Average prob

If you have seven integers, the median (m) MUST be the 4th integer in the set of integers. This means that 3 integers must be less than m and 3 must be great. No integers can be equal to m because the problem specifies that there are "seven DISTINCT integers."

We want a high average which means we want ever integer to be as high as possible.

Let's start by finding the lowest three values.

First, if m is the median, the integer just less than m is m-1, the next integer is m-2, and the lowest integer will be m-3. Our first four integers are now (in order)

m-3, m-2, m-1, m

For our 3 integers greater than m we know that the highest we can go is double m, which is 2m. So our GREATEST integer in the set is 2m. We still need two more integers to be as high as possible so we choose 2m-1 and 2m-2.

Now we have all of our integers: m-3, m-2, m-1, m, 2m-2, 2m-1, 2m.

Sum all the integers and divide by seven

(m-3 + m-2 + m-1 + m + 2m-2 + 2m-1 + 2m)/7

Simplify and get (10m-9)/7 which equals 10m/10 - 9/7, answer C

selango wrote:Set S contains seven distinct integers. The median of set S is the integer m, and all values in set S are equal to or less than 2m. What is the highest possible average (arithmetic mean) of all values in set S ?

A. m

B. 10m/7

C. 10m/7 – 9/7

D.5m/7 + 3/7

E.5m

OA C

Since there's a variable in the answer choices, we can plug in.

Plug in m=5.

For the median to be 5, we need to choose 3 distinct integers that are smaller than 5 and 3 distinct integers that are larger than 5. Since we're trying to maximize the average, we want each integer to be as large as possible.

The largest allowed value is 2m = 2*5 = 10. So the 3 larger integers will have to be 8, 9, 10.
For the 3 distinct integers smaller than 5, the largest possible are 2, 3, 4.
So our 7 integers are 2, 3, 4, 5, 8, 9, 10.
The average = (2 + 3 + 4 + 5 + 8 + 9 + 10)/7 = 41/7.

Only answer choice C works: 10m/7 – 9/7 = 10*5/7 - 9/7 = 50/7 - 9/7 = 41/7.

The correct answer is C.