### Relationships In Word Problems

Comparison: Children solve addition and subtraction problems by focusing either on the action (joining or separating) or on the relationship in a problem. Many problems involve relationships between amounts.
COMPARISON problems are the type of problems looked at this week, which involve figuring our similarities or differences between sets.

Difference Unknown: One type of compare problem involves finding out how many more are in one set than another. For example, James has 6 mice. Joy has 11 mice. How many more mice does Joy have than James?

Set Unknown Problems: Another type of compare problem involves finding out how many are in the one set by knowing both the number of items in one set and the difference between the two sets. For example, James has 6 mice. Joy has 5 more mice than James. How many mice does Joy have? or Joy has 11 mice. She has 5 more mice than James. How many mice does James have?

## Types of Addition and Subtraction Problems

 Join: 7 + 9 = ? 7 + ? = 16 ? + 9 = 16 Result Unknown Change Unknown Start Unknown Byron has 7 shells.Then Melissa gavehim 9 more shells.How many shellsdoes Byron havenow? Byron has 7 shells.Melissa give Byronsome shells.Now Byron has 16 shells.How many shellsdid Mellisa give him? Byron has some shells.Melissa gives him9 more.Now Byron has 16 shells.How many shells didByron start with?

 Separate: 8 - 3 = ? 8 - ? = 5 ? - 3 = 5 Result Unknown Change Unknown Start Unknown Colleen has 8 guppies.She gave 3guppies to Roger.How manyguppies doesColleen have left? Colleen has 8 guppies.She gave some guppiesto Roger. Then she had5 guppies left.How many guppiesdid Colleen give Roger? Colleen hassome guppies.She gave 3 guppiesto Roger. Then she had5 guppies left.How many guppiesdid Colleen have tostart with?

 Compare: Difference Unknown Set Unknown James has 6 mice.Joy has 11 mice.How many more micedoes Joy have than James? James has 6 mice.Joy has 5 more micethan James.How many mice doesJoy have? Joy has 11 mice.She has 5 more micethan James.How many micedoes James have?

### Strategies for Solving Joining Problems

Problems involving the difference between the sets are easier than those that ask about either of the two sets. The reason can be found in children's solution strategies.

Compare-Difference-Unknown problems can be solved by a physical (direct) modeling strategy involving matching. For the problem involving the unknown difference (James has 6 mice. Joy has 11 mice.), children use a matching strategy. Here children count out a set of six cubes for James, and another set of 11 cubes for Joy. Then they put the 6 cubes in a row, followed by making another row of 11 cubes next to the row of 6 cubes. Children then count the 5 cubes not matched with a cube in the first row. These 5 cubes represent the difference.

Compare-Set-Unknown problems are solved with a variety of counting strategies. For the problem involving the unknown larger set ( James has 5 mice etc.) children will count up: "6 (pause) 7 (1), 8 (2), 9 (3),10 (4), 11 (5). For the problem involving the unknown smaller set (Joy has 11 mice. She has 5 more mice than James.), children may count down to find the smaller set: "11 . . . 10(1), 9(2), 8(3), 7(4), 6(5). James has 6 mice." For example, to solve the problem with the unknown larger set (James has 6 mice. Joy has 5 more mice than James), children will count up.