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So, lets say, that you have the following numbers (4, 15, 3, 22) and you have to add the fifth number (which I would call X) and they must average out to say 20.

What do you have to do?

I figured the formula is something like (4+15+3+22+X)/5=20.

But for some reason I can't solve it. It just doesnt click in. Can anyone tell me if at least I have the formula right?

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Closest I can get this down to is (44+X)/5=20 – Liger86 May 20 '11 at 9:22
Well, i'm bad with math, like really bad. There must be something you can do to both sides to bring that X on one side. If my formula is even right... – Liger86 May 20 '11 at 9:24
$(4+15+3+22+X)/5 = 20$, then $4 + 15 + 3 + 22 + X = 100$, so $X = 56$. – Alexander Thumm May 20 '11 at 9:25
Oh, so you multiply both sides by 5, which cancels out the "divide by 5" and turns 20 into 100, right? – Liger86 May 20 '11 at 9:28
Correct... and some more characters to make this a comment :D – Alexander Thumm May 20 '11 at 9:33
up vote 2 down vote accepted

For such $X$ you have $$4+15+3+22+X = 5\cdot 20,$$ that is $$X = 100 -4 - 15-3-22=... $$

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For some reason subtraction makes this weird, I like the way Alexander did it in one of the comments. But this works too. I guess I better get used to doing things and moving things to both sides of equation. Thanks! – Liger86 May 20 '11 at 9:35
Yes - you free $X$ using first the inverse operation "times 5" and then the inverse operation "subtract (4+5+3+22)". – AD. May 20 '11 at 10:22

AD's answer (and the comments that follow) as well as comments on the original question cover the algebraic solution pretty well. This question, however, is one of the archetypal "algebra" problems that, out of context, most people wouldn't use algebra to do. Here's a non-algebraic line of thinking that leads to the answer.

You're going to have 5 numbers that have an average of 20. That means that when you add up these 5 numbers, it's like adding up 5 20s, or multiplying $5\times20$, which gives 100. So the sum of all five of the numbers is $5\times20=100$.

The numbers you have so far add up to $4+15+3+22=44$. To end up with a total of 100 with one more number, that number has to be $100-44=56$.

These are the same operations that fall out of doing the algebra, used in the same way. The difference is in using formal symbols versus an informal chain of reasoning. If this problem is posed to an elementary school student who has appropriate knowledge of averages, they will likely be able to solve it by thinking along the lines of what I've shown. If this problem is posed to a student who has done more symbolic manipulation and variable-use, they will much more likely write out an equation and formally solve it.

edit: For a bit more on what I'm talking about with regard to non-algebraic "algebra" word problems, see "What Should Not Be in the Algebra and Geometry Curricula of Average College-Bound Students?" by Zalman Usiskin (Mathematics Teacher 73 (September 1980), pdf), particularly the section titled "Deletion 1: The traditional word problems," starting on the page numbered 70 in the linked PDF.

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Thanks for the your solution Issac, it looks like something I can use, I will look into it more now. – Liger86 May 22 '11 at 6:30

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