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Chapter 15: Ratios, Rates, and Unit Rates

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.

15.1 Meaning of ratio

A ratio compares two quantities in a fixed order. In this section, the focus is Meaning of ratio.

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 the ratio 4:6.

  1. Find the GCF: 2.
  2. Divide both terms by 2.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 2:3

Worked Example 2

Problem: Simplify the ratio 8:12.

  1. Find the GCF: 4.
  2. Divide both terms by 4.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 2:3

Worked Example 3

Problem: Simplify the ratio 15:25.

  1. Find the GCF: 5.
  2. Divide both terms by 5.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 3:5

Worked Example 4

Problem: Simplify the ratio 21:28.

  1. Find the GCF: 7.
  2. Divide both terms by 7.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 3:4

Worked Example 5

Problem: Simplify the ratio 18:30.

  1. Find the GCF: 6.
  2. Divide both terms by 6.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 3:5

Worked Example 6

Problem: Simplify the ratio 12:20.

  1. Find the GCF: 4.
  2. Divide both terms by 4.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 3:5

Worked Example 7

Problem: Simplify the ratio 9:15.

  1. Find the GCF: 3.
  2. Divide both terms by 3.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 3:5

Worked Example 8

Problem: Simplify the ratio 16:24.

  1. Find the GCF: 8.
  2. Divide both terms by 8.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 2:3

Worked Example 9

Problem: Simplify the ratio 25:35.

  1. Find the GCF: 5.
  2. Divide both terms by 5.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 5:7

Worked Example 10

Problem: Simplify the ratio 14:49.

  1. Find the GCF: 7.
  2. Divide both terms by 7.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 2:7

Practice exercise

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

15.2 Part-to-part ratios

A ratio compares two quantities in a fixed order. In this section, the focus is Part-to-part ratios.

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 the ratio 4:6.

  1. Find the GCF: 2.
  2. Divide both terms by 2.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 2:3

Worked Example 2

Problem: Simplify the ratio 8:12.

  1. Find the GCF: 4.
  2. Divide both terms by 4.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 2:3

Worked Example 3

Problem: Simplify the ratio 15:25.

  1. Find the GCF: 5.
  2. Divide both terms by 5.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 3:5

Worked Example 4

Problem: Simplify the ratio 21:28.

  1. Find the GCF: 7.
  2. Divide both terms by 7.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 3:4

Worked Example 5

Problem: Simplify the ratio 18:30.

  1. Find the GCF: 6.
  2. Divide both terms by 6.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 3:5

Worked Example 6

Problem: Simplify the ratio 12:20.

  1. Find the GCF: 4.
  2. Divide both terms by 4.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 3:5

Worked Example 7

Problem: Simplify the ratio 9:15.

  1. Find the GCF: 3.
  2. Divide both terms by 3.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 3:5

Worked Example 8

Problem: Simplify the ratio 16:24.

  1. Find the GCF: 8.
  2. Divide both terms by 8.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 2:3

Worked Example 9

Problem: Simplify the ratio 25:35.

  1. Find the GCF: 5.
  2. Divide both terms by 5.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 5:7

Worked Example 10

Problem: Simplify the ratio 14:49.

  1. Find the GCF: 7.
  2. Divide both terms by 7.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 2:7

Practice exercise

Create one new Grade 8 problem involving Part-to-part ratios. Show the important steps, include units when needed, and explain how you checked the answer.

15.3 Part-to-whole ratios

A ratio compares two quantities in a fixed order. In this section, the focus is Part-to-whole ratios.

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 the ratio 4:6.

  1. Find the GCF: 2.
  2. Divide both terms by 2.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 2:3

Worked Example 2

Problem: Simplify the ratio 8:12.

  1. Find the GCF: 4.
  2. Divide both terms by 4.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 2:3

Worked Example 3

Problem: Simplify the ratio 15:25.

  1. Find the GCF: 5.
  2. Divide both terms by 5.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 3:5

Worked Example 4

Problem: Simplify the ratio 21:28.

  1. Find the GCF: 7.
  2. Divide both terms by 7.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 3:4

Worked Example 5

Problem: Simplify the ratio 18:30.

  1. Find the GCF: 6.
  2. Divide both terms by 6.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 3:5

Worked Example 6

Problem: Simplify the ratio 12:20.

  1. Find the GCF: 4.
  2. Divide both terms by 4.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 3:5

Worked Example 7

Problem: Simplify the ratio 9:15.

  1. Find the GCF: 3.
  2. Divide both terms by 3.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 3:5

Worked Example 8

Problem: Simplify the ratio 16:24.

  1. Find the GCF: 8.
  2. Divide both terms by 8.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 2:3

Worked Example 9

Problem: Simplify the ratio 25:35.

  1. Find the GCF: 5.
  2. Divide both terms by 5.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 5:7

Worked Example 10

Problem: Simplify the ratio 14:49.

  1. Find the GCF: 7.
  2. Divide both terms by 7.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 2:7

Practice exercise

Create one new Grade 8 problem involving Part-to-whole ratios. Show the important steps, include units when needed, and explain how you checked the answer.

15.4 Simplifying ratios

A ratio compares two quantities in a fixed order. In this section, the focus is Simplifying ratios.

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 the ratio 4:6.

  1. Find the GCF: 2.
  2. Divide both terms by 2.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 2:3

Worked Example 2

Problem: Simplify the ratio 8:12.

  1. Find the GCF: 4.
  2. Divide both terms by 4.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 2:3

Worked Example 3

Problem: Simplify the ratio 15:25.

  1. Find the GCF: 5.
  2. Divide both terms by 5.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 3:5

Worked Example 4

Problem: Simplify the ratio 21:28.

  1. Find the GCF: 7.
  2. Divide both terms by 7.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 3:4

Worked Example 5

Problem: Simplify the ratio 18:30.

  1. Find the GCF: 6.
  2. Divide both terms by 6.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 3:5

Worked Example 6

Problem: Simplify the ratio 12:20.

  1. Find the GCF: 4.
  2. Divide both terms by 4.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 3:5

Worked Example 7

Problem: Simplify the ratio 9:15.

  1. Find the GCF: 3.
  2. Divide both terms by 3.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 3:5

Worked Example 8

Problem: Simplify the ratio 16:24.

  1. Find the GCF: 8.
  2. Divide both terms by 8.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 2:3

Worked Example 9

Problem: Simplify the ratio 25:35.

  1. Find the GCF: 5.
  2. Divide both terms by 5.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 5:7

Worked Example 10

Problem: Simplify the ratio 14:49.

  1. Find the GCF: 7.
  2. Divide both terms by 7.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 2:7

Practice exercise

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

15.5 Equivalent ratios

A ratio compares two quantities in a fixed order. In this section, the focus is Equivalent ratios.

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 the ratio 4:6.

  1. Find the GCF: 2.
  2. Divide both terms by 2.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 2:3

Worked Example 2

Problem: Simplify the ratio 8:12.

  1. Find the GCF: 4.
  2. Divide both terms by 4.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 2:3

Worked Example 3

Problem: Simplify the ratio 15:25.

  1. Find the GCF: 5.
  2. Divide both terms by 5.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 3:5

Worked Example 4

Problem: Simplify the ratio 21:28.

  1. Find the GCF: 7.
  2. Divide both terms by 7.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 3:4

Worked Example 5

Problem: Simplify the ratio 18:30.

  1. Find the GCF: 6.
  2. Divide both terms by 6.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 3:5

Worked Example 6

Problem: Simplify the ratio 12:20.

  1. Find the GCF: 4.
  2. Divide both terms by 4.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 3:5

Worked Example 7

Problem: Simplify the ratio 9:15.

  1. Find the GCF: 3.
  2. Divide both terms by 3.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 3:5

Worked Example 8

Problem: Simplify the ratio 16:24.

  1. Find the GCF: 8.
  2. Divide both terms by 8.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 2:3

Worked Example 9

Problem: Simplify the ratio 25:35.

  1. Find the GCF: 5.
  2. Divide both terms by 5.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 5:7

Worked Example 10

Problem: Simplify the ratio 14:49.

  1. Find the GCF: 7.
  2. Divide both terms by 7.

Very beginner explanation: Simplifying a ratio is like simplifying a fraction: divide both parts by the same common factor.

Answer: 2:7

Practice exercise

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

15.6 Meaning of rate

A rate compares quantities measured in different units. In this section, the focus is Meaning of 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 Meaning of rate. Show the important steps, include units when needed, and explain how you checked the answer.

15.7 Unit rate

A unit rate compares a quantity with exactly one unit of another quantity. In this section, the focus is Unit 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 Unit rate. Show the important steps, include units when needed, and explain how you checked the answer.

15.8 Speed

Speed is an important Grade 8 concept in Ratios, Rates, and Unit Rates. 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: 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 Speed. Show the important steps, include units when needed, and explain how you checked the answer.

15.9 Price per item

Price per item is an important Grade 8 concept in Ratios, Rates, and Unit Rates. 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: 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 Price per item. Show the important steps, include units when needed, and explain how you checked the answer.

15.10 Cost per kilogram

Cost per kilogram is an important Grade 8 concept in Ratios, Rates, and Unit Rates. 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: 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 Cost per kilogram. Show the important steps, include units when needed, and explain how you checked the answer.

15.11 Comparing rates

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

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 Comparing rates. Show the important steps, include units when needed, and explain how you checked the answer.

15.12 Best-buy problems

Best-buy problems is an important Grade 8 concept in Ratios, Rates, and Unit Rates. 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: 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 Best-buy problems. Show the important steps, include units when needed, and explain how you checked the answer.

15.13 Rate tables

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

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 Rate tables. Show the important steps, include units when needed, and explain how you checked the answer.

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Chapter 15 Review Questions and Answers

Q1. What is important to remember about Meaning of ratio?

Answer: A ratio compares two quantities in a fixed order. In this section, the focus is Meaning of ratio.

Q2. What is important to remember about Part-to-part ratios?

Answer: A ratio compares two quantities in a fixed order. In this section, the focus is Part-to-part ratios.

Q3. What is important to remember about Part-to-whole ratios?

Answer: A ratio compares two quantities in a fixed order. In this section, the focus is Part-to-whole ratios.

Q4. What is important to remember about Simplifying ratios?

Answer: A ratio compares two quantities in a fixed order. In this section, the focus is Simplifying ratios.

Q5. What is important to remember about Equivalent ratios?

Answer: A ratio compares two quantities in a fixed order. In this section, the focus is Equivalent ratios.

Q6. What is important to remember about Meaning of rate?

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

Q7. What is important to remember about Unit rate?

Answer: A unit rate compares a quantity with exactly one unit of another quantity. In this section, the focus is Unit rate.

Q8. What is important to remember about Speed?

Answer: Speed is an important Grade 8 concept in Ratios, Rates, and Unit Rates. 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 Price per item?

Answer: Price per item is an important Grade 8 concept in Ratios, Rates, and Unit Rates. 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 Cost per kilogram?

Answer: Cost per kilogram is an important Grade 8 concept in Ratios, Rates, and Unit Rates. 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 Comparing rates?

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

Q12. What is important to remember about Best-buy problems?

Answer: Best-buy problems is an important Grade 8 concept in Ratios, Rates, and Unit Rates. 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 Rate tables?

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

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.