The average (arithmetic mean) of the integers from 200 to 400, inclusive, is how much greater than the average of the integers from 50 to 100, inclusive?
(A) 150
(B) 175
(C) 200
(D) 225
(E) 300
OA: D
Please, anyone, help in explaining the approach to solving this problem. TY.
OG2015 PS The average (arithmetic mean)
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- lionsshare
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Since the integers from 200 to 400 are equally spaced (200, 201, 202, ... 398, 399, 400), the average (arithmetic mean) of the integers from 200 to 400, inclusive would be the average of the first number (200) and the last number (400).lionsshare wrote:The average (arithmetic mean) of the integers from 200 to 400, inclusive, is how much greater than the average of the integers from 50 to 100, inclusive?
(A) 150
(B) 175
(C) 200
(D) 225
(E) 300
OA: D
Please, anyone, help in explaining the approach to solving this problem. TY.
Thus, the average (arithmetic mean) of the integers from 200 to 400, inclusive = (200 + 400)/2 = 300.
Similarly, the average (arithmetic mean) of the integers from 50 to 100, inclusive = (50 + 100)/2 = 75.
The average (arithmetic mean) of the integers from 200 to 400, inclusive is greater than the average (arithmetic mean) of the integers from 50 to 100, inclusive by 300 - 75 = 225.
The correct answer: D
Hope this helps!
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Since we have an evenly spaced set of integers, the average of the integers from 200 to 400, inclusive, is (200 + 400)/2 = 300.lionsshare wrote: ↑Fri Sep 08, 2017 3:17 pmThe average (arithmetic mean) of the integers from 200 to 400, inclusive, is how much greater than the average of the integers from 50 to 100, inclusive?
(A) 150
(B) 175
(C) 200
(D) 225
(E) 300
OA: D
Please, anyone, help in explaining the approach to solving this problem. TY.
Similarly, the average of the integers from 50 to 100, inclusive, is (50 + 100)/2 = 75.
Thus, the difference is 300 - 75 = 225.
Answer: D
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