# Lesson 15

Problem Solving with Line Plots

## Warm-up: Number Talk: Multiply by 18 (10 minutes)

### Narrative

The purpose of this Number Talk is to for students to demonstrate strategies and understandings they have for multiplying whole numbers by fractions. These understandings help students develop fluency and will be helpful later in this lesson when students will need to be able to solve problems involving multiplication of a whole number by a fraction.

### Launch

- Display one expression.
- “Give me a signal when you have an answer and can explain how you got it.”
- 1 minute: quiet think time

### Activity

- Record answers and strategy.
- Keep expressions and work displayed.
- Repeat with each expression.

### Student Facing

Find the value of each expression mentally.

- \(\frac{1}{3}\times18\)
- \(\frac{2}{3}\times18\)
- \(\frac{4}{3}\times18\)
- \(\frac{5}{3}\times18\)

### Student Response

For access, consult one of our IM Certified Partners.

### Activity Synthesis

- “What patterns do you notice in the products?” (There is an 18 in one factor for all of them and it’s always some number of thirds. Once I know the value of \(\frac{1}{3} \times 18\), I can find the rest by multiplying by the number of thirds.)

## Activity 1: Info Gap: Picking Fruit (25 minutes)

### Narrative

This Info Gap activity gives students an opportunity solve problems about data represented on line plots. In both sets of cards, there is a partially complete line plot and some missing data.

For the first set of cards, the problem card has the missing data and the data card has a partially complete line plot. Monitor for students who:

- request all the information on the data card and create a complete line plot which they may use to answer the question
- only request the information they need to answer the question about the heaviest apricot and the most common weight

For the second set of cards, the problem card has the partially complete line plot and the data card has information to determine the missing data. Here students will likely need to communicate with each other as the information about the most common weight is vital to solve the problem, but the student with the problem card may not think to ask about this.

The Info Gap allows students to refine the language they use and ask increasingly more precise questions until they get the information they need (MP6).

*Representation: Access for Perception.*Provide appropriate reading accommodations and supports to ensure student access to written directions, word problems, and other text-based content.

*Supports accessibility for: Conceptual Processing, Memory*

### Required Materials

Materials to Copy

- Info Gap: Picking Fruit

### Required Preparation

- Create a set of cards from the blackline master for each group of 2.

### Launch

- Groups of 2

**MLR4 Information Gap **

- Recall, if necessary, the steps of the info gap routine.
- “I will give you either a problem card or a data card. Silently read your card. Do not read or show your card to your partner.”
- Distribute the first set of cards.
- Remind students that after the person with the problem card asks for a piece of information, the person with the data card should respond with: “Why do you need to know (restate the information requested)?”

### Activity

- 8–10 minutes: partner work time
- After students solve the first problem, distribute the next set of cards. Students switch roles and repeat the process with Problem Card 2 and Data Card 2.

### Student Facing

Your teacher will give you either a problem card or a data card. Do not show or read your card to your partner.

Pause here so your teacher can review your work.

Ask your teacher for a new set of cards and repeat the activity, trading roles with your partner.

### Student Response

For access, consult one of our IM Certified Partners.

### Activity Synthesis

- “What kinds of questions were the most useful to ask?”
- (Card 1: I asked for the heaviest apricot and then realized it was one of the ones on my card. Then I asked for the most common weight and could solve the problem.
- Card 2: I needed to find out what the rest of the apricot weights were. I tried asking for that but my partner did not have that information. My partner told me she had some information about the heaviest apricot and the most common weight and once I found that out I was able to solve the problem.)

- Invite students to share their strategy for solving one of the problems.
- Consider asking “Did anyone solve the problem in a different way?”

## Activity 2: Mathematical Questions [OPTIONAL] (10 minutes)

### Narrative

The purpose of this optional activity is for students to answer questions about a line plot using the same context as the previous activity. Students relate repeated addition of the same fraction to multiplication which they studied in a previous unit. They also address a question about the sum of all of the data. Because there is a lot of data, there are many viable strategies to answer this question and the synthesis focuses on sharing these strategies.

When students solve problems about the apricot weights using the line plot, they reason abstractly and quantitatively (MP2).

### Launch

- Groups of 2

### Activity

- 1–2 minutes: quiet think time
- 5–6 minutes: partner work time
- When students decide whether or not the apricots altogether weigh more than 1 pound, monitor for these strategies:
- adding up all the weights, in ounces, to see if they weigh more than 1 pound
- estimating the sum but not calculating it exactly
- calculating the total weight of the apricots that weigh \(1\frac{5}{8}\) ounces and then using the structure of the graph

### Student Facing

This line plot shows the weights of some apricots that Mai picked.

- What fraction of the apricots weigh less than \(1\frac{1}{2}\) ounces? Explain or show your reasoning.
- Write a multiplication equation that represents the total weight of the apricots that each weigh \(1\frac{5}{8}\) ounces.
- Do all of Mai’s apricots together weigh more or less than a pound? Explain or show your reasoning.

### Student Response

For access, consult one of our IM Certified Partners.

### Activity Synthesis

- Invite previously selected students to share their equation for the total weight of the \(1\frac{5}{8}\) ounce apricots.
- Display the equation: \(5 \times 1 \frac{5}{8} = 8 \frac{1}{8}\).
- “Can you use this information to help decide whether or not the apricots weigh more than a pound?”
- (Yes. There are 10 more apricots, and except for one, they all weigh more than an ounce so that will be more than 16 ounces for sure.
- Yes. I added up all the weights and it was more than 16 ounces.
- Yes. The 5 apricots that weigh \(1\frac{5}{8}\) ounces together are more than \(\frac{1}{2}\) pound. So the 5 heavier apricots are also more than \(\frac{1}{2}\) pound so together they weigh more than 1 pound.)

## Lesson Synthesis

### Lesson Synthesis

“We have added, subtracted, and multiplied fractions to solve problems about line plots.”

“In what ways did we use these operations to help us solve problems about line plots?” (Line plots have a lot of different data and the data had fractions so when we answered questions about the data we had to add, subtract, or multiply.)

“Which was your favorite problem about line plots?” (The eggs because I thought the picture was really interesting.)

## Cool-down: Reflect (5 minutes)

### Cool-Down

For access, consult one of our IM Certified Partners.

## Student Section Summary

### Student Facing

In this section we learned to add and subtract fractions. When the denominators are the same, such as \(\frac{7}{10} + \frac{4}{10}\), we can just add the tenths and see that there are 11 of them so \(\frac{7}{10} + \frac{4}{10} = \frac{11}{10}\). When the denominators are not the same, such as \(\frac{1}{6} + \frac{3}{8}\), we look for a common denominator so that we can add parts of the same size. One way to find a common denominator is to use the product of the two denominators, \(6 \times 8\), because that’s always a multiple of both denominators. Using 48 as a denominator we find \(\frac{1}{6} + \frac{3}{8} = \frac{1 \times 8}{6 \times 8} + \frac{3 \times 6}{8 \times 6}\). This means \(\frac{1}{6} + \frac{3}{8} = \frac{26}{48}\). For the expression \(\frac{1}{6} + \frac{3}{8}\) we can also use a smaller common denominator. Since \(24\) is a multiple of 6 and 8 we can also rewrite \(\frac{1}{6} + \frac{3}{8}\) as \(\frac{4}{24} + \frac{9}{24}\) which is \(\frac{13}{24}\).