MGU Mathematics Series — Arithmetic & Pre-Algebra Foundations
Part III: Ratios, Percents, Exponents, and Scientific Notation
Ratios · Rates · Proportions · Percents · Simple Interest
Chapter 6 of 14
MGU Mathematics Series  ·  Arithmetic & Pre-Algebra Foundations  ·  Library Textbook Edition

Chapter 6:
Ratios, Proportions,
and Percents

Proportional reasoning as a unified way of thinking — how quantities compare, how they scale, and how they change in consistent, predictable ways

■ Math Level: Pre-Algebra ■

■ 14 Sections  •  Part III of 5 ■

Follows Chapters 1–5  ·  Prepares students for Exponents, Scientific Notation, and Algebra I

Chapter 6 — Front Matter

Purpose and Overview

This chapter introduces proportional reasoning, one of the most important mathematical ideas connecting arithmetic to algebra, science, finance, and everyday decision-making. Students who understand ratios, rates, proportions, and percents are equipped to read a nutrition label, compare prices, calculate a discount, interpret a map, analyze data, and manage money. These are not separate tricks to memorize. They are related expressions of a single idea: that quantities can be compared and scaled in consistent, predictable ways.

By the end of this chapter, students should not only be able to compute with ratios and percents. They should be able to reason proportionally — asking whether two quantities are in balance, whether a change is fair, and whether an answer makes sense in context.

Opening Question

A store sells a 12-ounce bag of trail mix for $3.60 and a 20-ounce bag for $5.80. Which bag is the better value? Before you can answer that question confidently, you need to know how to compare quantities that are not the same size. You need proportional reasoning. This chapter will give you the tools to answer questions like this one, and many others like it.

Why This Chapter Matters

Proportional reasoning appears in nearly every area of applied mathematics. Recipes scale up and down using ratios. Maps use scales that are proportions. Science experiments involve rates. Shopping decisions involve unit prices. Tax, discount, and interest are all forms of percent. Every time you ask "how much of one thing per one unit of another," you are thinking proportionally.

In algebra, proportional reasoning becomes even more important. Linear equations, slope, direct variation, and dimensional analysis are all built on the ideas introduced here. This chapter is not a detour from algebra preparation. It is one of the most direct paths toward it.

14Sections
15Objectives
20Key Terms
20+Worked Examples

Learning Objectives

  1. 1Define a ratio and write it in multiple forms.
  2. 2Distinguish part-to-part ratios from part-to-whole ratios.
  3. 3Identify and generate equivalent ratios.
  4. 4Build and extend ratio tables.
  5. 5Calculate rates and unit rates.
  6. 6Find unit price and use it to compare values.
  7. 7Use unit multipliers to convert between units of measurement.
  8. 8Define a proportion and determine whether a proportion is true.
  9. 9Solve a proportion using cross multiplication.
  10. 10Express a percent as a ratio out of 100.
  11. 11Solve the three types of percent problems.
  12. 12Calculate percent increase and percent decrease.
  13. 13Apply the simple interest formula.
  14. 14Connect proportional reasoning to real-world contexts in science, finance, and measurement.
  15. 15Recognize proportional reasoning as a unified way of thinking, not a collection of isolated procedures.

Key Terms

ratio, equivalent ratio, part-to-part ratio, part-to-whole ratio, rate, unit rate, unit price, unit multiplier, proportion, cross products, percent, base, percentage, percent increase, percent decrease, simple interest, principal, interest rate, proportional reasoning

Full definitions appear in the Key Terms Review at the end of this chapter and in the MGU Library Glossary.

Chapter 6 — Contents

Sections at a Glance

Sections 6.1 – 6.4
Ratios: Comparing and Organizing Quantities
What Ratios Compare · Part-to-Part and Part-to-Whole · Equivalent Ratios · Ratio Tables
Section 6.1

What Ratios Compare

A ratio is a comparison of two quantities by division. It answers the question: how much of one thing is there compared to another? When you say there are 3 red marbles and 5 blue marbles in a bag, you can express this comparison in three equivalent forms.

FormExampleRead as
Using the word to3 to 5"three to five"
Using a colon3 : 5"three to five"
Using a fraction3/5"three to five"

All three forms mean the same thing. The fraction form is especially useful for computation. The two quantities in a ratio are called terms. Order matters: a ratio of 3 to 5 is not the same as 5 to 3. When both quantities share the same unit, the units cancel in the simplified form. When they have different units, keep the units.

Like fractions, ratios can be simplified by dividing both terms by their greatest common factor (GCF).

Example 6.1.1
Simplifying a Ratio
  • ProblemA bag contains 15 red candies and 10 green candies. Write the ratio of red to green in simplest form.
  • ThinkThe ratio is 15/10. Find the GCF of 15 and 10.
  • SolveGCF(15, 10) = 5 15/10 = (15 ÷ 5)/(10 ÷ 5) = 3/2
  • Check3 and 2 share no common factor other than 1. The ratio is fully simplified.
Answer: The ratio of red to green is 3 to 2.
Student Warning

A ratio of 3 to 2 does not mean that 3/2 of something exists. It means for every 3 of the first quantity, there are 2 of the second. The fraction form is a notation choice, not a claim about parts of a whole, unless the ratio is specifically written as part-to-whole (see Section 6.2).

Practice 6.1Ratios and Simplification
  1. A class has 18 students: 10 wearing glasses and 8 not. Write the ratio of students wearing glasses to students not wearing glasses in all three forms.
  2. Why does the order of terms matter in a ratio?
  3. Simplify each ratio: (a) 6 to 9   (b) 20 : 12   (c) 25/15   (d) 14 to 21
  4. A recipe uses 4 cups of oats and 2 cups of nuts. Write the ratio of oats to nuts in simplest form.
  5. A parking lot has 36 compact cars and 24 large vehicles. Write the ratio of compact to large in simplest form.
  6. A sports team won 9 games and lost 3. Write the ratio of wins to losses. Then write the ratio of wins to total games played.
  7. Error Analysis: A student writes the ratio of 10 boys to 15 girls as 15/10. What mistake did the student make?
Section 6.2

Part-to-Part and Part-to-Whole Ratios

A part-to-part ratio compares one group to another group within the same whole. If a class has 14 girls and 11 boys, the ratio of girls to boys is 14 to 11. Neither group includes the other.

A part-to-whole ratio compares one group to the entire total. If a class has 14 girls out of 25 students, the ratio of girls to total students is 14 to 25. This form connects directly to fractions, decimals, and percents. Read each problem carefully to determine which type you need.

Example 6.2.1
Part-to-Part vs. Part-to-Whole
  • ProblemA fruit bowl has 6 apples, 4 bananas, and 2 oranges. Write (a) the ratio of apples to bananas, and (b) the ratio of apples to all pieces of fruit.
  • Think(a) Apples to bananas is part-to-part. (b) Apples to all fruit is part-to-whole. Total = 6 + 4 + 2 = 12.
  • Solvea) 6/4 = 3/2 (part-to-part) b) 6/12 = 1/2 (part-to-whole)
  • CheckIn (b), 1/2 means apples are half of all the fruit. Since 6 is half of 12, this is correct.
Answer: (a) 3 to 2    (b) 1 to 2
Student Tip

When a problem asks "what fraction of the total," you need a part-to-whole ratio. This is the form that connects most directly to percents and probability.

Practice 6.2Part-to-Part and Part-to-Whole
  1. Explain in your own words how a part-to-part ratio differs from a part-to-whole ratio.
  2. A jar has 5 red beads, 3 blue beads, and 2 yellow beads. Which ratio type answers "what fraction of the beads are blue?"
  3. A survey asked 30 students whether they prefer math or reading. 18 chose math. Write (a) the part-to-part ratio of math to reading and (b) the part-to-whole ratio of math to total students.
  4. A parking lot has 40 cars, 15 trucks, and 5 motorcycles. Write (a) the ratio of cars to trucks and (b) the ratio of motorcycles to all vehicles.
  5. A flower garden has 12 red roses, 8 white roses, and 4 yellow roses. Write the ratio of yellow roses to all roses. Then write the ratio of red roses to non-red roses.
  6. Error Analysis: A student is asked for the part-to-whole ratio of boys in a class of 20 where 8 are boys. The student writes 8/12. Identify and correct the error.
Section 6.3

Equivalent Ratios

Two ratios are equivalent if they represent the same comparison at different scales. To create an equivalent ratio, multiply or divide both terms by the same nonzero number. To test equivalence, simplify both ratios to lowest terms and compare, or use cross multiplication: if a/b = c/d, then a × d = b × c.

Example 6.3.1
Testing Whether Two Ratios Are Equivalent
  • ProblemAre the ratios 4 to 6 and 10 to 15 equivalent?
  • Method 1Simplify both: 4/6 = 2/3 10/15 = 2/3 ✓ Both equal 2/3.
  • Method 2Cross multiply: 4 × 15 = 60 6 × 10 = 60 ✓ Cross products are equal.
Answer: Yes. 4 to 6 and 10 to 15 are equivalent ratios.
Example 6.3.2
Creating an Equivalent Ratio
  • ProblemA recipe uses 2 cups of flour for every 3 cups of water. How many cups of flour are needed for 12 cups of water?
  • ThinkRatio is 2/3. Find the multiplier: 3 × ? = 12, so multiply both terms by 4.
  • Solve2/3 = (2 × 4)/(3 × 4) = 8/12
  • Check8/12 = 2/3. Correct.
Answer: 8 cups of flour are needed for 12 cups of water.
Practice 6.3Equivalent Ratios
  1. How do you create an equivalent ratio? What must remain unchanged?
  2. Determine whether each pair is equivalent: (a) 3 to 5 and 9 to 15   (b) 4 to 7 and 8 to 15   (c) 6 : 10 and 9 : 15
  3. Write three equivalent ratios for each: (a) 1 : 4   (b) 5 to 2   (c) 3/8
  4. A car travels 150 miles in 3 hours. Write an equivalent ratio to find how far it travels in 5 hours at the same speed.
  5. Error Analysis: A student says 3 to 4 and 6 to 9 are equivalent because "you double both numbers." Is the student correct? Explain.
Section 6.4

Ratio Tables

A ratio table is an organized display of equivalent ratios. It has at least two rows, one for each quantity. Each column holds one equivalent pair. Every column must reflect the same ratio relationship. To build a table, start with the original ratio and multiply both quantities by 1, 2, 3, 4, and so on. To find missing values, locate the multiplier that connects two known entries and apply it to the missing quantity.

Example 6.4.1
Building a Ratio Table
  • ProblemA car uses 3 gallons of fuel for every 90 miles. Complete a ratio table for 1 through 5 gallons.
  • ThinkFind the unit rate: 90 ÷ 3 = 30 miles per gallon. Then multiply.
  • Solve
    Gallons12345
    Miles306090120150
  • CheckEach miles value divided by its gallons value equals 30. The ratio is consistent.
Answer: Unit rate is 30 miles per gallon. Table shown above.
Example 6.4.2
Using a Ratio Table to Find a Missing Value
  • ProblemA store packs 5 notebooks per bag. How many bags are needed for 35 notebooks?
  • SolveMultiplier: 35 ÷ 5 = 7 Bags: 1 × 7 = 7 bags
  • Check7 × 5 = 35. Correct.
Answer: 7 bags are needed.
Practice 6.4Ratio Tables
  1. What is a ratio table, and what does each column represent?
  2. Why must every column in a ratio table be equivalent to the original ratio?
  3. Complete this ratio table for a 4 to 1 relationship.
    First Quantity481216?
    Second Quantity1???5
  4. A printer prints 8 pages per minute. Build a ratio table for 1, 2, 3, 4, and 5 minutes.
  5. A recipe needs 3 eggs for every 12 cookies. Use a ratio table to find how many eggs are needed for 48 cookies.
Sections 6.5 – 6.7
Rates, Unit Price, and Unit Multipliers
Rates and Unit Rates · Unit Price · Unit Multipliers for Conversion
Section 6.5

Rates and Unit Rates

A rate is a ratio that compares two quantities with different units. The word "per" is the key signal — it means "for each one" and indicates division. A unit rate is a rate in which the second quantity is exactly 1. Unit rates make it easy to compare, estimate, and scale because everything is expressed "per one unit." To find a unit rate, divide the first quantity by the second.

Unit Rate Formula

Unit rate = first quantity ÷ (1 unit of second quantity)

Examples: 60 miles per hour, $4.50 per pound, 24 heartbeats per minute.

Example 6.5.1
Finding a Unit Rate
  • ProblemA car travels 240 miles in 4 hours. Find the unit rate in miles per hour.
  • Solve240 ÷ 4 = 60 miles per hour
  • Check60 × 4 = 240. Correct.
Answer: 60 miles per hour.
Example 6.5.2
Comparing Unit Rates
  • ProblemRunner A completes 5 miles in 40 minutes. Runner B completes 8 miles in 72 minutes. Who runs faster?
  • SolveRunner A: 5 ÷ 40 = 0.125 miles per minute Runner B: 8 ÷ 72 ≈ 0.111 miles per minute
  • CheckBoth rates are less than 1 mile per minute, which is reasonable since neither runner is sprinting at track-record pace.
Answer: Runner A is faster at 0.125 miles per minute versus 0.111.
Student Tip

Always label unit rates with both units. Writing "60" is incomplete. Writing "60 miles per hour" is correct. Units tell you what is being compared and prevent confusion later.

Practice 6.5Rates and Unit Rates
  1. What is the difference between a rate and a unit rate?
  2. In the rate "45 students per classroom," what are the two quantities being compared?
  3. Find the unit rate: (a) 180 miles in 3 hours   (b) $12.50 for 5 pounds   (c) 560 words in 8 minutes   (d) 126 heartbeats in 3 minutes
  4. Car A travels 270 miles in 5 hours. Car B travels 320 miles in 6 hours. Which car is traveling faster?
  5. A factory produces 480 items in 8 hours. What is the unit rate of production?
  6. A cyclist rides 15 miles in 1.5 hours. What is the unit rate in miles per hour?
  7. Error Analysis: A student divides 3 hours by 180 miles to find a unit rate. What error did the student make? What is the correct unit rate?
Section 6.6

Unit Price

Unit price is a specific, practical kind of unit rate. It tells you the cost per single unit of a product and is the most reliable way to compare the value of products sold in different quantities. Stores often sell the same product in multiple sizes. Always calculate unit price rather than assuming the larger size is always cheaper.

Unit Price Formula

Unit price = Total price ÷ Number of units

The result is a price per single unit — per ounce, per count, per liter, etc.

Example 6.6.1
Calculating and Comparing Unit Prices
  • ProblemA 12-oz bag of trail mix costs $3.60. A 20-oz bag costs $5.80. Which is the better value?
  • Solve12-oz: $3.60 ÷ 12 = $0.30 per oz 20-oz: $5.80 ÷ 20 = $0.29 per oz
  • CheckThe 20-oz bag costs less per ounce even though it costs more in total.
Answer: The 20-ounce bag is the better value at $0.29 per ounce.
Student Warning

"Better value" means lower unit price — not lower total price. A product with a higher sticker price can still be the better deal if it contains significantly more quantity. Always calculate; never assume.

Practice 6.6Unit Price
  1. What formula is used to calculate unit price?
  2. Why is unit price a more reliable comparison tool than total price?
  3. Find the unit price: (a) 8 bottles of water for $4.00   (b) 15 pens for $4.50   (c) 16-oz peanut butter for $3.84   (d) 24-oz peanut butter for $5.04
  4. Which is the better value? (a) 6 granola bars for $4.20 or 10 for $6.50   (b) 32-oz sports drink for $2.56 or 20-oz for $1.60
  5. A grocery store sells orange juice: 64-oz for $4.48 and 32-oz for $2.40. Which has the lower unit price? How much do you save per ounce by choosing the better value?
Section 6.7

Unit Multipliers

A unit multiplier is a fraction equal to 1 used to convert a measurement from one unit to another. Because multiplying by 1 does not change a quantity's value, unit multipliers let you change the unit without changing what the measurement means. Every conversion fact yields two unit multipliers. Since 12 inches = 1 foot, both 12 in/1 ft and 1 ft/12 in equal 1. Place the unit you want to cancel in the denominator so it divides out. This technique is called dimensional analysis in science courses.

Steps for Using a Unit Multiplier

Step 1. Write the starting measurement as a fraction over 1.

Step 2. Multiply by the unit multiplier with the unwanted unit in the denominator.

Step 3. Cancel the unwanted unit. Simplify the numbers. Write the answer with the new unit.

Example 6.7.1
Converting Inches to Feet
  • ProblemConvert 84 inches to feet.
  • Think12 in = 1 ft. Cancel inches by placing them in the denominator.
  • Solve84 in × (1 ft / 12 in) = 84/12 ft = 7 ft
  • Check7 × 12 = 84. Correct.
Answer: 84 inches = 7 feet.
Example 6.7.2
Two-Step Unit Conversion
  • ProblemConvert 3 hours to seconds.
  • ThinkUse two multipliers in sequence: hours → minutes, then minutes → seconds.
  • Solve3 hr × (60 min / 1 hr) × (60 sec / 1 min) = 3 × 60 × 60 sec = 10,800 sec
  • Check3 × 60 = 180 min. 180 × 60 = 10,800 sec. Correct.
Answer: 3 hours = 10,800 seconds.
Student Tip

If you set up a unit multiplier and the units do not cancel, you have used the wrong orientation. Flip the fraction and try again.

Practice 6.7Unit Multipliers
  1. Why is a unit multiplier equal to 1?
  2. Which unit belongs in the denominator — the unit you are starting with or the unit you want to end with?
  3. Convert using unit multipliers. Show your work: (a) 5 feet to inches   (b) 180 minutes to hours   (c) 4 yards to feet   (d) 4,400 yards to miles (1 mile = 1,760 yd)
  4. Convert 2 days to minutes using two unit multipliers.
  5. A road is 3.5 miles long. How many feet is that? (1 mile = 5,280 ft)
  6. A swimming pool holds 12,000 gallons. Convert to quarts. (1 gallon = 4 quarts)
Sections 6.8 – 6.9
Proportions: Setting Ratios Equal
Testing Proportions · Cross Products · Solving Proportions · Word Problems and Map Scales
Section 6.8

Proportions

A proportion is a statement that two ratios are equal: a/b = c/d. When two ratios form a proportion, the quantities are in balance. If you scale one ratio up by a factor, the other must scale by the same factor. Proportions appear in recipes, map scales, enlarged photographs, tax rates, and scientific measurements — anywhere a consistent relationship must be maintained.

Cross Products

In the proportion a/b = c/d, the cross products are a × d and b × c.

If the cross products are equal, the proportion is true.

If the cross products are unequal, the proportion is false.

Example 6.8.1
Testing a True Proportion
  • ProblemIs 3/4 = 9/12 a true proportion?
  • Solve3 × 12 = 36 4 × 9 = 36 ✓ Cross products are equal.
Answer: Yes. 3/4 = 9/12 is a true proportion.
Example 6.8.2
Identifying a False Proportion
  • ProblemIs 5/8 = 15/25 a true proportion?
  • Solve5 × 25 = 125 8 × 15 = 120 ✗ 125 ≠ 120.
Answer: No. 5/8 = 15/25 is not a true proportion.
Practice 6.8Testing Proportions
  1. What is a proportion?
  2. How do cross products tell you whether a proportion is true or false?
  3. Determine whether each is a true proportion: (a) 2/5 = 6/15   (b) 4/7 = 12/22   (c) 9/6 = 12/8   (d) 3/11 = 9/33
  4. A map uses a scale of 1 inch to represent 25 miles. A student claims 4 inches represents 100 miles. Write this as a proportion and verify whether it is true.
Section 6.9

Solving Proportions

When one term in a proportion is unknown, solve for it using cross multiplication. Write the proportion, cross multiply, set the products equal, and divide to isolate the variable. This connects directly to solving one-step equations in Chapter 10. When setting up a proportion from a word problem, verify that both ratios compare the same quantities in the same order.

Example 6.9.1
Solving a Basic Proportion
  • ProblemSolve for x: 3/5 = x/20
  • Solve3 × 20 = 5 × x 60 = 5x x = 60 ÷ 5 = 12
  • Check3/5 = 12/20. Cross products: 3 × 20 = 60, 5 × 12 = 60. Equal. ✓
Answer: x = 12
Example 6.9.2
Proportion from a Word Problem
  • ProblemIf 4 pounds of apples cost $5.20, how much would 10 pounds cost?
  • ThinkCost to pounds = cost to pounds. Same order on both sides.
  • Solve$5.20 / 4 lb = x / 10 lb 5.20 × 10 = 4 × x 52 = 4x x = 52 ÷ 4 = $13.00
  • Check$5.20 ÷ 4 = $1.30/lb. $1.30 × 10 = $13.00. Correct.
Answer: 10 pounds of apples cost $13.00.
Example 6.9.3
Map Scale Proportion
  • ProblemA map scale is 1 inch : 40 miles. Two cities are 3.5 inches apart on the map. What is the actual distance?
  • Solve1 in / 40 mi = 3.5 in / x mi 1 × x = 40 × 3.5 x = 140 miles
  • Check3.5 × 40 = 140. Correct.
Answer: The actual distance between the cities is 140 miles.
Student Warning

When setting up a proportion from a word problem, always verify that both ratios compare the same quantities in the same order. A proportion with flipped terms will produce a wrong answer that may still look plausible at first glance.

Practice 6.9Solving Proportions
  1. What is cross multiplication, and why does it work?
  2. When setting up a proportion from a word problem, what must be true about both ratios?
  3. Solve each proportion: (a) 2/3 = x/12   (b) x/5 = 9/15   (c) 7/x = 14/10   (d) 4/9 = 20/x
  4. A car travels 200 miles on 8 gallons of gas. How far can it travel on 11 gallons?
  5. A recipe calls for 2.5 cups of flour to make 24 cookies. How much flour is needed to make 60 cookies?
  6. On a scale drawing, 1 cm represents 5 meters. A hallway is 8 cm on the drawing. How long is the actual hallway?
  7. Error Analysis: A student solves 3/4 = x/20 by computing x = 3 × 4 ÷ 20, getting x = 0.6. Find the error and solve correctly.
Sections 6.10 – 6.12
Percents: Ratios Out of 100
Percent as Per Hundred · Three Types of Percent Problems · Percent Increase and Decrease
Section 6.10

Percent as Per Hundred

The word percent comes from the Latin per centum, meaning "per hundred." A percent is a ratio that always compares a quantity to 100. So 45% simply means 45/100, and 7% means 7/100. Percent is a way of standardizing comparison by fixing the whole at 100. Because percent is always out of 100, it connects directly to decimals and fractions through reliable conversion rules.

ConversionMethodExamples
Percent → DecimalDivide by 100 (move decimal two places left)45% = 0.45   7% = 0.07   125% = 1.25
Decimal → PercentMultiply by 100 (move decimal two places right)0.38 = 38%   0.06 = 6%   1.5 = 150%
Percent → FractionWrite over 100 and simplify60% = 3/5   25% = 1/4
Fraction → PercentDivide numerator by denominator, then × 1003/4 = 75%   2/5 = 40%

Percents are not limited to the range 0 to 100. A percent greater than 100% means more than the whole. A price increase of 150% means the new price is 2.5 times the original. A percent less than 1% is also valid: 0.5% = 0.005 as a decimal.

Example 6.10.1
Converting Percent to Decimal and Fraction
  • ProblemConvert 35% to a decimal and a fraction in simplest form.
  • SolveDecimal: 35% = 35 ÷ 100 = 0.35 Fraction: 35/100 = 7/20
  • Check7 ÷ 20 = 0.35. Consistent.
Answer: 35% = 0.35 = 7/20
Example 6.10.2
Converting a Fraction to a Percent
  • ProblemConvert 3/8 to a percent.
  • Solve3 ÷ 8 = 0.375 0.375 × 100 = 37.5%
Answer: 3/8 = 37.5%
Practice 6.10Percent Conversions
  1. What does the word "percent" mean in its original Latin sense?
  2. Why can percents be greater than 100?
  3. Convert each percent to a decimal and a simplified fraction: (a) 50%   (b) 8%   (c) 72%   (d) 120%   (e) 0.5%
  4. Convert each decimal to a percent: (a) 0.25   (b) 0.09   (c) 1.4   (d) 0.003
  5. Convert each fraction to a percent: (a) 1/2   (b) 3/5   (c) 7/8   (d) 5/4
  6. Error Analysis: A student converts 6% to a decimal by writing 0.6. What error did the student make? What is the correct answer?
Section 6.11

Percent and Percentage Problems

Every percent problem involves three quantities: the base (the whole), the rate (the percent as a decimal), and the percentage (the part). All three types of percent problems use the same formula — only the unknown changes.

The Percent Formula

Percentage = Rate × Base    (P = r × B)

Type 1 — Find the part: P = r × B   (given rate and base)

Type 2 — Find the rate: r = P ÷ B   (given part and base; convert result to %)

Type 3 — Find the base: B = P ÷ r   (given part and rate)

Example 6.11.1
Type 1: Finding the Percentage
  • ProblemA backpack is priced at $65. Sales tax is 8%. How much is the tax?
  • SolveP = 0.08 × 65 = $5.20
  • Check10% of $65 is $6.50. 8% should be slightly less. $5.20 is reasonable.
Answer: The sales tax is $5.20.
Example 6.11.2
Type 2: Finding the Rate
  • ProblemA student answered 42 out of 56 questions correctly. What percent is that?
  • Solver = 42 ÷ 56 = 0.75 = 75%
  • Check0.75 × 56 = 42. Correct.
Answer: The student answered 75% correctly.
Example 6.11.3
Type 3: Finding the Base
  • ProblemA tip of $9 is 15% of the restaurant bill. What was the total bill?
  • SolveB = 9 ÷ 0.15 = $60
  • Check0.15 × 60 = 9. Correct.
Answer: The total restaurant bill was $60.
Student Tip

Identify the missing value before writing the equation. If you are given the rate and base, find the part. If given part and whole, find the rate. If given part and rate, find the base.

Practice 6.11Percent Problems
  1. Name the three quantities involved in every percent problem.
  2. If you know the percentage and the percent rate, which quantity are you solving for?
  3. Solve: (a) What is 40% of 90?   (b) What is 12% of 250?   (c) 35 is what % of 140?   (d) 72 is what % of 90?   (e) 24 is 60% of what number?   (f) 18 is 36% of what number?
  4. A jacket costs $120. There is a 15% off sale. What is the discount? What is the sale price?
  5. A class has 32 students. On one day, 4 are absent. What percent are absent?
  6. A scholarship covers 40% of tuition. If the scholarship is $3,200, what is the full tuition?
  7. Error Analysis: A student solves "What is 25% of 60?" by writing 60 ÷ 25 = 2.4. Identify and correct the error.
Section 6.12

Percent Increase and Decrease

When a quantity changes, percent change expresses that change as a percent of the original. This captures how significant the change is relative to where the quantity started. A raise of $1,000 means different things at a salary of $10,000 versus $100,000. In both cases the original value — not the new value — appears in the denominator.

Percent Change Formulas

Percent Increase = (New − Original) / Original × 100

Percent Decrease = (Original − New) / Original × 100

Finding a new value directly:

New value after increase = Original × (1 + r)

New value after decrease = Original × (1 − r)

Where r is the percent written as a decimal.

Example 6.12.1
Calculating Percent Increase
  • ProblemA store raised the price of a book from $12.00 to $15.00. What is the percent increase?
  • SolveIncrease = 15 − 12 = 3 Percent Increase = 3/12 × 100 = 25%
  • Check25% of $12 = $3. $12 + $3 = $15. Correct.
Answer: The price increased by 25%.
Example 6.12.2
Calculating Percent Decrease
  • ProblemA jacket originally priced at $80 is now on sale for $60. What is the percent decrease?
  • SolveDecrease = 80 − 60 = 20 Percent Decrease = 20/80 × 100 = 25%
Answer: The price decreased by 25%.
Example 6.12.3
Finding a Sale Price Directly
  • ProblemA television costs $450 and is marked down 30%. What is the sale price?
  • SolveSale Price = 450 × (1 − 0.30) = 450 × 0.70 = $315
  • Check30% of 450 = 135. 450 − 135 = $315. Correct.
Answer: The sale price is $315.
Student Warning

Always divide by the original value when calculating percent change. Dividing by the new value gives a different and incorrect result. This is the most common error in percent change problems.

Practice 6.12Percent Increase and Decrease
  1. In the percent change formula, what value is always in the denominator?
  2. A price increases from $50 to $75. Is the percent increase 33% or 50%? Show your work.
  3. Find the percent change. State increase or decrease: (a) 40 to 50   (b) 120 to 90   (c) 25 to 28   (d) 200 to 150
  4. Find the new value: (a) Original $60; Increase 20%   (b) Original $150; Decrease 35%   (c) Original 80; Increase 15%
  5. A car's value drops from $18,000 to $13,500 after one year. What is the percent decrease?
  6. A student's test score improved from 68 to 85. What is the percent increase?
  7. A store increases a price by 12%, then later decreases the new price by 12%. Is the final price equal to, less than, or greater than the original? Explain and calculate.
  8. Error Analysis: A price rises from $80 to $100. A student calculates the percent increase as (20/100) × 100 = 20%. Is this correct? Provide the correct answer and explain the error.
Sections 6.13 – 6.14
Interest and Real-Life Applications
Simple Interest · Shopping · Science · Cooking · Personal Finance
Section 6.13

Simple Interest

When money is borrowed or deposited in a bank, the amount paid or earned for using that money over time is called interest. Simple interest is calculated on the original amount alone, making it the most straightforward interest model. It applies to many savings accounts, short-term loans, and basic financial instruments.

Simple Interest Formula

I = P × r × t

I = interest earned or owed

P = principal (the original amount)

r = annual interest rate written as a decimal

t = time in years (convert months to years by dividing by 12)

Total amount after interest: A = P + I

Example 6.13.1
Finding Simple Interest and Total Amount
  • ProblemA student deposits $500 in a savings account that pays 4% simple annual interest. How much interest is earned after 3 years? What is the total balance?
  • SolveP = 500, r = 0.04, t = 3 I = 500 × 0.04 × 3 = $60 A = 500 + 60 = $560
  • Check$500 × 0.04 = $20 per year. $20 × 3 = $60. Correct.
Answer: Interest earned: $60. Total balance: $560.
Example 6.13.2
Time Given in Months
  • ProblemA loan of $1,200 is borrowed at 6% annual simple interest for 8 months. How much interest is owed?
  • ThinkConvert 8 months to years: t = 8/12 = 2/3
  • SolveI = 1200 × 0.06 × (2/3) = 1200 × 0.04 = $48
  • Check$1,200 × 0.06 = $72/year. For 2/3 year: $72 × 2/3 = $48. Correct.
Answer: The interest owed is $48.
Example 6.13.3
Solving for Principal
  • ProblemAn account earned $90 in simple interest at 5% per year over 3 years. What was the principal?
  • SolveP = I ÷ (r × t) = 90 ÷ (0.05 × 3) = 90 ÷ 0.15 = $600
  • Check600 × 0.05 × 3 = 90. Correct.
Answer: The principal was $600.
Student Tip — Simple vs. Compound Interest

Simple interest is calculated only on the original principal. In later mathematics, you will encounter compound interest, where interest is calculated on both the principal and the accumulated interest. Compound interest grows faster, which is good for savings but more expensive for loans.

MGU Library Connection
  • For an introduction to compound interest and exponential growth, see the future MGU Algebra I textbook, Chapter on Exponential Functions.
  • For a reference table of common interest types, see Appendix H of this textbook.
Practice 6.13Simple Interest
  1. What do P, r, and t represent in the simple interest formula?
  2. If an interest rate is 7%, what decimal value do you use in the formula?
  3. Why must time be expressed in years when using I = Prt?
  4. Find the simple interest and total amount: (a) P = $800, r = 3%, t = 2 yr   (b) P = $2,000, r = 5%, t = 4 yr   (c) P = $450, r = 6%, t = 6 months   (d) P = $1,500, r = 4.5%, t = 3 yr
  5. Solve for the missing variable: (a) I = $120, r = 4%, t = 5 yr. Find P.   (b) I = $180, P = $1,200, t = 3 yr. Find r.
  6. A student takes a loan of $3,000 at 5% simple interest for 2 years. How much interest is owed? What is the total repayment amount?
  7. A grandparent deposits $5,000 at 2.5% simple interest. How much will be in the account after 10 years?
  8. Error Analysis: A student calculates interest on a 9-month loan of $600 at 6% as I = 600 × 0.06 × 9 = $324. What error was made? What is the correct answer?
Section 6.14

Real-Life Applications of Proportional Reasoning

The ideas in this chapter do not exist in separate boxes. Ratios, rates, proportions, percents, and interest all express the same underlying idea: how quantities compare and how they scale. This section brings these tools together in realistic situations. The problem-solving method applies throughout: Read → Represent → Solve → Check → Explain.

Example 6.14.1
Shopping Application
  • ProblemA store is selling a $90 coat at 30% off. Sales tax is 7.5%. What is the final price?
  • RepresentFind the discount → sale price → apply tax → final price.
  • SolveDiscount = 0.30 × 90 = $27 Sale price = 90 − 27 = $63 Tax = 0.075 × 63 = $4.73 Final price = 63 + 4.73 = $67.73
  • CheckA 30% discount on $90 gives $63. A small tax raises it to $67.73. Reasonable.
Answer: The final cost of the coat is approximately $67.73.
Example 6.14.2
Science Application
  • ProblemA solution contains 15 g of salt in 200 mL of water. A larger batch uses 500 mL. How many grams of salt maintain the same concentration?
  • Represent15 g / 200 mL = x g / 500 mL
  • Solve15 × 500 = 200 × x 7,500 = 200x x = 37.5 g
  • Check37.5/500 = 0.075 and 15/200 = 0.075. Same concentration.
Answer: 37.5 grams of salt are needed for 500 mL of water.
Example 6.14.3
Recipe Scaling Application
  • ProblemA cookie recipe calls for 1.5 cups of sugar for 36 cookies. How much sugar is needed for 60 cookies?
  • Solve1.5 / 36 = x / 60 1.5 × 60 = 36 × x 90 = 36x x = 2.5 cups
Answer: 2.5 cups of sugar are needed for 60 cookies.
Example 6.14.4
Budget Application
  • ProblemA person earns $3,200 per month and budgets 28% for rent and 15% for food. How many dollars go to each category? What percent remains?
  • SolveRent: 0.28 × 3,200 = $896 Food: 0.15 × 3,200 = $480 Used: 28 + 15 = 43% Left: 57% → 0.57 × 3,200 = $1,824
Answer: Rent $896, food $480, $1,824 (57%) remains for other expenses.
Practice 6.14Mixed Applications
  1. A car is priced at $18,500. The dealer offers a 5% discount. Local sales tax is 6.5%. What is the total price after discount and tax?
  2. A cyclist rides 40 km in 2 hours. At the same rate, how long would it take to ride 110 km?
  3. A map has a scale of 2 cm = 30 km. Two landmarks are 7.5 cm apart on the map. What is the actual distance?
  4. A store sells cereal in two sizes: 18 oz for $3.78 and 26 oz for $5.46. Which has the lower unit price? How much cheaper per ounce is the better option?
  5. A student invests $2,500 at 4.5% simple annual interest. How long will it take for the account to reach $3,025?
  6. A class survey found that 18 out of 25 students owned a smartphone. If the school has 600 students, how many would be expected to own a smartphone?
Chapter 6 — Back Matter

Chapter Summary

Key Terms Review

ratio
A comparison of two quantities by division. Written as a to b, a : b, or a/b. Order of terms matters.
equivalent ratios
Ratios that represent the same comparison after simplification. Created by multiplying or dividing both terms by the same nonzero number.
part-to-part ratio
A ratio comparing one group to another group within the same whole. Neither group includes the other.
part-to-whole ratio
A ratio comparing one group to the total of all groups. Connects directly to fractions, decimals, and percents.
rate
A ratio comparing two quantities with different units. Must always include both units in its label.
unit rate
A rate with a second quantity of 1. Calculated by dividing the first quantity by the second.
unit price
The cost per single unit of measurement. Calculated as total price divided by number of units.
unit multiplier
A fraction equal to 1 used to convert between units. The unwanted unit goes in the denominator to cancel.
proportion
An equation stating that two ratios are equal: a/b = c/d. Solved using cross multiplication.
cross products
In a/b = c/d, the products a × d and b × c. Equal cross products confirm a true proportion.
percent
A ratio comparing a quantity to 100. Written with the % symbol. Equivalent to n/100 as a fraction.
base
The whole quantity in a percent problem. The denominator in a part-to-whole comparison.
percentage
The part of the whole in a percent problem. Found using P = rB.
percent increase
The ratio of the amount of increase to the original value, expressed as a percent. Original is the denominator.
percent decrease
The ratio of the amount of decrease to the original value, expressed as a percent. Original is the denominator.
simple interest
Interest computed only on the original principal using I = Prt. Does not compound over time.
principal
The original amount of money borrowed or deposited. The P in the simple interest formula.
interest rate
The percent charged or earned per year. Converted to a decimal before use in formulas.
proportional reasoning
The ability to think about how quantities compare and scale. The unifying idea behind ratios, rates, proportions, percents, and interest.
dimensional analysis
The scientific name for using unit multipliers to convert between units of measurement.

Concept Review Questions

Answer in complete sentences.

  1. What is a ratio, and what are the three ways to write it?
  2. How do you simplify a ratio?
  3. What is the difference between a part-to-part ratio and a part-to-whole ratio?
  4. How do you test whether two ratios are equivalent?
  5. What is a ratio table, and how is it used to solve problems?
  6. What is a unit rate, and how does it differ from a regular rate?
  7. Why is unit price a better comparison tool than total price?
  8. Describe how to use a unit multiplier to convert a unit of measurement.
  9. What is a proportion? How do you solve a proportion with an unknown term?
  10. What does it mean for two quantities to be proportional?
  11. What does percent mean, and how is it related to fractions and decimals?
  12. What are the three types of percent problems, and how do you identify which one you are solving?
  13. In percent change, why is the original value always the denominator?
  14. State the simple interest formula and explain what each variable represents.
  15. Give one example each of proportional reasoning used in cooking, shopping, and science.

Ratio and Rate Practice

Part AWriting and Simplifying Ratios
  1. 18 to 24
  2. 35 : 15
  3. 48 to 36
  4. 60/100
  5. 16 tomatoes : 12 peppers (simplify)
  6. 450 students : 30 teachers
Part BRates and Unit Rates

Find each unit rate.

  1. 250 miles in 5 hours
  2. $8.75 for 5 pounds
  3. 840 words in 7 minutes
  4. 312 miles on 12 gallons
  5. Train A: 480 mi in 6 hr
  6. Train B: 350 mi in 4 hr (which is faster?)
Part CUnit Price
  1. Find the unit price and determine the better value: (a) 4 cans for $3.20 or 7 cans for $5.46   (b) 18 oz of cereal for $4.50 or 24 oz for $5.76
Part DUnit Multipliers

Show your unit multiplier setup.

  1. 7 feet to inches
  2. 360 seconds to minutes
  3. 5 miles to feet (1 mi = 5,280 ft)
  4. 3.5 hours to minutes

Percent Practice

Part AConverting

Convert each percent to a decimal and a simplified fraction.

  1. 45%
  2. 2%
  3. 130%
  4. 37.5%
  5. 0.64 → percent
  6. 7/20 → percent
  7. 1.08 → percent
  8. 5/8 → percent
Part BPercent Problems
  1. What is 18% of 300?
  2. What is 6% of 1,500?
  3. 45 is what % of 180?
  4. 84 is what % of 120?
  5. 56 is 70% of what?
  6. 30 is 12% of what?
Part CPercent Change

Find percent change; state increase or decrease.

  1. 80 to 100
  2. 200 to 170
  3. 45 to 54
  4. 60 to 45
  5. $75 increased 20%
  6. $240 decreased 25%

Proportion Word Problems

  1. If 3 notebooks cost $4.50, how much do 8 notebooks cost?
  2. A recipe for 4 servings requires 2.5 cups of broth. How much broth is needed for 10 servings?
  3. On a map, 1.5 inches represents 45 miles. How many miles does 5 inches represent?
  4. A car travels 330 miles on 11 gallons of fuel. How many gallons are needed to travel 480 miles?
  5. A factory produces 720 parts in 9 hours. How many parts are produced in 15 hours at the same rate?
  6. A student reads 35 pages in 50 minutes. At that rate, how long would it take to read 98 pages?
  7. A photograph that is 4 inches wide and 6 inches tall is enlarged so the width becomes 10 inches. What is the new height?
  8. A school surveys 40 students and finds that 28 prefer math over reading. If the school has 600 students, how many would be expected to prefer math?

Financial Applications

  1. A laptop costs $899. Sales tax is 8.25%. What is the total price?
  2. A jacket was originally $120 and is on sale for 35% off. What is the discount? What is the sale price?
  3. A worker earns $18.50 per hour and works 38 hours per week. What is the weekly gross pay? If the worker receives a 4% raise, what is the new hourly rate?
  4. Calculate simple interest and total amount: (a) P = $1,000, r = 4%, t = 5 yr   (b) P = $2,500, r = 3.5%, t = 2 yr   (c) P = $800, r = 6%, t = 18 months
  5. An account earns $150 in simple interest over 3 years at 5% per year. What was the principal?
  6. A car valued at $22,000 depreciated to $15,400 over three years. What was the percent decrease in value?

Error Analysis

For each problem, identify the error and provide the correct solution.

Problem 1 — Ratio Order
Ratio of 14 girls to 21 boys written as 21/14 = 3/2; "ratio of girls to boys is 3 to 2"

What error did the student make? Write the correct ratio of girls to boys in simplest form.

Problem 2 — Unit Rate Direction
Unit rate for 210 miles in 3.5 hours: 3.5 ÷ 210 ≈ 0.017 "miles per hour"

What is the error, and what is the correct unit rate?

Problem 3 — Proportion Setup
"5 lb cost $8; find cost of 12 lb" → proportion set up as 5/8 = x/12 → x = 7.5

Identify the setup error and solve the problem correctly.

Problem 4 — Percent Conversion
15% converted to decimal: 15 ÷ 10 = 1.5

Find and correct the error. What is the correct decimal?

Problem 5 — Percent of a Number
"What is 40% of 120?" solved as 120 ÷ 40 = 3

Identify the error and solve correctly.

Problem 6 — Percent Change Denominator
Population increases from 1,000 to 1,250. Percent increase = (250/1,250) × 100 = 20%

Is this correct? Explain and provide the correct percent increase.

Problem 7 — Simple Interest Time
$400 loan at 5% for 6 months: I = 400 × 0.05 × 6 = $120

Find the error and solve correctly.

Chapter 6 Checkpoint

Chapter 6 Checkpoint
Complete without assistance — assesses all key skills and concepts from Chapter 6
Part 1 — Ratios and Rates (6.1–6.7)
  1. Simplify the ratio 36 to 24.
  2. A class has 13 boys and 17 girls. Write the part-to-whole ratio of boys to all students.
  3. Are the ratios 4/6 and 10/15 equivalent? Show your work using two methods.
  4. A car travels 210 miles in 3.5 hours. Find the unit rate in miles per hour.
  5. A 15-oz can of soup costs $2.25. A 24-oz can costs $3.36. Which has the lower unit price?
  6. Convert 450 minutes to hours using a unit multiplier.
  7. Complete the ratio table for a 2 : 5 relationship.
    First246?10
    Second5??20?
Part 2 — Proportions (6.8–6.9)
  1. Is 3/5 = 9/16 a true proportion? Use cross products.
  2. Solve: x/8 = 9/12
  3. If 6 cans of paint cover 840 square feet, how many cans are needed to cover 2,100 square feet?
Part 3 — Percents (6.10–6.12)
  1. Convert 62% to a decimal and a simplified fraction.
  2. Convert 3/8 to a percent.
  3. What is 35% of 160?
  4. 45 is what percent of 75?
  5. 72 is 48% of what number?
  6. A price rises from $60 to $78. What is the percent increase?
  7. A store marks down a $200 item by 15%. What is the sale price?
Part 4 — Simple Interest (6.13)
  1. Find the simple interest and total amount: P = $1,200, r = 4%, t = 3 years.
  2. A loan of $800 is taken for 9 months at 6% annual simple interest. How much interest is owed?
  3. An account earns $112.50 in simple interest over 2.5 years at 3% per year. What was the principal?
Part 5 — Mixed Applications (6.14)
  1. A map scale is 1 inch = 35 miles. Two cities are 4.5 inches apart on the map. What is the actual distance?
  2. A jacket costs $85. It is on sale for 20% off and the sales tax is 7%. What is the final price?
  3. A student invests $4,000 at 3.5% simple interest. How long will it take to earn $420 in interest?
  4. A car traveled 468 miles and used 13 gallons of gas. How far would it travel on 20 gallons at the same rate?
  5. A population of 2,400 decreased to 1,800 over five years. What was the percent decrease?
Bridge Note: Looking Ahead
Connecting Chapter 6 to Algebra I and Beyond

The work of this chapter does not end at the checkpoint. The ideas here grow into some of the most important structures in mathematics.

Proportional relationships become linear equations in algebra. When two quantities are always in a constant ratio, their relationship can be written as y = kx, where k is the unit rate. This is the simplest linear function and is the starting point for studying slope, direct variation, and graphing in Algebra I.

Percent problems become equation-solving problems. Every percent problem in this chapter used P = rB, which is a multiplication equation. As equations grow more complex, the same structure continues to appear.

Dimensional analysis using unit multipliers is the foundation of scientific measurement in chemistry, physics, and engineering. The same technique of canceling units through fraction multiplication will be essential in every science course you take.

Simple interest is the entry point to the study of exponential functions. Compound interest, population growth and decay, and radioactive half-life all develop from the basic question this chapter introduced: how does a quantity change over time at a consistent rate?

Proportional reasoning itself is not a chapter topic. It is a way of thinking. Students who develop this habit — asking "how do these quantities relate?" and "what stays the same when I scale?" — will find later mathematics more accessible, more connected, and more meaningful.

MGU Library Connection
  • MGU Algebra I textbook: direct variation, slope, and linear equations extend proportional reasoning from this chapter.
  • MGU Consumer Mathematics module: taxes, budgeting, and loans build directly on Sections 6.11 through 6.13.
  • MGU Science Series: unit analysis and concentration problems use the unit multipliers of Section 6.7.
  • Appendix E of this textbook — Formula Sheet: includes the percent, simple interest, and proportion formulas.
  • Appendix D of this textbook — Fraction-Decimal-Percent Conversion Chart.