[This article begins a series rescued from my old blog. Moving has been a long process, but I’m finally unpacking the last cardboard box! To read the entire series, click here: elementary problem solving series.]
Most young students solve story problems by the flash of insight method: When they read the problem, they know almost instinctively how to solve it. This is fine for problems like:
There are 7 children. 2 of them are girls. How many boys are there?
As problems get more difficult, however, that flash of insight becomes less reliable, so we find our students staring blankly at their paper or out the window. They complain, “I don’t know what to do. It’s too hard!”
We need to give our students a tool that will help them when insight fails.
Three Steps to Solving Word Problems
In the process of solving a word problem, the student must work his way through three steps:
- Translate the words into a mathematical calculation or algebraic equation.
- Do the calculation or solve the equation.
- Interpret the resulting number in the context of the original problem.
When our students struggle with solving problems, most of the time it is step one that gives them trouble. They do not know how to translate the problem from English into “mathish.” If we want to help our students with their math problems, we need to teach them how to do this sort of translation.
One approach teaches key or signal words. For instance, we can tell our students that a problem asking, “How many more…?” is probably going to require subtraction. The question asks for the difference between two quantities, and difference is the answer when you subtract. Unfortunately, this technique only works for the most simplistic story problems, and some textbook (or test) writers enjoy using key words in ways that will mislead a careless student.
If you would like to teach key words to your students, here is a list of the most useful ones:
I do teach a few key words to my students. I particularly like the translation “of” = “multiply” when dealing with fraction and percent problems. But I also want my students to read a math problem and analyze what is happening, no matter what words are used to describe the situation. For this, they need a more powerful tool than key words.
Tool #1: Algebra with Words
Advantages:
- efficient, getting to an answer quickly
- flexible, applicable to almost any situation
- useful in higher mathematics
Disadvantages:
- abstract, thus difficult for children to understand at a glance
- abstraction may lead to mental block for some students
- takes plenty of practice to learn to use well
We can introduce young children to algebra by using whole words, not letter variables. For instance, for the problem in the first paragraph above, we might write:
Children = 7
Girls = 2
Boys = Children – Girls = ?
With older children, we proceed to using initials, but it is important for the student’s understanding that we still say the whole word as we write:
C = 7
G = 2
B = C – G = ?
This is an easy approach when the relationships are simple, as in this situation. As the relationships between quantities in the problem get more complex, the algebraic equations must get correspondingly more complex. Our students must learn to use some basic but important rules to use as they work through their equations:
The Rule of Inverse Operations
If a mathematical operation has been done on some quantity, you can “undo” it by using the inverse of that operation. For example, if you know that N + 5 = 8, then you can “undo” the addition by subtracting 5.
The Balance Rule
Whatever you want to do to one side of the equation, you must also do the same thing to the other side, to keep the equation in balance. So in the example N + 5 = 8, if you want to subtract 5 from the “N + 5” side of the equal sign, you must also subtract 5 from the 8.
The Rule of Substitution
If one quantity is equal to another quantity, then it may be substituted for the other quantity in any equation. Thus, if you have determined that N = 3 in this particular problem, you can then use that information in any other equation within the same problem.
The Rule of Parentheses (Distributive Property)
If you add (or subtract, or multiply) a group of quantities, usually set apart from the rest of the equation by parentheses, you must add (or subtract, or multiply) everything within that group, one at a time. And you must do this very carefully, especially when subtracting, because this is one of the easiest places to make a mistake in algebra.
Tool #2: Bar Diagrams
Advantages:
- pictorial representation of abstract relationships
- fewer (and less abstract) rules to learn
- intuitive, easy to understand by students who have played with blocks
- flexible, applying in many situations
- diagrams eventually serve as a lead-in to algebra, giving students a way to visualize equations
Disadvantages:
- pictorial approach may not be intuitive to some students (or teachers)
- may not work in every situation
- takes plenty of practice to learn to use well
In bar diagrams, quantities (both known and unknown) are represented by block-like rectangles. The student imagines moving these blocks around or cutting them into smaller pieces in order to find a useful relationship between the known and unknown quantities. In this way, the abstract puzzle of the word problem becomes a 2-D visualization puzzle: How can we fit these blocks together? For many students, this pictorial approach makes the problem much easier to understand.
In the above problem, we would draw a rectangular bar to represent all the children. Then we would divide it into two parts, representing the boys and the girls:
As problems get more complex, the bar may be split into more than two parts. Also, the parts may be related to each other in ways that require a more involved diagram. However complicated the story, though, you usually begin by drawing a long bar to represent one whole thing and then dividing it into parts.
Again, the student must learn some basic but important rules:
The Whole Is the Sum of Its Parts
All bar diagrams descend from one very basic diagram showing the inverse relationship between addition and subtraction: The whole is the sum of its parts. If you know the value of both parts, you can add them up to get the whole. If you know the whole total and one of the parts, you subtract the part you know in order to find the other part.
In a picture:
Simplify to a Single Unknown Part (called a “unit”)
You cannot solve for two unknown numbers at once, so you must use the facts given in your problem and manipulate the blocks in your drawing until you can connect one unknown unit (or a group of same-size units) to a number. Once you find that single unknown unit, all the other quantities in your problem will fall into place.
Which Tool Should We Teach?
Because algebra is flexible and efficient in almost any problem-solving situation, it is our ultimate goal. We want all our students to learn to use it well. A friend has argued (on a now-defunct homeschooling forum) that this goal is best served by teaching algebra early and using it often. On the other hand, my students have done well with the Singapore Math approach, using bar diagrams in elementary school and making the transition to algebra in junior high.
I am not sure which of these tools will work best with your student, but I suggest you choose one of them and stick with it. Use the tool with your child over and over, until it becomes almost automatic. Start with simple story problems that are easy enough to solve with a flash of insight. Discuss how your chosen tool can model the situations, translating the English of the stories into math calculations or algebra equations—but don’t force your child to fully solve each problem. Just practice the tough part: the translation. Gradually work your way up to more challenging problems. If your math program doesn’t give your student enough problems to practice on, try the Challenging Word Problems series from Singapore Math.
In future posts, I plan to discuss typical word problems at different elementary grade levels and show how these problems can be translated into algebra and bar diagrams. I hope that seeing the two tools in action will help you decide which you prefer.
Denise Gaskins' Let's Play Math 
I have what seems to be a very straight forward math question, but it has haunted me for years. This is a question one of my daughters had a homework years back when in grade 4 or 5. It has “haunted” me, stuck in my mind so clearly because it was the very first time I was unable to help. Can you explain it to me.
There are 3 traveling salesmen, traveling together to a convention. It is late at night and they are very tired and decide to stop at a country inn for a few hours sleep. The inn only has one room left with a double bed and an couch. salesmen are tired and take the room as a few hours sleep is all they want. The room is $30.00, each of the 3 contribute $10.00.
In the morning the manager of the inn arrives. He feels bad about the uncomfortable room provided these 3 men and hoping to attract them back again decides to give them a partial refund. He gives the bell boy $5.00 to take to the room and give to these salesmen.
The bell boy realizes the men are not expecting any sort of refund. So he decides to give them back only $3.00 and keeps $2.00 for himself.
He gives the 3 men the $3.00 refund, each man takes a dollar. So as each of the men had originally paid $10.00, but as each had received a dollar back. It ended up costing each man $9.00. They are happy with this and the bell boy is happy as he has $2.00 in his pocket.
Question: each of the 3 men ended up paying $9.00. 3X9=27+2(money in bellboys pocket)=29. We started with $30.00 what has happened to the extra $1.00. All my numbers are correct, why doesn’t this math work?
I know it is easy to say just work the equation back ie. 29-2=27+3 and we are back to the 30. But why doesn’t the math work in the equation as I presented it.
I hope I have explained it in such a way as you understand my question. I hope you can clear up this 25 year old question for me
This is a mathematical sleight-of-hand problem. You know how a stage magician waves one hand while talking fast, so that you fail to notice what he is doing with the other hand. In the same way, the writer of this problem waves numbers around without meaning. The individual numbers are not difficult, but the barrage of them distracts the reader so that you fail to notice what has really happened.
Each man pays a net $9 for the room. Total payment = 3x$9 =$27.
The manager gets $25 = the original $30 payment, minus a $5 refund.
The bellboy takes for himself a tip of $2.
True equation: 3 x 9 = 25 + 2. What was paid equals what was received.
Sorry, I don’t mean to be a bother over something so trivial, but I still don’t understand. We start with $30.00, all of the transactions take place and when the dust settles. We still have 3X9=27+2=29 all are true and the math is correct. How is it we don’t end up back at 30 the number from which we started?
Hope I am not being a bother, I would just like to get it.
You are not being a bother, Bill, but you have been dazzled by the stage magician. You are trying to add the bellboy’s tip to the amount the men paid for the room, as if the tip came out of nowhere. The tip is part of what the men paid—it is already included in that $27, so when you add 27 + 2, it is like trying to pay the tip twice. Any number you get from that (in this case, the number 29) has nothing to do with what really happened in the story.
Let’s look at exactly where the money went:
$30 original payment
-$5 refund
+$2 tip confiscated by bellboy without permission
=$27 net payment for the room.
Thanks for this post. I teach elementary math to 9th graders, and I decided to start using the Singapore bar-model method with them this fall (I just posted about this). We’ve tried different approaches in the past – mainly focused on getting them to translate the problems into algebraic sentences, as you discuss in your first option. But this doesn’t work for many of our students. My gut tells me that the visual approach of the bar model will help them to make more sense out of the problems, especially when we work on ratios and proportions. I hope it works! Anyway, I’m interested in your further posts on this topic. It would be great if you could also talk about the common problems students have with bar modeling, and how you help them overcome them.
I will be interested to follow how your class works out, Dan. The main problem that I have had with trying to introduce bar diagrams to older students is that I want to move them into multi-step problems too quickly. They don’t want to draw the simple addition/subtraction bars, and I think the simple bars are so… well… simple, that it ought to be intuitive, so I don’t push them to do enough practice at that very basic level. But with my homeschooled son, that repeated practice is exactly what gave him enough familiarity with the bars as a thinking tool that he is able now to use them whenever he needs to work through a problem.
On thing I did for my son to make all that practice more palatable was that he didn’t have to finish every problem. He drew the bars and told me how to find the answer, and then he was allowed to skip the actual calculation. We went through all of the 2nd grade Challenging Word Problems book that way, and most of the way through the 3rd grade book, too. It was only when I knew he could model almost any problem that I started requiring him to calculate the answers—and at that point, I quit requiring the drawings. Now he uses the bars only if they help him think things through, which means mostly on ratio/proportion problems.
hi really like this site cause people learn on it, like me im really interested in math. and i want to know more in algebra. tanx to this site and people beyond this.thank you very much.
If you are interested in math, there is plenty of help on the internet. For instance, I have included a list of algebra tutorials and resources at the bottom of my The game of algebra post. And there are plenty of great links on my resource page.
The best way I know to learn math is to play around with it. Enjoy!
I understood Algebra when I actually began to use it. The problem is that my children are now learning things I have never heard of in Algebra. Tonight, my son is to explain the Singapore Box Model to me and I won’t have a clue if he actually understands it himself. Please help!?
Have you read through the examples in Pre-algebra problem solving: 2nd grade and Pre-algebra problem solving: 3rd grade? Those should give you a pretty good overview of how simple problems are solved with bar models. For more help — and especially if you want help with a particular problem — you might try the Singapore math Yahoo group.
i am also encouraged by this interactions to share my insight.. i hope i add something to your analysis. First i break down 3o in terms of the refund and 25 which is paid by the three costumers. 30= 25 + 5. then 30 = 25+3 (refunded by the bell boy)+ 2 (kept by the bellboy) so all is well. then to verify the three costumers actually paid 3(9) = 27 + the 3(refund) and that’s 30.
Bill, the trick lies in the 2nd line of the question (“3X9=27+2(money in bellboys pocket)=29”), where 2 is swiftly added to 27 without any explanation and then the sum is claimed to be equal to ‘the amount previously paid to innkeeper’.
The relationship should be like this
net payment by travellers(27) = amount in bellboy’s pocket(2) + net amount paid to innkeeper(25)
so 2 should be deducted from 27 rather than adding if you want to get any entity that makes sense(in this case ‘net amount paid to innkeeper’).
Otherwise it will just lead to a meaningless entity –> net amount paid to innkeeper+ twice the amount in bellboy’s pocket = 25+4=29 which is nowhere near to the entity ‘amount previously paid to innkeeper'(30)
Original Amount paid By each person=30/3=$10
After refunding, Manager got $25.Therefore each person gave=$(25/3)
Each person gets back $1(=3/3)
Total Money(spent+got Back)of each person =(25/3)+(3/3)=$(28/3)——–1st equation
Now,The bell boy has $2
If it was given back too,Each person gets=$(2/3),right?
So total money cycled=1st equation+(2/3)
=(28/3)+(2/3)
=30/3=$10 each.
Hence Proved………….!!