is the middle term, i.e., m + 6. There isn't an option that matches with our answer. If m + 2 = n as we derived in the previous step, m + 6 will be n + 4(adding 4 on both sides to get the required new average).

Option E matches with n + 4.
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The average of 5 consecutive integers starting with m as the first int 09 Dec 2022, 02:08
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