Ratios, proportions, and percents help you solve many different practical problems, from adjusting a recipe’s ingredients to comparing the time you spend on the phone at work to the time you spend in meetings. In this next hour, you will learn additional ways percents are useful when analyzing change, whether it’s the change in the price of an item that’s now on sale or the change in your utility bills from one month to the next. You will also be able to calculate not only the amount of different items needed to make a certain mixture, but also the amount of time it will take to get a job done if others help you. Algebra is the key to all these real-life problems, even though they are very different situations.

The proportion gives you an efficient way to work with many percent problems. But it doesn’t apply in some situations where percents often occur, such as when you are comparing prices, enrollment numbers, business expenses, or other statistics and want to know the percent of change.

Consider the following problem:

Your cable TV bill is now $90 per month. Last year you lived in another city and paid only $72 for the same service. What is the percent of increase for your cable service?

Solution: This percent problem is about change, and even though you can use the proportion from Hour 13 for many percent calculations, the following proportion helps you find the percent of change from an original amount to a new amount:

In the cable TV example here, you know the original price, $72, and the new price, $90. The proportion uses the original price, but it also uses the change in price—not the new price. Subtraction gives you the change in price:

You can now put all the numbers into the proportion, using x for the unknown % of change:

The cross-product process helps you create the next equation:

Dividing both sides by 72:

Your cable bill has increased 25%. you want to calculate percent of change, whether it is an increase or a decrease.

If you always find the change in price by using “new price” minus “old price,” the percent answer will be positive if it was an increase, and it will be negative if it was a decrease, as shown in the next example:

Your employer had to cut back everyone’s hours this month to save on expenses. You now work 20 hours a week, and you were working 25 hours each week. What is the percent of change in your hours per week?

Solution: This problem is about work schedule change, not price change, but the change proportion still works. It can be shortened so that it applies to all percent-of-change problems:

Put your numbers in the proportion. Use 25 for “original” because you were working 25 hours a week before the slowdown. What is “change”? Remember to use “new” minus “old” to find “change.” Your “new” schedule is 20 hours, and your “old” was 25.

The numbers can now be put into the proportion:

The cross-product creates the equation:

Your hours per week have decreased 20% because a negative percent shows a decrease, just as a positive percent shows an increase in your cable bill.

Let’s look at a utility bill comparison:

Last month your electricity bill was $68. The month before that it was $80. What is the percent of change?

Solution: Find the change by subtracting “new” minus “old.” Careful with this one—the words make it a bit difficult to know which bill is new and which is old. Your “new” bill is $68, and your “old” bill was $80:

Put the numbers into the proportion:

The cross-product gives the following equation:

Your utility bill decreased 15%, from $80 to $62.

The two proportions for all percent problems are very easy to use, and you should memorize them. It helps that they both have the after the equals signs. You will need to remember the is/of and the change/original ratios before the equals signs, but writing both proportions a few times will help keep them in your memory.

The percent-of-change proportion can lead to wrong answers if you forget that the first denominator must be the original number. Many people mistakenly use the new number. If you remember to always use the original, this proportion will be invaluable in finding percent of change, as well as finding other terms of the proportion if you already know the percent of change, as in the following example:

Enrollment at a local college increased 20% this year compared to a year ago. There are 640 more students this year than last. How many students were enrolled last year?

Solution: Use the percent-of-change proportion. change %

You already know the “change” is 640, the percent is 20, and x is the unknown percent, so these terms can be put into the proportion:

The cross-product creates the equation:

There were 3,200 students enrolled last year.

Now try these percent of change problems, checking your answers on page 174.

Proportions have played a large role in the percent work so far this hour. The mixture problems that follow show that algebra comes in handy in lots of places, from labs to grocery stores.

When different items are combined to make a mixture, whether it’s a saline solution or granola cereal, percents are often an important part of finding the equation you need to solve.

Let’s look first at a saline solution problem. A chemist in a lab has 20 liters of a 25% saline solution. How many liters of water should be added to make a solution that is 20% saline?

Drawing the liters of solution as barrels is helpful. In each mixture problem, the percents are changed to decimal numbers. The lid of each barrel shows the percent of that barrel’s salinity, written in decimal form. The lower part of each barrel shows the number of liters in that barrel.

In this problem, the 25% saline solution (in the first barrel, with .25 on the lid) will have water (in the second barrel) added to make a new barrel. Water is 0% saline, so 0 is on the lid of the second barrel. The new barrel needs to be 20% saline, so .20 is on the lid. The chemist has 20 liters in the first barrel and will be adding x liters in the second barrel to make a new barrel that will hold 20 + x liters. The barrels are shown here:

This illustration lets you see the problem clearly, and the equation is easily written from the barrels. You multiply the numbers in each barrel because “of” always indicates multiplication:

5 liters of water need to be added to make a 20% saline solution.

The following mixture problem shows how to increase the acidity in a solution. The method is similar to the saline problem, and the barrels will help you write the equation.

A chemist in a lab has 30 liters of a 20% acidic solution. How many liters of acid should be added to make a solution that is 40% acidic?

The 20% acid solution (in the first barrel, with .20 on the lid) will have acid (in the second barrel) added to make a new barrel. Acid is 100% acidic, and 100% = 1 in decimal form, so 1 is on the lid of the second barrel. The new barrel needs to be 40% acid, so .40 is on the lid. The chemist has 30 liters in the first barrel and will be adding x liters in the second barrel to make a new barrel that will hold 30 + x liters. The barrels are shown here:

The equation can now be written from the barrels’ labels:

Adding 10 liters of acid will give a new solution that is 40% acidic.

The barrel system is an efficient way to organize the information in a rather complex real-life situation to be used in an equation. When percents are involved, they are changed to decimals. In the next example, dollars are the units, along with pounds instead of liters, but the barrels help you see that it’s really just an algebra problem that is easily solved.

Consider a granola mixture problem:

A store’s produce department sells granola in bulk and makes its own custom mixture from a combination of granola and nut clusters. The granola costs $2 per pound, and the nut clusters cost $5 per pound. They want to make 60 pounds of the custom mixture that will cost $3 per pound. How much of the granola and nut clusters will it take?

Solution: The first barrel is for the granola, and the second barrel is for nut clusters. The new mixture is in the barrel after the equals sign. This problem doesn’t give an amount for either of the ingredients, but you do know there will be a total of 60 pounds in the new mixture barrel. If you let x be the pounds of granola, then 60 - x is the pounds of nut clusters. (Adding x and 60 - x equals 60, which is the total you want in the new barrel.) The barrels can be labeled as shown here:

Write the equation from the labels on the barrels:

Because x is the pounds of granola, it will take 40 pounds of granola and 20 pounds (from 60 - x) of the nut clusters to make 60 pounds of the custom mixture.

Now try the following mixture problems, checking your answers on pages 174-175.

Mixture problems can most efficiently be shown as barrel drawings that lead you directly to the algebraic equation. Work problems also have a specific method that helps you write the equation. Let’s finish this hour learning about the effect of people working together to get a job done.

Setting up work schedules for a large business or a small shop has many elements that need to be considered. Algebraic equations are a big part of the decision making that goes into analyzing how long it takes more than one person to do a certain job.

In this last part of Hour 14, you may be surprised when you see how long it takes two people to do a job that they usually do separately.

Consider the following problem:

If it takes you 20 minutes to load packages onto the delivery truck and it takes Joe 30 minutes to do the same loading job, how long will it take if you both work together?

Solution: The method used to solve this type of problem is based on work rates. The following general work equation helps you find the time (t) it takes for two people—a and b—to do a job when they work together.

The variables a and b represent the time it takes each person to do the given job when they are working alone. Putting in the example’s numbers creates the following equation:

To solve for t, you need to move the 20 and the 30 to the other side of the equals sign. This problem isn’t a proportion, so the cross-product method won’t help solve this equation.

The key to isolating the variable t is the multiplication property of equality, which happens to be the same property that helps you solve proportions. If you multiply both sides of the equation by the least common denominator (LCD) for 20 and 30, the equation will be in a much simpler form.

30 is not the LCD because 20 is not a factor of 30. The easiest way to find the LCD for 20 and 30 is to multiply the larger number, 30, by 2. This equals 60, and 20 is a factor of 60, so the LCD is 60. The solution steps for the equation are shown here:

Working together, it will take only 12 minutes for you and Joe to load the delivery truck, as long as you both work at your regular speeds.

The answer to this type of work-together problem is usually a fraction, as in the following example:

You work for the maintenance department of a large apartment building. Before new tenants move in, you paint the walls of each apartment. If you can paint all bedroom walls and ceiling in a two-bedroom apartment in three hours and another employee can paint the same number of walls and the ceiling in the same apartment in four hours, how long will it take if you both work together to paint one apartment?

Solution: The general work equation is used for this problem:

To find the LCD, multiply 4 times 2, which is 8. Since 3 is not a factor of 8, multiply 4 times 3, which is 12 and the LCD. Multiplying both sides of the equation by 12 will help you isolate the variable:

Using a calculator to divide 12 by 7 gives a decimal answer of about 1.7. The times in the problem are in hours, as is the answer, so it will take about 1.7 hours if you both paint the apartment together, or a little less than 2 hours.

The work equation can also be used if three or more people are all working together—the LCD may be a little harder to find, but it can be done. The equation for three workers would be as follows:

This problem-solving technique applies only to a job that can be done by more than one person at the same time, not an assembly-line type job where many lines of work are in progress. The ability to calculate the amount of time it takes for people to do a job working together instead of separately relies heavily on algebraic skills, and you have mastered them.

Now try these work problems, checking your answers on page 175.

Based on what you’ve learned in Hour 14, do the following problems. Check your answers with the solution key in Appendix B.