Algebra Word Problems: Learning from Mistakes

A recent series of questions from an insightful high school student about word problems, provided a number of opportunities to discuss how to find and correct your mistakes – or the book’s! We’ll look at five.

1: Averages; failing to rethink the plan

Here’s the first question (all are from Ivka):

Could you please help me solve this problem? I believe the solution might be incorrect. I solved it a few times but kept getting the same answer. The givens are:

There are eight students in the class. Seven of the eight grades are:

71, 76, 82, 87, 89, 92, 95.

If the class average of all eight students is 82, and the student with the lowest grade drops the class, what is the average of the remaining students? (Calculate to the nearest tenth.)

My solution is:

mean = 82 = (71 + 76 + 82 + 87 + 89 + 92 + 95 + x)/8 = 64

new mean = (64 + 76 + 82 + 87 + 89 + 92 + 95)/7 = 83.6

According to the solution, the correct answer is 84.6. Could you explain why I am not arriving at the answer?

Thank you very much.

What she wrote isn’t quite what she meant; the 64 at the end of the first line of work is not equal to the expression before it, but is the solution for x, which comes from simplifying the equation to \((592+x)/8=82\) and solving to get \(592+x=656\), and then \(x=64\). Then Ivka put that in place of x, dropped the lowest grade (71), and found the new average. Do you see the error?

I answered:

Hi, Ivka.

Your work is good until the last step.

You found that the 8th student has a grade of 64, which is correct; but then you put that grade in place of the lowest of the initially given grades, 71.

You should be dropping the lowest grade in the class, which is 64, not 71! This will give you the expected answer.

I suspect that you may have decided that 71 would be dropped before putting the 64 into the list, and didn’t modify your opinion.

This is a nice example where planning ahead (too much) can be a problem: Don’t decide what the lowest grade is until you know all of them!

2: Coins; interpreting tricky grammar

Here’s the next problem:

I am supposed to solve the word problem on coins. However, it’s missing an important piece of information. Here are the givens:

Brian paid for an item at the store and received eleven dollars in change. He received an equal number of dimes and nickels and two fewer than four times the number of nickels in quarters. How many nickels, dimes and quarters did he receive?

What does the problem mean when it says: received ‘two fewer than’?  Does it mean that two fewer dimes and nickels were received altogether than four times the number of nickels and quarters?

I answered:

Hi, Ivka.

The problem isn’t faulty, but it is very easy to misread statements like these, with somewhat complicated grammar. Let’s break it apart:

Brian paid for an item at the store and received eleven dollars in change.

He received an equal number of dimes and nickels

and two fewer than four times the number of nickels in quarters.

How many nickels, dimes and quarters did he receive?

We are told that the numbers of dimes and nickels are equal:

D = N

Then we are told that the number of quarters is 2 less than 4 times the number of nickels:

Q = 4N – 2

That last phrase, “in quarters” is a little awkward, and easy to miss.

Together with

25Q + 10D + 5N = 1100 (in terms of cents),

This is enough to solve the problem.

What I did here is first to break the problem into its separate statements, and then rephrase each into a more straightforward statement about how the different quantities are related, before writing equations. The tricky part, “two fewer than four times the number of nickels in quarters”, had to be straightened out by bringing “the number of quarters is …” before the calculation part. Even so, the phrase “in quarters” seems a little odd to me, as it is more typically used of a value (“I have two dollars in quarters“). So I think the problem is poorly phrased — and if someone said this to me face to face, I would probably ask them to rephrase it themselves!

When the grammar gets complicated, I often have to rewrite it, making little changes to make sure it still means the same thing, to smooth out the relationships. We touched on that in A Fraction Word Problem – Algebra or Not?

Does that help?

I wasn’t sure what algebraic methods were in Ivka’s toolbox, so I gave three equations in three variables as a way to display the meaning of the problem without implying a way to solve it.

Ivka replied, showing her work:

Dear Dr. Peterson,

With your help, I arrived to (hopefully) the right answers. I solved the problem this way:

n … x

d … x

q … y = 4(x) – 2

Equation 1:

0.05(x) + 0.10(x) + 0.25(y) = $11

0.15x + 0.25y = $11

Now I substituted y = 4x – 2, into equation 1:

0.15x + 0.25(4x – 2) = $11

0.15x + x – 0.5 = $11

1.15x – 0.5 = $11

1.15x = $11 + 0.50

1.15x = $11.50

x = 10


n = x = 10 nickels

d = x = 10 dimes

q = y = 4x – 2 = 4(10) – 2 = 40 – 2 = 38 quarters


10 nickels = 0.05(10) = $0.50

10 dimes = 0.10(10) = $1

38 quarters = 0.25(38) = $9.50

Together, they add up to $11, which matches the given of the problem that all coins have a value of $11.

That was written up very clearly! She made a “problem-to-algebra dictionary”, showing a variable or expression for each quantity in the problem, then turned my ideas into a single equation in one variable. The work is good, including the check. The only thing I’d do differently is to omit the dollar sign in the work, because the entire equation is about numbers of dollars.

I concluded:

Good work. (And I like that you didn’t just take it from where I left it, but backed up and did it all your own way, so I know you’re thinking well.)

I often tell students, the hardest part of math is often the English!

Ivka closed:

Thank you so much! I love your website and service. It’s so supportive!

3: More coins; catching your error

Shortly after that, she sent this similar problem:

I solved a new word problem involving coins. After checking the solution, my answers for dimes, nickels and quarters vary.

Could you let me know where I made a mistake? I cannot find it!

Thank you!

The givens are:

Stacey paid for an item at the store and received two dollars in change. She received three fewer dimes than nickels and two fewer quarters than the number of nickels. How many nickels, dimes and quarters did she receive?

My reasoning is:

d … x – 3

n … x

q … x – 2


0.10(x – 3) + 0.05(x) + 0.25(x – 2) = $2

0.1x – 0.3x + 0.05x + 0.25x – 0.5 = $2

-0.2x + 0.3x – 0.5 =$2

0.1x = $2 + 0.50

0.1x = $2.50

x = 25


d = x – 3 = 25 – 3 = 22 dimes

n = x = 25 nickels

q = x – 2 = 25 – 2 = 23 quarters

So this is all I could come up with. However, their answers are: 7 nickels, 4 dimes, 5 quarters.

Have you found the error? These coins will add up to far more than $2 (namely, $9.20).

Ten minutes later, she added this:

My apologies. I figured out what I did wrong.

When solving the equation, I made a mistake when cross-multiplying 0.1(x – 3). I said 0.1 times 3 is 0.3x, which I corrected to 0.3!!!

I am getting the correct number of dimes, nickels, and quarters.

Thank you for your help (for making me rethink the problem all over again!).

Yes, sometimes just trying to explain your work to someone else reveals your error!

Ivka distributed incorrectly, going from \(0.10(x-3)\) to \(0.1x-0.3x\). Here is the correct work: $$0.10(x-3)+0.05(x)+0.25(x-2)=2\\0.10x-0.3+0.05x+0.25x-0.5=2\\0.4x-0.8=2\\0.4x=2.8\\x=7\\d=7-3=4\\q=7-2=5$$

Doctor Rick answered with some things to learn from this:

Hi, Ivka.

Good work in catching your error! All of us make errors like this from time to time; you demonstrate the importance of checking your solution as part of the work.

If you had made some other little error (like I did when I tried to do it in my head), you might get something like

0.4x – 0.75 = 2

x = 6.875

At this point you would think: That can’t be right! The number of nickels should be a whole number. It just happened that the incorrect answer you got was a whole number, so you had to do the full check to see that your solution was incorrect.

When I did this problem myself, I actually wrote the equation in cents rather than dollars:

10(x – 3) + 5x + 25(x – 2) = 200

This way, I am only working with whole numbers, rather than decimals. This reduces the chance for errors, as we generally find integers easier to work with. What you did is perfectly valid; this is just a suggestion.

I demonstrated using cents rather than dollars in my explanation of the previous problem.

Ivka’s error was not directly due to missing a decimal point, but could have indirectly resulted from the increased complexity of the equation.

Ivka replied:

Thank you for explaining why I should use integers when setting up equations. The reason I set up the equation in dollars is because in the past, when I was learning the word problems for the first time, I noticed that when a solution is given in dollars, I set up my equation in dollars so I do not need to convert to cents later. But in this case, I could have solved it with integers to find my error faster.

Yes, this is a common trade-off, between simplifying the work of solving, and having to remember to change the total to cents as well (or, in other cases, to do an extra step at the end).

Doctor Rick confirmed:

As I said, there is nothing wrong with what you did! Sometimes it may be easier to work with integers, but that is a personal preference. It’s just good to be aware that you have options.

4: Speed and distance; ambiguous wording

Three days later, Ivka was on to a new type of problem:

Good evening,

I tried calculating this problem in three different ways. Each way I arrived to the same answer except my answer does not match the textbook’s answer. Could you help me figure out why, please?

Suppose you leave your house by bicycle and travel 10 miles per hour. One hour later, your brother leaves and travels down the same road by car at 45 miles per hour. How long, in hours to the nearest tenth, does it take for the car to overtake the bicycle?

Method 1:

10(x + 1) = 45x

x = 0.29 hrs

Method 2:

t_a – t_b = 1 hour

(D/v_a) – (D/v_b) = 1

D/10 – D/45 = 1

D = 12.86 miles

t_b = D/v_b = 12.86/45 = 0.29 hours

Method 3:

v_a (t_a_first hour) + v_aT = v_bT

10(1) +10T = 45T

T = 0.29 hours

Thank you for your help.

The textbook’s possible listed answer choices to choose from are 1.1, 1.2, 1.3, and 1.4. Mine is not listed.

Method 1 takes x as the time the car drives, and sets the two distances equal.

Method 2 takes D as the distance both go, and makes the times for the two people differ by 1 hour; then she uses the distance to find how long the car drives.

Method 3 takes T as the time the car drives, and sets the bike’s distance for the first hour, plus its distance while the car drives, equal to the distance the car drives.

All are excellent methods, and all give the same (correct!) answer. Why does the book not even list 0.3 hours (rounded) as an option they expect students to get by mistake?

Doctor Rick answered:

Hi, Ivka.

Your work looks good, and you’ve got two different approaches (the first and third being rather similar) that give the same answer. What went wrong?

I think that the problem is in how to interpret the question, “how long … does it take for the car to overtake the bicycle?”.

You interpreted that question the same way I would: “What is the length of time that the car must drive?” In other words, “How much time passes from the time the car starts to when it overtakes the bicycle?”

However, looking at the book’s options, I think the writer of the problem meant, “How much time passes from the time the bicycle started?” That is, what is the total amount of time for everything that happens?

The question is not clearly stated; your interpretation is the one that makes the most sense to me, but the other interpretation is also reasonable. That’s my best idea as to why your answer isn’t what the author had in mind.

We take the question as, “How long does the car take to overtake the bike?”; the book takes it to mean, “How long does it take for the bicycle to be overtaken?”

Since the bike was ridden for one hour longer, this makes sense; their answer is \(0.3+1=1.3\) hours.

What the book said

Ivka replied:

Good morning, Dr. Rick,

I see what you are saying. I watched the videos on a website, solving exact same problems and solved this problem the same way.

I am sorry, I do not see why I would calculate the time when they meet (when the car catches up to the bike) starting with the time the biker starts riding. I am just not following that idea. Could you elaborate, please?

Here is the solution they gave me:

When the car overtakes the bicycle, the distances each travel are the same. Thus we can use the formula, where D is the distance, t is the time, b is the bicycle, and c is the car: D = sb(t) = sc(t − 1).

Substitute the numbers from the problem into the formula:

10t = 45(t − 1)

10t = 45t − 45

−35t = −45

t = 1.285714 ≈ 1.3

Answer: 1.3 hours

Doctor Rick responded:

Thanks for all this.

An important part of problem solving is to define the variable(s) precisely. This was not done in the solution you show; it just says “t is the time”. Exactly which time is it?

From the work, I can see that t is the total time that the bicycle is moving, because 10t clearly comes from speed × time = distance for the bicycle. The car is moving for one hour less time, or (t – 1) hours. The solution should have made this clear.

This is not a matter of math, and it’s not something you did wrong. It’s a matter of ambiguity in the English statement of the problem.

Just as the solution should define the variable t clearly, so also the problem should have likewise made clear what time it was asking for. The solution assumes it was the total time, but we couldn’t know that just from the wording of the problem.

Evidently the people who wrote the problem and the solution think that everything is clear. If you continue working with this material, now you know something of their thinking, which may help you to interpret the problems as they intend. You will not necessarily find that problems from other sources should be interpreted in the same way.

I generally find that authors interpret this sort of wording the same way Doctor Rick (and I) do. But sometimes they may not!

Our general advice when there is any ambiguity in a problem is to clearly state your interpretation, so the grader knows what problem you are solving. The habits of defining your variables, and of stating the answer in a full sentence (“The car takes 0.3 hours to catch up to the bike”) can do this without any special planning.

5: Speed; error in the book

The last problem we’ll look at today is similar, and was asked at the same time as the last:

Here is the third problem that I am not getting the answer to:

Two people leave from two towns that are 240 miles apart and travel toward each other along the same road. Person “A” drives 12 miles/hour (mph) slower than person “B”. If they meet in 5 hours, at what speed did each person travel?

I did this:

d_a + d_b = 240 miles

((x – 12) miles /1 hour) * 5 hours + ((x miles)/1hour) * 5 hours) = 240 miles

(x – 12)5 + 5x = 240

x = 30 mi/hour = v_b

Thus, v_a = 30 – 12 = 18 miles/hour

The textbook’s answer choices are: 30 and 42; 28 and 46; 32 and 44; 25 and 37.

Thank you again for your help!

Doctor Rick answered this, too:

This time the problem, options, and your work all match up reasonably. You can see that one of the options is most like yours: option (a) has one car at a speed of 30 mph, but the other is 12 mph faster than this, rather than slower.

You would do well to write down explicitly what your variable “x” stands for, but I can see that it is the speed of car B, the faster car. Your work is correct. It’s easy to imagine a student calculating x as you did, but then thinking that x is the speed of car A; that student would give answer (a).

I can only wonder if whoever wrote the answer choices made exactly this mistake! What is the source – is it from a reputable textbook? (Even those do have the occasional error, especially in problems.) Can you ask a teacher about errors in the textbook?

We can check your answer. I person “A” drives at 18 miles/hour for 5 hours, and person “B” drives at 30 miles/hour for 5 hours, then the total distance they drove is 5(18 + 30) = 240 miles. That’s correct.

When your answer passes the check, and “their” answer does not, you can move on, knowing you were right.

Ivka replied:

Dear Dr. Rick,

Thank you very much for checking my answer!  Yes, I looked at the solution and it gives 18 and 30 miles/hour. The correct answer choice was unlisted! Here is their solution:

Two people leave from two towns that are 240 miles apart and travel toward each other along the same road. Person “A” drives 12 miles/hour (mph) slower than person “B”. If they meet in 5 hours, at what speed did each person travel?

Using the variable x as the speed of the faster person, we can write and solve the following equation:

240 = (x − 12)(5) + (x)(5)

240 = 5x − 60 + 5x

240 = 10x − 60

300 = 10x

x = 30

Answer: Person A’s speed: 18 mph; Person B’s speed: 30 mph

So they give the right answer; the only error is in the list of choices.

Doctor Rick responded:

In this solution, the variable x is defined as “the speed of the faster person”, who is Person “B”. That’s good.

If the options listed in the problem did not include the answer that was provided by this same source, that is not good! It makes me reluctant to trust the source.

At any rate, you know now not to be quite so worried when you get an answer that isn’t listed. You need to check the solutions. If you got a different answer, look in the solution for how they are interpreting the problem, not just how they solve it.

If this should happen on a test, you would be justified in (a) raising your hand to ask about the problem, and (b) objecting if your answer is marked wrong.

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