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For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8, the mean
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16 Aug 2021, 01:50
16 Aug 2021, 01:50
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For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8, the mean is how much greater than the median?
(A) 0
(B) 1
(C) n+l
(D) n+2
(E) n+3
(A) 0
(B) 1
(C) n+l
(D) n+2
(E) n+3
Retired Moderator
Joined: 19 Nov 2020
Posts: 326
Given Kudos: 64
GRE 1: Q160 V152
Re: For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8, the mean
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16 Aug 2021, 02:17
16 Aug 2021, 02:17
the most stupid and awkward way I used under timer: find the difference between (n+4)-(n+3)=1 and (n+8)-(n+4)=4
that is suppose, mean=median for consecutive integers, then the above stated difference is 5 spread over five elements and should increase mean by 1
answer is B
please post other ways to apply for the solution under 1 min, I am interested in learning
that is suppose, mean=median for consecutive integers, then the above stated difference is 5 spread over five elements and should increase mean by 1
answer is B
please post other ways to apply for the solution under 1 min, I am interested in learning
Carcass wrote:
For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8, the mean is how much greater than the median?
(A) 0
(B) 1
(C) n+l
(D) n+2
(E) n+3
(A) 0
(B) 1
(C) n+l
(D) n+2
(E) n+3
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8, the mean
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17 Aug 2021, 06:49
17 Aug 2021, 06:49
1
Carcass wrote:
For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8, the mean is how much greater than the median?
(A) 0
(B) 1
(C) n+l
(D) n+2
(E) n+3
(A) 0
(B) 1
(C) n+l
(D) n+2
(E) n+3
To calculate the median, arrange the numbers in ascending order: {n, n + 1, n + 2, n + 4, n + 8}
Since we have an ODD number of values, the median is the middlemost term
Median = n + 2
The mean = [n + (n+1) + (n+2) + (n+4) + (n+8)]/5
= (5n + 15)/5
= n + 3
The mean is how much greater than the median?
Difference = (n + 3) - (n + 2)
= 1
Answer: B
Cheers,
Brent
