Learn on PengiSaxon Math, Course 2Chapter 6: Lessons 51-60, Investigation 6

Lesson 53: Ratio Word Problems

In Saxon Math Course 2 for Grade 7, Lesson 53 teaches students how to solve ratio word problems using proportions and ratio boxes. Students learn to organize given ratio numbers and actual count numbers into a two-column table, then set up and solve a proportion to find an unknown quantity. This skill builds a foundation for applying proportional reasoning to more complex problems in later lessons.

Section 1

πŸ“˜ Welcome to Saxon Math

New Concept

Math concepts are introduced incrementally. Simple tools you master now, like ratio boxes, will be expanded to solve more complex problems in later lessons.

What’s next

Our journey begins with a foundational skill. Next, you will see worked examples on how to use ratio boxes to solve word problems.

Section 2

Increasing Knowledge

Property

A ratio box organizes information into two columns (Ratio, Actual Count) and rows for the groups being compared, like parrots and macaws.

Examples

Parrots to macaws is 3 to 5. With 45 parrots: 35=45m\frac{3}{5} = \frac{45}{m}, so m=75m=75 macaws. Boys to girls is 5 to 4. With 200 girls: 54=B200\frac{5}{4} = \frac{B}{200}, so B=250B=250 boys.

Explanation

Think of a ratio box as a cheat sheet for word problems. It helps you sort the 'ratio' numbers from the 'actual count' numbers, making it a piece of cake to write a proportion correctly.

Section 3

Analyze

Property

A proportion is a statement that two ratios are equal, written as an equation like ab=cd\frac{a}{b} = \frac{c}{d}.

Examples

From 54=B200\frac{5}{4} = \frac{B}{200}, we get 4B=5β‹…2004B = 5 \cdot 200, so B=250B=250. From girl-boy ratio 9 to 7 with 63 girls: 97=63B\frac{9}{7} = \frac{63}{B}, giving 9B=4419B = 441, so B=49B=49.

Explanation

Solving is a cross-multiplication adventure! Multiply diagonally across the equals sign to form a simple equation. This is the key action step where you finally reveal the unknown number in your problem.

Book overview

Jump across lessons in the current chapter without opening the full course modal.

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Chapter 6: Lessons 51-60, Investigation 6

  1. Lesson 1

    Lesson 51: Scientific Notation for Large Numbers

  2. Lesson 2

    Lesson 52: Order of Operations

  3. Lesson 3Current

    Lesson 53: Ratio Word Problems

  4. Lesson 4

    Lesson 54: Rate Word Problems

  5. Lesson 5

    Lesson 55: Average and Rate Problems with Multiple Steps

  6. Lesson 6

    Lesson 56: Plotting Functions

  7. Lesson 7

    Lesson 57: Negative Exponents, Scientific Notation for Small Numbers

  8. Lesson 8

    Lesson 58: Symmetry

  9. Lesson 9

    Lesson 59: Adding Integers on the Number Line

  10. Lesson 10

    Lesson 60: Fractional Part of a Number, Part 1, Percent of a Number, Part 1

  11. Lesson 11

    Investigation 6: Classifying Quadrilaterals

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

πŸ“˜ Welcome to Saxon Math

New Concept

Math concepts are introduced incrementally. Simple tools you master now, like ratio boxes, will be expanded to solve more complex problems in later lessons.

What’s next

Our journey begins with a foundational skill. Next, you will see worked examples on how to use ratio boxes to solve word problems.

Section 2

Increasing Knowledge

Property

A ratio box organizes information into two columns (Ratio, Actual Count) and rows for the groups being compared, like parrots and macaws.

Examples

Parrots to macaws is 3 to 5. With 45 parrots: 35=45m\frac{3}{5} = \frac{45}{m}, so m=75m=75 macaws. Boys to girls is 5 to 4. With 200 girls: 54=B200\frac{5}{4} = \frac{B}{200}, so B=250B=250 boys.

Explanation

Think of a ratio box as a cheat sheet for word problems. It helps you sort the 'ratio' numbers from the 'actual count' numbers, making it a piece of cake to write a proportion correctly.

Section 3

Analyze

Property

A proportion is a statement that two ratios are equal, written as an equation like ab=cd\frac{a}{b} = \frac{c}{d}.

Examples

From 54=B200\frac{5}{4} = \frac{B}{200}, we get 4B=5β‹…2004B = 5 \cdot 200, so B=250B=250. From girl-boy ratio 9 to 7 with 63 girls: 97=63B\frac{9}{7} = \frac{63}{B}, giving 9B=4419B = 441, so B=49B=49.

Explanation

Solving is a cross-multiplication adventure! Multiply diagonally across the equals sign to form a simple equation. This is the key action step where you finally reveal the unknown number in your problem.

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 6: Lessons 51-60, Investigation 6

  1. Lesson 1

    Lesson 51: Scientific Notation for Large Numbers

  2. Lesson 2

    Lesson 52: Order of Operations

  3. Lesson 3Current

    Lesson 53: Ratio Word Problems

  4. Lesson 4

    Lesson 54: Rate Word Problems

  5. Lesson 5

    Lesson 55: Average and Rate Problems with Multiple Steps

  6. Lesson 6

    Lesson 56: Plotting Functions

  7. Lesson 7

    Lesson 57: Negative Exponents, Scientific Notation for Small Numbers

  8. Lesson 8

    Lesson 58: Symmetry

  9. Lesson 9

    Lesson 59: Adding Integers on the Number Line

  10. Lesson 10

    Lesson 60: Fractional Part of a Number, Part 1, Percent of a Number, Part 1

  11. Lesson 11

    Investigation 6: Classifying Quadrilaterals