Learn on PengiSaxon Algebra 1Chapter 5: Inequalities and Linear Systems

Lesson 47: Solving Problems Involving the Percent of Change

In this Grade 9 Saxon Algebra 1 lesson, students learn how to calculate the percent of change by dividing the amount of increase or decrease by the original amount and expressing the result as a percent. The lesson covers real-world applications including markups and discounts, where students find new prices after applying a percent of increase or decrease to an original cost. Part of Chapter 5 on Inequalities and Linear Systems, Lesson 47 builds percent reasoning skills through practical examples involving salaries, retail pricing, and fundraising totals.

Section 1

📘 Solving Problems Involving the Percent of Change

New Concept

Algebra is a mathematical language that uses variables and symbols to represent unknown quantities, allowing us to model and solve complex, real-world problems.

What’s next

To start, we'll apply these powerful ideas to a practical concept. You'll soon see worked examples and challenge problems on calculating the percent of change.

Section 2

Finding the Percent of Increase or Decrease

Property

amount of increase or decreaseoriginal amount=Percent of change\frac{\text{amount of increase or decrease}}{\text{original amount}} = \text{Percent of change}

Examples

  • From 200 students to 250: The increase is 250−200=50250 - 200 = 50. The percent increase is 50200=0.25\frac{50}{200} = 0.25, or 25%25\%.
  • From 80 dollars to 60 dollars: The decrease is 80−60=2080 - 60 = 20. The percent decrease is 2080=0.25\frac{20}{80} = 0.25, or 25%25\%.

Explanation

First, find the difference between the new and original numbers to find the amount of change. If the new number is bigger, you've got an increase! Then, just divide that change by the original amount and multiply by 100 to get your percentage. It's the perfect way to track growth or shrinkage in numbers!

Section 3

Application: Finding Markups and Discounts

Property

A markup results when a percent of increase is applied to a cost. A discount results when a percent of decrease is applied to a cost.
New Price = Original Price + Markup
New Price = Original Price - Discount

Examples

  • A store buys a keyboard for 50 dollars and marks it up by 80%. The markup is 0.80â‹…50=400.80 \cdot 50 = 40 dollars. The new price is 50+40=9050 + 40 = 90 dollars.
  • A 30 dollars hoodie is on sale for 20% off. The discount is 0.20â‹…30=60.20 \cdot 30 = 6 dollars. The new price is 30−6=2430 - 6 = 24 dollars.

Explanation

Stores use markups to make a profit—they buy an item cheap and sell it for more! For you, a discount is awesome because it means a sale! You calculate the percentage of the original price and either add it (markup) or subtract it (discount) to find the final price. This is the math behind every price tag.

Book overview

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

Continue this chapter

Chapter 5: Inequalities and Linear Systems

  1. Lesson 1

    Lesson 41: Finding Rates of Change and Slope

  2. Lesson 2

    Lesson 42: Solving Percent Problems

  3. Lesson 3

    Lesson 43: Simplifying Rational Expressions

  4. Lesson 4

    Lesson 44: Finding Slope Using the Slope Formula

  5. Lesson 5

    Lesson 45: Translating Between Words and Inequalities

  6. Lesson 6

    Lesson 46: Simplifying Expressions with Square Roots and Higher-Order Roots

  7. Lesson 7Current

    Lesson 47: Solving Problems Involving the Percent of Change

  8. Lesson 8

    Lesson 48: Analyzing Measures of Central Tendency

  9. Lesson 9

    Lesson 49: Writing Equations in Slope-Intercept Form

  10. Lesson 10

    Lesson 50: Graphing Inequalities

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

📘 Solving Problems Involving the Percent of Change

New Concept

Algebra is a mathematical language that uses variables and symbols to represent unknown quantities, allowing us to model and solve complex, real-world problems.

What’s next

To start, we'll apply these powerful ideas to a practical concept. You'll soon see worked examples and challenge problems on calculating the percent of change.

Section 2

Finding the Percent of Increase or Decrease

Property

amount of increase or decreaseoriginal amount=Percent of change\frac{\text{amount of increase or decrease}}{\text{original amount}} = \text{Percent of change}

Examples

  • From 200 students to 250: The increase is 250−200=50250 - 200 = 50. The percent increase is 50200=0.25\frac{50}{200} = 0.25, or 25%25\%.
  • From 80 dollars to 60 dollars: The decrease is 80−60=2080 - 60 = 20. The percent decrease is 2080=0.25\frac{20}{80} = 0.25, or 25%25\%.

Explanation

First, find the difference between the new and original numbers to find the amount of change. If the new number is bigger, you've got an increase! Then, just divide that change by the original amount and multiply by 100 to get your percentage. It's the perfect way to track growth or shrinkage in numbers!

Section 3

Application: Finding Markups and Discounts

Property

A markup results when a percent of increase is applied to a cost. A discount results when a percent of decrease is applied to a cost.
New Price = Original Price + Markup
New Price = Original Price - Discount

Examples

  • A store buys a keyboard for 50 dollars and marks it up by 80%. The markup is 0.80â‹…50=400.80 \cdot 50 = 40 dollars. The new price is 50+40=9050 + 40 = 90 dollars.
  • A 30 dollars hoodie is on sale for 20% off. The discount is 0.20â‹…30=60.20 \cdot 30 = 6 dollars. The new price is 30−6=2430 - 6 = 24 dollars.

Explanation

Stores use markups to make a profit—they buy an item cheap and sell it for more! For you, a discount is awesome because it means a sale! You calculate the percentage of the original price and either add it (markup) or subtract it (discount) to find the final price. This is the math behind every price tag.

Book overview

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

Continue this chapter

Chapter 5: Inequalities and Linear Systems

  1. Lesson 1

    Lesson 41: Finding Rates of Change and Slope

  2. Lesson 2

    Lesson 42: Solving Percent Problems

  3. Lesson 3

    Lesson 43: Simplifying Rational Expressions

  4. Lesson 4

    Lesson 44: Finding Slope Using the Slope Formula

  5. Lesson 5

    Lesson 45: Translating Between Words and Inequalities

  6. Lesson 6

    Lesson 46: Simplifying Expressions with Square Roots and Higher-Order Roots

  7. Lesson 7Current

    Lesson 47: Solving Problems Involving the Percent of Change

  8. Lesson 8

    Lesson 48: Analyzing Measures of Central Tendency

  9. Lesson 9

    Lesson 49: Writing Equations in Slope-Intercept Form

  10. Lesson 10

    Lesson 50: Graphing Inequalities