The mean will always be 1 greater than the median in this list of numbers. Hence, B

Try to solve this by substituting values of n.

irfan_m1973 wrote:For the positive numbers, n, n + 1, n + 2, n + 4, and n+ 8, the mean is how much greater the median?

(A) 0
(B) 1
(C) n + 1
(D) n + 2
(E) n + 3

Picking numbers is a great way to attack this question; we can also solve algebraically.

The median of a set with an odd number of terms is simply the middle term in the set, in this case n+2.

To solve for the mean, we use the average formula:

Average = sum of terms / # of terms

= (n + (n + 1) + (n + 2) + (n + 4) + (n+ 8)) / 5
= (5n + 15)/5
= n + 3

We want to know by how much the mean is greater than the median, so:

mean - median = (n + 3) - (n + 2) = 1... choose (B).

irfan_m1973 wrote:For the positive numbers, n, n + 1, n + 2, n + 4, and n+ 8, the mean is how much greater the median?

(A) 0
(B) 1
(C) n + 1
(D) n + 2
(E) n + 3

Let's first calculate the mean (arithmetic average).

mean = sum/quantity

mean = (n + n + 1 + n + 2 + n + 4 + n + 8)/5

mean = (5n + 15)/5

mean = n + 3

Next, we determine the median. The median is the middle value when the terms are ordered from least to greatest. The terms ordered from least to greatest are as follows:

n, n + 1, n + 2, n + 4, n + 8

The median is n + 2.

Finally we are asked how much greater the mean is than the median. To determine the difference we can subtract the smaller value (the median) from the larger value (the mean) and we get:

n + 3 - (n + 2) = n + 3 - n - 2 = 1

Answer: B

Average = sum of terms / # of terms

= (n + (n + 1) + (n + 2) + (n + 4) + (n+ 8)) / 5
= (5n + 15)/5
= n + 3

We want to know by how much the mean is greater than the median, so:

mean - median = (n + 3) - (n + 2) = 1... choose (B).