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About this book series.

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More (almost) impossible integrals, sums, and series.

A New Collection of Fiendish Problems and Surprising Solutions

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  • Copyright: 2023

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All of what teachers have loved for over twenty years and more! The most popular, classroom-tested problem solving materials have been updated and are now available in Spanish! The updated edition of The Problem Solver has been aligned to NCTM standards providing complete coverage of all national content standards.

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Problem Solvers: A Free Early Math Curriculum

A free early math curriculum for children aged 30 to 48 months, research tells us early math skills are a powerful predictor of overall school achievement. but what does “early math” look like for toddlers and young preschoolers it looks like problem solvers..

Problem Solvers  is a free, downloadable early math curriculum that includes:

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Algebra Topics  - Distance Word Problems

Algebra topics  -, distance word problems, algebra topics distance word problems.

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Algebra Topics: Distance Word Problems

Lesson 10: distance word problems.


What are distance word problems?

Distance word problems are a common type of algebra word problems. They involve a scenario in which you need to figure out how fast , how far , or how long one or more objects have traveled. These are often called train problems because one of the most famous types of distance problems involves finding out when two trains heading toward each other cross paths.

In this lesson, you'll learn how to solve train problems and a few other common types of distance problems. But first, let's look at some basic principles that apply to any distance problem.

The basics of distance problems

There are three basic aspects to movement and travel: distance , rate , and time . To understand the difference among these, think about the last time you drove somewhere.

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The distance is how far you traveled. The rate is how fast you traveled. The time is how long the trip took.

The relationship among these things can be described by this formula:

distance = rate x time d = rt

In other words, the distance you drove is equal to the rate at which you drove times the amount of time you drove. For an example of how this would work in real life, just imagine your last trip was like this:

  • You drove 25 miles—that's the distance .
  • You drove an average of 50 mph—that's the rate .
  • The drive took you 30 minutes, or 0 .5 hours—that's the time .

According to the formula, if we multiply the rate and time , the product should be our distance.

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And it is! We drove 50 mph for 0.5 hours—and 50 ⋅ 0.5 equals 25 , which is our distance.

What if we drove 60 mph instead of 50? How far could we drive in 30 minutes? We could use the same formula to figure this out.

60 ⋅ 0.5 is 30, so our distance would be 30 miles.

Solving distance problems

When you solve any distance problem, you'll have to do what we just did—use the formula to find distance , rate , or time . Let's try another simple problem.

On his day off, Lee took a trip to the zoo. He drove an average speed of 65 mph, and it took him two-and-a-half hours to get from his house to the zoo. How far is the zoo from his house?

First, we should identify the information we know. Remember, we're looking for any information about distance, rate, or time. According to the problem:

  • The rate is 65 mph.
  • The time is two-and-a-half hours, or 2.5 hours.
  • The distance is unknown—it's what we're trying to find.

You could picture Lee's trip with a diagram like this:

problem solver math book

This diagram is a start to understanding this problem, but we still have to figure out what to do with the numbers for distance , rate , and time . To keep track of the information in the problem, we'll set up a table. (This might seem excessive now, but it's a good habit for even simple problems and can make solving complicated problems much easier.) Here's what our table looks like:

We can put this information into our formula: distance = rate ⋅ time .

We can use the distance = rate ⋅ time formula to find the distance Lee traveled.

The formula d = rt looks like this when we plug in the numbers from the problem. The unknown distance is represented with the variable d .

d = 65 ⋅ 2.5

To find d , all we have to do is multiply 65 and 2.5. 65 ⋅ 2.5 equals 162.5 .

We have an answer to our problem: d = 162.5. In other words, the distance Lee drove from his house to the zoo is 162.5 miles.

Be careful to use the same units of measurement for rate and time. It's possible to multiply 65 miles per hour by 2.5 hours because they use the same unit: an hour . However, what if the time had been written in a different unit, like in minutes ? In that case, you'd have to convert the time into hours so it would use the same unit as the rate.

Solving for rate and time

In the problem we just solved we calculated for distance , but you can use the d = rt formula to solve for rate and time too. For example, take a look at this problem:

After work, Janae walked in her neighborhood for a half hour. She walked a mile-and-a-half total. What was her average speed in miles per hour?

We can picture Janae's walk as something like this:

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And we can set up the information from the problem we know like this:

The table is repeating the facts we already know from the problem. Janae walked one-and-a-half miles or 1.5 miles in a half hour, or 0.5 hours.

As always, we start with our formula. Next, we'll fill in the formula with the information from our table.

The rate is represented by r because we don't yet know how fast Janae was walking. Since we're solving for r , we'll have to get it alone on one side of the equation.

1.5 = r ⋅ 0.5

Our equation calls for r to be multiplied by 0.5, so we can get r alone on one side of the equation by dividing both sides by 0.5: 1.5 / 0.5 = 3 .

r = 3 , so 3 is the answer to our problem. Janae walked 3 miles per hour.

In the problems on this page, we solved for distance and rate of travel, but you can also use the travel equation to solve for time . You can even use it to solve certain problems where you're trying to figure out the distance, rate, or time of two or more moving objects. We'll look at problems like this on the next few pages.

Two-part and round-trip problems

Do you know how to solve this problem?

Bill took a trip to see a friend. His friend lives 225 miles away. He drove in town at an average of 30 mph, then he drove on the interstate at an average of 70 mph. The trip took three-and-a-half hours total. How far did Bill drive on the interstate?

This problem is a classic two-part trip problem because it's asking you to find information about one part of a two-part trip. This problem might seem complicated, but don't be intimidated!

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You can solve it using the same tools we used to solve the simpler problems on the first page:

  • The travel equation d = rt
  • A table to keep track of important information

Let's start with the table . Take another look at the problem. This time, the information relating to distance , rate , and time has been underlined.

Bill took a trip to see a friend. His friend lives 225 miles away. He drove in town at an average of 30 mph , then he drove on the interstate at an average of 70 mph . The trip took three-and-a-half hours total. How far did Bill drive on the interstate?

If you tried to fill in the table the way we did on the last page, you might have noticed a problem: There's too much information. For instance, the problem contains two rates— 30 mph and 70 mph . To include all of this information, let's create a table with an extra row. The top row of numbers and variables will be labeled in town , and the bottom row will be labeled interstate .

We filled in the rates, but what about the distance and time ? If you look back at the problem, you'll see that these are the total figures, meaning they include both the time in town and on the interstate. So the total distance is 225 . This means this is true:

Interstate distance + in-town distance = Total distance

Together, the interstate distance and in-town distance are equal to the total distance. See?

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In any case, we're trying to find out how far Bill drove on the interstate , so let's represent this number with d . If the interstate distance is d , it means the in-town distance is a number that equals the total, 225 , when added to d . In other words, it's equal to 225 - d .

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We can fill in our chart like this:

We can use the same technique to fill in the time column. The total time is 3.5 hours . If we say the time on the interstate is t , then the remaining time in town is equal to 3.5 - t . We can fill in the rest of our chart.

Now we can work on solving the problem. The main difference between the problems on the first page and this problem is that this problem involves two equations. Here's the one for in-town travel :

225 - d = 30 ⋅ (3.5 - t)

And here's the one for interstate travel :

If you tried to solve either of these on its own, you might have found it impossible: since each equation contains two unknown variables, they can't be solved on their own. Try for yourself. If you work either equation on its own, you won't be able to find a numerical value for d . In order to find the value of d , we'll also have to know the value of t .

We can find the value of t in both problems by combining them. Let's take another look at our travel equation for interstate travel.

While we don't know the numerical value of d , this equation does tell us that d is equal to 70 t .

Since 70 t and d are equal , we can replace d with 70 t . Substituting 70 t for d in our equation for interstate travel won't help us find the value of t —all it tells us is that 70 t is equal to itself, which we already knew.

But what about our other equation, the one for in-town travel?

When we replace the d in that equation with 70 t , the equation suddenly gets much easier to solve.

225 - 70t = 30 ⋅ (3.5 - t)

Our new equation might look more complicated, but it's actually something we can solve. This is because it only has one variable: t . Once we find t , we can use it to calculate the value of d —and find the answer to our problem.

To simplify this equation and find the value of t , we'll have to get the t alone on one side of the equals sign. We'll also have to simplify the right side as much as possible.

Let's start with the right side: 30 times (3.5 - t ) is 105 - 30 t .

225 - 70t = 105 - 30t

Next, let's cancel out the 225 next to 70 t . To do this, we'll subtract 225 from both sides. On the right side, it means subtracting 225 from 105 . 105 - 225 is -120 .

- 70t = -120 - 30t

Our next step is to group like terms—remember, our eventual goal is to have t on the left side of the equals sign and a number on the right . We'll cancel out the -30 t on the right side by adding 30 t to both sides. On the right side, we'll add it to -70 t . -70 t + 30 t is -40 t .

- 40t = -120

Finally, to get t on its own, we'll divide each side by its coefficient: -40. -120 / - 40 is 3 .

So t is equal to 3 . In other words, the time Bill traveled on the interstate is equal to 3 hours . Remember, we're ultimately trying to find the distanc e Bill traveled on the interstate. Let's look at the interstate row of our chart again and see if we have enough information to find out.

It looks like we do. Now that we're only missing one variable, we should be able to find its value pretty quickly.

To find the distance, we'll use the travel formula distance = rate ⋅ time .

We now know that Bill traveled on the interstate for 3 hours at 70 mph , so we can fill in this information.

d = 3 ⋅ 70

Finally, we finished simplifying the right side of the equation. 3 ⋅ 70 is 210 .

So d = 210 . We have the answer to our problem! The distance is 210 . In other words, Bill drove 210 miles on the interstate.

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Solving a round-trip problem

It might have seemed like it took a long time to solve the first problem. The more practice you get with these problems, the quicker they'll go. Let's try a similar problem. This one is called a round-trip problem because it describes a round trip—a trip that includes a return journey. Even though the trip described in this problem is slightly different from the one in our first problem, you should be able to solve it the same way. Let's take a look:

Eva drove to work at an average speed of 36 mph. On the way home, she hit traffic and only drove an average of 27 mph. Her total time in the car was 1 hour and 45 minutes, or 1.75 hours. How far does Eva live from work?

If you're having trouble understanding this problem, you might want to visualize Eva's commute like this:

problem solver math book

As always, let's start by filling in a table with the important information. We'll make a row with information about her trip to work and from work .

1.75 - t to describe the trip from work. (Remember, the total travel time is 1.75 hours , so the time to work and from work should equal 1.75 .)

From our table, we can write two equations:

  • The trip to work can be represented as d = 36 t .
  • The trip from work can be represented as d = 27 (1.75 - t ) .

In both equations, d represents the total distance. From the diagram, you can see that these two equations are equal to each other—after all, Eva drives the same distance to and from work .

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Just like with the last problem we solved, we can solve this one by combining the two equations.

We'll start with our equation for the trip from work .

d = 27 (1.75 - t)

Next, we'll substitute in the value of d from our to work equation, d = 36 t . Since the value of d is 36 t , we can replace any occurrence of d with 36 t .

36t = 27 (1.75 - t)

Now, let's simplify the right side. 27 ⋅(1.75 - t ) is 47.25 .

36t = 47.25 - 27t

Next, we'll cancel out -27 t by adding 27 t to both sides of the equation. 36 t + 27 t is 63 t .

63t = 47.25

Finally, we can get t on its own by dividing both sides by its coefficient: 63 . 47.25 / 63 is .75 .

t is equal to .75 . In other words, the time it took Eva to drive to work is .75 hours . Now that we know the value of t , we'll be able to can find the distance to Eva's work.

If you guessed that we were going to use the travel equation again, you were right. We now know the value of two out of the three variables, which means we know enough to solve our problem.

First, let's fill in the values we know. We'll work with the numbers for the trip to work . We already knew the rate : 36 . And we just learned the time : .75 .

d = 36 ⋅ .75

Now all we have to do is simplify the equation: 36 ⋅ .75 = 27 .

d is equal to 27 . In other words, the distance to Eva's work is 27 miles . Our problem is solved.

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Intersecting distance problems

An intersecting distance problem is one where two things are moving toward each other. Here's a typical problem:

Pawnee and Springfield are 420 miles apart. A train leaves Pawnee heading to Springfield at the same time a train leaves Springfield heading to Pawnee. One train is moving at a speed of 45 mph, and the other is moving 60 mph. How long will they travel before they meet?

This problem is asking you to calculate how long it will take these two trains moving toward each other to cross paths. This might seem confusing at first. Even though it's a real-world situation, it can be difficult to imagine distance and motion abstractly. This diagram might help you get a sense of what this situation looks like:

problem solver math book

If you're still confused, don't worry! You can solve this problem the same way you solved the two-part problems on the last page. You'll just need a chart and the travel formula .

Pawnee and Springfield are 420 miles apart. A train leaves Pawnee heading toward Springfield at the same time a train leaves Springfield heading toward Pawnee. One train is moving at a speed of 45 mph , and the other is moving 60 mph . How long will they travel before they meet?

Let's start by filling in our chart. Here's the problem again, this time with the important information underlined. We can start by filling in the most obvious information: rate . The problem gives us the speed of each train. We'll label them fast train and slow train . The fast train goes 60 mph . The slow train goes only 45 mph .

problem solver math book

We can also put this information into a table:

We don't know the distance each train travels to meet the other yet—we just know the total distance. In order to meet, the trains will cover a combined distance equal to the total distance. As you can see in this diagram, this is true no matter how far each train travels.

problem solver math book

This means that—just like last time—we'll represent the distance of one with d and the distance of the other with the total minus d. So the distance for the fast train will be d , and the distance for the slow train will be 420 - d .

Because we're looking for the time both trains travel before they meet, the time will be the same for both trains. We can represent it with t .

The table gives us two equations: d = 60 t and 420 - d = 45 t . Just like we did with the two-part problems, we can combine these two equations.

The equation for the fast train isn't solvable on its own, but it does tell us that d is equal to 60 t .

The other equation, which describes the slow train, can't be solved alone either. However, we can replace the d with its value from the first equation.

420 - d = 45t

Because we know that d is equal to 60 t , we can replace the d in this equation with 60 t . Now we have an equation we can solve.

420 - 60t = 45t

To solve this equation, we'll need to get t and its coefficients on one side of the equals sign and any other numbers on the other. We can start by canceling out the -60 t on the left by adding 60 t to both sides. 45 t + 60 t is 105 t .

Now we just need to get rid of the coefficient next to t . We can do this by dividing both sides by 105 . 420 / 105 is 4 .

t = 4 . In other words, the time it takes the trains to meet is 4 hours . Our problem is solved!

If you want to be sure of your answer, you can check it by using the distance equation with t equal to 4 . For our fast train, the equation would be d = 60 ⋅ 4 . 60 ⋅ 4 is 240 , so the distance our fast train traveled would be 240 miles. For our slow train, the equation would be d = 45 ⋅ 4 . 45 ⋅ 4 is 180 , so the distance traveled by the slow train is 180 miles . Remember how we said the distance the slow train and fast train travel should equal the total distance? 240 miles + 180 miles equals 420 miles , which is the total distance from our problem. Our answer is correct.

Practice problem 1

Here's another intersecting distance problem. It's similar to the one we just solved. See if you can solve it on your own. When you're finished, scroll down to see the answer and an explanation.

Jon and Dani live 270 miles apart. One day, they decided to drive toward each other and hang out wherever they met. Jon drove an average of 65 mph, and Dani drove an average of 70 mph. How long did they drive before they met up?

Problem 1 answer

Here's practice problem 1:

Jon and Dani live 270 miles apart. One day, they decided to drive toward each other and hang out wherever they met. Jon drove an average of 65 mph, and Dani drove 70 mph. How long did they drive before they met up?

Answer: 2 hours .

Let's solve this problem like we solved the others. First, try making the chart. It should look like this:

Here's how we filled in the chart:

  • Distance: Together, Dani and Jon will have covered the total distance between them by the time they meet up. That's 270 . Jon's distance is represented by d , so Dani's distance is 270 - d .
  • Rate: The problem tells us Dani and Jon's speeds. Dani drives 65 mph , and Jon drives 70 mph .
  • Time: Because Jon and Dani drive the same amount of time before they meet up, both of their travel times are represented by t .

Now we have two equations. The equation for Jon's travel is d = 65 t . The equation for Dani's travel is 270 - d = 70 t . To solve this problem, we'll need to combine them.

The equation for Jon tells us that d is equal to 65 t . This means we can combine the two equations by replacing the d in Dani's equation with 65 t .

270 - 65t = 70t

Let's get t on one side of the equation and a number on the other. The first step to doing this is to get rid of -65 t on the left side. We'll cancel it out by adding 65 t to both sides: 70 t + 65 t is 135 t .

All that's left to do is to get rid of the 135 next to the t . We can do this by dividing both sides by 135 : 270 / 135 is 2 .

That's it. t is equal to 2 . We have the answer to our problem: Dani and Jon drove 2 hours before they met up.

Overtaking distance problems

The final type of distance problem we'll discuss in this lesson is a problem in which one moving object overtakes —or passes —another. Here's a typical overtaking problem:

The Hill family and the Platter family are going on a road trip. The Hills left 3 hours before the Platters, but the Platters drive an average of 15 mph faster. If it takes the Platter family 13 hours to catch up with the Hill family, how fast are the Hills driving?

You can picture the moment the Platter family left for the road trip a little like this:

problem solver math book

The problem tells us that the Platter family will catch up with the Hill family in 13 hours and asks us to use this information to find the Hill family's rate . Like some of the other problems we've solved in this lesson, it might not seem like we have enough information to solve this problem—but we do. Let's start making our chart. The distance can be d for both the Hills and the Platters—when the Platters catch up with the Hills, both families will have driven the exact same distance.

Filling in the rate and time will require a little more thought. We don't know the rate for either family—remember, that's what we're trying to find out. However, we do know that the Platters drove 15 mph faster than the Hills. This means if the Hill family's rate is r , the Platter family's rate would be r + 15 .

Now all that's left is the time. We know it took the Platters 13 hours to catch up with the Hills. However, remember that the Hills left 3 hours earlier than the Platters—which means when the Platters caught up, they'd been driving 3 hours more than the Platters. 13 + 3 is 16 , so we know the Hills had been driving 16 hours by the time the Platters caught up with them.

Our chart gives us two equations. The Hill family's trip can be described by d = r ⋅ 16 . The equation for the Platter family's trip is d = ( r + 15) ⋅ 13 . Just like with our other problems, we can combine these equations by replacing a variable in one of them.

The Hill family equation already has the value of d equal to r ⋅ 16. So we'll replace the d in the Platter equation with r ⋅ 16 . This way, it will be an equation we can solve.

r ⋅ 16 = (r + 15) ⋅ 13

First, let's simplify the right side: r ⋅ 16 is 16 r .

16r = (r + 15) ⋅ 13

Next, we'll simplify the right side and multiply ( r + 15) by 13 .

16r = 13r + 195

We can get both r and their coefficients on the left side by subtracting 13 r from 16 r : 16 r - 13 r is 3 r .

Now all that's left to do is get rid of the 3 next to the r . To do this, we'll divide both sides by 3: 195 / 3 is 65 .

So there's our answer: r = 65. The Hill family drove an average of 65 mph .

You can solve any overtaking problem the same way we solved this one. Just remember to pay special attention when you're setting up your chart. Just like the Hill family did in this problem, the person or vehicle who started moving first will always have a greater travel time.

Practice problem 2

Try solving this problem. It's similar to the problem we just solved. When you're finished, scroll down to see the answer and an explanation.

A train moving 60 mph leaves the station at noon. An hour later, a train moving 80 mph leaves heading the same direction on a parallel track. What time does the second train catch up to the first?

Problem 2 answer

Here's practice problem 2:

Answer: 4 p.m.

To solve this problem, start by making a chart. Here's how it should look:

Here's an explanation of the chart:

  • Distance: Both trains will have traveled the same distance by the time the fast train catches up with the slow one, so the distance for both is d .
  • Rate: The problem tells us how fast each train was going. The fast train has a rate of 80 mph , and the slow train has a rate of 60 mph .
  • Time: We'll use t to represent the fast train's travel time before it catches up. Because the slow train started an hour before the fast one, it will have been traveling one hour more by the time the fast train catches up. It's t + 1 .

Now we have two equations. The equation for the fast train is d = 80 t . The equation for the slow train is d = 60 ( t + 1) . To solve this problem, we'll need to combine the equations.

The equation for the fast train says d is equal to 80 t . This means we can combine the two equations by replacing the d in the slow train's equation with 80 t .

80t = 60 (t + 1)

First, let's simplify the right side of the equation: 60 ⋅ ( t + 1) is 60 t + 60 .

80t = 60t + 60

To solve the equation, we'll have to get t on one side of the equals sign and a number on the other. We can get rid of 60 t on the right side by subtracting 60 t from both sides: 80 t - 60 t is 20 t .

Finally, we can get rid of the 20 next to t by dividing both sides by 20 . 60 divided by 20 is 3 .

So t is equal to 3 . The fast train traveled for 3 hours . However, it's not the answer to our problem. Let's look at the original problem again. Pay attention to the last sentence, which is the question we're trying to answer.

Our problem doesn't ask how long either of the trains traveled. It asks what time the second train catches up with the first.

The problem tells us that the slow train left at noon and the fast one left an hour later. This means the fast train left at 1 p.m . From our equations, we know the fast train traveled 3 hours . 1 + 3 is 4 , so the fast train caught up with the slow one at 4 p.m . The answer to the problem is 4 p.m .



Top 10 Problem Solving Books

Gus's Garage

Whenever my son encounters a problem—be it building block pieces that won't fit together the way he wants them to, a door he can't open, or a bucket on the playground his friend won't share—my mom heart immediately leaps to help him. I want to solve his problems for him, to help him be happy and make life easy . . . but the truth I know deep down is that if I always help him, I'm not helping him at all. By allowing him opportunities to problem solve himself when a problem of appropriate difficulty arises, while it may be painful for both of us at the moment, I know he's developing crucial problem-solving skills, and problem-solving is one of those essential skills that, once developed, will serve children their entire lifetime. To help showcase different techniques for problem-solving, and hone metacognition for kids, we've collected here on this list the very best books for teaching problem solving through children's literature! Reading these problem-solving books with your child provides an unparalleled opportunity to have shared references to help you as a team through a learning moment when it arises, plus you'll get to enjoy the bonding moment of reading together! Some books are absolute classics, such as "The Little Mouse, the Red Ripe Strawberry and the Big Hungry Bear," that take a more humorous approach to problem-solving; others are popular titles you may be familiar with that take a more direct approach to flexible thinking techniques, such as New York Times Bestseller "What To Do With a Problem"; and some are hidden gems you may be discovering for the first time. There are books that teach social problem solving, highlight out-of-the-box thinking in innovation, speak to the role of teamwork in overcoming obstacles, and address the very real possibility that problem-solving may be needed to cope with failure at many stages of the process. Because problem solving is important in all of life's stages, this list includes board, picture and chapter books. Board books are best for infants and toddlers. Picture books are excellent for toddlers and also include stories for kindergarten and early elementary students (although we think picture books are great for all ages!). Chapter books are great for elementary- and middle school-age readers. If you know your target age group, feel free to filter to a single category, or just browse the entire list. Without further ado, enjoy this problem-solving list, and let us know what titles you would include!

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Math scores stink in America. Other countries teach it differently - and see higher achievement.

problem solver math book

American students struggle in math. 

The latest results of an international exam given to teenagers ranked the USA ninth in reading and 31st in math literacy out of 79 countries and economies. America has a smaller-than-average share of top-performing math students, and scores have essentially been flat for two decades.

One likely reason: U.S. high schools teach math differently than other countries. 

Classes here often focus on formulas and procedures rather than teaching students to think creatively about solving complex problems involving all sorts of mathematics, experts said. That makes it harder for students to compete globally, be it on an international exam or in colleges and careers that value sophisticated thinking and data science. 

There is a  growing chorus of math experts who recommend ways to bring America’s math curriculum into the 21st century to make it more reflective of what children in higher-performing countries learn. Some schools experiment with ways to make math more exciting, practical and inclusive. 

“There’s a lot of research that shows when you teach math in a different way, kids do better, including on test scores,” said Jo Boaler, a mathematics professor at Stanford University who is behind a major push to remake America's math curriculum.

Standardized tests: How many exams should kids have to take?

Here are some ideas for improving it:

Stop teaching the ‘geometry sandwich’

Most American high schools teach algebra I in ninth grade, geometry in 10th grade and algebra II in 11th grade – something Boaler calls “the geometry sandwich.”

Other countries teach three straight years of integrated math – I, II and III — in which concepts of algebra, geometry, probability, statistics and data science are taught together, allowing students to take deep dives into complex problems.

Geographic disparity: States with the best (and worst) schools

In higher-performing countries, statistics or data science – the computer-based analysis of data, often coupled with coding – is a larger part of the math curriculum, Boaler said. Most American classes focus on teaching rote procedures, she said. 

Next year, Boaler and a research team plan to recommend that California phase out the algebra-geometry pathway in favor of integrated math for all students – something she pitched to education leaders across the state. 

Some states, such as Utah, have made the switch. The Common Core academic standards, a version of which most states adopted, say high school math can be taught in either format.

Does Common Core work?  Despite new standards and more testing, reading and math scores haven't budged in a decade

The move requires extra time and resources to train teachers . Georgia mandated high schools teach integrated math starting in 2008. After pushback from teachers and parents, it gave schools the option to go back to the old sequence in 2016. In one large survey , Georgia teachers said they didn’t want to specialize in more than one math area.

The podcast  Freakonomics  featured an  episode in October about the peculiarities of America’s math curriculum. Hosted by University of Chicago economist Steve Levitt, it highlighted Boaler's work and garnered significant feedback, given the specificity of the topic, Levitt told USA TODAY.

Levitt is engaged in the movement to upend traditional math instruction. He said high schools could consider whittling down the most useful elements of geometry and the second year of algebra into a one-year course. Then students would have more room in their schedules for more applicable math classes.

“When you talk to people within the math education industry, they call that insanely radical,” Levitt said. “I think most parents would not consider it radical to teach only the best of two subjects that most people don’t like that much.”

Make more room for data science

“Ninety percent of the data we have in the world right now was created in the past two years,” Boaler said. “We’re at a point in this world where things are changing, and we need to help students navigate that new world.”

Other countries act more quickly on that idea. Estonia students ranked first among European countries in mathematics, as well as reading and science, on the 2018 Programme for International Student Assessment . Many factors may have helped: The country offers high-quality early childhood education to all kids, class sizes are small, and there's little high-stakes testing, leaving more time for instruction. 

Unlike other countries, Estonia teaches computer programming at all grade levels – a  strategy started in the upper grades in the late '90s and extended to elementary schools around 2012. The country is experimenting with adopting a new computer-based math curriculum .

Computer-based math: What it looks like, and why it's important

In the USA, about 3,300 students this year in 15 Southern California school districts are taking a new Introduction to Data Science course that features data and statistics, real-life data collection and coding to analyze the data. The course was developed by the University of California-Los Angeles and the LA Unified School District, and it counts as a statistics credit.

The class features a scripted curriculum with engaging exercises, such as having students record how much time they spend grooming themselves, then comparing that with national data collected for the American Time Use survey.

Teachers are trained to teach the class, as many haven’t been exposed to programming before, said Suyen Machado, director of the Introduction to Data Science project.

Students who took the new course showed significant growth in their statistical understanding over the year, studies show. Students said they felt learning to code was a valuable skill.

“A lot of students report that they find the content is more applicable to real life,” Machado said. “One of the more challenging things about the course is learning programming. They say it’s hard, but they want to do it.”

Stop splitting up students so much, and don’t hasten the curriculum

Over the years, some schools have sought to raise math achievement by pushing algebra down to eighth grade. High-flying students may adapt and have room to take more advanced high school classes. Hastening the curriculum can widen the gulf in achievement between lower-performing students, including those who are economically disadvantaged and racial minorities.

The practice reflects a long-standing feature of American math education: As early as middle school, students are often split into "tracks"  in ways that predetermine who will take advanced classes in high school. The advanced classes are often full of students who are  white or Asian and attend suburban schools – while black and Latino students continue to be underrepresented, research shows.

About six years ago, San Francisco’s school leaders sought to tackle the problem. They halted teaching algebra I in eighth grade. Students take the same three-year sequence of math courses in middle school, and everyone is enrolled in mixed-ability classrooms, said Lizzy Hull Barnes, math supervisor at the San Francisco Unified School District.

In high school, all students take ninth grade algebra and 10th grade geometry. After that, students can choose their path : Some may pick algebra II, others may choose a course combining algebra II and pre-calculus. Some may accelerate to AP statistics.

Before the change, 40% of graduating seniors in San Francisco had to repeat algebra I in their academic careers. For the Class of 2019, the first cohort of students to follow the new sequence, just 8% of students had to repeat the course.

The changes led to a major increase in disadvantaged students enrolling in higher-level math classes as juniors and seniors, Barnes said. Boosting the success of black and Latino students did not harm the progress of high-achieving white and Asian students. 

“It’s been a seismic shift,” Barnes said.

In New York, an uproar over axing gifted tracks: This school is doing it anyway

Change the way elementary teachers think about math

Improving the math aptitude of older students in the USA is connected to messages students hear about why math is important and who's good at it when they're younger.

Those messages often come from their elementary school teachers, many of whom didn’t like math as students themselves.

"Math phobia is real. Math anxiety is real," said DeAnn Huinker, a professor of mathematics education at the University of Wisconsin-Milwaukee who teaches future elementary and middle school teachers.

New research suggests that when teachers improve their attitude toward math , it can help to raise student test scores. At Stanford, Boaler and her team designed an  online course for teachers  featuring research showing anyone can learn math with enough practice, intelligence isn’t fixed and math is connected to all sorts of everyday activities.

They recruited fifth grade teachers from a county in central California to take and discuss the course. Within a year, the participating teachers' students posted significantly higher state math scores compared with previous years. The jumps were particularly significant for girls and low-income students, Boaler said.

“They thought they had to teach procedures, and then realized they could teach in this open, visual, creative way," Boaler said. "A lot of research studies suggest that it takes a long time for changes to come about. In this one, it was quick.”

Make high school math reflect real life

Beyond data science, some districts design courses that include more real-world math and topics such as financial algebra and mathematical modeling.

The approach has led other countries to success. Teens in the Netherlands post some of the strongest math scores in the world on the PISA assessment. That's largely because the exam prioritizes the application of mathematical concepts to real-life situations, and the Dutch teach math rooted in reality and relevant to society. 

Some longtime Dutch math experts were involved in the design of PISA, which began in 2000 and is given every three years to a sample of 15-year-old students in developed countries and economies.

At Sweetwater High School in Chula Vista, California, math teacher Melody Morris teaches a new 12th grade course that explores topics such as two-player games, graph theory, sequences and series and cryptography. The course, called Discrete Math, was developed through a partnership with San Diego State University.

In one exercise, Morris teaches students to play a capture-the-flag style game featured on the television show "Survivor." They learn that by using math, they can win every time.

'Survivor: Winners at War': Previous champs compete in Season 40

“Their typical response is: ‘This is math?’ ” Morris said. “They think it’s about playing games and having fun. But what they’re really learning is how to break down large problems into small ones and how to make hypotheses and test them.”

Students at Sweetwater still rise up through the traditional “geometry sandwich” from ninth through 11th grades. Morris said many who choose her class senior year find themselves much more engaged in the material. They develop a toolkit that will allow them to approach any problem in life, Morris said.

 “A lot of what we’re building is habits of mind,” she said.

Who's best at tech and engineering?  Girls outperform boys on exam, 'whether they take a class or not'

Education coverage at USA TODAY is made possible in part by a grant from the Bill & Melinda Gates Foundation. The Gates Foundation does not provide editorial input.

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Graduate Physics Problems Books

Need to brush up on my late-undergrad and early-grad physics and was wondering if anyone can recommend books or lecture notes (hard copy, or on-line) that also have solutions.

Two that I have come across are:

Princeton Problems in Physics with Solutions - Nathan Newbury

University of Chicago Graduate Problems in Physics with Solutions - Jeremiah A. Cronin

Spacetime Physics - Taylor & Wheeler (favorite book on special relativity; has a lot of problems with solutions at the back; a lot of the problems really enforce the material and discuss paradoxes)

If possible, please also provide a reason why you like the books as opposed to just listing them.

  • soft-question
  • resource-recommendations
  • 1 $\begingroup$ Take it you've thrown away all your old lecture notes, problem sheets and past papers then? They surely would be the best place to start.. I've given the question +1 all-the-same as I would also be interested in others' suggestions. $\endgroup$ –  qftme May 24, 2011 at 16:39
  • $\begingroup$ Yes (lol). I unfortunately threw away my notes. $\endgroup$ –  Kamil Sindi May 24, 2011 at 17:26
  • 1 $\begingroup$ This post is old, but I want to add my two cents-it matters what you want the material for. If you are just interested in physics then any of the answers below are fine resources. However, if you are practicing for a particular exam, I cant suggest these sorts of resources as the primary study tool since you are just going to spend huge amounts of time on problems that might not have a significant impact on your exam performance. That is, if you are studying for a qualification exam, its in your best interest to completely exhaust all old exams before moving onto resources like ones below. $\endgroup$ –  DJBunk Nov 14, 2012 at 21:04
  • $\begingroup$ Roger Blandford & Kip Thorne, Modern Classical Physics: Optics, Fluids, Plasmas, Elasticity, Relativity, and Statistical Physics. Lecture notes: Ph 136: Applications of classical physics, cns.gatech.edu/PHYS-4421/caltech136/index.html $\endgroup$ –  Qmechanic ♦ Nov 27, 2018 at 12:38

11 Answers 11

(I have a suggestion to make this question a CW.)

General Physics : (Early undergrad and advanced high school)

  • Problems in Physics I.E Irodov - (Highly recommended)
  • Problems in Physics S S Krotov - (Once again, highly recommended but out of print)
  • Physics Olympiad Books - (Haven't read but saw some olympiad problems back in the day)
  • Physics by example (like this book a lot, lower undergrad)
  • Feynman's Tips on Physics (Exercises to accompany the famous lectures )

General Qualifying exam books: The following books are a part of a series dedicated to the qualifying exams in American Universities and has a large compilation of problems of all levels. Others in the series include Mechanics , Electromagnetism , Quantum Mechanics , Thermodynamics , Optics and Solid State Physics . Unlike other compilation of exercises for qualifiers (such as Princeton or Chicago Problems, or the one mentioned below), they make no excuse for economy and include as many problems from all levels for each subtopic.

Another good book that I read recently for my exam is the two volume series: A Guide to Physics Problems (Part 2 has some relatively easy but interesting problems. I haven't gone through the first part, which is much much more challenging.)

Thinking Like a Physicist: Physics Problems for Undergrads : I love this book because it fosters a real sense of physical understanding, so it's not just mathematics, but actual physical reasoning. Plus, I found the problems challenging and interesting.

Then, there is always the MIT Open Course Ware in Phsyics , which has undergrad and graduate courses with assignments, lecture notes, tests, problems, solutions, etc.

If its qualifying exam questions and problems that you're after, a lot of universities will post examples of past versions online, its just a matter of looking at the department website hard enough.

  • $\begingroup$ +1 @Jen: Thanks. If you know any particularly good qualifying exam questions with answers, do you mind posting them? $\endgroup$ –  Kamil Sindi May 24, 2011 at 20:06
  • $\begingroup$ It's pretty google-able - the first page of results for "Physics qualifying exams with answers" turn up a number of past exams and solution banks. $\endgroup$ –  Jen May 24, 2011 at 20:19
  • $\begingroup$ Also, the GRE subject test in physics is pretty decent for quickly getting up to speed. There are practice tests available: ets.org/gre/subject/about/content/physics $\endgroup$ –  Jen May 24, 2011 at 20:22

Some review/problem set books that I like are:

Solid State Physics: problems and solutions - Laszalo Mihaly and Michael C. Martin

Problems in Quantum Mechanics with Solutions - Gordon Leslie Squires (The reviews are bare bones but I find the questions to be very good at making you think)

Hope this helps. I am interested to see what other people come up with.

"Problem Book on Relativity and Gravitation" - A. Lightman, W. Press, R. Price, S. Teukolsky

The two you have were my favorites.

For completeness, a couple of others, which also have worked solutions:

L.A. Sena, A Collection of Questions and Problems in Physics

Constantinescu, Problems in Quantum Mechanics

There are many problems and solutions sets available for Caltech's comprehensive "Applications of Classical Physics" course here: http://www.pma.caltech.edu/Courses/ph136/yr2008/

It might be worth it to look at some past problems from physics olympiads. In particular:

International Physics Olympiad website http://www.jyu.fi/ipho/ for high-school students, all problems with solutions. Used to be tricky problems, not its more step-by-step, but still challenging and interesting

Moscow City Physics Olympiad http://olympiads.mccme.ru/mfo/ there is a pdf in russian, but I managed to understand almost every single problem with google translator

And of course BAUPC http://liquids.seas.harvard.edu/oleg/competition/ which has few, but (in my opinion) wonderful problems. Some of them are repeated in the classical mechanics book by David Morin ( http://www.amazon.com/Introduction-Classical-Mechanics-Problems-Solutions/dp/0521876222 )

I'll add absolutely great book with problems in analytical mechanics: G.L. Kotkin and V.G. Serbo , "Problems in Classical Mechanics" It is basically a problem-driven approach to learning Landau's first tome.

Also I really like the book: Zhong-Qi Ma and Xiao-Yan Gu , "Problems & Solutions in Group theory for physicists."

Problems and Solutions on Electromagnetism (Major American Universities Ph.D. by Yung-Kuo Lim

Problems and Solutions on Mechanics: Major American Universities Ph.D. by Yung-Kuo Lim

Problems and Solutions on Thermodynamics (Major American Universities Ph.D. by U.S.T. of China Physics Coaching Class and Yung-Kuo Lim

I think the Y K Lim series are the best to follow, especially for last couple week of preparations. These books have all sort of good problems (with solutions) collected from reputed US universities.

You also can dig the internet for various universities solved exams but that is quite time consuming.

Some problem books on quantum field theory:

T.P. Cheng, L.F. Li, "Gauge Theories of Elementary Particle Physics: Problems and Solutions" , Oxford University Press (2000)

V. Radovanovi, "Problem Book in Quantum Field Theory" , Springer (2008).

  • D. Atkinson, P.W. Johnson, "Exercises in Quantum Field Theory: A Self-Contained Book of Questions and Answers" , Rinton Press (2003).

More can be found in this page: http://physics.library.wisc.edu/practicebooks.html

One of the best books for IIT mains and advance exams 2013 is Advanced Problems in School Physics Vol-I & II Cengage Learning

  • $\begingroup$ Hi Arvind. Welcome to Physics.SE. Whilst this may theoretically answer the question, it would be preferable to include the essential parts of the answer here, and provide the link for reference. $\endgroup$ –  Waffle's Crazy Peanut Feb 10, 2013 at 12:14

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