Learn on PengiSaxon Math, Course 3Chapter 4: Algebra • Measurement

Lesson 34: Proportions and Ratio Word Problems

In this Grade 8 Saxon Math Course 3 lesson, students learn how to identify proportional relationships by testing whether two ratios are equal and solving for missing terms in a proportion using constant factors. The lesson also introduces ratio word problems, teaching students to organize ratio numbers and actual counts in a ratio table to find unknown quantities.

Section 1

📘 Proportions and Ratio Word Problems

New Concept

A proportion is a statement that two ratios are equal.

What’s next

Next, you'll learn to solve for missing terms in proportions and apply this skill to organize and solve complex ratio word problems.

Section 2

Proportions

Property

A proportion is a statement that two ratios are equal. Equal ratios reduce to the same value or are connected by a constant multiplier. For example,

24=612\frac{2}{4} = \frac{6}{12}
because both sides can be simplified to
12\frac{1}{2}
.

Examples

  • The ratios
    24\frac{2}{4}
    and
    612\frac{6}{12}
    form a proportion because multiplying the first fraction by
    33\frac{3}{3}
    gives you the second.
  • A car lot charging 3 dollars for 2 hours and 4 dollars for 3 hours is not proportional because
    3243\frac{3}{2} ≠ \frac{4}{3}
    .

Explanation

Think of proportions as two fractions playing dress-up! They look different, but they're the same value in disguise. You can reveal their secret identity by finding the magic number that connects them. This 'constant factor' proves that the relationship between the numbers is consistent and fair.

Section 3

Solving Proportions

Property

To find a missing number in a proportion, use a variable and find the multiplier between the terms. In the proportion

28=6x\frac{2}{8} = \frac{6}{x}
, you can solve for the unknown value, x.

Examples

  • To solve
    24m=85\frac{24}{m} = \frac{8}{5}
    , notice that 8×3=248 \times 3 = 24. Apply this to the denominator: 5×3=155 \times 3 = 15, so m=15m=15.
  • To solve
    45=16y\frac{4}{5} = \frac{16}{y}
    , see that 4×4=164 \times 4 = 16. Apply this to the denominator: 5×4=205 \times 4 = 20, so y=20y=20.

Explanation

Put on your detective hat! When a number is missing from a proportion, find the relationship between the numbers you do know. If you see 2×3=62 \times 3 = 6, you can apply that same secret rule to the other pair of numbers. It’s a simple trick to solve the mystery every time!

Section 4

Ratio Word Problems

Property

Use a ratio table with two columns, 'Ratio' and 'Actual Count', and two rows for the items being compared. This table helps organize the numbers from a word problem to set up a proportion.

Examples

  • The ratio of boys to girls is 3 to 4. If there are 12 girls, how many boys? Set up
    34=b12\frac{3}{4} = \frac{b}{12}
    . Since 4×3=124 \times 3 = 12, then 3×3=93 \times 3 = 9. There are 9 boys.
  • A triangle's base-to-height ratio is 3 to 2. If the base is 24 inches, find the height. Set up
    32=24h\frac{3}{2} = \frac{24}{h}
    . Since 3×8=243 \times 8 = 24, then 2×8=162 \times 8 = 16. The height is 16 inches.

Explanation

Word problems can be messy, but a ratio table is your secret weapon for tidiness! It neatly sorts the 'recipe' numbers (the ratio) from the 'real-world' numbers (the actual count). This little grid makes it super easy to build a proportion and solve for the missing piece of the puzzle.

Book overview

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

Continue this chapter

Chapter 4: Algebra • Measurement

  1. Lesson 1

    Lesson 31: Adding Integers and Collecting Like Terms

  2. Lesson 2

    Lesson 32: Probability

  3. Lesson 3

    Lesson 33: Subtracting Integers

  4. Lesson 4Current

    Lesson 34: Proportions and Ratio Word Problems

  5. Lesson 5

    Lesson 35: Similar and Congruent Polygons

  6. Lesson 6

    Lesson 36: Multiplying and Dividing Integers and Multiplying and Dividing Terms

  7. Lesson 7

    Lesson 37: Areas of Combined Polygons

  8. Lesson 8

    Lesson 38: Using Properties of Equality to Solve Equations

  9. Lesson 9

    Lesson 39: Circumference of a Circle

  10. Lesson 10

    Lesson 40: Area of a Circle

  11. Lesson 11

    Lesson 11: Investigation 4: Drawing Geometric Solids

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

📘 Proportions and Ratio Word Problems

New Concept

A proportion is a statement that two ratios are equal.

What’s next

Next, you'll learn to solve for missing terms in proportions and apply this skill to organize and solve complex ratio word problems.

Section 2

Proportions

Property

A proportion is a statement that two ratios are equal. Equal ratios reduce to the same value or are connected by a constant multiplier. For example,

24=612\frac{2}{4} = \frac{6}{12}
because both sides can be simplified to
12\frac{1}{2}
.

Examples

  • The ratios
    24\frac{2}{4}
    and
    612\frac{6}{12}
    form a proportion because multiplying the first fraction by
    33\frac{3}{3}
    gives you the second.
  • A car lot charging 3 dollars for 2 hours and 4 dollars for 3 hours is not proportional because
    3243\frac{3}{2} ≠ \frac{4}{3}
    .

Explanation

Think of proportions as two fractions playing dress-up! They look different, but they're the same value in disguise. You can reveal their secret identity by finding the magic number that connects them. This 'constant factor' proves that the relationship between the numbers is consistent and fair.

Section 3

Solving Proportions

Property

To find a missing number in a proportion, use a variable and find the multiplier between the terms. In the proportion

28=6x\frac{2}{8} = \frac{6}{x}
, you can solve for the unknown value, x.

Examples

  • To solve
    24m=85\frac{24}{m} = \frac{8}{5}
    , notice that 8×3=248 \times 3 = 24. Apply this to the denominator: 5×3=155 \times 3 = 15, so m=15m=15.
  • To solve
    45=16y\frac{4}{5} = \frac{16}{y}
    , see that 4×4=164 \times 4 = 16. Apply this to the denominator: 5×4=205 \times 4 = 20, so y=20y=20.

Explanation

Put on your detective hat! When a number is missing from a proportion, find the relationship between the numbers you do know. If you see 2×3=62 \times 3 = 6, you can apply that same secret rule to the other pair of numbers. It’s a simple trick to solve the mystery every time!

Section 4

Ratio Word Problems

Property

Use a ratio table with two columns, 'Ratio' and 'Actual Count', and two rows for the items being compared. This table helps organize the numbers from a word problem to set up a proportion.

Examples

  • The ratio of boys to girls is 3 to 4. If there are 12 girls, how many boys? Set up
    34=b12\frac{3}{4} = \frac{b}{12}
    . Since 4×3=124 \times 3 = 12, then 3×3=93 \times 3 = 9. There are 9 boys.
  • A triangle's base-to-height ratio is 3 to 2. If the base is 24 inches, find the height. Set up
    32=24h\frac{3}{2} = \frac{24}{h}
    . Since 3×8=243 \times 8 = 24, then 2×8=162 \times 8 = 16. The height is 16 inches.

Explanation

Word problems can be messy, but a ratio table is your secret weapon for tidiness! It neatly sorts the 'recipe' numbers (the ratio) from the 'real-world' numbers (the actual count). This little grid makes it super easy to build a proportion and solve for the missing piece of the puzzle.

Book overview

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

Continue this chapter

Chapter 4: Algebra • Measurement

  1. Lesson 1

    Lesson 31: Adding Integers and Collecting Like Terms

  2. Lesson 2

    Lesson 32: Probability

  3. Lesson 3

    Lesson 33: Subtracting Integers

  4. Lesson 4Current

    Lesson 34: Proportions and Ratio Word Problems

  5. Lesson 5

    Lesson 35: Similar and Congruent Polygons

  6. Lesson 6

    Lesson 36: Multiplying and Dividing Integers and Multiplying and Dividing Terms

  7. Lesson 7

    Lesson 37: Areas of Combined Polygons

  8. Lesson 8

    Lesson 38: Using Properties of Equality to Solve Equations

  9. Lesson 9

    Lesson 39: Circumference of a Circle

  10. Lesson 10

    Lesson 40: Area of a Circle

  11. Lesson 11

    Lesson 11: Investigation 4: Drawing Geometric Solids