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Chapter 58: Simple Interest

Learn Grade 8 Math from beginner foundations through advanced Grade 8 problem solving with detailed explanations, at least five examples per topic, practice exercises, and review questions.

Grade 8Beginner Friendly5+ Examples Per TopicPractice30 Q&A
Estimated reading time0% read
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What this chapter covers

This chapter contains 13 topics. Technical terms are followed by plain-language meanings where they first appear. Each topic includes at least five worked examples. Coding is included only where it naturally supports the Grade 8 coding expectations.

58.1 Principal

Principal is the starting amount of money saved, invested, or borrowed. In this section, the focus is Principal.

For a beginner, focus on the meaning before memorizing a procedure. Start with a small example, name the quantities and units, apply one rule at a time, and connect the result back to the question.

10 Worked Examples with Answers and Very-Beginner Explanations

Worked Example 1

Problem: Find simple interest on $500 at 5% per year for 2 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 500×0.05×2.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $50.00

Worked Example 2

Problem: Find simple interest on $1000 at 4% per year for 3 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 1000×0.04×3.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $120.00

Worked Example 3

Problem: Find simple interest on $750 at 6% per year for 1 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 750×0.06×1.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $45.00

Worked Example 4

Problem: Find simple interest on $1200 at 3.5% per year for 4 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 1200×0.035×4.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $168.00

Worked Example 5

Problem: Find simple interest on $2000 at 2.5% per year for 5 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 2000×0.025×5.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $250.00

Worked Example 6

Problem: Find simple interest on $1500 at 4.5% per year for 2 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 1500×0.045×2.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $135.00

Worked Example 7

Problem: Find simple interest on $2500 at 3% per year for 4 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 2500×0.03×4.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $300.00

Worked Example 8

Problem: Find simple interest on $800 at 7.5% per year for 1 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 800×0.075×1.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $60.00

Worked Example 9

Problem: Find simple interest on $3000 at 2% per year for 5 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 3000×0.02×5.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $300.00

Worked Example 10

Problem: Find simple interest on $600 at 8% per year for 3 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 600×0.08×3.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $144.00

Practice exercise

Create one new Grade 8 problem involving Principal. Show the important steps, include units when needed, and explain how you checked the answer.

58.2 Interest

Interest is an important Grade 8 concept in Simple Interest. Start by identifying the quantities, relationships, units, or rules involved. Work with a small example first, show the important steps, and check whether the result is reasonable.

For a beginner, focus on the meaning before memorizing a procedure. Start with a small example, name the quantities and units, apply one rule at a time, and connect the result back to the question.

10 Worked Examples with Answers and Very-Beginner Explanations

Worked Example 1

Problem: Find simple interest on $500 at 5% per year for 2 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 500×0.05×2.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $50.00

Worked Example 2

Problem: Find simple interest on $1000 at 4% per year for 3 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 1000×0.04×3.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $120.00

Worked Example 3

Problem: Find simple interest on $750 at 6% per year for 1 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 750×0.06×1.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $45.00

Worked Example 4

Problem: Find simple interest on $1200 at 3.5% per year for 4 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 1200×0.035×4.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $168.00

Worked Example 5

Problem: Find simple interest on $2000 at 2.5% per year for 5 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 2000×0.025×5.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $250.00

Worked Example 6

Problem: Find simple interest on $1500 at 4.5% per year for 2 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 1500×0.045×2.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $135.00

Worked Example 7

Problem: Find simple interest on $2500 at 3% per year for 4 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 2500×0.03×4.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $300.00

Worked Example 8

Problem: Find simple interest on $800 at 7.5% per year for 1 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 800×0.075×1.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $60.00

Worked Example 9

Problem: Find simple interest on $3000 at 2% per year for 5 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 3000×0.02×5.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $300.00

Worked Example 10

Problem: Find simple interest on $600 at 8% per year for 3 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 600×0.08×3.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $144.00

Practice exercise

Create one new Grade 8 problem involving Interest. Show the important steps, include units when needed, and explain how you checked the answer.

58.3 Interest rate

A rate compares quantities measured in different units. In this section, the focus is Interest rate.

For a beginner, focus on the meaning before memorizing a procedure. Start with a small example, name the quantities and units, apply one rule at a time, and connect the result back to the question.

10 Worked Examples with Answers and Very-Beginner Explanations

Worked Example 1

Problem: A quantity of 80 units is used over 2 units of time. Find the unit rate.

  1. Divide the first quantity by the second.
  2. 80 ÷ 2 = 40.

Very beginner explanation: A unit rate answers “how much for one?”

Answer: 40 per unit

Worked Example 2

Problem: A quantity of 120 units is used over 3 units of time. Find the unit rate.

  1. Divide the first quantity by the second.
  2. 120 ÷ 3 = 40.

Very beginner explanation: A unit rate answers “how much for one?”

Answer: 40 per unit

Worked Example 3

Problem: A quantity of 190 units is used over 4 units of time. Find the unit rate.

  1. Divide the first quantity by the second.
  2. 190 ÷ 4 = 47.5.

Very beginner explanation: A unit rate answers “how much for one?”

Answer: 47.5 per unit

Worked Example 4

Problem: A quantity of 250 units is used over 5 units of time. Find the unit rate.

  1. Divide the first quantity by the second.
  2. 250 ÷ 5 = 50.

Very beginner explanation: A unit rate answers “how much for one?”

Answer: 50 per unit

Worked Example 5

Problem: A quantity of 220 units is used over 2 units of time. Find the unit rate.

  1. Divide the first quantity by the second.
  2. 220 ÷ 2 = 110.

Very beginner explanation: A unit rate answers “how much for one?”

Answer: 110 per unit

Worked Example 6

Problem: A quantity of 160 units is used over 3 units of time. Find the unit rate.

  1. Divide the first quantity by the second.
  2. 160 ÷ 3 = 53.3333.

Very beginner explanation: A unit rate answers “how much for one?”

Answer: 53.3333 per unit

Worked Example 7

Problem: A quantity of 130 units is used over 4 units of time. Find the unit rate.

  1. Divide the first quantity by the second.
  2. 130 ÷ 4 = 32.5.

Very beginner explanation: A unit rate answers “how much for one?”

Answer: 32.5 per unit

Worked Example 8

Problem: A quantity of 200 units is used over 5 units of time. Find the unit rate.

  1. Divide the first quantity by the second.
  2. 200 ÷ 5 = 40.

Very beginner explanation: A unit rate answers “how much for one?”

Answer: 40 per unit

Worked Example 9

Problem: A quantity of 290 units is used over 2 units of time. Find the unit rate.

  1. Divide the first quantity by the second.
  2. 290 ÷ 2 = 145.

Very beginner explanation: A unit rate answers “how much for one?”

Answer: 145 per unit

Worked Example 10

Problem: A quantity of 180 units is used over 3 units of time. Find the unit rate.

  1. Divide the first quantity by the second.
  2. 180 ÷ 3 = 60.

Very beginner explanation: A unit rate answers “how much for one?”

Answer: 60 per unit

Practice exercise

Create one new Grade 8 problem involving Interest rate. Show the important steps, include units when needed, and explain how you checked the answer.

58.4 Time

Time is an important Grade 8 concept in Simple Interest. Start by identifying the quantities, relationships, units, or rules involved. Work with a small example first, show the important steps, and check whether the result is reasonable.

For a beginner, focus on the meaning before memorizing a procedure. Start with a small example, name the quantities and units, apply one rule at a time, and connect the result back to the question.

10 Worked Examples with Answers and Very-Beginner Explanations

Worked Example 1

Problem: Time: Explain Time in one simple sentence.

  1. Look at the words in “Time”.
  2. State what the idea is used for.
  3. Give one small mathematical example.

Very beginner explanation: A beginner should first understand the meaning before memorizing a rule.

Answer: Time is a Grade 8 math idea used to describe or solve a specific mathematical relationship.

Worked Example 2

Problem: Time: A student says, “I can use Time without checking units or labels.” Is that a good method?

  1. Read the statement.
  2. Ask whether units, signs, labels, or conditions matter.
  3. Decide whether the method is safe.

Very beginner explanation: Many math mistakes come from using a correct calculation on the wrong quantity.

Answer: No. Important units, labels, signs, and conditions must be checked.

Worked Example 3

Problem: Time: What is the first step when solving a problem about Time?

  1. Read the whole question.
  2. Underline the known information.
  3. Identify what must be found.

Very beginner explanation: This prevents you from calculating before you understand the problem.

Answer: Identify the given information and the unknown before choosing a rule.

Worked Example 4

Problem: Time: After solving a Time problem, what should you do before accepting the answer?

  1. Estimate the expected size or direction.
  2. Check units and signs.
  3. Use an inverse method if possible.

Very beginner explanation: Checking is part of solving, not an optional extra.

Answer: Check whether the answer is reasonable and consistent with the problem.

Worked Example 5

Problem: Time: Give one way Time could appear outside a textbook.

  1. Think about money, measurements, data, maps, geometry, or technology.
  2. Connect the topic to one of those situations.

Very beginner explanation: Connecting math to real situations makes the rule easier to remember.

Answer: Time can be used in a real-life situation where quantities must be compared, measured, predicted, or calculated.

Worked Example 6

Problem: Time: Which representation could help explain Time: a table, graph, diagram, equation, or number line?

  1. Choose the representation that makes the relationship easiest to see.
  2. Label it clearly.

Very beginner explanation: Different representations show different features of the same mathematics.

Answer: Use the representation that best matches the problem; more than one may be valid.

Worked Example 7

Problem: Time: A student gets an answer for Time but cannot explain the steps. What should be improved?

  1. Rewrite the solution one step at a time.
  2. Name the rule used at each important step.

Very beginner explanation: A correct final number without reasoning may hide an error.

Answer: The reasoning should be shown so the solution can be checked.

Worked Example 8

Problem: Time: Why can estimation help before a detailed Time calculation?

  1. Round or use benchmark values.
  2. Predict the approximate answer.
  3. Compare the exact result with the estimate.

Very beginner explanation: If the exact answer is far from the estimate, recheck the work.

Answer: Estimation gives a target range for the final answer.

Worked Example 9

Problem: Time: How can you test whether your rule for Time works?

  1. Choose a very small easy example.
  2. Apply the rule.
  3. Check the result another way.

Very beginner explanation: Small examples expose mistakes quickly.

Answer: Test the rule on a simple case whose answer can be verified.

Worked Example 10

Problem: Time: How would you teach Time to someone seeing it for the first time?

  1. Define the idea.
  2. Show one easy example.
  3. Explain every step.
  4. Then let the learner try a similar problem.

Very beginner explanation: This sequence reduces memorization without understanding.

Answer: Teach meaning first, then a small worked example, then guided practice.

Practice exercise

Create one new Grade 8 problem involving Time. Show the important steps, include units when needed, and explain how you checked the answer.

58.5 Simple-interest formula I = Prt

A formula is a rule written with symbols to show how quantities are related. In this section, the focus is Simple-interest formula I = Prt.

For a beginner, focus on the meaning before memorizing a procedure. Start with a small example, name the quantities and units, apply one rule at a time, and connect the result back to the question.

10 Worked Examples with Answers and Very-Beginner Explanations

Worked Example 1

Problem: Solve 2x + 5 = 17.

  1. Simplify each side if needed.
  2. Use inverse operations to move variable terms and constants.
  3. Keep both sides balanced.
  4. Check by substitution when a single solution exists.

Very beginner explanation: An equation is like a balance: whatever operation is done to one side must preserve equality.

Answer: x = 6

Worked Example 2

Problem: Solve 3x - 4 = 11.

  1. Simplify each side if needed.
  2. Use inverse operations to move variable terms and constants.
  3. Keep both sides balanced.
  4. Check by substitution when a single solution exists.

Very beginner explanation: An equation is like a balance: whatever operation is done to one side must preserve equality.

Answer: x = 5

Worked Example 3

Problem: Solve 5x + 7 = 2x + 22.

  1. Simplify each side if needed.
  2. Use inverse operations to move variable terms and constants.
  3. Keep both sides balanced.
  4. Check by substitution when a single solution exists.

Very beginner explanation: An equation is like a balance: whatever operation is done to one side must preserve equality.

Answer: x = 5

Worked Example 4

Problem: Solve 4(x + 2) = 24.

  1. Simplify each side if needed.
  2. Use inverse operations to move variable terms and constants.
  3. Keep both sides balanced.
  4. Check by substitution when a single solution exists.

Very beginner explanation: An equation is like a balance: whatever operation is done to one side must preserve equality.

Answer: x = 4

Worked Example 5

Problem: Solve 7x - 3 = 4x + 12.

  1. Simplify each side if needed.
  2. Use inverse operations to move variable terms and constants.
  3. Keep both sides balanced.
  4. Check by substitution when a single solution exists.

Very beginner explanation: An equation is like a balance: whatever operation is done to one side must preserve equality.

Answer: x = 5

Worked Example 6

Problem: Solve 0.5x + 2 = 7.

  1. Simplify each side if needed.
  2. Use inverse operations to move variable terms and constants.
  3. Keep both sides balanced.
  4. Check by substitution when a single solution exists.

Very beginner explanation: An equation is like a balance: whatever operation is done to one side must preserve equality.

Answer: x = 10

Worked Example 7

Problem: Solve 2(x-3)+4=10.

  1. Simplify each side if needed.
  2. Use inverse operations to move variable terms and constants.
  3. Keep both sides balanced.
  4. Check by substitution when a single solution exists.

Very beginner explanation: An equation is like a balance: whatever operation is done to one side must preserve equality.

Answer: x = 6

Worked Example 8

Problem: Solve 9 - 2x = 1.

  1. Simplify each side if needed.
  2. Use inverse operations to move variable terms and constants.
  3. Keep both sides balanced.
  4. Check by substitution when a single solution exists.

Very beginner explanation: An equation is like a balance: whatever operation is done to one side must preserve equality.

Answer: x = 4

Worked Example 9

Problem: Solve 3x + 8 = 3x + 8.

  1. Simplify each side if needed.
  2. Use inverse operations to move variable terms and constants.
  3. Keep both sides balanced.
  4. Check by substitution when a single solution exists.

Very beginner explanation: An equation is like a balance: whatever operation is done to one side must preserve equality.

Answer: all real numbers

Worked Example 10

Problem: Solve 4x + 1 = 4x + 9.

  1. Simplify each side if needed.
  2. Use inverse operations to move variable terms and constants.
  3. Keep both sides balanced.
  4. Check by substitution when a single solution exists.

Very beginner explanation: An equation is like a balance: whatever operation is done to one side must preserve equality.

Answer: no solution

Practice exercise

Create one new Grade 8 problem involving Simple-interest formula I = Prt. Show the important steps, include units when needed, and explain how you checked the answer.

58.6 Converting percent rate to decimal

A decimal uses place value to represent parts of one, including tenths, hundredths, thousandths, and smaller values. In this section, the focus is Converting percent rate to decimal.

For a beginner, focus on the meaning before memorizing a procedure. Start with a small example, name the quantities and units, apply one rule at a time, and connect the result back to the question.

10 Worked Examples with Answers and Very-Beginner Explanations

Worked Example 1

Problem: Which is greater: 3.6 or 0.4?

  1. Compare whole-number parts first.
  2. Then compare tenths, hundredths, and later decimal places.

Very beginner explanation: Decimal comparison is place-value comparison.

Answer: 3.6

Worked Example 2

Problem: Which is greater: 8.25 or 1.5?

  1. Compare whole-number parts first.
  2. Then compare tenths, hundredths, and later decimal places.

Very beginner explanation: Decimal comparison is place-value comparison.

Answer: 8.25

Worked Example 3

Problem: Which is greater: 12.75 or 2.4?

  1. Compare whole-number parts first.
  2. Then compare tenths, hundredths, and later decimal places.

Very beginner explanation: Decimal comparison is place-value comparison.

Answer: 12.75

Worked Example 4

Problem: Which is greater: 0.96 or 0.3?

  1. Compare whole-number parts first.
  2. Then compare tenths, hundredths, and later decimal places.

Very beginner explanation: Decimal comparison is place-value comparison.

Answer: 0.96

Worked Example 5

Problem: Which is greater: 5.04 or 1.2?

  1. Compare whole-number parts first.
  2. Then compare tenths, hundredths, and later decimal places.

Very beginner explanation: Decimal comparison is place-value comparison.

Answer: 5.04

Worked Example 6

Problem: Which is greater: 14.8 or 2?

  1. Compare whole-number parts first.
  2. Then compare tenths, hundredths, and later decimal places.

Very beginner explanation: Decimal comparison is place-value comparison.

Answer: 14.8

Worked Example 7

Problem: Which is greater: 7.125 or 0.5?

  1. Compare whole-number parts first.
  2. Then compare tenths, hundredths, and later decimal places.

Very beginner explanation: Decimal comparison is place-value comparison.

Answer: 7.125

Worked Example 8

Problem: Which is greater: 20.05 or 3.75?

  1. Compare whole-number parts first.
  2. Then compare tenths, hundredths, and later decimal places.

Very beginner explanation: Decimal comparison is place-value comparison.

Answer: 20.05

Worked Example 9

Problem: Which is greater: 2.4 or 0.06?

  1. Compare whole-number parts first.
  2. Then compare tenths, hundredths, and later decimal places.

Very beginner explanation: Decimal comparison is place-value comparison.

Answer: 2.4

Worked Example 10

Problem: Which is greater: 100.5 or 4.02?

  1. Compare whole-number parts first.
  2. Then compare tenths, hundredths, and later decimal places.

Very beginner explanation: Decimal comparison is place-value comparison.

Answer: 100.5

Practice exercise

Create one new Grade 8 problem involving Converting percent rate to decimal. Show the important steps, include units when needed, and explain how you checked the answer.

58.7 Calculating simple interest

Simple interest is calculated only on the original principal using I = Prt. In this section, the focus is Calculating simple interest.

For a beginner, focus on the meaning before memorizing a procedure. Start with a small example, name the quantities and units, apply one rule at a time, and connect the result back to the question.

10 Worked Examples with Answers and Very-Beginner Explanations

Worked Example 1

Problem: Find simple interest on $500 at 5% per year for 2 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 500×0.05×2.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $50.00

Worked Example 2

Problem: Find simple interest on $1000 at 4% per year for 3 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 1000×0.04×3.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $120.00

Worked Example 3

Problem: Find simple interest on $750 at 6% per year for 1 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 750×0.06×1.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $45.00

Worked Example 4

Problem: Find simple interest on $1200 at 3.5% per year for 4 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 1200×0.035×4.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $168.00

Worked Example 5

Problem: Find simple interest on $2000 at 2.5% per year for 5 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 2000×0.025×5.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $250.00

Worked Example 6

Problem: Find simple interest on $1500 at 4.5% per year for 2 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 1500×0.045×2.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $135.00

Worked Example 7

Problem: Find simple interest on $2500 at 3% per year for 4 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 2500×0.03×4.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $300.00

Worked Example 8

Problem: Find simple interest on $800 at 7.5% per year for 1 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 800×0.075×1.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $60.00

Worked Example 9

Problem: Find simple interest on $3000 at 2% per year for 5 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 3000×0.02×5.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $300.00

Worked Example 10

Problem: Find simple interest on $600 at 8% per year for 3 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 600×0.08×3.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $144.00

Practice exercise

Create one new Grade 8 problem involving Calculating simple interest. Show the important steps, include units when needed, and explain how you checked the answer.

58.8 Calculating final amount

Calculating final amount is an important Grade 8 concept in Simple Interest. Start by identifying the quantities, relationships, units, or rules involved. Work with a small example first, show the important steps, and check whether the result is reasonable.

For a beginner, focus on the meaning before memorizing a procedure. Start with a small example, name the quantities and units, apply one rule at a time, and connect the result back to the question.

10 Worked Examples with Answers and Very-Beginner Explanations

Worked Example 1

Problem: Find 10% of 80.

  1. Convert 10% to decimal 0.1.
  2. Multiply 0.1 × 80.

Very beginner explanation: “Of” usually means multiply in a percent-of-a-number problem.

Answer: 8

Worked Example 2

Problem: Find 25% of 60.

  1. Convert 25% to decimal 0.25.
  2. Multiply 0.25 × 60.

Very beginner explanation: “Of” usually means multiply in a percent-of-a-number problem.

Answer: 15

Worked Example 3

Problem: Find 15% of 120.

  1. Convert 15% to decimal 0.15.
  2. Multiply 0.15 × 120.

Very beginner explanation: “Of” usually means multiply in a percent-of-a-number problem.

Answer: 18

Worked Example 4

Problem: Find 5% of 250.

  1. Convert 5% to decimal 0.05.
  2. Multiply 0.05 × 250.

Very beginner explanation: “Of” usually means multiply in a percent-of-a-number problem.

Answer: 12.5

Worked Example 5

Problem: Find 13% of 75.

  1. Convert 13% to decimal 0.13.
  2. Multiply 0.13 × 75.

Very beginner explanation: “Of” usually means multiply in a percent-of-a-number problem.

Answer: 9.75

Worked Example 6

Problem: Find 35% of 140.

  1. Convert 35% to decimal 0.35.
  2. Multiply 0.35 × 140.

Very beginner explanation: “Of” usually means multiply in a percent-of-a-number problem.

Answer: 49

Worked Example 7

Problem: Find 8% of 500.

  1. Convert 8% to decimal 0.08.
  2. Multiply 0.08 × 500.

Very beginner explanation: “Of” usually means multiply in a percent-of-a-number problem.

Answer: 40

Worked Example 8

Problem: Find 60% of 45.

  1. Convert 60% to decimal 0.6.
  2. Multiply 0.6 × 45.

Very beginner explanation: “Of” usually means multiply in a percent-of-a-number problem.

Answer: 27

Worked Example 9

Problem: Find 2.5% of 320.

  1. Convert 2.5% to decimal 0.025.
  2. Multiply 0.025 × 320.

Very beginner explanation: “Of” usually means multiply in a percent-of-a-number problem.

Answer: 8

Worked Example 10

Problem: Find 125% of 64.

  1. Convert 125% to decimal 1.25.
  2. Multiply 1.25 × 64.

Very beginner explanation: “Of” usually means multiply in a percent-of-a-number problem.

Answer: 80

Practice exercise

Create one new Grade 8 problem involving Calculating final amount. Show the important steps, include units when needed, and explain how you checked the answer.

58.9 Comparing simple-interest options

Comparing simple-interest options is an important Grade 8 concept in Simple Interest. Start by identifying the quantities, relationships, units, or rules involved. Work with a small example first, show the important steps, and check whether the result is reasonable.

For a beginner, focus on the meaning before memorizing a procedure. Start with a small example, name the quantities and units, apply one rule at a time, and connect the result back to the question.

10 Worked Examples with Answers and Very-Beginner Explanations

Worked Example 1

Problem: Find simple interest on $500 at 5% per year for 2 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 500×0.05×2.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $50.00

Worked Example 2

Problem: Find simple interest on $1000 at 4% per year for 3 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 1000×0.04×3.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $120.00

Worked Example 3

Problem: Find simple interest on $750 at 6% per year for 1 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 750×0.06×1.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $45.00

Worked Example 4

Problem: Find simple interest on $1200 at 3.5% per year for 4 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 1200×0.035×4.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $168.00

Worked Example 5

Problem: Find simple interest on $2000 at 2.5% per year for 5 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 2000×0.025×5.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $250.00

Worked Example 6

Problem: Find simple interest on $1500 at 4.5% per year for 2 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 1500×0.045×2.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $135.00

Worked Example 7

Problem: Find simple interest on $2500 at 3% per year for 4 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 2500×0.03×4.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $300.00

Worked Example 8

Problem: Find simple interest on $800 at 7.5% per year for 1 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 800×0.075×1.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $60.00

Worked Example 9

Problem: Find simple interest on $3000 at 2% per year for 5 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 3000×0.02×5.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $300.00

Worked Example 10

Problem: Find simple interest on $600 at 8% per year for 3 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 600×0.08×3.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $144.00

Practice exercise

Create one new Grade 8 problem involving Comparing simple-interest options. Show the important steps, include units when needed, and explain how you checked the answer.

58.10 Savings examples

Savings examples is an important Grade 8 concept in Simple Interest. Start by identifying the quantities, relationships, units, or rules involved. Work with a small example first, show the important steps, and check whether the result is reasonable.

For a beginner, focus on the meaning before memorizing a procedure. Start with a small example, name the quantities and units, apply one rule at a time, and connect the result back to the question.

10 Worked Examples with Answers and Very-Beginner Explanations

Worked Example 1

Problem: Monthly income is $1500.00 and expenses are $1100.00. Find the budget balance.

  1. Subtract expenses from income.
  2. 1500.00 - 1100.00 = 400.00.

Very beginner explanation: A positive balance is a surplus; a negative balance is a deficit.

Answer: $400.00 surplus

Worked Example 2

Problem: Monthly income is $2000.00 and expenses are $1600.00. Find the budget balance.

  1. Subtract expenses from income.
  2. 2000.00 - 1600.00 = 400.00.

Very beginner explanation: A positive balance is a surplus; a negative balance is a deficit.

Answer: $400.00 surplus

Worked Example 3

Problem: Monthly income is $1750.00 and expenses are $1350.00. Find the budget balance.

  1. Subtract expenses from income.
  2. 1750.00 - 1350.00 = 400.00.

Very beginner explanation: A positive balance is a surplus; a negative balance is a deficit.

Answer: $400.00 surplus

Worked Example 4

Problem: Monthly income is $2200.00 and expenses are $1800.00. Find the budget balance.

  1. Subtract expenses from income.
  2. 2200.00 - 1800.00 = 400.00.

Very beginner explanation: A positive balance is a surplus; a negative balance is a deficit.

Answer: $400.00 surplus

Worked Example 5

Problem: Monthly income is $3000.00 and expenses are $2600.00. Find the budget balance.

  1. Subtract expenses from income.
  2. 3000.00 - 2600.00 = 400.00.

Very beginner explanation: A positive balance is a surplus; a negative balance is a deficit.

Answer: $400.00 surplus

Worked Example 6

Problem: Monthly income is $2500.00 and expenses are $2100.00. Find the budget balance.

  1. Subtract expenses from income.
  2. 2500.00 - 2100.00 = 400.00.

Very beginner explanation: A positive balance is a surplus; a negative balance is a deficit.

Answer: $400.00 surplus

Worked Example 7

Problem: Monthly income is $3500.00 and expenses are $3100.00. Find the budget balance.

  1. Subtract expenses from income.
  2. 3500.00 - 3100.00 = 400.00.

Very beginner explanation: A positive balance is a surplus; a negative balance is a deficit.

Answer: $400.00 surplus

Worked Example 8

Problem: Monthly income is $1800.00 and expenses are $1400.00. Find the budget balance.

  1. Subtract expenses from income.
  2. 1800.00 - 1400.00 = 400.00.

Very beginner explanation: A positive balance is a surplus; a negative balance is a deficit.

Answer: $400.00 surplus

Worked Example 9

Problem: Monthly income is $4000.00 and expenses are $3600.00. Find the budget balance.

  1. Subtract expenses from income.
  2. 4000.00 - 3600.00 = 400.00.

Very beginner explanation: A positive balance is a surplus; a negative balance is a deficit.

Answer: $400.00 surplus

Worked Example 10

Problem: Monthly income is $1600.00 and expenses are $1200.00. Find the budget balance.

  1. Subtract expenses from income.
  2. 1600.00 - 1200.00 = 400.00.

Very beginner explanation: A positive balance is a surplus; a negative balance is a deficit.

Answer: $400.00 surplus

Practice exercise

Create one new Grade 8 problem involving Savings examples. Show the important steps, include units when needed, and explain how you checked the answer.

58.11 Loan examples

Loan examples is an important Grade 8 concept in Simple Interest. Start by identifying the quantities, relationships, units, or rules involved. Work with a small example first, show the important steps, and check whether the result is reasonable.

For a beginner, focus on the meaning before memorizing a procedure. Start with a small example, name the quantities and units, apply one rule at a time, and connect the result back to the question.

10 Worked Examples with Answers and Very-Beginner Explanations

Worked Example 1

Problem: Find simple interest on $500 at 5% per year for 2 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 500×0.05×2.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $50.00

Worked Example 2

Problem: Find simple interest on $1000 at 4% per year for 3 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 1000×0.04×3.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $120.00

Worked Example 3

Problem: Find simple interest on $750 at 6% per year for 1 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 750×0.06×1.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $45.00

Worked Example 4

Problem: Find simple interest on $1200 at 3.5% per year for 4 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 1200×0.035×4.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $168.00

Worked Example 5

Problem: Find simple interest on $2000 at 2.5% per year for 5 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 2000×0.025×5.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $250.00

Worked Example 6

Problem: Find simple interest on $1500 at 4.5% per year for 2 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 1500×0.045×2.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $135.00

Worked Example 7

Problem: Find simple interest on $2500 at 3% per year for 4 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 2500×0.03×4.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $300.00

Worked Example 8

Problem: Find simple interest on $800 at 7.5% per year for 1 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 800×0.075×1.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $60.00

Worked Example 9

Problem: Find simple interest on $3000 at 2% per year for 5 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 3000×0.02×5.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $300.00

Worked Example 10

Problem: Find simple interest on $600 at 8% per year for 3 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 600×0.08×3.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $144.00

Practice exercise

Create one new Grade 8 problem involving Loan examples. Show the important steps, include units when needed, and explain how you checked the answer.

58.12 Spreadsheet simple-interest calculations

Spreadsheet simple-interest calculations is an important Grade 8 concept in Simple Interest. Start by identifying the quantities, relationships, units, or rules involved. Work with a small example first, show the important steps, and check whether the result is reasonable.

For a beginner, focus on the meaning before memorizing a procedure. Start with a small example, name the quantities and units, apply one rule at a time, and connect the result back to the question.

10 Worked Examples with Answers and Very-Beginner Explanations

Worked Example 1

Problem: Find simple interest on $500 at 5% per year for 2 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 500×0.05×2.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $50.00

Worked Example 2

Problem: Find simple interest on $1000 at 4% per year for 3 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 1000×0.04×3.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $120.00

Worked Example 3

Problem: Find simple interest on $750 at 6% per year for 1 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 750×0.06×1.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $45.00

Worked Example 4

Problem: Find simple interest on $1200 at 3.5% per year for 4 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 1200×0.035×4.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $168.00

Worked Example 5

Problem: Find simple interest on $2000 at 2.5% per year for 5 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 2000×0.025×5.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $250.00

Worked Example 6

Problem: Find simple interest on $1500 at 4.5% per year for 2 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 1500×0.045×2.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $135.00

Worked Example 7

Problem: Find simple interest on $2500 at 3% per year for 4 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 2500×0.03×4.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $300.00

Worked Example 8

Problem: Find simple interest on $800 at 7.5% per year for 1 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 800×0.075×1.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $60.00

Worked Example 9

Problem: Find simple interest on $3000 at 2% per year for 5 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 3000×0.02×5.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $300.00

Worked Example 10

Problem: Find simple interest on $600 at 8% per year for 3 years.

  1. Convert the percent rate to a decimal.
  2. Use I = Prt.
  3. I = 600×0.08×3.

Very beginner explanation: Simple interest is calculated only on the original principal.

Answer: $144.00

Practice exercise

Create one new Grade 8 problem involving Spreadsheet simple-interest calculations. Show the important steps, include units when needed, and explain how you checked the answer.

58.13 Long-term planning with simple interest

A term is one part of an algebraic expression separated by addition or subtraction signs. In this section, the focus is Long-term planning with simple interest.

For a beginner, focus on the meaning before memorizing a procedure. Start with a small example, name the quantities and units, apply one rule at a time, and connect the result back to the question.

10 Worked Examples with Answers and Very-Beginner Explanations

Worked Example 1

Problem: Simplify 3x + 2x.

  1. 3x and 2x are like terms.
  2. Add the coefficients: 3 + 2 = 5.

Very beginner explanation: Like terms have exactly the same variable part. Distribution multiplies every term inside parentheses.

Answer: 5x

Worked Example 2

Problem: Simplify 7y - 4y + 3.

  1. 7y and -4y are like terms.
  2. Combine them; keep the constant 3.

Very beginner explanation: Like terms have exactly the same variable part. Distribution multiplies every term inside parentheses.

Answer: 3y + 3

Worked Example 3

Problem: Simplify 4(a + 3).

  1. Multiply 4 by a.
  2. Multiply 4 by 3.

Very beginner explanation: Like terms have exactly the same variable part. Distribution multiplies every term inside parentheses.

Answer: 4a + 12

Worked Example 4

Problem: Simplify 2(3x - 5) + x.

  1. Distribute 2.
  2. 2(3x - 5)=6x-10.
  3. Combine 6x+x.

Very beginner explanation: Like terms have exactly the same variable part. Distribution multiplies every term inside parentheses.

Answer: 7x - 10

Worked Example 5

Problem: Simplify 5m + 8 - 2m - 3.

  1. Combine variable terms.
  2. Combine constant terms.

Very beginner explanation: Like terms have exactly the same variable part. Distribution multiplies every term inside parentheses.

Answer: 3m + 5

Worked Example 6

Problem: Simplify -3(2p + 4).

  1. Multiply -3 by both terms.
  2. Keep signs carefully.

Very beginner explanation: Like terms have exactly the same variable part. Distribution multiplies every term inside parentheses.

Answer: -6p - 12

Worked Example 7

Problem: Simplify 6x + 4 + x - 9.

  1. Combine x-terms.
  2. Combine constants.

Very beginner explanation: Like terms have exactly the same variable part. Distribution multiplies every term inside parentheses.

Answer: 7x - 5

Worked Example 8

Problem: Simplify 0.5x + 1.5x.

  1. Both terms have x.
  2. Add decimal coefficients 0.5 + 1.5.

Very beginner explanation: Like terms have exactly the same variable part. Distribution multiplies every term inside parentheses.

Answer: 2x

Worked Example 9

Problem: Simplify 3(2a + 1) - a.

  1. Distribute 3.
  2. Combine 6a-a.

Very beginner explanation: Like terms have exactly the same variable part. Distribution multiplies every term inside parentheses.

Answer: 5a + 3

Worked Example 10

Problem: Simplify 8q - 2(q + 3).

  1. Distribute -2.
  2. Combine 8q-2q.

Very beginner explanation: Like terms have exactly the same variable part. Distribution multiplies every term inside parentheses.

Answer: 6q - 6

Practice exercise

Create one new Grade 8 problem involving Long-term planning with simple interest. Show the important steps, include units when needed, and explain how you checked the answer.

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Chapter 58 Review Questions and Answers

Q1. What is important to remember about Principal?

Answer: Principal is the starting amount of money saved, invested, or borrowed. In this section, the focus is Principal.

Q2. What is important to remember about Interest?

Answer: Interest is an important Grade 8 concept in Simple Interest. Start by identifying the quantities, relationships, units, or rules involved. Work with a small example first, show the important steps, and check whether the result is reasonable.

Q3. What is important to remember about Interest rate?

Answer: A rate compares quantities measured in different units. In this section, the focus is Interest rate.

Q4. What is important to remember about Time?

Answer: Time is an important Grade 8 concept in Simple Interest. Start by identifying the quantities, relationships, units, or rules involved. Work with a small example first, show the important steps, and check whether the result is reasonable.

Q5. What is important to remember about Simple-interest formula I = Prt?

Answer: A formula is a rule written with symbols to show how quantities are related. In this section, the focus is Simple-interest formula I = Prt.

Q6. What is important to remember about Converting percent rate to decimal?

Answer: A decimal uses place value to represent parts of one, including tenths, hundredths, thousandths, and smaller values. In this section, the focus is Converting percent rate to decimal.

Q7. What is important to remember about Calculating simple interest?

Answer: Simple interest is calculated only on the original principal using I = Prt. In this section, the focus is Calculating simple interest.

Q8. What is important to remember about Calculating final amount?

Answer: Calculating final amount is an important Grade 8 concept in Simple Interest. Start by identifying the quantities, relationships, units, or rules involved. Work with a small example first, show the important steps, and check whether the result is reasonable.

Q9. What is important to remember about Comparing simple-interest options?

Answer: Comparing simple-interest options is an important Grade 8 concept in Simple Interest. Start by identifying the quantities, relationships, units, or rules involved. Work with a small example first, show the important steps, and check whether the result is reasonable.

Q10. What is important to remember about Savings examples?

Answer: Savings examples is an important Grade 8 concept in Simple Interest. Start by identifying the quantities, relationships, units, or rules involved. Work with a small example first, show the important steps, and check whether the result is reasonable.

Q11. What is important to remember about Loan examples?

Answer: Loan examples is an important Grade 8 concept in Simple Interest. Start by identifying the quantities, relationships, units, or rules involved. Work with a small example first, show the important steps, and check whether the result is reasonable.

Q12. What is important to remember about Spreadsheet simple-interest calculations?

Answer: Spreadsheet simple-interest calculations is an important Grade 8 concept in Simple Interest. Start by identifying the quantities, relationships, units, or rules involved. Work with a small example first, show the important steps, and check whether the result is reasonable.

Q13. What is important to remember about Long-term planning with simple interest?

Answer: A term is one part of an algebraic expression separated by addition or subtraction signs. In this section, the focus is Long-term planning with simple interest.

Q14. Why should you show your steps?

Answer: Showing steps makes reasoning easier to verify and helps locate mistakes.

Q15. Why is estimation useful?

Answer: Estimation helps you judge whether a calculated answer is reasonable.

Q16. Why do units matter?

Answer: Units tell what a value measures and prevent incompatible quantities from being mixed.

Q17. When should you round?

Answer: Usually round near the end unless the question specifically asks for earlier rounding.

Q18. How can you check an answer?

Answer: Use an inverse operation, substitution, estimation, a graph, or a second method.

Q19. What should you identify first in a word problem?

Answer: Identify what is known, what is unknown, and the mathematical relationship connecting them.

Q20. What makes a final answer complete?

Answer: Give the value, correct units when needed, and a brief interpretation in context.

Q21. What should you do if an answer seems unreasonable?

Answer: Re-read the question and check copied values, signs, units, formulas, and calculations.

Q22. Why are diagrams useful?

Answer: A labelled diagram can reveal relationships among lengths, angles, areas, and coordinates.

Q23. Why are tables useful?

Answer: Tables organize values and help reveal patterns, rates, relationships, and missing information.

Q24. Why should you explain your reasoning?

Answer: Reasoning shows why a method works, not just what answer was obtained.

Q25. How do mistakes help learning?

Answer: Analyzing a mistake identifies the misunderstood rule or step and helps prevent repetition.

Q26. When is a calculator useful?

Answer: A calculator helps with lengthy arithmetic after the mathematical setup is understood and estimated.

Q27. Why compare more than one strategy?

Answer: Different strategies may make a problem easier and provide a way to verify the result.

Q28. What is mathematical communication?

Answer: It is presenting ideas clearly with words, symbols, diagrams, tables, graphs, and justified steps.

Q29. What is mathematical modelling?

Answer: It is representing a real situation with mathematics, testing the model, and revising it if needed.

Q30. Why should assumptions be stated?

Answer: Assumptions show the conditions the model depends on and help readers judge whether it is reasonable.