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This post is about rate problems. There are two main types of rate problems that you would probably see in an Algebra class: rate of travel and rate of work. Rate of travel problems involve distance, rate, and time for variables. Rate of work problems involve two people or objects that work on a job or complete a task at different rates. We will cover rate of work problems in the next post.

Before looking at these specific types of word problems, I want to say a few things about word problems in general. You must This post is about rate problems. There are two main types of rate problems that you would probably see in an Algebra class: rate of travel and rate of work. Rate of travel problems involve distance, rate, and time for variables. Rate of work problems involve two people or objects that work on a job or complete a task at different rates. We will cover rate of work problems in the next post.

**read**the problem. I’ve had students come into tutoring for help on a word problem when they haven’t even read it yet! How do you know you can’t do the problem if you don’t even know what it says? Here are my tips for working word problems of any kind:

- Read the problem as many times as necessary until you fully understand the scenario and the question you are asked.
- Find (circle, highlight, box) all relevant information such as given values, variables you don't know, and anything that relates two variables together.
- Make a list of all information, using variables to identify each item. Use subscripts to differentiate between two variables of a single type, such as two rates.
- Set up equations to relate the information you are given.
- Solve the equations for the variable you are asked to find using any technique for solving systems of equations that you have learned.
- Practice! Work all of the word problems in the section of your textbook, even if they are not all assigned for homework. Practice is the only way to feel more comfortable solving word problems.

Now let’s look specifically at rate of travel word problems. The easiest way to solve these problems is to use a table for your information. Set up the table with four columns: person/object, distance, rate, and time; and a row for each acting person or object plus a “totals” row, which you may or may not use. A sample table is below, with variables in place of numbers.

Then, fill the table with the values you are given and the variables you are trying to find. Use the distance formula across the rows and sum any columns with given totals. Then, solve your equations for the variables you need.

That brings us to the distance formula.

Here, D is the distance, R is the average rate, and T is the time. An important thing to note is that your units need to all be the same. If the rate is given in miles per hour, then your distance needs to be in miles and your time needs to be in hours. If you are given other units, you need to convert them before using the formula. Let’s work a few example problems.

**Example 1:**Plane A leaves Los Angeles for New York City at 500 mph at the same time that Plane B leaves New York City for Los Angeles on the same path traveling 650 mph. The distance from LA to NYC is 3000 miles. Assuming the plans do not make any stops on the trip, how long until the planes meet each other?

Start with a blank table (below) with the two planes in two different rows.

What information are we given? Well, plane A is traveling at 500 mph and plane B at 650 mph. Those are the rates of the two planes, so we can put those into the table.

We are also told that the distance between the two cities is 3000 miles. It may be tempting to input 3000 into each of the distance boxes, but wait! Think about this first. The two planes are traveling on the same path towards each other. The total distance is 3000 miles. Then each plane, when they meet, will have already traveled some portion of the total distance, say D. The sum of the distances that each plane has already traveled will be the total. Does that make sense? Think of it this way: if plane A meets plane B after traveling 1000 miles, then plane B has already traveled 2000 miles. So, we can input two variables in the distance boxes and input the sum, 3000 miles, into the total box in that column.

We can’t solve this problem until we input something into the time boxes. What is the time? The question asks “how long until,” which indicates that we are trying to find a time. The planes leave each airport at the same time, so when they meet, they will have traveled for the same amount of time. Let’s call this time T. We don’t need subscripts here because the time for each plane is the same.

Now we are ready to set up some equations. For each plane row, use the distance formula to obtain two equations. Then, since we have a total in the distance column, add the two distances and set them equal to the total.

The only other thing to do is solve this system of equations. There are many ways to do this. Notice that the first two equations can be substituted into the third. Let’s do that. Below are my steps. The blue arrows show where to substitute these equations. Then, collect like terms, and then divide to get T by itself. I obtained the decimal answer of 2.61 hours. To write it in terms of hours and minutes, bring the whole number down and multiply the decimal portion by 60, which leaves me with 2 hours, 36.5 minutes for the planes to meet.

Let’s try another one.

**Example 2:**A bus traveling at 50 kilometers per hour (km/hr) made the trip to town in 6 hours. If it had to travel 45 km/hr instead, how many more minutes would the trip to town have taken?

Start with your empty table and fill in information. Here, you are given two scenarios: either the bus goes fast or the bus goes slow. The fast rate is 50 km/hr and the slow is 45 km/hr. The time when the bus went fast was 6 hours. We are asked to find how much longer the trip would take if the bus drove more slowly. This means we are looking for a difference in time between fast and slow. In order to find that, we first need to find the slow time, so let that variable be T.

Now, what about these distances? In order to solve, we need to have some distance there. The question is asking about the same trip to town that the bus originally made, just at a different rate. Therefore, the distance whether the bus goes fast or slow is the same. We are not given a number for this distance, so let’s just let it be D.

We are not given any totals information, so set up equations by using the distance formula across each row. Then solve the equations. The first equation gives us a number, D = 300 km. The second equation has two variables, D and T. Notice that we already know that D is 300, so substitute 300 for D. Then solve for T to get that the trip to town would have taken 6 hours, 40 minutes if the bus had gone slow. The question asks for how much more time, so we need to subtract the time going fast from the time going slow. Our solution is that the bus would have taken 40 minutes longer had it gone 45 km/hr instead of 50 km/hr.

**Example 3:**A bus and car leave the same restaurant at 1 PM on a straight road traveling in opposite directions. The bus travels at 50 mph and the car at 55 mph. At what time will the bus and car be 210 miles apart?

To set up the table, we need to figure out what information is given. It is obvious that the rates of the bus and car are given, so input those first. They are traveling in opposite directions and will travel two different distances. We want to know when the total distance between the vehicles is 210 miles. Now, how about their times? Since the two vehicles left the restaurant at the same time, they will be 210 miles apart after some common amount of time. So, let this variable be T and it is the same for both vehicles.

Now, set up equations and solve them. This one is solved just like the first example. At the end, you find that the time it takes for the vehicles to be 210 miles apart is 2 hours. Since they left the restaurant at 1 PM, we know that they will be 210 miles apart at 3 PM.

Let’s work one final problem like this.

**Example 4:**Mary leaves her house at 3 PM traveling at 30 mph. Mary’s brother, Mark, leaves their house one hour later traveling at 60 mph. What time will Mary’s brother catch up to her?

Set this one up just like all the others. Here, we know the two rates. And we also know the distance because when they are at the same place, they will each be the same distance from their house. Let T be the time that has passed since one of the two (here, I’ve used Mark’s time) left the house. Then Mary’s time is T + 1 because she left an hour

**earlier**. Think of it as Mary getting a head-start on Mark.
Now we obtain the equations and solve for the variable we need. Here, we have two equations that are equal to the same thing (D). So, we can set these equations equal to each other. Then, solve for T—don’t forget to distribute in the parentheses on the left! We get that T = 1 hour, so Mark meets Mary on the road 1 hour after he left their house. He left 1 hour

**after**Mary. Since Mary left at 3 PM, Mark left at 4 PM, and they meet on the road one hour later, at 5 PM.
I think we’ve done enough problems that you should have a good handle on how to work these. In the next post, we will be looking at rate of work problems, so keep an eye out for that. To get email updates every time I post something new, enter your email address in the subscription box on the right. If you have any comments or questions, feel free to leave them below.