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

Answer:

$33\frac{1}{3} \%$

Step-by-step 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 = $100+ 100\times50\%$

                                                                                                $=100+100\times\frac{50}{100} \\= 100+\frac{5000}{100} \\= 100+50 = 150$

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

$=\frac{100-50}{150} \times100\\\ =\frac{50}{150} \times100\\=\frac{5000}{150} = 33\frac{1}{3} \%$

Thus, $33\frac{1}{3} \%$ greater than the average of the numbers in list L is the average of the numbers in list M.