problem solving distance rate and time formulas lesson 6.5 answers

Pardon Our Interruption

As you were browsing something about your browser made us think you were a bot. There are a few reasons this might happen:

• You've disabled JavaScript in your web browser.
• You're a power user moving through this website with super-human speed.
• You've disabled cookies in your web browser.
• A third-party browser plugin, such as Ghostery or NoScript, is preventing JavaScript from running. Additional information is available in this support article .

To regain access, please make sure that cookies and JavaScript are enabled before reloading the page.

Solving problems with the formula for distance, rate, and time

Formula that relates distance, rate, and time

This lesson looks at how to find distance, rate and time given two out of three of these values.

Distance, rate and time are related by the equation

???\text{Distance}=\text{Rate} \cdot \text{Time}???

Hi! I'm krista.

I create online courses to help you rock your math class. Read more.

Let’s talk about the units of each of these values.

Distance has units in inches, feet, miles, or centimeters, meters, kilometers etc.

Time has units in seconds, minutes, hours, etc.

Rate has units in distance/time, for example inches/second, miles/hour, or kilometers/hours.

Before you can use the formula ???D=RT??? you need to make sure that your units for the distance and time are the same units in your rate. If they aren’t, you’ll need to change them so you’re working with the same units.

How to solve distance, rate, and time problems

Take the course

Want to learn more about algebra 2 i have a step-by-step course for that. :), finding average rate given distance and time.

Heather ran ???56??? km in ???5??? hours. What was Heather’s average rate in km/hr?

We’ll use the formula for distance.

Let’s write down what we know.

???D=56??? km

???T=5??? hr

If we plug these into the distance formula, we get

???56\text{ km} = R\cdot 5\text{ hr}???

Now solve for the Rate.

???\frac{56\ \text{km}}{5\ \text{hr}} = \frac{R \cdot 5\ \text{hr}}{5\ \text{hr}}???

???R=11.2\ \frac{\text{km}}{\text{hr}}???

Before you can use the formula D=RT you need to make sure that your units for the distance and time are the same units in your rate.

Distance, rate, and time problems with two people who leave at different times

Susan and Benjamin were ???60??? miles apart on a straight trail. Susan started walking toward Benjamin at a rate of ???5??? mph at 7:30 a.m. Benjamin left three hours later and they met on the trail at 3:30 p.m. How fast is Benjamin?

We’ve been given information about distance, rate and time, so we’ll use the formula

where ???D??? is the distance traveled, ???R??? is the rate, and ???T??? is the time. We can use subscripts to create unique equations for Susan and Benjamin.

Susan: ???D_{S} =R_{S} T_{S}???

Benjamin: ???D_{B} =R_{B} T_{B}???

We know that in order to meet each other, they must have covered a distance of ???60??? miles between them. Therefore,

???D_{S}+D_{B}=60???

Since we know that ???D_{S}=R_{S}T_{S}??? and ???D_{B}=R_{B}T_{B}???, we can make substitutions into this equation for their rates and times, instead of their distances.

???R_{S}T_{S}+R_{B}T_{B}=60???

The problem tells us that Susan was walking at ???5??? mph, and that she walked for ???8??? hours, since she walked from 7:30 a.m. until 3:30 p.m. So

???(5)(8)+R_{B}T_{B}=60???

???40+R_{B}T_{B}=60???

???R_{B}T_{B}=20???

Benjamin left three hours after Susan, which means he started walking at 10:30 p.m., and he kept walking until they met at 3:30 p.m., which means he walked for ???5??? hours. So

???R_{B}(5)=20???

???R_{B}=4???

Which means that Benjamin walks at ???4??? mph.

Get access to the complete Algebra 2 course

Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices.

Chapter 8: Rational Expressions

8.8 Rate Word Problems: Speed, Distance and Time

Distance, rate and time problems are a standard application of linear equations. When solving these problems, use the relationship rate (speed or velocity) times time equals distance .

[latex]r\cdot t=d[/latex]

For example, suppose a person were to travel 30 km/h for 4 h. To find the total distance, multiply rate times time or (30km/h)(4h) = 120 km.

The problems to be solved here will have a few more steps than described above. So to keep the information in the problem organized, use a table. An example of the basic structure of the table is below:

The third column, distance, will always be filled in by multiplying the rate and time columns together. If given a total distance of both persons or trips, put this information in the distance column. Now use this table to set up and solve the following examples.

Example 8.8.1

Joey and Natasha start from the same point and walk in opposite directions. Joey walks 2 km/h faster than Natasha. After 3 hours, they are 30 kilometres apart. How fast did each walk?

The distance travelled by both is 30 km. Therefore, the equation to be solved is:

[latex]\begin{array}{rrrrrrl} 3r&+&3(r&+&2)&=&30 \\ 3r&+&3r&+&6&=&30 \\ &&&-&6&&-6 \\ \hline &&&&\dfrac{6r}{6}&=&\dfrac{24}{6} \\ \\ &&&&r&=&4 \text{ km/h} \end{array}[/latex]

This means that Natasha walks at 4 km/h and Joey walks at 6 km/h.

Example 8.8.2

Nick and Chloe left their campsite by canoe and paddled downstream at an average speed of 12 km/h. They turned around and paddled back upstream at an average rate of 4 km/h. The total trip took 1 hour. After how much time did the campers turn around downstream?

The distance travelled downstream is the same distance that they travelled upstream. Therefore, the equation to be solved is:

[latex]\begin{array}{rrlll} 12(t)&=&4(1&-&t) \\ 12t&=&4&-&4t \\ +4t&&&+&4t \\ \hline \dfrac{16t}{16}&=&\dfrac{4}{16}&& \\ \\ t&=&0.25&& \end{array}[/latex]

This means the campers paddled downstream for 0.25 h and spent 0.75 h paddling back.

Example 8.8.3

Terry leaves his house riding a bike at 20 km/h. Sally leaves 6 h later on a scooter to catch up with him travelling at 80 km/h. How long will it take her to catch up with him?

The distance travelled by both is the same. Therefore, the equation to be solved is:

[latex]\begin{array}{rrrrr} 20(t)&=&80(t&-&6) \\ 20t&=&80t&-&480 \\ -80t&&-80t&& \\ \hline \dfrac{-60t}{-60}&=&\dfrac{-480}{-60}&& \\ \\ t&=&8&& \end{array}[/latex]

This means that Terry travels for 8 h and Sally only needs 2 h to catch up to him.

Example 8.8.4

On a 130-kilometre trip, a car travelled at an average speed of 55 km/h and then reduced its speed to 40 km/h for the remainder of the trip. The trip took 2.5 h. For how long did the car travel 40 km/h?

[latex]\begin{array}{rrrrrrr} 55(t)&+&40(2.5&-&t)&=&130 \\ 55t&+&100&-&40t&=&130 \\ &-&100&&&&-100 \\ \hline &&&&\dfrac{15t}{15}&=&\dfrac{30}{15} \\ \\ &&&&t&=&2 \end{array}[/latex]

This means that the time spent travelling at 40 km/h was 0.5 h.

Distance, time and rate problems have a few variations that mix the unknowns between distance, rate and time. They generally involve solving a problem that uses the combined distance travelled to equal some distance or a problem in which the distances travelled by both parties is the same. These distance, rate and time problems will be revisited later on in this textbook where quadratic solutions are required to solve them.

For Questions 1 to 8, find the equations needed to solve the problems. Do not solve.

• A is 60 kilometres from B. An automobile at A starts for B at the rate of 20 km/h at the same time that an automobile at B starts for A at the rate of 25 km/h. How long will it be before the automobiles meet?
• Two automobiles are 276 kilometres apart and start to travel toward each other at the same time. They travel at rates differing by 5 km/h. If they meet after 6 h, find the rate of each.
• Two trains starting at the same station head in opposite directions. They travel at the rates of 25 and 40 km/h, respectively. If they start at the same time, how soon will they be 195 kilometres apart?
• Two bike messengers, Jerry and Susan, ride in opposite directions. If Jerry rides at the rate of 20 km/h, at what rate must Susan ride if they are 150 kilometres apart in 5 hours?
• A passenger and a freight train start toward each other at the same time from two points 300 kilometres apart. If the rate of the passenger train exceeds the rate of the freight train by 15 km/h, and they meet after 4 hours, what must the rate of each be?
• Two automobiles started travelling in opposite directions at the same time from the same point. Their rates were 25 and 35 km/h, respectively. After how many hours were they 180 kilometres apart?
• A man having ten hours at his disposal made an excursion by bike, riding out at the rate of 10 km/h and returning on foot at the rate of 3 km/h. Find the distance he rode.
• A man walks at the rate of 4 km/h. How far can he walk into the country and ride back on a trolley that travels at the rate of 20 km/h, if he must be back home 3 hours from the time he started?

Solve Questions 9 to 22.

• A boy rides away from home in an automobile at the rate of 28 km/h and walks back at the rate of 4 km/h. The round trip requires 2 hours. How far does he ride?
• A motorboat leaves a harbour and travels at an average speed of 15 km/h toward an island. The average speed on the return trip was 10 km/h. How far was the island from the harbour if the trip took a total of 5 hours?
• A family drove to a resort at an average speed of 30 km/h and later returned over the same road at an average speed of 50 km/h. Find the distance to the resort if the total driving time was 8 hours.
• As part of his flight training, a student pilot was required to fly to an airport and then return. The average speed to the airport was 90 km/h, and the average speed returning was 120 km/h. Find the distance between the two airports if the total flying time was 7 hours.
• Sam starts travelling at 4 km/h from a campsite 2 hours ahead of Sue, who travels 6 km/h in the same direction. How many hours will it take for Sue to catch up to Sam?
• A man travels 5 km/h. After travelling for 6 hours, another man starts at the same place as the first man did, following at the rate of 8 km/h. When will the second man overtake the first?
• A motorboat leaves a harbour and travels at an average speed of 8 km/h toward a small island. Two hours later, a cabin cruiser leaves the same harbour and travels at an average speed of 16 km/h toward the same island. How many hours after the cabin cruiser leaves will it be alongside the motorboat?
• A long distance runner started on a course, running at an average speed of 6 km/h. One hour later, a second runner began the same course at an average speed of 8 km/h. How long after the second runner started will they overtake the first runner?
• Two men are travelling in opposite directions at the rate of 20 and 30 km/h at the same time and from the same place. In how many hours will they be 300 kilometres apart?
• Two trains start at the same time from the same place and travel in opposite directions. If the rate of one is 6 km/h more than the rate of the other and they are 168 kilometres apart at the end of 4 hours, what is the rate of each?
• Two cyclists start from the same point and ride in opposite directions. One cyclist rides twice as fast as the other. In three hours, they are 72 kilometres apart. Find the rate of each cyclist.
• Two small planes start from the same point and fly in opposite directions. The first plane is flying 25 km/h slower than the second plane. In two hours, the planes are 430 kilometres apart. Find the rate of each plane.
• On a 130-kilometre trip, a car travelled at an average speed of 55 km/h and then reduced its speed to 40 km/h for the remainder of the trip. The trip took a total of 2.5 hours. For how long did the car travel at 40 km/h?
• Running at an average rate of 8 m/s, a sprinter ran to the end of a track and then jogged back to the starting point at an average of 3 m/s. The sprinter took 55 s to run to the end of the track and jog back. Find the length of the track.

Answer Key 8.8

Intermediate Algebra Copyright © 2020 by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

Share This Book

Distance, Rate, and Time Worksheets

• Pre Algebra & Algebra
• Math Tutorials
• Exponential Decay
• Worksheets By Grade

In math, distance, rate, and time are three important concepts you can use to solve many problems if you know the formula. Distance is the length of space traveled by a moving object or the length measured between two points. It is usually denoted by  d  in math problems.

The rate is the speed at which an object or person travels. It is usually denoted by  r  in equations. Time is the measured or measurable period during which an action, process, or condition exists or continues. In distance, rate, and time problems , time is measured as the fraction in which a particular distance is traveled. Time is usually denoted by  t  in equations.

Use these free, printable worksheets to help students learn and master these important math concepts. Each slide provides the student worksheet, followed by an identical worksheet that includes the answers for ease of grading. Each worksheet provides three distance, rate, and time problems for students to solve.

Worksheet No. 1

Print the PDF: Distance, Rate, and Time Worksheet No. 1

When solving distance problems, explain to students that they will use the formula:

or rate (speed) times time equals distance. For example, the first problem states:

The Prince David ship headed south at an average speed of 20 mph. Later the Prince Albert traveled north with an average speed of 20 mph. After the Prince David ship had traveled for eight hours, the ships were 280 miles apart. How many hours did the Prince David Ship Travel?

Students should find that the ship traveled for six hours.

Worksheet No. 2

Print the PDF: Distance, Rate, and Time Worksheet No. 2

If students are struggling, explain that to solve these problems, they will apply the formula that solves distance, rate, and time, which is  distance = rate x tim e. It is abbreviated as:

The formula can also be rearranged as:

r = d/t or  t = d/r

Let students know that there are many examples where you might use this formula in real life. For example, if you know the time and rate a person is traveling on a train, you can quickly calculate how far he traveled. And if you know the time and distance a passenger traveled on a plane, you could quickly figure the distance she traveled simply by reconfiguring the formula.

Worksheet No. 3

Print the PDF: Distance, Rate, Time Worksheet No. 3

On this worksheet, students will solve problems such as:

Two sisters Anna and Shay left the home at the same time. They headed out in opposite directions toward their destinations. Shay drove 50 mph faster than her sister Anna. Two hours later, they were 220 mph apart from each other. What was Anna’s average speed?

The students should find that Anna's average speed was 30 mph.

Worksheet No. 4

Print the PDF: Distance, Rate, Time Worksheet No. 4

Ryan left home and drove to his friend's house driving 28 mph. Warren left an hour after Ryan traveling at 35 mph hoping to catch up with Ryan. How long did Ryan drive before Warren caught up to him?

Students should find that Ryan drove for five hours before Warren caught up to him.

Worksheet No. 5

Print the PDF: Distance, Rate, and Time Worksheet No. 5

On this final worksheet, students will solve problems including:

Pam drove to the mall and back. It took one hour longer to go there than it did to come back home. The average speed she was traveling on the trip there was 32 mph. The average speed on the way back was 40 mph. How many hours did the trip there take?

They should find that Pam's trip took five hours.

• Solving Problems Involving Distance, Rate, and Time
• Realistic Math Problems Help 6th-graders Solve Real-Life Questions
• Newspaper Printables
• 4th-Grade Math Word Problems
• Basic Subtraction Fact Worksheets to 20
• Timestable Facts to 10
• Find the Main Idea Worksheets and Practice Questions
• Rate of Change Worksheet with Solutions
• What Is Velocity in Physics?
• Addition and Multiplication Printables
• Help Kids Calculate the Area and Circumference of Circles
• Money Worksheets for Counting Change
• Maple Syrup Printables
• 7th Grade Science Fair Projects
• Simple Interest Worksheets With Answers
• What Speed Actually Means in Physics

Distance Word Problems - Given the Total Time

These lessons, with videos, examples and step-by-step solutions, explain how to solve time-distance-rate problems.

Related Pages Rate, Time, Distance Solving Speed, Time, Distance Problems Using Algebra More Algebra Lessons Math Worksheets

Distance problems are word problems that involve the distance an object will travel at a certain average rate for a given period of time.

The formula for distance problems is: distance = rate × time or d = r × t .

Things to watch out for: Make sure that you change the units when necessary. For example, if the rate is given in miles per hour and the time is given in minutes then change the units appropriately.

It would be helpful to use a table to organize the information for distance problems. A table helps you to think about one number at a time instead being confused by the question.

The following diagram shows how to set up a table to help solve time-distance problems. Scroll down the page for more examples and solutions on how to solve distance problems.

Distance Problems: Given Total Time

Example: John took a drive to town at an average rate of 40 mph. In the evening, he drove back at 30 mph. If he spent a total of 7 hours traveling, what is the distance traveled by John?

Solution: Step 1: Set up a rtd table.

Step 2: Fill in the table with information given in the question.

John took a drive to town at an average rate of 40 mph. In the evening, he drove back at 30 mph. If he spent a total of 7 hours traveling, what is the distance traveled by John?

Let t = time to travel to town.

7 – t = time to return from town.

Step 3: Fill in the values for d using the formula d = rt

Step 4: Since the distances traveled in both cases are the same, we get the equation:

40 t = 120 The distance traveled by John to go back is also 120 So, the total distance traveled by John is 240

Distance, Rate, Time Word Problems Two examples of distance, rate, and time. One involves adding the distances in our chart, where as the other example involves setting the distances equal to each other.

• Two truck drivers leave a cafe at the same time, traveling in opposite directions. On truck goes 7 mph faster than the other one. After 4 hr, they are 404 miles apart. How fast is each truck going?
• Ryan left the science museum and drove south at a rate of 28 km/h. Jenna left three hours later driving 42 km/h faster in an effort to catch up to him. How long did Jenna have to travel to catch up with Ryan?

Distance-time word problem where the total time is given

Example: Gordon rode his bike at 15 mph to go get his car. He then drove back at 45 mph. If the entire trip took him 8 hours, how far was his car?

We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.

• HW Guidelines
• Study Skills Quiz
• Find Local Tutors
• Demo MathHelp.com
• Join MathHelp.com

Select a Course Below

• ACCUPLACER Math
• Math Placement Test
• PRAXIS Math
• + more tests
• 5th Grade Math
• 6th Grade Math
• Pre-Algebra
• College Pre-Algebra
• Introductory Algebra
• Intermediate Algebra
• College Algebra

"Distance" Word Problems

Explanation More Examples

What is a "distance" word problem?

"Distance" word problems, often also called "uniform rate" problems, involve something travelling at some fixed and steady ("uniform") pace ("rate", "velocity", or "speed"), or else you are told to regard to object as moving at some average speed.

Content Continues Below

MathHelp.com

Distance Word Problems

Whenever you read a problem that involves "how fast", "how far", or "for how long", you should think of the distance equation, d = rt , where d stands for distance, r stands for the (constant or average) rate of speed, and t stands for time.

Make sure that the units for time and distance agree with the units for the rate. For instance, if they give you a rate of feet per second, then your time must be in seconds and your distance must be in feet. Sometimes they try to trick you by using mis-matched units, and you have to catch this and convert to the correct units.

In case you're wondering, this type of exercise requires that the rate be fixed and steady (that is, unchanging) for the d = rt formula to work. The only way you can deal with a speed that might be changing over time is to take the average speed over the time or distance in question. Working directly with changing speeds will be something you'll encounter in calculus, as it requires calculus-based (or more advanced) methods.

What is the difference between a fixed speed and an average speed?

A fixed-speed exercise is one in which the car, say, is always going exactly sixty miles an hour; in three hours, the car, on cruise-control, will have gone 180 miles. An average-speed exercise is one in which the car, say, averaged forty miles an hour, but this average includes the different speeds related to stop lights, highways, and back roads; in three hours the car went 120 miles, though the car's speed was not constant. Most of the exercises you'll see will be fixed-speed exercises, but obviously they're not very "real world". It's a simplification they do in order to make the situation feasible using only algebraic methods.

What is an example of a "distance" word problem?

• A 555 -mile, 5 -hour plane trip was flown at two speeds. For the first part of the trip, the average speed was 105 mph. Then the tailwind picked up, and the remainder of the trip was flown at an average speed of 115 mph. For how long did the plane fly at each speed?

There is a method for setting up and solving these exercises that I first encountered well after I'd actually been doing them while taking a class as an undergraduate. But, as soon as I was introduced to the method, I switched over, because it is *so* way easier.

First I set up a grid, with the columns being labelled with the variables from the "distance" formula, and the rows being labelled with the "parts" involved:

In the first part, the plane covered some distance. I don't know how much, so I'll need a variable to stand for this unknown value. I'll use the variable they give me in the distance equation:

They gave me the speed, or rate, for this part, so I'll add this to my table:

The plane flew for some amount of time during this first part, but I don't know how long that was. So I need a variable to stand for this unknown value; I'll use the variable from the distance equation:

For the second part, the plane travelled the rest of the total distance. I don't know the exact distance that was flown during this second part, but I do know that it was "however much was left of the 555 miles, after the first d miles were flown in the first part. "How much was left after [some amount] was taken out" is expressed with subtraction: I take the amount that has been taken care of already, and subtract this from whatever was the total. Adding this to my table, I get:

They've given me the speed, or rate, for the second part, and I can use the same "How much is left?" construction for whatever was the time for this second part. So now my table looks like this:

For the "total" row, I add down (or take info from the exercise statement):

Advertisement

Why did I not add down in the "rate" column? Because I cannot add rates! In this exercise, adding the rates would have said that the average rate for the entire trip was 105 + 115 = 220 miles per hour. But obviously this makes no sense.

The genius of this table-based method of set-up is that I can now create equations from the rows and columns. In this exercise, there is more than one way to proceed. I'll work with the "distance" equation to create expressions for the distances covered in each part.

Multiplying across, the first row tells me that the distance covered in the first part of the flight was:

1st part distance: 105 t

Again multiplying across, the second row tells me that the distance covered in the second part of the flight was:

2nd part distance: 115(5 − t )

I can add these two partial-distance expressions, and set them equal to the known total distance:

105 t + 115(5 − t ) = 555

This is an equation in one variable, which I can solve:

105 t + 115(5 − t ) = 555 105 t + 575 − 115 t = 555 575 − 10 t = 555 20 = 10 t 2 = t

Looking back at my table, I see the I had defined t to be the time that the plane spent in the air on the first part of its journey. Looking back at the original exercise, I see that they want to know the times that the plane spent at each of the two speeds.

I now have the time for the first part of the flight; the time was two hours. The exercise said that the entire trip was five hours, so the second part must have taken three hours (found by subtracting the first-part time from the total time). They haven't asked for the partial distances, so I now have all the information I need; no further computations are necessary. My answer is:

first part: 2 hours second part: 3 hours

When I was setting up my equation, I mentioned that there was more than one way to proceed. What was the other way? I could have used the table to create an expression for each of the two partial times, added, set the result equal to the given total, and solved for the variable d . Since the distance equation is d  =  rt , then the expressions for the partial times would be created by solving the equation for t  = . My work would have looked like this:

first part: d /105

second part: (555 − d )/115

adding: d /105 + (555 − d )/115 = 5 23 d + 11,655 − 21 d = 12,075 2 d = 420 d = 210

Looking back at my table, I would have seen that this gives me the distance covered in the first half of the flight. Looking back at the exercise, I would have seeing that they are wanting times, not distances. So I would have back-solved for the time for the first part, and then done the subtraction to find the time for the second part. My work would have had more steps, but my answer would have been the same.

There are three things that I hope you take from the above example:

• Using a table or grid to keep track of what you're doing can be incredibly helpful.
• It is important to clearly define your variables, so you know (by the end) what you'd meant (back in the beginning), so you can apply your results correctly.
• You should always check the original exercise, so you can be sure that you're answering the question that they'd actually asked.

(My value for the distance, found above, is correct, but was not what they'd asked for.) But even more important to understand is this:

NEVER TRY TO ADD RATES! Think about it: If you drive 20 mph on one street, and 40 mph on another street, does that mean you averaged 60 mph? Of course not.

Can I even average the rates? If I drove at 20 mph for one hour, and then drove 60 mph for two hours, then I would have travelled 140 miles in three hours, or a little under 47 mph. But 47 is not the average of 20 and 60 .

As you can see, the actual math involved in solving this type of exercise is often quite simple. It's the set-up that's the hard part. So what follows are some more examples, but with just the set-up displayed. Try your hand at solving, and click on the links to get pop-ups from which to check your equations and solutions.

• An executive drove from home at an average speed of 30 mph to an airport where a helicopter was waiting. The executive boarded the helicopter and flew to the corporate offices at an average speed of 60 mph. The entire distance was 150 miles; the entire trip took three hours. Find the distance from the airport to the corporate offices.

I will start in the usual way, by setting up my table:

I have labelled my rows so it's clear how they relate to the exercise. Now I need to fill in the rows. As before, I don't know the distance or the time for the part where the executive was driving, so I'll use variables for these unknowns, along with the given rate.

For the flying portion of the trip, I'll use the "how much is left" construction, along with the given rate, to fill in my second row. I'll also fill in the totals.

The first row gives me the equation d  = 30 t . The second row is messier, giving me the equation:

150 − d = 60(3 − t )

There are various ways I can go from here; I think I'll solve this second equation for the variable d , and then set the results equal to each other.

150 − d = 60(3 − t ) 150 − 60(3 − t ) = d

Setting equal these two expressions for d , I get:

30 t = 150 − 60(3 − t )

Solve for t ; interpret the value; state the final answer.

• A car and a bus set out at 2 p.m. from the same point, headed in the same direction. The average speed of the car is 30 mph slower than twice the speed of the bus. In two hours, the car is 20 miles ahead of the bus. Find the rate of the car.

Both vehicles travelled for the same amount of time.

The car's values are expressed in terms of the bus' values, so I'll use variables for the bus' unknowns, and then define the car in terms of the bus' variables. This gives me:

(As it turns out, I won't need the "total" row this time.) The car's row gives me:

d + 20 = 2(2 r − 30)

This is not terribly helpful. The second row gives me:

I'll use the second equation to simplify the first equation by substituting " 2 r " from the second equation in for the " d " in the first equation. Then I'll solve the equation for the value of " r ". Finally, I'll need to interpret this value within the context of the exercise, and then I'll state the final answer.

(Remember that the expression for the car's speed, from the table, was 2 r  − 30 , so all you need to do is find the numerical value of this expression. Just evaluate; don't try to solve — again — for the value of r .)

URL: https://www.purplemath.com/modules/distance.htm

Page 1 Page 2

Standardized Test Prep

College math, homeschool math, share this page.

• Terms of Use
• Privacy / Cookies
• About Purplemath
• About the Author
• Tutoring from PM
• Advertising
• Linking to PM
• Site licencing

Visit Our Profiles

Module 10: Linear Equations

Problems involving formulas i, learning outcomes.

• Solve a formula for a specific variable
• Use the distance, rate, and time formula
• Apply for the steps for solving word problems to interest rate problems

There is often a well-known formula or relationship that applies to a word problem. For example, if you were to plan a road trip, you would want to know how long it would take you to reach your destination. [latex]d=rt[/latex] is a well-known relationship that associates distance traveled, the rate at which you travel, and how long the travel takes.

Distance, Rate, and Time

If you know two of the quantities in the relationship [latex]d=rt[/latex], you can easily find the third using methods for solving linear equations. For example, if you know that you will be traveling on a road with a speed limit of [latex]30\frac{\text{ miles }}{\text{ hour }}[/latex] for [latex]2[/latex] hours, you can find the distance you would travel by multiplying rate times time or [latex]\left(30\frac{\text{ miles }}{\text{ hour }}\right)\left(2\text{ hours }\right)=60\text{ miles }[/latex].

We can generalize this idea depending on what information we are given and what we are looking for. For example, if we need to find time, we could solve the [latex]d=rt[/latex] equation for [latex]t[/latex] using division:

[latex]d=rt[/latex]

[latex]\frac{d}{r}=t[/latex]

Likewise, if we want to find rate, we can isolate [latex]r[/latex] using division:

[latex]\frac{d}{t}=r[/latex]

In the following examples, you will see how this formula is applied to answer questions about ultra marathon running.

Ultra marathon running (defined as anything longer than [latex]26.2[/latex] miles) is becoming very popular among women even though it remains a male-dominated niche sport. Ann Trason has broken twenty world records in her career. One such record was the American River [latex]50[/latex]-mile Endurance Run, which begins in Sacramento, California, and ends in Auburn, California. [1] In 1993, Trason finished the run with a time of [latex]6:09:08[/latex] .  The men’s record for the same course was set in 1994 by Tom Johnson, who finished the course with a time of  [latex]5:33:21[/latex]. [2]

In the next examples, we will use the [latex]d=rt[/latex] formula to answer the following questions about the two runners.

• What was each runner’s rate for their record-setting runs?

By the time Johnson had finished, how many more miles did Trason have to run?

• How much further could Johnson have run if he had run as long as Trason?

What was each runner’s time for running one mile?

To make answering the questions easier, we will round the two runners’ times to [latex]6[/latex] hours and [latex]5.5[/latex] hours.

What was each runner’s rate for their record-setting runs? Round to two decimal places.

Read and Understand:  We are looking for rate and we know distance and time, so we can use the idea: [latex]d=rt[/latex], [latex]\frac{d}{t}=r[/latex]. Let’s solve one runner’s rate using the original formula, and one using the rearranged formula.

Define and Translate: Because there are two runners, making a table to organize this information helps. Note how we keep units to help us keep track of how all the terms are related to each other.

Write and Solve:

Trason’s rate:

[latex]50\text{ miles }=r\left(6\text{ hours}\right)[/latex]

[latex]50=6r[/latex]

[latex]r\approx 8.33[/latex]

Johnson’s rate:

[latex]r=\frac{d}{t}[/latex]

[latex]r=\frac{50\text{ miles}}{5.5\text{ hours}}[/latex]

[latex]r=\frac{50}{5.5}[/latex]

[latex]r\approx 9.10[/latex]

Check and Interpret:

We can fill in our table with this information.

Now that we know each runner’s rate, we can answer the second question.

Here is the table we created for reference:

Read and Understand:  We are looking for how many miles Trason still had on the trail when Johnson had finished after [latex]5.5[/latex] hours. This is a distance, and we know rate and time.

Define and Translate:  We can use the formula [latex]d=rt[/latex] again. This time the unknown is [latex]d[/latex], and the time Trason had run is [latex]5.5[/latex] hours.

[latex]\begin{array}{l}d=rt\\\\d=8.33\frac{\text{ miles }}{\text{ hour }}\left(5.5\text{ hours}\right)\\\\d=45.82\text{ miles}\end{array}[/latex]

Have we answered the question? We were asked to find how many more miles she had to run after [latex]5.5[/latex] hours.  What we have found is how long she had run after [latex]5.5[/latex] hours. We need to subtract [latex]d=45.82\text{ miles }[/latex] from the total distance of the course.

[latex]50\text{ miles }-45.82\text{ miles }=4.18\text{ miles }[/latex]

The third question is similar to the second. Now that we know each runner’s rate, we can answer questions about individual distances or times.

How much further could Johnson have run if he had run for the same amount of time as Trason?

Read and Understand:  The word further implies we are looking for a distance.

Define and Translate:  We can use the formula [latex]d=rt[/latex] again. This time the unknown is [latex]d[/latex], the time is [latex]6[/latex] hours, and Johnson’s rate is [latex]9.1\frac{\text{ miles }}{\text{ hour }}[/latex]

[latex]\begin{array}{l}d=rt\\\\d=9.1\frac{\text{ miles }}{\text{ hour }}\left(6\text{ hours }\right)\\\\d=54.6\text{ miles }\end{array}[/latex].

Have we answered the question? We were asked to find how many more miles Johnson would have run if he had run at his rate of [latex]9.1\frac{\text{ miles }}{\text{ hour }}[/latex] for [latex]6[/latex] hours.

Johnson would have run [latex]54.6[/latex] miles, so that’s [latex]4.6[/latex] more miles than he ran during the race.

Now we will tackle the last question, where we are asked to find a time for each runner.

Read and Understand:  we are looking for time, and this time our distance has changed from  [latex]50[/latex] miles to [latex]1[/latex] mile, so we can use

[latex]d=rt\\\frac{d}{r}=t[/latex]

Define and Translate: we can use the formula [latex]d=rt[/latex] again. This time the unknown is [latex]t[/latex], the distance is [latex]1[/latex] mile, and we know each runner’s rate. It may help to create a new table:

We will need to divide to isolate time.

[latex]\begin{array}{c}d=rt\\\\1\text{ mile }=8.33\frac{\text{ miles }}{\text{ hour }}\left(t\text{ hours }\right)\\\\\frac{1\text{ mile }}{\frac{8.33\text{ miles }}{\text{ hour }}}=t\text{ hours }\\\\0.12\text{ hours }=t\end{array}[/latex].

[latex]0.12[/latex]  hours is about  [latex]7.2[/latex] minutes, so Trason’s time for running one mile was about [latex]7.2[/latex] minutes. WOW! She did that for  [latex]6[/latex] hours!

[latex]\begin{array}{c}d=rt\\\\1\text{ mile }=9.1\frac{\text{ miles }}{\text{ hour }}\left(t\text{ hours }\right)\\\\\frac{1\text{ mile }}{\frac{9.1\text{ miles }}{\text{ hour }}}=t\text{ hours }\\\\0.11\text{ hours }=t\end{array}[/latex].

[latex]0.11[/latex] hours is about [latex]6.6[/latex] minutes, so Johnson’s time for running one mile was about  [latex]6.6[/latex] minutes. WOW! He did that for [latex]5.5[/latex] hours!

Have we answered the question? We were asked to find how long it took each runner to run one mile given the rate at which they ran the whole  [latex]50[/latex]-mile course.  Yes, we answered our question.

Trason’s mile time was [latex]7.2\frac{\text{minutes}}{\text{mile}}[/latex] and Johnsons’ mile time was [latex]6.6\frac{\text{minutes}}{\text{mile}}[/latex]

In the following video, we show another example of answering many rate questions given distance and time.

Simple Interest

In order to entice customers to invest their money, many banks will offer interest-bearing accounts. The accounts work like this: a customer deposits a certain amount of money (called the Principal, or [latex]P[/latex]), which then grows slowly according to the interest rate ([latex]r[/latex], measured in percent) and the length of time ([latex]t[/latex], usually measured in months) that the money stays in the account. The amount earned over time is called the interest ([latex]I[/latex]), which is then given to the customer.

The simplest way to calculate interest earned on an account is through the formula [latex]\displaystyle I=P\,\cdot \,r\,\cdot \,t[/latex]

If we know any of the three amounts related to this equation, we can find the fourth. For example, if we want to find the time it will take to accrue a specific amount of interest, we can solve for [latex]t[/latex] using division:

[latex]\displaystyle\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,I=P\,\cdot \,r\,\cdot \,t\\\\ \frac{I}{{P}\,\cdot \,r}=\frac{P\cdot\,r\,\cdot \,t}{\,P\,\cdot \,r}\\\\\,\,\,\,\,\,\,\,\,\,\,{t}=\frac{I}{\,r\,\cdot \,t}\end{array}[/latex]

Below is a table showing the result of solving for each individual variable in the formula.

In the next examples, we will show how to substitute given values into the simple interest formula, and decipher which variable to solve for.

If a customer deposits a principal of [latex]$2000[/latex] at a monthly rate of  [latex]0.7\%[/latex], what is the total amount that she has after [latex]24[/latex] months?

Substitute in the values given for the Principal, Rate, and Time.

[latex]\displaystyle\begin{array}{l}I=P\,\cdot \,r\,\cdot \,t\\I=2000\cdot 0.7\%\cdot 24\end{array}[/latex]

Rewrite  [latex]0.7\%[/latex] as the decimal  [latex]0.007[/latex], then multiply.

[latex]\begin{array}{l}I=2000\cdot 0.007\cdot 24\\I=336\end{array}[/latex]

Add the interest and the original principal amount to get the total amount in her account.

[latex] \displaystyle 2000+336=2336[/latex]

She has  [latex]$2336[/latex] after  [latex]24[/latex] months.

The following video shows another example of finding an account balance after a given amount of time, principal invested, and a rate.

In the following example, you will see why it is important to make sure the units of the interest rate match the units of time when using the simple interest formula.

Alex invests [latex]$600[/latex] at [latex]3.5\%[/latex] monthly interest for [latex]3[/latex] years. What amount of interest has Alex earned?

Read and Understand: The question asks for an amount, so we can substitute what we are given into the simple interest formula [latex]I=P\,\cdot \,r\,\cdot \,t[/latex]

Define and Translate:  we know [latex]P[/latex], [latex]r[/latex], and [latex]t[/latex] so we can use substitution. [latex]r[/latex]  =[latex]0.035[/latex], [latex]P[/latex] =  [latex]$600[/latex], and [latex]t[/latex] =[latex]3[/latex] years. We have to be careful, because [latex]r[/latex] is in months, and [latex]t[/latex] is in years.  We need to change [latex]t[/latex] into months, because we can’t change the rate—it is set by the bank.

[latex]{t}=3\text{ years }\cdot12\frac{\text{ months }}{\text {year }}=36\text{ months }[/latex]

Substitute the given values into the formula.

[latex]\begin{array}{l} I=P\,\cdot \,r\,\cdot \,t\\\\I=600\,\cdot \,0.035\,\cdot \,36\\\\{I}=756\end{array}[/latex]

We were asked what amount Alex earned, which is the amount provided by the formula. In the previous example, we were asked the total amount in the account, which included the principal and interest earned.

Alex has earned [latex]$756[/latex].

In the following video, we show another example of how to find the amount of interest earned after an investment has been sitting for a given monthly interest.

After  [latex]10[/latex] years, Jodi’s account balance has earned  [latex]$1080[/latex] in interest. The rate on the account is  [latex]0.09\%[/latex] monthly. What was the original amount she invested in the account?

Read and Understand: The question asks for the original amount invested, the principal. We are given a length of time in years, and an interest rate in months, and the amount of interest earned.

Define and Translate:  we know [latex]I[/latex] =  [latex]$1080[/latex], [latex]r[/latex] =[latex]0.009[/latex], and [latex]t[/latex] =[latex]10[/latex] years, so we can use [latex]{P}=\frac{I}{{r}\,\cdot \,t}[/latex]

We also need to make sure the units on the interest rate and the length of time match, and they do not. We need to change time into months again.

[latex]{t}=10\text{ years }\cdot12\frac{\text{ months }}{\text{ year }}=120\text{ months }[/latex]

Substitute the given values into the formula

[latex]\begin{array}{l}{P}=\frac{I}{{R}\,\cdot \,T}\\\\{P}=\frac{1080}{{0.009}\,\cdot \,120}\\\\{P}=\frac{1080}{1.08}=1000\end{array}[/latex]

We were asked to find the principal given the amount of interest earned on an account.  If we substitute [latex]P[/latex] =[latex]$1000[/latex] into the formula [latex]I=P\,\cdot \,r\,\cdot \,t[/latex] we get

[latex]I=1000\,\cdot \,0.009\,\cdot \,120\\I=1080[/latex]

Our solution checks out. Jodi invested  [latex]$1000[/latex].

In the next section, we will apply our problem-solving method to problems involving dimensions of geometric shapes.

• "Ann Trason." Wikipedia. Accessed May 05, 2016. https://en.wikipedia.org/wiki/Ann_Trason . ↵
•  "American River [latex]50[/latex] Mile Endurance Run." Wikipedia. Accessed May 05, 2016. https://en.wikipedia.org/wiki/American_River_50_Mile_Endurance_Run . ↵
• Image: Ann Trason Trail Running. Authored by : Lumen Learning. License : CC BY: Attribution
• Rates. Provided by : Lumen Learning. License : CC BY: Attribution
• Problem Solving Using Distance, Rate, Time (Running). Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/3WLp5mY1FhU . License : CC BY: Attribution
• Simple Interest - Determine Account Balance (Monthly Interest). Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/XkGgEEMR_00 . License : CC BY: Attribution
• Simple Interest - Determine Interest Balance (Monthly Interest). Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/mRV5ljj32Rg . License : CC BY: Attribution
• Simple Interest - Determine Principal Balance (Monthly Interest). Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/vbMqN6lVoOM . License : CC BY: Attribution
• Ann Trason. Provided by : Wikipedia. Located at : https://en.wikipedia.org/wiki/Ann_Trason . License : CC BY-SA: Attribution-ShareAlike
• American River 50 Mile Endurance Run. Provided by : Wikipedia. Located at : https://en.wikipedia.org/wiki/American_River_50_Mile_Endurance_Run . License : CC BY-SA: Attribution-ShareAlike
• Unit 10: Solving Equations and Inequalities, from Developmental Math: An Open Program. Provided by : Monterey Institute of Technology and Education. Located at : http://nrocnetwork.org/resources/downloads/nroc-math-open-textbook-units-1-12-pdf-and-word-formats/ . License : CC BY: Attribution

• Pre-algebra lessons
• Pre-algebra word problems
• Algebra lessons
• Algebra word problems
• Algebra proofs
• Advanced algebra
• Geometry lessons
• Geometry word problems
• Geometry proofs
• Trigonometry lessons
• Consumer math
• Baseball math
• Math for nurses
• Statistics made easy
• High school physics
• Basic mathematics store
• SAT Math Prep
• Math skills by grade level
• Ask an expert
• Other websites
• K-12 worksheets
• Worksheets generator
• Algebra worksheets
• Geometry worksheets
• Free math problem solver
• Pre-algebra calculators
• Algebra Calculators
• Geometry Calculators
• Math puzzles
• Math tricks
• Member login

Distance rate time problems

Distance rate time problems involve object moving at a constant rate and this is called uniform motion. The formula d = r × t is the formula to use to solve problems related to distance, rate, and time.

Examples showing how to solve distance rate time problems

It may be obvious that for this type of problem, the distance is the same since you left the same place (your house) and are going to the same location.

Suppose that you went to no other places, the distance then to go back to your house is again the same.

Therefore, when solving distance rate time problems involving opposite direction travel, you can add the distances to get the total distance.

Study also the distance rate time problem in the figure below carefully

Mixture word problems

Recent Articles

How to divide any number by 5 in 2 seconds.

Feb 28, 24 11:07 AM

Math Trick to Square Numbers from 50 to 59

Feb 23, 24 04:46 AM

Sum of Consecutive Odd Numbers

Feb 22, 24 10:07 AM

100 Tough Algebra Word Problems. If you can solve these problems with no help, you must be a genius!

 Recommended

About me :: Privacy policy :: Disclaimer :: Donate   Careers in mathematics  

Copyright © 2008-2021. Basic-mathematics.com. All right reserved

Mr. Math Blog

Convert Units of Length - Lesson 6.1

Convert Units of Capacity - Lesson 6.2

Convert Units of Weight and Mass - Lesson 6.3

Transform Units - Lesson 6.4

Problem Solving - Distance, Rate, and Time - Lesson 6.5

Fractions and Decimals - Lesson 2.1

Compare and order Fractions and Decimals - Lesson 2.2

Multiply Fractions - Lesson 2.3

Simplify Factors - Lesson 2.4

Model Fraction Division - Lesson 2.5

Estimate Quotients - Lesson 2.6

Dividing Fractions - Lesson 2.7

Model Mixed Number Division - Lesson 2.8

Divide Mixed Numbers - Lesson 2.9

Problem Solving - Fraction Operations - Lesson 2.10

Understanding Positive and Negative Integers - Lesson 3.1

Compare and Order Integers - Lesson 3.2

Rational Numbers and the Number Line - Lesson 3.3

Compare and Order Rational Numbers - Lesson 3.4

Absolute Value - Lesson 3.5

Compare Absolute Value - Lesson 3.6

Rational Numbers and the Coordinate Plane - Lesson 3.7

Ordered Pair Relationships - Lesson 3.8

Distance on the Coordinate Plane - Lesson 3.9

Problem Solving - The Coordinate Plane - Lesson 3.10

Solutions of Equations - Lesson 8.1

Writing Equations - Lesson 8.2

Model and Solve Addition Equations - Lesson 8.3

Solve Addition & Subtraction Equations - Lesson 8.4

Sixth Grade

     Math

Model Ratios - Lesson 4.1

Ratios and Rates - Lesson 4.2

Equivalent Ratios and the Multiplication Table - Lesson 4.3

Use Tables to Compare Ratios - Lesson 4.4

Use Equivalent Ratios - Lesson 4.5

Find Unit Rates - Lesson 4.6

Use Unit Rates - Lesson 4.7

Equivalent Ratios and Graphs - Lesson 4.8

Three Dimensional Figures and Nets - Lesson 11.1

Surface Area of Prisms - Lesson 11.3

Surface Area of Pyramids - Lesson 11.4

Volume of a Rectangular Prism - Lesson 11.6

Area of Parallelograms - Lesson 10.1

Explore Area of Triangles - Lesson 10.2

Area of Triangles - Lesson 10.3

Explore Area of Trapezoids - Lesson 10.4

Area of Trapezoids - Lesson 10.5

Area of Regular Polygons - Lesson 10.6

​Composite Figures - Lesson 10.7

Divide Multi-Digit Numbers - Lesson 1.1

Prime Factorization - Lesson 1.2

Least Common Multiple (LCM) - Lesson 1.3

Greatest Common Factor (GCF) - Lesson 1.4

Problem Solving: Apply the GCF - Lesson 1.5

Add and Subtract Decimals - Lesson 1.6

Multiply Decimals - Lesson 1.7

Divide Decimals by Whole Numbers - Lesson 1.8

Divide with Decimals - Lesson 1.9

Chapter 1 Review for Test

Please Donate, if you're a regular!

The donate link is below. Thanks so much!!

Exponents - Lesson 7.1

Evaluating Expressions Involving Exponents - Lesson 7.2

Write Algebraic Expressions - Lesson 7.3

Identify Parts of Expressions - Lesson 7.4

Evaluate Algebraic Expressions and Formulas - Lesson 7.5

Use Algebraic Expressions - Lesson 7.6

Problem Solving - Combining Like Terms - Lesson 7.7

Generate Equivalent Expressions - Lesson 7.8

Identify Equivalent Expressions - Lesson 7.9

Thanks for your donation!

​Every little bit helps me help you!  :-)

Independent and Dependent Variables - Lesson 9.1

Equations and Tables - Lesson 9.2

Model Percents - Lesson 5.1

Write Percents as Fractions and Decimals - Lesson 5.2

Write Fractions and Decimals as Percents - Lesson 5.3

Percent of a Quantity - Lesson 5.4

Problem Solving - Percents - Lesson 5.5

Find the Whole From a Percent - Lesson 5.6

• AP Calculus
• AP Statistics
• Independent Study
• Second Grade Math
• Third Grade Math
• Fourth Grade Math
• Fifth Grade Math
• Sixth Grade Math
• Sixth Grade Math (CA)
• Seventh Grade Math (CA)
• Eighth Grade Math (CA)
• Integrated Math 1
• Integrated Math 2
• Integrated Math 3
• PreCalculus
• AP Statistics Exam Prep
• Elementary Statistics
• ELM Practice
• Percents and Decimals
• Sixth Grade Math (Big Ideas)

Online Math Class

Copyright 2013. All rights reserved.

• 888-309-8227
• 732-384-0146

New User Registration

Forgot Password

Go Math! 6 Common Core Edition, Grade: 6 Publisher: Houghton Mifflin Harcourt

Go math 6 common core edition, title : go math 6 common core edition, publisher : houghton mifflin harcourt, isbn : 547587783, isbn-13 : 9780547587783, use the table below to find videos, mobile apps, worksheets and lessons that supplement go math 6 common core edition..

textbook resources

• Call us toll-free
• FAQs – Frequently Asked Questions
• Contact Lumos Learning – Proven Study Programs by Expert Teachers

Follow us: Lumos Learning -->

• 2024 © Lumos Learning
• Privacy Policy - Terms of Service - Disclaimers

PARCC® is a registered trademark of PARCC, Inc. Lumos Learning, is not owned by or affiliated in any fashion with PARCC, Inc... Read More

PARCC® is a registered trademark of PARCC, Inc. Lumos Learning, is not owned by or affiliated in any fashion with PARCC, Inc., the Partnership for the Assessment of Readiness for College and Careers, nor any state of the Union. Neither PARCC, Inc., nor The Partnership for the Assessment of Readiness for College and Careers, nor any member state has endorsed this product. No portion of any fees or charges paid for any products or services Lumos Learning offers will be paid or inure to the benefit of PARCC, Inc., or any state of the Union

SBAC is a copyright of The Regents of the University of California – Smarter Balanced Assessment Consortium, which is not aff... Read More

SBAC is a copyright of The Regents of the University of California – Smarter Balanced Assessment Consortium, which is not affiliated to Lumos Learning. The Regents of the University of California – Smarter Balanced Assessment Consortium, was not involved in the production of, and does not endorse these products or this site.

ACT® Aspire™ is a registered trademark of ACT Aspire LLC., which is not affiliated to Lumos Learning. ACT Aspire LLC, was not... Read More

ACT® Aspire™ is a registered trademark of ACT Aspire LLC., which is not affiliated to Lumos Learning. ACT Aspire LLC,was not involved in the production of, and does not endorse these products or this site.

Florida Department of Education is not affiliated to Lumos Learning. Florida department of education, was not involved in the... Read More

Florida Department of Education is not affiliated to Lumos Learning. Florida department of education, was not involved in the production of, and does not endorse these products or this site.

Indiana Department of Education is not affiliated to Lumos Learning. Indiana department of education, was not involved in the... Read More

Indiana Department of Education is not affiliated to Lumos Learning. Indiana department of education, was not involved in the production of, and does not endorse these products or this site.

Mississippi Department of Education is not affiliated to Lumos Learning. Mississippi department of education, was not involved... Read More

Mississippi Department of Education is not affiliated to Lumos Learning. Mississippi department of education, was not involved in the production of, and does not endorse these products or this site.

Ohio Department of Education is not affiliated to Lumos Learning. Ohio department of education, was not involved in the prod... Read More

Ohio Department of Education is not affiliated to Lumos Learning. Ohio department of education, was not involved in the production of, and does not endorse these products or this site.

Tennessee Department of Education is not affiliated to Lumos Learning. Tennessee department of education, was not involved... Read More

Tennessee Department of Education is not affiliated to Lumos Learning. Tennessee department of education, was not involved in the production of, and does not endorse these products or this site.

Georgia Department of Education is not affiliated to Lumos Learning. Georgia department of education, was not involved... Read More

Georgia Department of Education is not affiliated to Lumos Learning. Georgia department of education, was not involved in the production of, and does not endorse these products or this site.

Missouri Department of Education is not affiliated to Lumos Learning. Missouri department of education, was not involved... Read More

Missouri Department of Education is not affiliated to Lumos Learning. Missouri department of education, was not involved in the production of, and does not endorse these products or this site.

Louisiana Department of Education is not affiliated to Lumos Learning. Louisiana department of education, was not involved... Read More

Louisiana Department of Education is not affiliated to Lumos Learning. Louisiana department of education, was not involved in the production of, and does not endorse these products or this site.

• Create a ShowMe
• Community Guidelines
• Edit Profile
• Edit ShowMes
• Students' ShowMes
• My Subscriptions
• ShowMe Ambassadors

I'm an elementary school teacher who...

• Elementary Math
• Ratios and Proportional Reasoning
• 6th grade go math

You must be logged into ShowMe

Are you sure you want to remove this ShowMe? You should do so only if this ShowMe contains inappropriate content.