Rate Problems

 

Learning Objective(s)

·         Use systems of equations to describe and solve problems involving rates.

 

 

A rate is a mathematical way of relating two quantities, which are usually measured in different units. A favorite type of a rate problem in algebra courses sends two hypothetical trains rushing towards each other at different speeds, and asks you to determine when they will meet. Less exciting, but also common, rate situations involve calculating wages or determining the time it takes for a container to fill or empty. The secret to solving all rate problems is creating a system of equations to represent the situation.

 

 

Speed is a familiar kind of rate. Common units used for speed include miles per hour, feet per second, and—in the case of very fast things, like light—miles per second. Sometimes it is helpful to think of a rate as a ratio of two quantities. You could think of speed, then, as the ratio of distance traveled to time taken.

 

Are you ready to try a rate problem? Here goes:

 

Two trains leave stations 60 miles apart at the same time heading toward one another on parallel tracks. Train A is traveling 30 miles per hour, while Train B is going 50 miles per hour. When do they pass each other?

 

The key to solving rate problems is to figure out the context of the problem and then identify a formula that relates all of the information in the problem. In this problem, our context is distance: we have two objects traveling at different rates and in opposite directions. We can relate all of the information using the simple formula distance = rate • time, or d = rt.

 

One technique that can help us keep track of the information in the problem is to make a table. This will help us organize our data and create a system that we can solve. Let’s make a table with columns for “distance,” “rate,” and “time,” and a row for each train. We can fill in the speed at which each train is going, as this was provided in the problem.

 

 

Notice that the “Distance” and “Time” columns are empty for now. We have left these empty because the specific distances and times that the trains traveled were not provided in the original problem. But we can use some logic to figure out what these are.

 

The problem tells us that the stations are 60 miles apart. The trains will meet at some point between the stations, so we can call the distance Train A travels d, and the distance Train B travels 60 – d. One way of thinking about these two distances is “Train A will travel some distance to the meeting point, and Train B will have to travel a different distance, and the total distance will be 60 miles.” Let’s put this information into the table.

 

 

Finally, there is the matter of time. The problem asks us to find when these trains will pass each other, so we know that this quantity is unknown. However, if the trains leave at the same time from different stations, then they must have traveled the same amount of time when they pass each other, no matter how fast or far each one goes. So because these times are equivalent, we can use the same variable, t, for both of them.

 

Based on this new information, our table now looks like this:

 

 

Now we have our system of equations. Train A is represented by the equation d = 30t, and Train B is represented by the equation 60 – d = 50t. We can use the substitution method to substitute 30t for d in the second equation, and then solve it for t.

 

 

It turns out that t = , so the trains will meet  of an hour (45 minutes) after they leave the stations.

 

Wages provide another example of a commonly used rate: they represent the ratio of money earned to a certain length of time worked.

 

Solving wage problems is similar to solving distance problems. Although the measurement units for the two problems are different, the fundamental ideas about how to solve these problems are the same. As in the previous problem, we will first identify the formula that relates all of the information in the problem, and then we will build a table to organize the data.

 

Two roommates saved their wages from weekend jobs to buy a television. One earns $10/hr, and the other earns $8/hr. Together they worked for 43 hours and earned exactly $400. How long did each of them work in order to save the money?

 

The question is asking how long each roommate worked in order to save $400. We can find out the amount of money they earned by using the simple formula earnings = wages • time.

 

 

The only roommate-specific information that the problem gives us is their hourly wage, which we’ve included in the table. We also know that they earned $400 together and also worked for 43 hours total. How do we include that information in the table?

 

Let’s start with the $400 total earnings. We know that one roommate earned a portion of that amount, and the other roommate earned the rest. So we can call Roommate 1’s earnings e and Roommate 2’s earnings 400 − e. (Note that this is an arbitrary assignment, and we could have called Roommate 2’s earnings e and used 400 − e to represent Roommate 1’s earnings.)

 

We can use the same logic for the hours worked. Together they worked for 43 hours—let’s call Roommate 1’s hours t, and Roommate 2’s hours 43 − t. (Assigning the variable t to Roommate 1 and 43 − t to Roommate 2 has the effect of making substitution a breeze, as you will see in a moment.)

 

When we put all this information into the table, it looks impressive:

 

 

We have had to introduce two variables into this problem, e and t. But that’s okay because we have used them to create a system of two equations: e = 10t and 400 − e = 8(43 − t). Now that we have our system, we can solve for the variable t, by substituting the value of e from the first equation into the second equation.

 

 

We find that t = 28. So, if Roommate 1 worked for 28 hours, then the time that Roommate 2 worked, 43 − t, must be 15 hours.

 

 

Rate problems can often be solved using systems of equations. One effective method is to identify a formula for the problem’s context, make a table to record information about the situation, and then use substitution to solve the system of two variables that results.