Lesson 5

Decimal Points in Products

5.1: Multiplying by 10 (5 minutes)

Warm-up

In this warm-up, students use the structure (MP7) of a set of multiplication equations to see the relationship between two numbers that differ by a factor of a power of 10. Students evaluate the expression $x\boldcdot 10$ by considering the effect of multiplication by 10. 

Launch

Ask students to answer the following questions without writing anything and to be prepared to explain their reasoning. Follow with whole-class discussion.

Student Facing

  1. In which equation is the value of \(x\) the largest?

    \(x \boldcdot 10 = 810\)

    \(x \boldcdot 10 = 81\)

    \(x \boldcdot 10 = 8.1\)

    \(x \boldcdot 10 = 0.81\)

  2. How many times the size of 0.81 is 810?

Student Response

Student responses to this activity are available at one of our IM Certified Partners

Activity Synthesis

Ask students to share what they noticed about the first four equations. Record student explanations that connect multiplying a number by 10 with moving the decimal place. 

Discuss how students could use their observations from the first question to multiply 0.81 by a number to get 810. 

5.2: Fractionally Speaking: Powers of Ten (15 minutes)

Activity

In grade 5, students recognize that multiplying a number by \(\frac{1}{10}\) is the same as dividing the number by 10, and multiplying by \(\frac{1}{100}\) is the same as dividing by 100. In this lesson, students will recognize and use the fact that multiplying by 0.1, 0.01, and 0.001 is equivalent to multiplying by \(\frac{1}{10}\), \(\frac{1}{100}\), and \(\frac{1}{1,000}\), respectively. In all cases, the essential point to understand is that in the base-ten system, the value of each place is \(\frac{1}{10}\) the value of the place immediately to its left. Writing the decimals 0.1, 0.01, and 0.001 in fraction form will help students recognize how the position of the decimal point in the product is affected by the positions of the decimal points in the factors.

The structure of the base-ten system can serve as a guide for calculating products of decimals, and the goal of this lesson is to begin to uncover that structure (MP7).

Students might need to see decimal and fraction names. If so, display a place-value chart for reference.

Launch

Arrange students in groups of 2. Ask one student in each group to complete the questions for Partner A, and have the other take the questions for Partner B. Then ask them to discuss their responses, answer the second question together, and pause for a brief class discussion.

Representation: Internalize Comprehension. Activate or supply background knowledge about multiplying a number by \(\frac{1}{10}\) or \(\frac{1}{100}\). Remind students that multiplying a number by \(\frac{1}{10}\) is the same as dividing by 10 and multiplying a number by \(\frac{1}{100}\) is the same as dividing by 100.
Supports accessibility for: Memory; Conceptual processing
Representing: MLR2 Collect and Display. Use this routine while students are working through the first two questions. As students work, circulate and listen for the connections students make between the problems. Write the students’ words and phrases on a visual display and update it throughout the remainder of the lesson. Listen for language like “the same,” reciprocal, and inverse operation. Remind students to borrow language from the display as needed. 
Design Principle(s): Optimize output (for explanation); Maximize meta-awareness

Student Facing

Work with a partner. One person solves the problems labeled “Partner A” and the other person solves those labeled “Partner B.” Then compare your results.

  1. Find each product or quotient. Be prepared to explain your reasoning.

    Partner A

    1. \(250 \boldcdot \frac{1}{10}\)
    2. \(250 \boldcdot \frac {1}{100}\)
    3. \(48 \div 10\)
    4. \(48 \div 100\)

    Partner B

    1. \(250 \div 10\)
    2. \(250 \div 100\)
    3. \(48\boldcdot \frac{1}{10}\)
    4. \(48 \boldcdot \frac{1}{100}\)
  2. Use your work in the previous problems to find \(720 \boldcdot (0.1)\) and \(720 \boldcdot (0.01)\). Explain your reasoning.

    Pause here for a class discussion.

  3. Find each product. Show your reasoning.

    1. \(36 \boldcdot (0.1)\)

    2. \((24.5) \boldcdot (0.1)\)

    3. \((1.8) \boldcdot (0.1)\)

    4. \(54 \boldcdot (0.01)\)

    5. \((9.2)\boldcdot (0.01)\)

  4. Jada says: “If you multiply a number by 0.001, the decimal point of the number moves three places to the left.” Do you agree with her? Explain your reasoning.

Student Response

Student responses to this activity are available at one of our IM Certified Partners

Anticipated Misconceptions

Students may readily see that \(36 \boldcdot (0.1)\) but be unsure about what to do when the factor being multiplied by the decimal 0.1 and 0.01 is also a decimal (e.g., \((24.5) \boldcdot (0.1)\)). Encourage them to try writing both decimals (e.g., \(24.5\) and \(0.1\)) as fractions, multiply, and convert the resulting fraction back to a decimal.

Activity Synthesis

The purpose of this discussion is to highlight the placement of the decimal point in a product. Consider asking some of the following questions:

  • “How does the size of a product compare to the size of the factor when the factor is multiplied by 0.1? How does the placement of the decimal point change?” (Multiplying by 0.1 makes the product ten times smaller than the factor. The decimal point moves to the left one place.)
  • “How does the size of a product compare to the size of the factor when the factor is multiplied by 0.01?” (Multiplying by 0.01 makes the factor one hundred times smaller. The decimal point moves to the left two places.)
  • “Can you predict the outcome of multiplying 750 by 0.1 or 0.01 without calculating? If so, how?” (Multiplying 750 by 0.1 would produce 75. Multiplying 750 by 0.01 would produce 7.5.)

5.3: Fractionally Speaking: Multiples of Powers of Ten (15 minutes)

Activity

In this activity, students continue to think about products of decimals using fractions. They use what they know about \(\frac{1}{10}\) and \(\frac{1}{100}\), as well as the commutative and associative properties, to identify and write multiplication expressions that could help them find the product of two decimals.

While students may be able to start by calculating the value of each decimal product, the goal is for them to look for and use the structure of equivalent expressions (MP7), and later generalize the process to multiply any two decimals.

As students work, listen for the different ways students decide on which expressions are equivalent to \((0.6) \boldcdot (0.5)\). Identify a few students or groups with differing approaches so they can share later.

Launch

Arrange students in groups of 2. Give groups 3–4 minutes to work on the first two questions, and then pause for a whole-class discussion. Discuss:

  • Why is \((0.6) \boldcdot (0.5)\) equivalent to \(6 \boldcdot \frac{1}{10} \boldcdot 5 \boldcdot \frac{1}{10}\)? (0.6 is 6 tenths, which is the same as \(6 \boldcdot \frac{1}{10}\), and 0.5 is 5 tenths, or \(5 \boldcdot \frac{1}{10}\))
  • Why is the expression \(6 \boldcdot \frac{1}{10} \boldcdot 5 \boldcdot \frac{1}{10}\) equivalent to \(6 \boldcdot 5 \boldcdot \frac{1}{10} \boldcdot \frac{1}{10}\)? (We can ‘switch’ the places of 5 and \(\frac{1}{10}\) in the multiplication and not change the product. This follows the commutative property of operations.)
  • How did you find the value of \(30 \boldcdot \frac{1}{100}\)? (Multiplying by \(\frac{1}{100}\) means dividing by 100, which moves the decimal point 2 places to the left, so the result is \(0.30\) or \(0.3\).
Action and Expression: Develop Expression and Communication. Maintain a display of important terms and vocabulary. During the launch take time to review terms that students will need to access for this activity. Invite students to suggest language or diagrams to include that will support their understanding of commutative and associative properties of operations.
Supports accessibility for: Memory; Language

Student Facing

  1. Select all expressions that are equivalent to \((0.6) \boldcdot (0.5)\). Be prepared to explain your reasoning.

    1. \(6 \boldcdot (0.1) \boldcdot 5 \boldcdot (0.1)\)
    2. \(6 \boldcdot (0.01) \boldcdot 5 \boldcdot (0.1)\)
    3. \(6 \boldcdot \frac{1}{10} \boldcdot 5 \boldcdot \frac{1}{10}\)
    4. \(6 \boldcdot \frac{1}{1,000} \boldcdot 5 \boldcdot \frac{1}{100}\)
    5. \(6 \boldcdot (0.001) \boldcdot 5 \boldcdot (0.01)\)
    6. \(6 \boldcdot 5 \boldcdot \frac{1}{10} \boldcdot \frac{1}{10}\)
    7. \(\frac{6}{10} \boldcdot \frac{5}{10}\)
  2. Find the value of \((0.6) \boldcdot (0.5)\). Show your reasoning.

  3. Find the value of each product by writing and reasoning with an equivalent expression with fractions.

    1. \((0.3) \boldcdot (0.02)\)
    2. \((0.7) \boldcdot (0.05)\)

Student Response

Student responses to this activity are available at one of our IM Certified Partners

Student Facing

Are you ready for more?

Ancient Romans used the letter I for 1, V for 5, X for 10, L for 50, C for 100, D for 500, and M for 1,000. Write a problem involving merchants at an agora, an open-air market, that uses multiplication of numbers written with Roman numerals.

Student Response

Student responses to this activity are available at one of our IM Certified Partners

Anticipated Misconceptions

If students try to use vertical calculation to find the products, ask them to instead do so by thinking of the decimals as fractions and about any patterns they observed.

Activity Synthesis

Select several students to share their responses and reasonings for the last question.

To conclude, ask students to consider how writing \(6 \boldcdot 5 \boldcdot \frac{1}{100}\) might be a favorable way to find \(0.6 \boldcdot 0.5\). Students may respond that using whole numbers and fractions makes multiplication simpler; even if there is division, it is division by a power of 10. In future lessons, students will apply this reasoning to find products of more elaborate decimals, such as \((0.24) \boldcdot (0.011)\).

Writing, Speaking: MLR3 Clarify, Critique, Correct. Present an incorrect justification for one of the expressions from the first problem that was not equivalent (such as B or E). For example, “The expression \(6 \boldcdot (0.01) \boldcdot 5 \boldcdot (0.1)\) is equivalent to \(0.6\boldcdot0.5\) because the same whole numbers are used and where you put the zeros in a decimal doesn’t matter.” Ask students, “Is this reasoning correct? Why or why not?” Give students 1–2 minutes to write a brief explanation about why the expressions are not equivalent. Student responses should show attention to the placement of the decimal point and the place values of the digits. For students who need additional support, provide a sentence frame, such as “_____ is (equivalent/not equivalent) because _____.” This will help students produce clearer justifications that demonstrate their reasoning about what equivalence means.
Design Principle(s): Optimize output (for justification); Cultivate conversation

Lesson Synthesis

Lesson Synthesis

We can use our understanding of fractions and place value in calculating the product of two decimals. Writing decimals in fraction form can help us determine the number of decimal places the product will have and place the decimal point in the product.

  • What are some ways to find \((0.4) \boldcdot (0.0007)\)? (We can think of 0.4 as \(\frac {4}{10}\) and 0.0007 as \(\frac {7}{10,000}\), multiply the fractions to get \(\frac {28}{100,000}\), and write the product as the decimal 0.00028. Or we can write 0.4 as \(4 \boldcdot  \frac {1}{10}\) and 0.0007 as \(7 \boldcdot \frac {1}{10,000}\), multiply the whole numbers and the fractions, and again convert the fractional product into a decimal.)
  • How might we tell which product will have a greater number of places to the right of the decimal point: \((0.03) \boldcdot (0.001)\) or \((0.3) \boldsymbol \boldcdot (0.0001)\)? (If we write the decimals as fractions and multiply them, we can see that both products equal \(\frac {3}{100,000}\) or 0.00003, so they would have the same number of places to the right of the decimal point.)

5.4: Cool-down - Placing Decimal Points in Products (5 minutes)

Cool-Down

Cool-downs for this lesson are available at one of our IM Certified Partners

Student Lesson Summary

Student Facing

We can use fractions like \(\frac{1}{10}\) and \(\frac{1}{100}\) to reason about the location of the decimal point in a product of two decimals.  

Let’s take \(24 \boldcdot (0.1)\) as an example. There are several ways to find the product:

  • We can interpret it as 24 groups of 1 tenth (or 24 tenths), which is 2.4.
  • We can think of it as \(24 \boldcdot \frac{1}{10}\), which is equal to \(\frac {24}{10}\) (and also equal to 2.4).
  • Multiplying by \(\frac {1}{10}\) has the same result as dividing by 10, so we can also think of the product as \(24 \div 10\), which is equal to 2.4.
     

Similarly, we can think of \((0.7) \boldcdot (0.09)\) as 7 tenths times 9 hundredths, and write:

\(\displaystyle \left(7 \boldcdot  \frac {1}{10}\right) \boldcdot \left(9 \boldcdot  \frac {1}{100}\right)\)

We can rearrange whole numbers and fractions:

\(\displaystyle (7 \boldcdot 9) \boldcdot \left( \frac {1}{10} \boldcdot  \frac {1}{100}\right)\)

This tells us that \((0.7) \boldcdot (0.09) = 0.063\).

\(\displaystyle 63 \boldcdot \frac {1}{1,\!000} = \frac {63}{1,\!000}\)


Here is another example: To find \((1.5) \boldcdot (0.43)\), we can think of 1.5 as 15 tenths and 0.43 as 43 hundredths. We can write the tenths and hundredths as fractions and rearrange the factors. \(\displaystyle \left(15 \boldcdot \frac{1}{10}\right) \boldcdot \left(43 \boldcdot \frac{1}{100}\right) = 15 \boldcdot 43 \boldcdot \frac{1}{1,\!000}\)

Multiplying 15 and 43 gives us 645, and multiplying \(\frac{1}{10}\) and \( \frac{1}{100}\) gives us \(\frac{1}{1,000}\). So \((1.5) \boldcdot (0.43)\) is \(645 \boldcdot \frac{1}{1,000}\), which is 0.645.