Marveling At The Historical

Math Oldies But Goodies

  • About This Blog

    This blog is mostly about math procedures in textbooks dated from about 1825-1900. I’m writing about them because some of the procedures are exquisite and much more powerful, and simpler, than some of the procedures in current text books. Really!

    I update this blog as frequently as possible ... every 2-3 days. And, if you are a lover of old texts and unique procedures, you might want to talk to me about them, at I’m not an antiquarian; the books I have are dusty, musty, brown-paged scribbled-in texts written by authors with insights into how math works. Unfortunately, most of their procedures have vanished. They’ve been overcome by more traditional perspectives, but you have to realize that at that time, they were teaching the traditional methods.

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Archive for the ‘find the average’ Category

A Different Formula for Average

Posted by mark schwartz on April 17, 2016


I came across this in one of my old texts but I unfortunately didn’t write down the citation. I don’t want to go through my 50-some old texts to find it, so this will have to go without knowing the precise source … sorry. I haven’t seen this presented in any current texts, but I think it should be considered.

The Story

A type of problem presented in Pre-Algebra and Algebra texts is: given three test scores (sometimes more), find the test score that is needed to find a specified average. For example, given test scores of 76, 84, and 88, what score would be needed for a student to get an 85 average?

Algebraically, we teach students the following procedure:

(76 + 84 + 88 + x)/4 = 85

Solving this equation, x = 92.

However, with many algebraic problems, if algebra is applied to the algebra problem, a general solution can be found. In this case, the outcome is a form of alligation. Generally, the principle of alligation addresses a problem by identifying “units” above and “units” below a target value. Look what happens.

Using the above example, let 76 = a, 84 = b, 88 = c, desired grade = x and the target average of 85 = d.

The equation is then: (a + b + c+ x)/4 = d

Solving this equation, using what can be called “slow motion math” meaning showing all the steps in the solution, gives:
(a + b + c+ x)/4 = d
a + b + c + x = 4d
x = 4d ─ a ─ b ─ c
x = d + d + d + d ─ a ─ b ─ c
x = d + d ─ a + d ─ b + d ─ c
x = d + (d ─ a) + (d ─ b) + (d ─ c)

Looking at the last statement it says: start with the target average, add to it the difference between the target average and each known score. Some of the differences will be positive, some negative. The result is the desired grade. This is somewhat awkward in words, but not awkward as a procedure.

Doing the above problem using this procedure, again in “slow motion”;

x = 85 + (85 ─ 76) + (85 ─ 84) + (85 ─ 88)
x = 85 + 9 + 1 ─ 3
x = 92

Both the traditional procedure and this procedure are quite straightforward and both are quite efficient. The difference is that with the “alligation” idea, there is no equation to solve and, if the numbers are somewhat small and you can do integer subtraction in your head, just add the numbers to get the desired grade.

For example, what grade is needed on a test to get an average of 85, if the student’s current grades are 82, 79, and 82?

Answer: 85 + 3 + 6 + 3 = 97 …. So, it’s possible with a lot of paying attention, hard work, tutoring, etc.

There is also the possibility that a student wants to know the minimum grade they can get on a test and yet have an average such that they will pass the course. For example, with 3 grades of 72, 75, and 81 what’s the lowest score on the fourth test which will allow her to pass? A passing grade is 65.

Using the formula above we have 65 + (65 ̶ 72) + (65 ̶ 75) + (65 ̶ 81) giving 65 ̶ 7 ̶ 10 ̶ 16, which is 32. I would still urge this student not to have 32 as a goal!

It’s possible that a student might want an unattainable target grade on a test. For example, with test scores of 85, 82, and 87, the student might desire an average of 90. Given that, the desired score would be 90 + (90 – 85) + (90 – 82) + (90 – 87), giving a needed test score of 106. (if the student is really motivated … maybe extra credit for a well done project, research, etc.)

In essence, every student could calculate or be told by the instructor exactly what is needed to get any desired target grade or average, with the caveat that some grades or averages may not be possible. But I do believe it’s reasonable for students to expect to know what is needed on a test to reach a goal … and they can readily calculate it themselves.


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