Lynne Zolli, a colleague with whom I’ve worked on different projects for many years, told me about one of her third-grade student’s complaints about this test item:

*There are 40 chairs and 10 tables. If each table has the same number of chairs, how many chairs will there be at each table? What do you need to do to solve this problem?
a) Add
b) Subtract
c) Multiply
d) Divide*

The “correct” answer was the last choice, to divide 40 by 10. But the student called Lynne over and argued that he could solve the problem with any of the operations.

Can You Solve the Problem with Any Operation?

After Lynne described this incident, I began to think about how the problem might be solved using each of the operations, and how I would “mathematize” the problem for each.

*Division*

Using division seemed straightforward, and I could represent this situation as **40 ÷ 10 = 4**. This problem situation is an example of division as sharing or partitioning, which involves dividing a collection of objects (here, 40 chairs) into a given number of equal parts (placed at 10 tables, each with the same number of chairs).

*Multiplication*

I thought of this as a missing-factor problem: If the 40 chairs were arranged at 10 tables with the same number of chairs at each, what number could I multiply by 10 to get 40? I could represent that as **10 × 4 = 40**.

*Addition*

Well, if I were actually setting the chairs out, I could first put one chair at each table, which would use up 10 chairs. Then I’d put another chair at each table to place another 10 chairs. I’d do this twice more until I had placed all 40 chairs: **10 + 10 + 10 + 10 = 40**. After four rounds, there were four chairs at each table.

*Subtraction*

I was initially stuck about how to solve the problem using subtraction. But then I decided to think about it similarly to how I thought about using addition. Instead of adding 10s to keep track of how many chairs I had put out, I’d subtract 10s to see how many chairs I had left. After placing the first round of 10, I’d be left with 30 chairs. After placing another 10 chairs, I’d be left with 20 chairs. After subtracting 10 four times, there would be 4 chairs at each table and no chairs left to place: **40 – 10 – 10 – 10 – 10 = 0**. As when I used addition, after four rounds, there were four chairs at each table.

There’s More Than One Way to Solve Most Problems

I think that the standardized test item had an important goal: to see if students could connect a word-problem situation to the appropriate operation. But I think the item didn’t recognize the possibility of interpreting the problem with more than one operation or embrace the idea that there’s more than one path to get to a solution. I realize that it’s hard to measure either of these concepts on a machine-scored multiple-choice test, but it’s reasonable to include these ideas in the context of classroom instruction. Word problems are useful for developing students’ understanding of the meanings of the basic arithmetic operations. Thinking about word problems in this way keeps the focus on developing understanding.

**Classroom Suggestion
**Present students the problem of putting 40 chairs at 10 tables so each table has the same number of chairs at it, and ask them to figure out ways to solve the problem with each of the four operations. Lead a class discussion for students to present their ideas.

The question should have read, “What is the most efficient thing you can do to solve the problem?”

Children have different ways of attacking or thinking about math problems. As a teacher any way is appropriate if the student can justify his reasoning. Standardized test makers do not take into account this factor. We are teaching our children to think, not repeat what someone else wants us to think. This child knows his math functions and can think in a variety of ways to solve a problem.

Very interesting! that kind of student’s response shows his conceptual understanding and creativity to solve any problem in his own way.

I also agree that these kinds of questions should be a part of classroom instruction rather than the part of an assessment. A discussion and practice of reflecting about different ways of solving one problem should be done in the classroom.

First, I would caution about the use of the word “efficient” with 3rd grade students. Regardless, I’m not sure it solves the problem. What is most efficient for one is not the most efficient for another. If I’m still struggling with division, the missing factor approach would be more efficient. Personally, this would be best if it was open-ended, leaving the decision of how to arrive at a solution entirely up to the student! It would also give the teacher more information about the level of understanding of the students.

My immediate thought as I read was similar to Maria’s. The question should have been “What is the most efficient way to solve the problem?”

As the previous comments have stated, it is a poor question for a test and, I would say, for learners’ view of mathematics as a subject since it rewards the idea that the Right Answer (see mathwithbaddrawings.com/2015/02/11/the-church-of-the-right-answer) is only available by using the Fixed Method. As Marylin points out, it can be solved using any of the four basic operators and some lateral thinking. For example, I gave this question to two degree educated (non-teaching) adults and both instantly answered 4 and, when asked, stated that they had used division. however, it may not be the method chosen by a child, or even the most efficient method for a child – for instance with a slightly different example, using bus passengers and a non-exact result, the children (age 9/10) rejected both multiplication and division on the grounds they missed the number, and used repeated subtraction. This implies that the original chairs/tables question was set by an adult, with an adult mindset.

Surely it’s lateral thinking that we should be instilling in our learners, even though it may be harder to teach and certainly harder to assess.

Maths, where walking on the grass is encouraged.

And actually, it never said every chair had to be placed at a table. Why couldn’t you have 0, 1, 2, or 3 chairs at each table?

As far as “fixing” the original question goes, “efficient” is subjective. If the testers really insist on having this sort of question (with vocabulary adjusted for the appropriate reading level):

Jamie is going to be setting forty chairs around ten tables so there are an equal number of chairs at each table. She performed an operation once on the two numbers she had and figured out how many chairs to put at each table. Which operation did she use?

Another weakness of the question is that it contains two questions.

On the tests we give our 3rd graders we present problems like this in different ways. One way is to offer equations rather than operations for the students to choose from (40+10, 40-10, 40×10 and 40/10). We also ask students to choose 1 operation, then justify their choice in writing. This helps use see their understanding and number sense.

I don’t understand why the problem doesn’t ask instead to work out the answer and explain reasoning. As written this is less a problem about mathematics than about semantics or weird metacognition, and implies that there is only one right *method*, which is way worse than one right *answer*!