Education

More Than One Way To Count Peanuts

May 31, 1995 3 min read
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The following vignette is excerpted from the chapter “Monitoring Students’ Progress” in “Assessment Standards for School Mathematics,” which was released last week by the National Council of Teachers of Mathematics.

Ms. Morris’s 1st-grade students are solving whole-number addition and sub~trac~tion problems. Rather than rely~ing on written tests or formal assessment procedures, Ms. Morris continually asks her students to describe the processes they used to solve a given problem, and students are encouraged to describe alternative solutions. In the following dialogue, the students solve a comparison problem.

Ms. Morris: The African elephant ate 37 peanuts. The Indian elephant ate 43 peanuts. How many fewer peanuts did the African elephant eat than the Indian elephant?

The children worked on the problem for two or three minutes. Some of the children used stacking cubes that had been joined together in stacks of 10 cubes. Others did not use any materials. ...

Ms. Morris: O.K.? How many fewer peanuts did the African elephant eat? Mike?

Mike: Six.

Ms. Morris: Does everyone agree with that? How did you figure it out, Mike?

Mike: Well, I had 43 here (pushing out four stacks of 10 cubes and three additional cubes joined together), and I had 37 here (pushing out three stacks of 10 cubes and a stack of seven). I put 30 on top of these 30. I took three, and I put them here; there were four left, so I took four off, and there were six left.

As he described what he did, he took three of the 10-stacks from the collection of 43 and put them on top of the three 10-stacks in the collection of 37. Then he took the three single cubes from the original set of 43 and put them on top of the seven cubes in the set of 37. Then he took the remaining stack of 10 cubes from the original 43 and broke off four cubes. He put these four cubes on the four cubes in the set of 37 that were not covered. He was left with six cubes from the set of 43 that did not match up with cubes in the set of 37.

Ms. Morris: What do you think of Mike’s solution?... Did anyone do it a different way?

Marci: I took 37, and I needed 43. So I counted up three more. That was 40. Then I took three more to 43.

Ms. Morris: Good. Does her way work out well? It sure does. Did anybody do it differently?

Linda: Well, first I got 37. Then I got 43 (pushes out collections of 37 and 43 cubes joined together in stacks of 10, with the extra cubes also connected together). See, I know it couldn’t be 10, because if you had 10, it would be 47 instead of 43. So I realized that it had to be less than 10. So what I did was I imagined three more cubes here (points to the top of the stack of seven cubes in the set of 37), and I imagined three more right here (pointing to a space next to the collection of 37 that corresponds to where the three cubes are in the collection of 43).

Ms. Morris gave each child in the group time to complete the problems, and she gave children who had a different solution an opportunity to explain their solutions. The children all listened attentively to other children’s solutions, so they had the chance to learn from one another. Ms. Morris also learned what each child could do, and she learned more than whether a child got the correct answer. The different solution strategies reflected quite different levels of understanding. Mike had to model the problem directly, whereas the solutions of Marci and Linda showed more flexibility in operating with numbers.

Copies of the report are available for $25 each, with discounts for quantities, from the National Council of Teachers of Mathematics. To order, call (800) 235-7566.

A version of this article appeared in the May 31, 1995 edition of Education Week as More Than One Way To Count Peanuts

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