Ben Franklin Math: Elementary Problem Solving 3rd Grade

The ability to solve word problems ranks high on any math teacher’s list of goals. How can I teach my students to solve math problems? I must help them develop the ability to translate “real world” situations into mathematical language.

In two previous posts, I introduced the problem-solving tools algebra and bar diagrams. These tools help our students organize the information in a word problem and translate it into a mathematical calculation.

Working Math Problems with Poor Richard

This time I will demonstrate these problem-solving tools in action with a series of 3rd-grade problems based on the Singapore Primary Math series, level 3A. For your reading pleasure, I have translated the problems into the life of Ben Franklin, inspired by the biography Poor Richard by James Daugherty.

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Boston, 1706

Ben helped his father make 650 tallow candles. After selling some, they had 39 candles left. How many candles did they sell?

Using algebra

This is much like a 2nd-grade problem, but with larger numbers. The main difficulty is to convert the common-sense equation that represents the story into its fact-family partner. If your student doesn’t recognize the logic of that switch, review how fact families work by using smaller numbers: 5 = 9 – 4 means also that 4 = 9 – 5.

Made = 650
Sold = ?
Leftover = 39

Leftover = Made – Sold
Which means also that:
Sold = Made – Leftover

Sold = 650 – 39 = 611 candles.

Using a bar diagram

In the bar diagram, the relationship is clear. This is a simple, one-step problem.

650 – 39 = 611 candles sold.

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1718

Ben sold 830 newspapers. His brother James sold 177 fewer newspapers than Ben.
(a) How many newspapers did James sell?
(b) How many newspapers did they sell altogether?

Using algebra

In early 2-step problems, the textbook asks for each step explicitly. Later, the student will be expected to figure out which intermediate steps are needed to find the answer. For this problem, the main danger is that your student will insert a mental period after “James sold 177” and not finish reading the sentence.

We will use each man’s name to represent the number of newspapers he sold.

Ben = 830
James = Ben – 177 = 830 – 177 = 653

Total = Ben + James = 830 + 653 = 1483 newspapers.

Using a bar diagram

This is a comparison problem. We need a bar for the newspapers Ben sold and a bar for the papers sold by James. To indicate that we need to find the total, we put a bracket to the right of our drawing, encompassing both bars.

830 – 177 = 653 newspapers James sold.
830 + 653 = 1483 newspapers sold altogether.

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London, 1724

Ben loved to visit the London book shops. In one small shop, there were 6 shelves of books. Each shelf held the same number of books. There were 30 books altogether. How many books were there on each shelf?

Using algebra
Words like “each” or “every” or “average” usually signal a this per that (or rate) problem. This per that problems require multiplication or division, and your student must be able to use logic to figure out which is needed. Since you are taking the total number of books and sharing them out among the shelves, it is division.

Shelves = 6
Books = 30

Books per shelf = 30 ÷ 6 = 5

Using a bar diagram

The basic “part + part = whole” diagram showed the inverse relationship between addition and subtraction. Now students can use the “(number of units) x (size of units) = total” diagram to represent both multiplication and division problems, because multiplication and division are also inverse operations. Remember that when the units are all the same size, we only need to write a number in the first one.

6 units = 30
1 unit = 30 ÷ 6 = 5 books on each shelf.

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Philadelphia, 1726

Ben and his friends made a club called “The Junto” to read books and discuss ideas. Ben read 7 science books. He read 5 times as many history books as science books. How many more history books than science books did he read?

Using algebra

Notice that the math book is no longer asking a question for each step in the problem. Students will tend to do one calculation and then move on to the next problem. Remind them to check that they have fully answered the question before going on.

Science = 7
History = 5 x Science = 5 x 7 = 35

How many more = History – Science = 35 – 7 = 28 books.

Using a bar diagram

Another comparison, with same-size units of books. We do not need the extra step of calculating how many history books Ben read. The bar diagram contains enough information to lead us directly to the answer.

1 unit = 7
4 units = 4 x 7 = 28 books.

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1732

Ben collected donations for many worthy organizations. He had 2467 pounds in a bank account to start a new hospital. A friend gave him another 133 pounds. How much more money must Ben collect if he needs 3000 pounds for the hospital?

Using algebra

Like the candles problem above, this problem requires a knowledge of how equations work — in this case, the inverse relationship between addition and subtraction. You may need to remind your students that 7 = 5 + 2 means also that 7 – 5 = 2.

Here we introduce our first algebra variable: I chose a question mark to stand in for our unknown value. Students may use an x or m or any symbol they like.

Bank = 2467
Friend = 133
Total needed = 3000
How much more money needed = ?

Total = Bank + Friend + ?
3000 = 2467 + 133 + ?
3000 = 2600 + ?

? = 3000 – 2600 = 400 pounds still needed

Using a bar diagram

The bar diagram makes the inverse relationship easy to see.

3000 – (2467 + 133) = 3000 – 2600 = 400 pounds needed.

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Paris, 1776-1785

While in France to negotiate a treaty, Ben went to a fancy party. There were 1930 women at the party. There were 859 fewer men than women. How many people were at the party altogether?

Using algebra

This is exactly like the newspaper problem earlier, but with bigger numbers. Again, the main danger is that your student will read the sentence about men as “There were 859 (fewer) men” and think, “Well, of course 859 is fewer!” but not notice the rest of the phrase.

Women = 1930
Men = Women – 859 = 1930 – 859 = 1071

Total = Women + Men = 1930 + 1071 = 3001 people.

Using a bar diagram

Again, there should be a bracket to the right, showing that we need to find the sum of the two bars.

(2 x 1930) – 859 = 3860 – 859 = 3001 people.

There are often more ways to find an answer than you might expect. Here, I have shown an alternate calculation. Although I imagine most 3rd-grade students would do the calculation exactly as is done in the algebra section above, the bar diagram reveals another way to look at the problem, if we wish. We could first find out what the total would be IF there had been the same number of men as women, and then subtract the extra men (the ones who weren’t really there). Sometimes the less obvious method will lead to a much easier calculation, as it does here: 860 – 859 = 1.

Which Approach Is Best for Your Student?

As I mentioned in the 2nd-grade article, the algebra approach required me to recognize on my own which operation was needed to solve the problem. Algebra offers an efficient way to write down my reasoning, which usually lets me move quickly from problem to solution. Students with strong reading and reasoning skills will appreciate this efficiency, but weaker students may have trouble deciding whether to add or subtract, multiply or divide. Even strong students may have difficulty when it becomes necessary to rearrange an equation or when a story requires several steps.

Bar diagrams often take up more space and require more pencil-to-the-paper work from the student in drawing them out. But in most cases, the bar diagrams offer more help for students who struggle with the question, “What do I do?” Diagrams put the inverse relationships on display, enabling the student to decide which arithmetical operation to use. On occasion, a bar diagram will surprise me by offering a more efficient solution than the algebra approach.

One clear advantage of bar diagrams, in my opinion, is how well they lead to understanding ratios. The problem with the science and history books is a good example. Students will meet many problems like this in Singapore math — one thing is some number times as many as another thing. The bar diagram shows this relationship: the bar really is 5 times as big.

The algebra approach leaves the relationship hidden in the abstraction of the numbers: 5 x 7 = 35, but what does that really mean? Problems like this are going to grow more challenging as students progress, until they become the dreaded ratios and proportions of pre-algebra. These topics are notoriously difficult for students [JSTOR access required, or try this article instead], but I believe the bar diagrams provide a much better foundation for understanding than any other method I have seen.

Teaching Tips

When using algebra with young children, keep the abstraction to a minimum. Do not introduce generic variables like x and y. Instead, use significant words from the story, like the names of the characters or their initials, or use words like Total and Leftover that name the relationship between quantities. And when you write or read an equation, emphasize the connection between the math and the story by saying the whole word, even if all you write is the initial.

Bar diagrams are normally drawn as rectangles, like blocks or Cuisenaire rods, and numbers or words may be written inside to label them. Brackets are used to group the bars together or to indicate a specific section of a bar.

When introducing bar diagrams, help your students recognize the meaning of the bar by saying, “Let’s imagine all the candles/newspapers/books set out in a row…” If your student has trouble figuring out where the numbers go in the diagram, you might ask, “Which is the big amount, the whole thing? What are the parts it is made of? Are we comparing one thing to another?”

For Tutoring or Homeschooling Situations

When I teach my students to draw bar diagrams, I do it apart from their daily homework. My students are allowed to work their daily homework by whatever method they choose — including doing it all in their heads and just writing an answer — as long as they can explain the logic of their solution. But in our story problem workbook, they have to draw the bars.

As a compensation for the extra pencil work of drawing, they do not have to actually calculate the answer. Once they show me how the bars are set up and tell me what would have to be done to solve it, they are done with that problem. They think they are pulling a fast one on me, because they aren’t doing the multiplication or subtraction or whatever, but I want them to focus on reasoning through to a solution. I want them to learn how the bar diagram tool works before they get to the really tough problems where they need it.

To get more practice creating bar diagrams, your students may enjoy these online tutorials:

• Thinking Blocks Word Problems