Learn on PengiOpenstax Elementary Algebra 2EChapter 3: Math Models

Lesson 3.2: Solve Percent Applications

In this lesson from OpenStax Elementary Algebra 2E, students learn to translate and solve basic percent equations, applying algebraic methods to real-world problems involving tips, sales tax, discounts, simple interest, and markups. Key skills include finding an unknown number, an unknown percent, and calculating percent increase and percent decrease using a structured problem-solving strategy. The content is designed for elementary algebra students building fluency with decimal conversions and equation solving.

Section 1

📘 Solve Percent Applications

New Concept

This lesson shows how to translate real-world percent problems—like tips, discounts, and interest—into simple algebraic equations. You'll learn to solve for unknown amounts, rates, or base values, making everyday math clear and manageable.

What’s next

Next up, you'll tackle interactive examples and practice cards to solidify your skills in solving these percent applications.

Section 2

Basic Percent Equations

Property

To solve percent problems, we translate English sentences into algebraic equations and then solve them. We must be sure to change the given percent to a decimal when we put it in the equation. The three basic types of percent equations are:

  1. Finding the amount: What number is 35% of 90? translates to n=0.35â‹…90n = 0.35 \cdot 90
  2. Finding the base: 6.5% of what number is 1.17 dollars? translates to 0.065â‹…n=1.170.065 \cdot n = 1.17
  3. Finding the percent: 144 is what percent of 96? translates to 144=pâ‹…96144 = p \cdot 96

Examples

  • What number is 25% of 160?

Translate the sentence into an equation: n=0.25â‹…160n = 0.25 \cdot 160. Solving this gives n=40n = 40. So, 40 is 25% of 160.

  • 4.5% of what number is 18 dollars?

Translate this as 0.045â‹…n=180.045 \cdot n = 18. To find n, divide both sides by 0.045: n=180.045=400n = \frac{18}{0.045} = 400. So, 4.5% of 400 dollars is 18 dollars.

Section 3

Percent Applications

Property

Many applications of percent occur in our daily lives. To solve these applications, we'll translate to a basic percent equation. Use this problem-solving strategy:

  1. Read the problem to understand it.
  2. Identify what you are looking for.
  3. Name the unknown quantity with a variable.
  4. Translate the problem into an equation.
  5. Solve the equation.
  6. Check if the answer makes sense.
  7. Answer with a complete sentence.

Examples

  • A dinner bill is 72.50 dollars, and you want to leave a 20% tip. How much is the tip?

Let tt be the tip. The problem is: What is 20% of 72.50 dollars? This translates to t=0.20â‹…72.50=14.50t = 0.20 \cdot 72.50 = 14.50. The tip should be 14.50 dollars.

  • A cereal serving provides 120 milligrams of sodium, which is 5% of the recommended daily amount. What is the total recommended amount?

Let aa be the total amount. The problem is: 120 is 5% of what amount? This translates to 120=0.05â‹…a120 = 0.05 \cdot a. Solving for aa gives a=2400a = 2400. The total recommended amount is 2400 mg.

Section 4

Simple Interest

Property

If an amount of money, PP, called the principal, is invested for a period of tt years at an annual interest rate rr, the amount of interest, II, earned is

I=PrtI = Prt

where

  • II = interest
  • PP = principal
  • rr = rate
  • tt = time

Interest earned according to this formula is called simple interest.

Examples

  • How much interest will be earned on a principal of 8,000 dollars invested at an interest rate of 3% for 5 years?

Using the formula I=PrtI = Prt, we substitute the values: I=(8000)(0.03)(5)=1200I = (8000)(0.03)(5) = 1200. The interest earned is 1,200 dollars.

  • A loan of 4,000 dollars was repaid with 640 dollars in interest after 2 years. What was the interest rate?

Using I=PrtI = Prt, we have 640=(4000)(r)(2)640 = (4000)(r)(2). This simplifies to 640=8000r640 = 8000r. Solving for rr gives r=0.08r = 0.08, so the interest rate was 8%.

Section 5

Discount

Property

The formulas for calculating a discount are:

amount of discount=discount rate×original price\text{amount of discount} = \text{discount rate} \times \text{original price}
sale price=original price−amount of discount\text{sale price} = \text{original price} - \text{amount of discount}

Keep in mind that the sale price should always be less than the original price.

Examples

  • A jacket with an original price of 220 dollars is on sale for 25% off. What is the amount of discount and the sale price?

The amount of discount is 0.25×220=550.25 \times 220 = 55 dollars. The sale price is 220−55=165220 - 55 = 165 dollars.

  • A pair of shoes originally priced at 80 dollars is on sale for 60 dollars. What is the amount of discount and the discount rate?

The amount of discount is 80−60=2080 - 60 = 20 dollars. To find the rate, solve 20=r⋅8020 = r \cdot 80, which gives r=0.25r = 0.25, or a 25% discount rate.

Section 6

Mark-Up

Property

The formulas for calculating a mark-up are:

amount of mark-up=mark-up rate×original cost\text{amount of mark-up} = \text{mark-up rate} \times \text{original cost}
list price=original cost+amount of mark up\text{list price} = \text{original cost} + \text{amount of mark up}

Keep in mind that the list price should always be more than the original cost.

Examples

  • An art gallery buys a sculpture for 400 dollars and marks the price up by 60%. What is the amount of mark-up and the list price?

The amount of mark-up is 0.60×400=2400.60 \times 400 = 240 dollars. The list price is 400+240=640400 + 240 = 640 dollars.

  • A music store buys a guitar for 1,500 dollars and sells it for 2,100 dollars. What is the amount of mark-up and the mark-up rate?

The amount of mark-up is 2100−1500=6002100 - 1500 = 600 dollars. To find the rate, solve 600=r⋅1500600 = r \cdot 1500, which gives r=0.4r = 0.4, or a 40% mark-up rate.

Book overview

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

Continue this chapter

Chapter 3: Math Models

  1. Lesson 1

    Lesson 3.1: Use a Problem-Solving Strategy

  2. Lesson 2Current

    Lesson 3.2: Solve Percent Applications

  3. Lesson 3

    Lesson 3.3: Solve Mixture Applications

  4. Lesson 4

    Lesson 3.4: Solve Geometry Applications: Triangles, Rectangles, and the Pythagorean Theorem

  5. Lesson 5

    Lesson 3.5: Solve Uniform Motion Applications

  6. Lesson 6

    Lesson 3.6: Solve Applications with Linear Inequalities

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

📘 Solve Percent Applications

New Concept

This lesson shows how to translate real-world percent problems—like tips, discounts, and interest—into simple algebraic equations. You'll learn to solve for unknown amounts, rates, or base values, making everyday math clear and manageable.

What’s next

Next up, you'll tackle interactive examples and practice cards to solidify your skills in solving these percent applications.

Section 2

Basic Percent Equations

Property

To solve percent problems, we translate English sentences into algebraic equations and then solve them. We must be sure to change the given percent to a decimal when we put it in the equation. The three basic types of percent equations are:

  1. Finding the amount: What number is 35% of 90? translates to n=0.35â‹…90n = 0.35 \cdot 90
  2. Finding the base: 6.5% of what number is 1.17 dollars? translates to 0.065â‹…n=1.170.065 \cdot n = 1.17
  3. Finding the percent: 144 is what percent of 96? translates to 144=pâ‹…96144 = p \cdot 96

Examples

  • What number is 25% of 160?

Translate the sentence into an equation: n=0.25â‹…160n = 0.25 \cdot 160. Solving this gives n=40n = 40. So, 40 is 25% of 160.

  • 4.5% of what number is 18 dollars?

Translate this as 0.045â‹…n=180.045 \cdot n = 18. To find n, divide both sides by 0.045: n=180.045=400n = \frac{18}{0.045} = 400. So, 4.5% of 400 dollars is 18 dollars.

Section 3

Percent Applications

Property

Many applications of percent occur in our daily lives. To solve these applications, we'll translate to a basic percent equation. Use this problem-solving strategy:

  1. Read the problem to understand it.
  2. Identify what you are looking for.
  3. Name the unknown quantity with a variable.
  4. Translate the problem into an equation.
  5. Solve the equation.
  6. Check if the answer makes sense.
  7. Answer with a complete sentence.

Examples

  • A dinner bill is 72.50 dollars, and you want to leave a 20% tip. How much is the tip?

Let tt be the tip. The problem is: What is 20% of 72.50 dollars? This translates to t=0.20â‹…72.50=14.50t = 0.20 \cdot 72.50 = 14.50. The tip should be 14.50 dollars.

  • A cereal serving provides 120 milligrams of sodium, which is 5% of the recommended daily amount. What is the total recommended amount?

Let aa be the total amount. The problem is: 120 is 5% of what amount? This translates to 120=0.05â‹…a120 = 0.05 \cdot a. Solving for aa gives a=2400a = 2400. The total recommended amount is 2400 mg.

Section 4

Simple Interest

Property

If an amount of money, PP, called the principal, is invested for a period of tt years at an annual interest rate rr, the amount of interest, II, earned is

I=PrtI = Prt

where

  • II = interest
  • PP = principal
  • rr = rate
  • tt = time

Interest earned according to this formula is called simple interest.

Examples

  • How much interest will be earned on a principal of 8,000 dollars invested at an interest rate of 3% for 5 years?

Using the formula I=PrtI = Prt, we substitute the values: I=(8000)(0.03)(5)=1200I = (8000)(0.03)(5) = 1200. The interest earned is 1,200 dollars.

  • A loan of 4,000 dollars was repaid with 640 dollars in interest after 2 years. What was the interest rate?

Using I=PrtI = Prt, we have 640=(4000)(r)(2)640 = (4000)(r)(2). This simplifies to 640=8000r640 = 8000r. Solving for rr gives r=0.08r = 0.08, so the interest rate was 8%.

Section 5

Discount

Property

The formulas for calculating a discount are:

amount of discount=discount rate×original price\text{amount of discount} = \text{discount rate} \times \text{original price}
sale price=original price−amount of discount\text{sale price} = \text{original price} - \text{amount of discount}

Keep in mind that the sale price should always be less than the original price.

Examples

  • A jacket with an original price of 220 dollars is on sale for 25% off. What is the amount of discount and the sale price?

The amount of discount is 0.25×220=550.25 \times 220 = 55 dollars. The sale price is 220−55=165220 - 55 = 165 dollars.

  • A pair of shoes originally priced at 80 dollars is on sale for 60 dollars. What is the amount of discount and the discount rate?

The amount of discount is 80−60=2080 - 60 = 20 dollars. To find the rate, solve 20=r⋅8020 = r \cdot 80, which gives r=0.25r = 0.25, or a 25% discount rate.

Section 6

Mark-Up

Property

The formulas for calculating a mark-up are:

amount of mark-up=mark-up rate×original cost\text{amount of mark-up} = \text{mark-up rate} \times \text{original cost}
list price=original cost+amount of mark up\text{list price} = \text{original cost} + \text{amount of mark up}

Keep in mind that the list price should always be more than the original cost.

Examples

  • An art gallery buys a sculpture for 400 dollars and marks the price up by 60%. What is the amount of mark-up and the list price?

The amount of mark-up is 0.60×400=2400.60 \times 400 = 240 dollars. The list price is 400+240=640400 + 240 = 640 dollars.

  • A music store buys a guitar for 1,500 dollars and sells it for 2,100 dollars. What is the amount of mark-up and the mark-up rate?

The amount of mark-up is 2100−1500=6002100 - 1500 = 600 dollars. To find the rate, solve 600=r⋅1500600 = r \cdot 1500, which gives r=0.4r = 0.4, or a 40% mark-up rate.

Book overview

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

Continue this chapter

Chapter 3: Math Models

  1. Lesson 1

    Lesson 3.1: Use a Problem-Solving Strategy

  2. Lesson 2Current

    Lesson 3.2: Solve Percent Applications

  3. Lesson 3

    Lesson 3.3: Solve Mixture Applications

  4. Lesson 4

    Lesson 3.4: Solve Geometry Applications: Triangles, Rectangles, and the Pythagorean Theorem

  5. Lesson 5

    Lesson 3.5: Solve Uniform Motion Applications

  6. Lesson 6

    Lesson 3.6: Solve Applications with Linear Inequalities