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30 Jul 2026, 09:33
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
36% (02:47) correct
64%
(02:58)
wrong
based on 36
sessions
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If the integer k is removed from the set of positive integers from 1 to n, inclusive, the average of the remaining integers is 25.8. What is the value of k?
A. 2
B. 11
C. 29
D. 36
E. 51
A. 2
B. 11
C. 29
D. 36
E. 51
31 Jul 2026, 02:04
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Original average = (n+1)/2
k = [n(n+1)]/2 - [129(n-1)]/5
(n-1) should be divisibly by 5
n should be of the form 5k+1
When k is removed, the average changes slightly, so we can roughly equate original average and new average just to get a close value of n
(n+1)/2 = 25.8
n = 50.6
Closest number to 50.6 of the form 5k+1 is 51, check for n=51
k = [n(n+1)]/2 - [129(n-1)]/5
k = [51(51+1)]/2 - [129(51-1)]/5
k = 36
Hence, we got option D
k = [n(n+1)]/2 - [129(n-1)]/5
(n-1) should be divisibly by 5
n should be of the form 5k+1
When k is removed, the average changes slightly, so we can roughly equate original average and new average just to get a close value of n
(n+1)/2 = 25.8
n = 50.6
Closest number to 50.6 of the form 5k+1 is 51, check for n=51
k = [n(n+1)]/2 - [129(n-1)]/5
k = [51(51+1)]/2 - [129(51-1)]/5
k = 36
Hence, we got option D
