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
The arithmetic mean of the 5 consecutive integers...
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Consecutive integers constitute an EVENLY SPACED SET.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
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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5 consecutive integres - {s,s + 1, s + 2, s + 3, s + 4}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
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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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.
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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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.
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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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).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
Answer: E
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