I have seen this problem so many times that I need to write about it, it is so easy that all of my students should get this one right. You should too. It goes like this example:
The average (arithmetic mean) age of a certain group of 20 people is 35. If five additional people join the group, the new average (arithmetic mean) is 36. What is the average age of the five new people?
Solution:
Average = (the sum of the things)/(the number of the things)
What do we know?
Our average is 35; AVG = 35 [notice that in any word problem the word "is" can be substituted with an equal sign, I always tell my students "IS is EQUAL" "IS" is "="]
Our number of things is 20
What don't we know?
The ages of our group of people. We don't need to know each one, just the sum total of them all.
Let's let S = [sum of the age of the 20 people]
fill in the known information into our equation
35 avg = (S sum of things)/ (20 number of things), solve for S by multiplying both sides by 20
S= 700
Next step the problem changes the average and the number of people. The unknown quantity it the sum of the new people's ages. No sweat, just a little algebra will get us an answer in a hurry. Let's call the new sum the variable N.
New Average = (Old Sum plus New Sum of things)/(Old number of things plus new number of things)
36 = (700 + N)/(20 + 5)
36 = (700 + N)/(25), multiply both sides by 25
36*25 = 700 + N
900 = 700 + N, subtract 700 from both sides
200 = N
To find the average age of the new people, we divide the sum of their ages, N, by their total number, 5.
200/5 = 40
Generically speaking, the question format goes like this:
The average (arithmetic mean) MEASUREMENT of a certain group of a NUMBER of THINGS is the AVERAGE.
IF SO MANY MORE new THINGS are added to the group, then the NEW AVERAGE of the THINGS is SECOND AVERAGE.
WHAT is the THIRD AVERAGE of the MEASUREMENT of the new THINGS?
I think that I shall write a computer program to fill in random measurements, things and numbers for practice.
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