Percents permeate our lives. And they perplex many students in their math classes. Therefore, I am writing this page in an attempt to bring some common sense help to this very important topic.

Problems with percents come in two basic styles:

• problems with only one step or operation, and
• those with two steps or operations.

We will explain these cases the way we do in our math classes.

### One Step Case

Problem statements in this category essentially come in three types of wordings (given here as samples with numbers in them):

1. 20% of 75 is what number?
2. 16 is 25% of what number?
3. 60 is what % of 40?

A big part of making these statements less troublesome is to learn to translate them into math symbols. To that end, know that in phrases like "X% of Y", the word of should be understood as meaning "multiply". So it is the same as "X% times Y", or "X% × Y".

Now if we approach this with an eye toward algebra, we will see that there is just one structure behind all three problems.

### number × number = product

Algebraically this is

### A × B = P

Now, if we have two of the three numbers given to us, the third one is always easy to find. Observe:

20 × 35 = P means "just multiply";

and

12 × B = 48 means "divide 48 by 12".

And that's all there is to it. Well, not quite. There is the matter of changing percents to decimals and vice versa. Also the other matter of how you will do the computations themselves -- by paper and pencil, mentally, or by calculator. But those are minor and easily managed. Let's get back to the three sample problems above.

First thing to be done, as explained above, is to "translate" them into algebra. Then do the appropriate "arithmetic".

```1.  20% of 75 is what number?   -->  20% × 75 = P

0.20 × 75 = P

15 = P

So, 15 is 20% of 75.

2.  16 is 25% of what number?  -->  16 = 25% × B

16 = 0.25 × B

64 = B

So, 25% of 64 is 16.

3.  60 is what % of 40?  -->	   60 = A% × 40

1.5 = A%

150 = A
```

### Two Step Case

Turning now to the two-step case, we see that this is really "where it's at", regarding percent problems. Every day when a person makes a purchase at a store and pays a sales tax, that person is involved with a two-step percent situation. So let's go shopping WITH PERCENTS!

```1.  Mary buys a sweater for \$50 and has to pay 8% in sales tax.  What
does she pay the clerk?

1st:  Find the amount of tax.

8% of \$50 is the tax (T).  So we write

8% × 50 = T

0.08 × 50 = T

4 = T

2nd:  Add the tax to the original price.

50 + 4 = P (price to pay)

54 = P

So she pays \$54.
```

For reasons that will soon be more clear, I like to put this process into one "formula". However I use words instead of abstract letters. In this real-world application, I let "old" stand for the original purchase price, and "new" stands for the price after taxes. Finally, "pct" stands for the percent figure, in this case the tax rate. My formula then is expressed as follows:

### new = old + old × pct

Another important situation arises when a person buys something "on sale" or "at a discount". And much the same thing happens here -- with one difference, of course. Observe:

```2.  Later Mary found a shoe sale, offering a 20% discount.  The shoes
she wants are regularly priced at \$35.  What would be the "sale
price"?

1st:  Find the amount of discount.

20% of \$35 is her savings (S).

20% × 35 = S

0.20 × 35 = S

7 = S

2nd:  Subtract the savings from the regular price.

35 - 7 = S.P. (sale price)

28 = S.P.

So she pays \$28 (before taxes).

```

We can derive a formula here as well. This time "new" stands for the "sale price"; everything else remains as before.

### new = old - old × pct

[Notice the important "minus sign" there.]

There is an old saying about "killing two birds with one stone" that has a unique application here. Our two formulas can be combined into one, like this:

### new = old ± old × pct

[In formulas using the plus-or-minus sign (±), you use whatever is necessary for a given situation.]

One final observation is important before proceeding with our discussion. The step-by-step procedure presented here is what I like to call the "clerk's" method. Because in filling out the paperwork for a customer, the clerk needs to show specifically what the amount of tax and/or the amount of discount was.

However, the customer is really only concerned with "the bottom line", the final amount that he or she must pay the store. So perhaps a short cut is in order now. If we use the distributive property from basic algebra, the formulas look like these:

### new = old(1 ± pct)

Applying these formulas to our story problems above, we see that for tax-add-on's, you merely add the percent rate to 1, then multiply by the price.

\$50 × (1 + 0.08) = \$50 × 1.08 = \$54.

And for the discounted sale price for the shoes, you have

\$35 × (1 - 0.20) = \$35 × 0.80 = \$28.

It's easily noted via "common sense" in the 2nd situation that if you receive 20% off, you must therefore be paying the other 80% of the regular price, right? And that sort of computation -- adding to or subtracting something from 1 -- often can be done mentally. Just be careful with your decimal points.

### Non-money examples

Now, do not get the idea that all this discussion applies ONLY to buying things in a store. The same formulas and procedures still are used in such problems as these:

 3. A school had 1200 students last year. A new factory opens in the community, and the enrollment increases by 5%. What is the new enrollment for this year? 4. The number of sea birds in a given area was estimated to be about 4,000. Due to an oil spill recently, their population was cut by an estimated 22%. How many birds are currently in the area?

### Percent Increase or Decrease

Another popular use of percents in two-step situations that often baffles students occurs in problems such as these:

 5. The hourly minimum wage was \$5.75 a few years ago. Later it was raised to \$7.50. What was the percent increase? 6. A person who once weighed 210 pounds went on a diet and reduced his weight so that now he weighs 185 pounds. What was the percent decrease?

Working these through ordinary arithmetic seems to be direct, at least for the first step. You must find out what was the increase or decrease for each situation. And that's done by simple subtraction.

For the minimum wage --> \$7.50 - \$5.75 = \$1.75.

For the weight loss --> 210 lb - 185 lb = 25 lb

But now where? Do we multiply, divide or just what? This is where the formulas and the idea of time [i.e. old vs. new] comes into play.

For the minimum wage problem, we have

old = \$5.75 & new = \$7.50

If we enter those values in our formula for tax-add-on above, since the concept is one of increasing something, and let p = the percent value we seek, we have

### 750 = 575 + 575p

Now it is a simple matter of using basic algebra.
```                        750  -  575  =  575p
175  =  575p
175  ÷  575  =  p
```
Ah ha! Now we see, it's divide! But in this problem, even I admit it's a little messy. So why not turn to our trusty little calculator to do it. It comes out to be

p = 0.304347826087

Here is where skill in rounding decimals is needed. A good answer for this number would about 30% or, if you prefer, 30.4%. Hence, tomorrow's news headlines would shout:

## Minimum Wage Rises 30%

Then how does all this apply to the weight loss problem? Happy to say, there's not much difference here either. Watch...

old = 210 & new = 185

Substituting these numbers in the "discount" formula [because this is a decreasing idea as well], we have

```                                 185  =  210  -  210p

185  -  210  =  -210p

-25  =  -210p

-25  ÷  -210  =  p
```
The division happens to be messy again, so our calculator comes to the rescue, giving us

p = 0.1190476190476.

Rounding the percent to the nearest 10th says that the individual had a weight decrease of about 11.9%. (And I'll bet he's happy about that!)

An Alternate Approach to Percent Increase/Decrease

When I need to do a lot of problems of this type at one time, I actually prefer to use an alternate mode of thinking. I decide to take my basic formula for tax-add-on and just "solve" it for pct once and for all. This gives me

```                                 new  -  old
pct  =  -------------
old
```
This makes my work decidedly more efficient. All I have to do is recall that when new is greater than old, I am doing a percent increase. On the other hand, if old is greater than new, there is a decrease happening. This is more evident as the numerator comes out negative this time. [Ya' gotta be careful with those negatives in algebra.]

Finding Original Values

I will now explain one final problem situation, mostly for completeness and because it appears in textbook exercises, yet I find it less frequently used outside of the classroom. Here is an example:

 Today in his exercise program John was able to do 120 pushups. This respresents a 50% increase over what he was able to do one month ago. How many was he able to do then?

If we think about this carefully, it's not so difficult to understand. At some point in the past he could do so many pushups. Now he can do more. His increase is half again as many. If he once did 50, he now can do "half of 50 more", or 75, because half of 50 is 25, etc. Our difficulty lies in the fact that we don't know how many he COULD do in the past, but rather what he CAN do now.

Where do we turn?

To our formulas, of course.

This time we have the new value and the pct value; we need the old value. Perhaps by now you would accept my advice and use a short cut approach on this. If we write the tax-add-on formula and solve it for old, our algebra looks like this:

```                old  +  old × pct  =  new

(1  +  pct)old  =  new

new
old  =  ---------
1 + pct
```

For our pushup problem, this gives us

```                                   120
old  =  ----------
1  + 0.5
```

or 80 pushups.

Suffice it to say that if our problem situation indicated that the new value represented a decrease over a past value, then the plus sign in the formula merely changes to a minus sign. Agreed?

I commented above that this situation is probably less frequently used in the "real-world". My reason for that statement is that in order to know what the percent of increase or decrease is, you probably had to know BOTH the old and new values in the first place! Hence, it is more of a puzzle problem in this regard. But its solution is easy if you let algebra be your helper. Hooray for algebra!!!