Question

Help pls!

Suppose ten distinct, positive integers have a median of 10. ("Distinct integers" means that no two integers are the same.)

What is the smallest the average of those ten integers could be?

Explain your answer in complete sentences.

Answer 1

8 (or 8.4)

Explanation:

First, the problem states the words: Distinct and positive.

That means that the smallest number we can use is 0.

We can make 10 blanks for 10 numbers for us to fill in.

_ _ _ _ _ _ _ _ _ _

Since 10 is an even number, and the median is not one single number, it will be the middle of the two numbers.

In this case, our two numbers are the 5th and 6th blank.

Any two numbers can be used, as long as they are the same actual value from 10.

Let us first put in the numbers we can, which are the numbers before the two middle blanks.

0,1,2,3, _ _ _ _ _ _

To find the median between two numbers, we can do:

(a+b) divided by 2 = median

We can use the smallest following number, 4. Then the 6th number will have to be 16 for 10 to be in the middle.

Also try the largest possible number for the 5th blank, 9. Then the 6th number will be 11.

4+16 and 9+11 both equal 20, and 20 divided by 2 is 10. So both of these work.

Now let's place the other numbers in for these two equations.

0,1,2,3,4,16,17,18,19,20

0,1,2,3,9,11,12,13,14,15

If we add the numbers of each together, we get:

0,1,2,3,4,16,17,18,19,20=100

0,1,2,3,9,11,12,13,14,15= 80

If we now divide each sum by 10 (to find the average) we get:

10

8

Since both of these were the most we could go, one with the 5th number as small as possible and one with the 5th number as large as possible.

Since the smaller answer we got was 8, the answer must be 8.

(I believe 0 is a positive integer because it doesn't carry a negative sign. If 0 is not a positive integer, the answer is 8.4. Use the same process

Answer 2

8.4

Explanation:

Since we want to minimize the average, the integers should be made as small as possible. So, we have 4 integers:

1, 2, 3, 4

However, the median must be 10. The next two integers must have an average of 10 - and since the integers are distinct, we should use:

9, 11

From there, we use the smallest integers (that are still above 11) possible:

12, 13, 14, 15

The average of these 10 integers is 84/10, or 8.4.

The median is still 10, because (9+11)/2 is 10.