Taxonomy of Word Problems
The Common Core uses a chart giving a taxonomy of one step addition and subtraction situations.25 Three main types are identified: change, comparison, and part-part-whole. The change and comparison are then split into two main subtypes: change plus and change minus, and more and less, respectively. Each of these types is further subdivided into three more categories based on the unknown. For the change problems, these categories are: result unknown, change unknown, start unknown. For the comparison problems, the categories are: difference unknown, bigger unknown, and smaller unknown. The third main type, the part-part-whole, differs because there are only two subtypes: total unknown (whole unknown) or addend unknown (part unknown). There are interchangeable names for the various types of problems because the Common Core standards use names in the chart, but in literature other names are used to describe the problem types. I refer to them with the various names to build teacher background knowledge and hopefully to avoid confusion. These types will be discussed below with examples. I do not plan to have the students name the types, but it is really important to expose the students to all of the different kinds of problems.
As you explore these word problems, the natural connection between addition and subtraction will become more evident. For example, you will see that there are equations in which addition represents the problem, but subtraction is required to solve it.
As the different types of problems are introduced, the same numbers and situation will be explored. The numbers chosen for the initial explanation of the problem types will be small whole numbers. This is done purposefully, because the focus at this point is on introducing the problem types and evoking the similarities and differences.
Addition and Subtraction - Change
“Add to” or “Change plus”
For this category, the problem could have a result unknown, a change unknown, or a start unknown.
Eight birds are sitting on the fence. Three more land on the fence. How many are on the fence now? The bar diagram for this problem is in Figure 1.
Figure 1
This is an add to (change plus)/ result unknown since there were eight birds on the fence and a positive change happened indicating addition. The total number of birds on the fence is unknown, therefore it is a result unknown. The equation to solve the problem is 8+3=∎.
The next example is an add to (change plus)/change unknown. Eight birds were sitting on the fence. Some more landed on the fence. There are now 11 birds on the fence. How many birds landed on the fence? The diagram for this is given in Figure 2.
Figure 2
In this situation, the number of birds originally on the fence is known. Then some more birds join them, but the problem does not state how many more came. The result is 11 birds on the fence. The equation is now 8+∎=11. While the equation shows addition, this problem is solved by subtraction 8+∎=11. If students were using a key word strategy, they would see the word more, thus students may be inclined to add eleven and eight, which does not answer the question.
The third type is add to (change plus)/start unknown.
Some birds are on the fence. Three more land on the fence. There are now 11 on the fence. How many were on the fence in the beginning? Figure 3 shows the bar model for this problem.
Figure 3
It would be expressed as ∎+3=11. The initial number of birds on the fence is unknown. The equation shows addition, but again, subtraction is required to solve for the answer; 11-3=∎. Once again, using the key word method, students would mistakenly be inclined to add eleven plus three since the key word more is used.
“Take from” or “Change minus”
These are also “add to” problems, but now the change is being “taken from.” There are the same three types: result unknown, change unknown, and start unknown.
The first problem is a Take from/result unknown.
Six squirrels were in a tree. Two ran away. How many were left in the tree? The equation is 6-2=∎ ,and it would be solved with subtraction. The bar model is shown in Figure 4.
Figure 4
The second problem of this kind is a Take from/change unknown.
Six squirrels were in a tree. Some ran away. There were four left in the tree. How many ran away? Mathematically, the equation is 6-∎=4, and the solution is ascertained by subtracting 6-4=2. The diagram for this is given in Figure 5.
Figure 5
The last one in this series is a Take from/ start unknown.
Some squirrels were in a tree. Two ran away. There were four left. How many were in the tree at the start? The bar model for this problem is shown in Figure 6.
Figure 6
The equation is ∎-2=4 , but the solution is found by computing 4+2=6. Once again, the key word method would be problematic. The problem says left, which typically means to subtract, but in fact addition is required.
Addition and Subtraction - “Put together” or “Part-part-whole”
For these, it is either an unknown part or unknown addend or an unknown whole or unknown total.
There are six green apples and four red apples are in a bowl. How many apples are in bowl?
The number sentence is 6+4=∎, and the answer is 10. The diagram for this problem is given in Figure 7.
Figure 7
An example of an unknown part or unknown addend is:
There are 10 apples in the bowl. There are six green apples and the rest are red apples. How many apples are red? Figure 8 shows the bar model for this problem.
Figure 8
The equation, 6+∎=10, is used to represent the problem, but to compute the answer students need to compute 10-6=4.
Addition and Subtraction - Compare
The compare problems are probably the trickiest of the addition and subtraction types. There are two of each kind: difference unknown, bigger unknown, and smaller unknown, for a total of six. Part of the confusion stems from the interchangeable vocabulary used to describe these types.
Difference unknown/more
Jaquan has six apples. Devin has two apples. How many more apples does Jaquan have than Devin? The model is shown in Figure 9.
Figure 9
It is a difference unknown, as the answer to the subtraction problem is unknown. It falls into the more category because the question asks, “How much more?” The number sentence 2+∎=6 represents this problem. However, the operation needed solve the problem is 6-2=∎.
Difference unknown/fewer (less)
Jaquan has six apples. Devin has two apples. How many fewer does Devin have?
The sentence 6-2=∎ will provide the solution. Figure 10 models this problem.
Figure 10
The two problems above sound quite different, but are answered with the same computation.
Bigger unknown/more
Jaquan has 4 more apples than Devin. Devin has 2 apples. How many apples does Jaquan have?
The equation 2+4=∎, leads to the correct answer of 6 apples. The bar model to represent this problem is shown in Figure 11.
Figure 11
Bigger unknown/fewer (less)
Devin has 4 fewer apples than Jaquan. Devin has 2 apples. How many apples does Jaquan have?
In order to determine the number of apples Jaquan has, use the equation 2+4=∎. Figure 12 provides the model for this problem.
Figure 12
The two problems above, again, sound quite different, but are answered with the same computation.
Smaller unknown/more
Jaquan has 4 more apples than Devin. Jaquan has 6 apples. How many apples does Devin have? The statement ∎+4=6 represents this scenario, but the subtraction sentence 6-4=∎ is needed to solve this problem. Once again, the word more that indicates addition when using the key word strategy would lead students astray. The model in Figure 13 represents this problem.
Figure 13
Smaller unknown/fewer (less)
Devin has 4 fewer apples than Jaquan. Jaquan has 6 apples. How many apples does Devin have? 6-4=∎ will solve this problem. Figure 14 shows the bar model for this problem.
Figure 14
The two comparison problems above are answered with the same computation, and sound quite different.
As with addition/subtraction problems, there are designated problem types for multiplication and division with three main categories: equal groups, comparison, and array/area. Just as the main types of addition/subtraction problems addressed different important contexts where addition and subtraction are used, the categories of multiplication/division problems point to the main interpretations or applications of these operations. The equal groups and comparison ones are closely related. The number of groups is analogous to the comparison factor. In fact, when a collection is made out of a number of equal groups, the comparison factor between the big collection and one of the groups is exactly the number of groups. Another similarity between the compare and the comparison problems are that they both can be represented through linear measurement. The equal groups and the comparison problems have three subcategories: unknown product, groups size unknown, and number of groups unknown. The arrays/area problems are different because they represent problems in two-dimensions. The Singapore bar models are a linear representation of a problem, and would not be an effective strategy for array/area models. As you read these examples, notice how multiplication equations are often solved with division and vice versa. I will now explain each type as I did for the addition and subtraction ones.
Multiplication/division – equal groups
Equal groups/unknown product
At the toy store, cars come in packs of six. If Jose buys four packs, how many toy cars will he buy in total? 6×4=∎ is the equation and the product is 24. The representation is shown in Figure 15.
Figure 15
Equal groups/group size unknown
At the toy store, Jose buys 24 cars. Jose buys four packs. How many are in each pack?
In this situation, 4×∎=24 the number of groups (packs) and the product is known, but the number in each group or the groups size is unknown. The equation sets up as a multiplication problem, but it uses division 24÷4=6 to find the solution. The bar model in Figure 16 shows this problem.
Figure 16
Equal groups/number of groups unknown
This next problem type is about equal groups. It is like the previous one but instead of group size, the number of groups is unknown.
At the toy store, Jose buys 24 cars. There are six in each pack. How many packs does Jose buy? The equation 6×∎=24 represents the scenario. The number in each group is known as is the product or total that Jose bought. The unknown is the number of groups or how many packs Jose buys. Once again division is used: the equation 24÷6=4 determines that the answer is four groups. The bar model in Figure 17 shows this problem. Notice the ellipses, which are used when the number of groups is unknown or when there are a large number of groups.
Figure 17
Multiplication/division – arrays/area
These problems use situations with objects arranged into rows or columns, representing an array, or they use measurement to show area. These are essentially equal groups problems, with the rows being the groups, and the columns being the number of groups. The point of making them a distinct category is to connect with the area model for multiplication. The arrays and area problems do not align well with Singapore bar models; therefore, I did not include images of bar models in this section.
Arrays/area – unknown product
Some chairs are arranged in the classroom. There are three rows of seven chairs. How many chairs are in there? The multiplication sentence 3×7=∎ corresponds to this problem, and the answer is 21 chairs.
Arrays/area – group size unknown
Twenty-one chairs are arranged into three equal rows. How many chairs are in each row? 3×∎=21 depicts this situation, but division is required to determine the group size of each row. So, 21÷3=7 results in the solution to this problem. In this problem, the number of equal groups is known, but the group size is unknown.
Arrays/area – number of groups unknown
Twenty-one chairs are arranged into equal rows. There are seven chairs in each row. How many rows will there be? The number sentence 7×∎=21 depicts this problem, but students must divide 21÷7=3 to ascertain the answer. In this similar situation, the number of chairs in each group is known, but the number of groups or rows is unknown.
Multiplication/division – comparison
The multiplication comparisons are also used to express measurement problems. An example and explanation of one with and without measurement is listed below.
Compare/unknown product
A pack of gummy bears cost $ 2 and a sandwich costs three times as much. How much does the sandwich cost? The equation $ 2×3=∎ denotes the scenario, and multiplication is used to determine that the cost of the sandwich is $ 6. The bar model in Figure 18 represents this problem.
Figure 18
Jamie wants to measure the length of her desk using a large paper clip. She lays 24 paper clips end to end. Each paper clip is two inches long. What is the length of her desk? This would be represented mathematically as 24×2=∎. Figure 19 models this problem.
Figure 19
Compare/smaller quantity unknown
A sandwich costs $6. It costs three times as much as gummy bears. How much do the gummy bears cost? Therefore 3 × ∎ = $ 6 , but the operation of division $ 6 ÷ 3 = $ 2 is necessary to find the answer. The smaller quantity, or in this case, the price of the gummy bears is unknown. Three groups of $2 equals $6. The bar model is shown in Figure 20.
Figure 20
Compare/ Comparison factor unknown
A sandwich costs $6 and gummy bears cost $2. How many times as much does the sandwich cost? The statement $ 2 × ∎ = $ 6 represents the situation, but the division sentence $ 6 ÷ $ 2 = 3 is needed to determine the comparison factor. The bar model for this problem is shown in Figure 21.
Figure 21
Multi-step word problems
All of the above-mentioned types of problems are one-step problems: they require only a single computation to arrive at an answer. One-step problems can be combined to make problems with several steps – two, three, or even more. One could say, they are the nuts and bolts that make up multi-step problems. I expect that, exposing my students to the component parts and developing a level of comfort with the various kinds will provide a strong foundation for them to be more successful with multi-step problems. Now, let’s look at a few examples of multi-step problems.
There are 37 children in the swimming pool. There are nine fewer adults than children in the pool. How many adults and children are in the pool? One way to solve this problem is to think 37-9=∎ to determine the number of adults in the pool. This first part is an addition/subtraction comparison, which will guide us to the number of adults in the pool. If 37-9=28, then there are 28 adults in the pool. The second step is to combine the number of adults and children with the equation 37+28=65. The second step is a put together or part-part-whole with the total unknown. As you can see, this two-step problem is comprised of situations that have already been discussed. Figure 22 shows the model for the first step in this problem.
Figure 22
37-9=28 adults, then a second bar model would represent the second step. The model below represents the second step which is an addition/subtraction part-part-whole problem with the whole unknown. The second step is modeled in figure 23.
Figure 23
Let’s look at one more example. Five children are on the playground. Then three times as many join them. How many children are on the playground now? The first step is multiplication/division comparison with the result unknown. This step is solved by calculating 5×3=∎. So, 15 children joined them. The second part is a change plus/result unknown which would use the addition statement 15+5=20 to figure out the result. Figure 24 shows both steps using a single bar model.
Figure 24
The one-step addition/subtraction and multiplication/division problems can be combined into about 400 types of two-step problems. It would be challenging to expose your students to all the different combinations of two-step problems. It would be impractical to teach all the combinations of three-step problems, as there are about 8,000 variations. Thus, the strategy of introducing students to the 23 parts that compose all multi-step problems is important to lay a strong foundation and should be taught with fidelity. An important takeaway, as demonstrated through the examples of word problems, is that inverse relationships between operations are not discrete. Therefore, addition and subtraction should be dealt with together, and likewise and multiplication and division should be dealt with at the same time.
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