Question

The average (arithmetic mean) of the 43 numbers in list L is a positive number. The average of all 48 numbers in both lists L and M is 50 percent greater than the average of the 43 numbers in list L. What percent greater than the average of the numbers in list L is the average of the numbers in list M?

Answer 1

`33(1)/(3) \%`

Explanation:

Given:

The average (arithmetic mean) of the 43 numbers in list L is a positive number.

The average of all 48 numbers in both lists L and M is 50 percent greater than the average of the 43 numbers in list L.

Question asked:

What percent greater than the average of the numbers in list L is the average of the numbers in list M?

Solution:

As the total number of observation in both list = 48

And the number of observation in list L = 43

Then, the number of observation in list M = 48 - 43 = 5

Let the average of the 43 numbers in list L = 100

Then the average of all 48 numbers in both lists L and M =
`100+ 100*50\%`

`=100+100*(50)/(100) \\= 100+(5000)/(100) \\= 100+50 = 150`

The average of the numbers in list M = 150 - 100 = 50

To find percent greater than the average of the numbers in list L in compare to average of the numbers in list M,

Average of the numbers in list L - average of the numbers in list M divided by the average of all 48 numbers in both lists L and M multiplied by 100

`=(100-50)/(150) *100\\\ =(50)/(150) *100\\=(5000)/(150) = 33(1)/(3) \%`

Thus,
`33(1)/(3) \%` greater than the average of the numbers in list L is the average of the numbers in list M.