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The average (arithmetic mean) of the even integers from 100 to 1000, i
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23 Sep 2021, 02:52
23 Sep 2021, 02:52
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26% (01:28) wrong based on 15 sessions
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The average (arithmetic mean) of the even integers from 100 to 1000, inclusive, is how much greater than the average of the even integers from 10 to 100, inclusive?
(A) 495
(B) 500
(C) 545
(D) 550
(E) 900
(A) 495
(B) 500
(C) 545
(D) 550
(E) 900
Re: The average (arithmetic mean) of the even integers from 100 to 1000, i
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23 Sep 2021, 11:59
23 Sep 2021, 11:59
Carcass wrote:
The average (arithmetic mean) of the even integers from 100 to 1000, inclusive, is how much greater than the average of the even integers from 10 to 100, inclusive?
(A) 495
(B) 500
(C) 545
(D) 550
(E) 900
(A) 495
(B) 500
(C) 545
(D) 550
(E) 900
The Arithmetic mean of all even integers from 100 to 1000 is (1000 + 100)/2 = 550
Similarly for even integers from 10 to 100, is (100 + 10)/2 = 55
So 550 - 55 = 495.
Answer is A.
Re: The average (arithmetic mean) of the even integers from 100 to 1000, i
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20 Jan 2022, 19:08
20 Jan 2022, 19:08
1
eskay1981 wrote:
Carcass wrote:
The average (arithmetic mean) of the even integers from 100 to 1000, inclusive, is how much greater than the average of the even integers from 10 to 100, inclusive?
(A) 495
(B) 500
(C) 545
(D) 550
(E) 900
(A) 495
(B) 500
(C) 545
(D) 550
(E) 900
The Arithmetic mean of all even integers from 100 to 1000 is (1000 + 100)/2 = 550
Similarly for even integers from 10 to 100, is (100 + 10)/2 = 55
So 550 - 55 = 495.
Answer is A.
The problem says the word "inclusive" twice, thus we must add 1 to each mean.
550 + 1 = 551
55 + 1 = 56
Difference = 551 - 56 = 495
The answer is still A, but for other problems, we might need the extra 1 for each mean.
