The arithmetic mean of the 5 consecutive integers...

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The arithmetic mean of the 5 consecutive integers starting with 's' is 'a'. What is the arithmetic mean of 9 consecutive integers that start with s + 2?


A 2 + s + a
B 22 + a
C 2s
D 2a + 2
E 4 + a

OA: E

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by akshay_gooner » Mon May 09, 2016 12:49 am
the answer should be 4+a

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by GMATGuruNY » Mon May 09, 2016 1:50 am
Musicat wrote:The arithmetic mean of the 5 consecutive integers starting with 's' is 'a'. What is the arithmetic mean of 9 consecutive integers that start with s + 2?


A 2 + s + a
B 22 + a
C 2s
D 2a + 2
E 4 + a
Consecutive integers constitute an EVENLY SPACED SET.
For any evenly spaced set, AVERAGE = MEDIAN.

The arithmetic mean of the 5 consecutive integers starting with 's' is 'a'.
Let s=1, implying that the 5 consecutive integers are as follows:
1, 2, 3, 4, 5.
Here, a = average = median = 3.

What is the arithmetic mean of 9 consecutive integers that start with s + 2?
Since s+2 = 1+2 = 3, the 9 consecutive integers starting with s+2 are as follows:
3, 4, 5, 6, 7, 8, 9, 10, 11.
Here, average = median = 7. This is our target.

Now plug s=1 and a=3 into the answers to see which yields our target value of 7.
Only E works:
4 + a = 4 + 3 = 7.

The correct answer is E.
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by OptimusPrep » Mon May 09, 2016 5:35 pm
Musicat wrote:The arithmetic mean of the 5 consecutive integers starting with 's' is 'a'. What is the arithmetic mean of 9 consecutive integers that start with s + 2?


A 2 + s + a
B 22 + a
C 2s
D 2a + 2
E 4 + a

OA: E
5 consecutive integres - {s,s + 1, s + 2, s + 3, s + 4}

Arithmetic mean = (5s + 10)/5 = s + 2 = a
Hence s = a - 2

9 consecutive integers - { s + 2, s + 3, s + 4, s + 5, s + 6, s + 7, s + 8, s + 9, s + 10}
Arithmetic mean = (9s + 54)/9 = s + 6

Putting the value of s,

Mean = a - 2 + 6 = a + 4

Correct Option: E

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by Matt@VeritasPrep » Wed May 11, 2016 11:32 pm
Option 1:

Pick numbers. Suppose s = 1. The five integers are 1 -> 5, and the mean is 3.

s + 2 = 3, and the 9 integers are 3 through 11. Their mean is 7, or 4 + a.

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by Matt@VeritasPrep » Wed May 11, 2016 11:35 pm
Option 2:

Use algebra. Our first set is {s, s+1, s+2, s+3, s+4}, so the mean is s + 2.

Our second set is {s+2, s+3, ..., s+10}. Since the set is evenly spaced, the mean and the median are the same. The median is s + 6, or (s + 2) + 4, or a + 4.

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by Scott@TargetTestPrep » Sat Mar 16, 2019 8:42 am
Musicat wrote:The arithmetic mean of the 5 consecutive integers starting with 's' is 'a'. What is the arithmetic mean of 9 consecutive integers that start with s + 2?


A 2 + s + a
B 22 + a
C 2s
D 2a + 2
E 4 + a

OA: E
that the mean and median of any number of consecutive integers are equal). For the 9 consecutive integers that start with (s + 2), the median (or mean) is s + 2 + 4 = s + 6. Since a = s + 2, then s + 6 = s + 2 + 4 = a + 4. So (a + 4) is the mean of the consecutive integers that start with (s + 2).

Answer: E

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