# The Craft of Word Problems

## CONTENTS OF CURRICULUM UNIT 04.05.02

- Purpose of Unit
- Introduction and Rationale
- New Haven Public Schools Mathematics Standards
- Unit Objectives
- Whole Class Interaction
- Discussion
- Guided Practice
- Independent Practice
- Lesson Plan Timeline
- Bibliography for Teachers
- Student Resources
- Materials for Classroom Use
- Appendix A
- Appendix B
- Appendix C
- Appendix D
- Appendix E
- Appendix F
- Appendix G
- Appendix H
- Appendix I
- Notes

### Unit Guide

## Developing Word Problems: A Student’s Task

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## Appendix C

A.5. Tony and I were comparing our baseball card collections. I have three times as many cards as Tony and 4 more besides. Tony has 9. How many do I have?

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A.6. Tony and I were comparing our baseball card collections. I have three times as many cards as Tony and 4 more besides. I have 31. How many does Tony have?

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First Step: Identify the givens and the question
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A.5. Givens:

Question:

- - I have 3 times the amount of cards that Tony has with 4 more besides
- - Tony has 9 cards

- - How many cards do I have?

A.6. Givens:

Question:

- - I have 3 times the amount of cards that Tony has with 4 more besides
- - I have 30 cards

- - How many cards does Tony have?

Second Step: Problem Analysis

A.5.

The problem first states that I have 3 times the amount of cards as Tony and that Tony has 9 cards. To find the first step of the problem 9 must be multiplied by 3. I then not only have 3 times the amount of Tony, but also 4 more cards besides. Since I have 4 more, then 4 must be added to the product of 9 times 3 to find my number of cards.

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A.6.

This problem states that I have 30 cards and that is 3 times the amount of Tony plus 4 more. If my total number of cards (30) is 3 times the amount and 4 more than Tony, then he must have 4 less than 30 and a third of that number of cards. So 3 must be subtracted from 30 and the difference must then be divided by 3 to find Tony’s total number of cards.

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Third Step: Identify the operation(s) needed to find the answer and then solve
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A.5.
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Multiplication and Addition

3 x 9 = 27

27 + 4 = 31 cards

A.6.

Subtraction and Division

31 - 4 = 27

27 / 3 = 9 cards

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Fourth Step: Identify the connection between both problems and reasoning.
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The problems are connected because they utilize the same numbers within both equations. The equations use the inverse operations of multiplication with addition and subtraction with division to solve. It can be see that when there are 3 groups of 9 cards that there are 27 cards plus 3 additional cards making 30 cards in all, and when those 30 cards have those 3 additional cards subtracted to make 27 cards which are then divided, into 3 groups there are 9 cards in each group. The operations are also undone in the opposite order (multiplication and addition inverted and opposite are then subtracted and divided).