Solving Word Problems: Comparisons (Beginning Algebra Study Notes)

Solving Word Problems: Comparisons

Introduction to Word Problems Involving Comparisons

Word problems that involve comparisons require translating real-world situations into algebraic equations. These problems often compare two or more quantities, such as scores, lengths, or costs, and ask you to find unknown values. The key steps are understanding the problem, defining variables, writing an equation, solving, and checking your answer.

Understanding and Setting Up Comparison Problems

Step 1: Understand the Problem

• Identify what is being compared: Look for phrases like "more than," "less than," or "longer than." These indicate a relationship between two quantities.

• Define variables: Assign a variable to the unknown quantity, usually the one mentioned second in the comparison.

• Restate the relationships algebraically: Express the other quantities in terms of your variable.

Step 2: Write an Equation

• Translate the relationships into an equation: Use the information given to set up an equation that models the situation.

• Combine all parts of the problem: Make sure your equation includes all relevant quantities and their relationships.

Step 3: Solve and State the Answer

• Solve the equation: Use algebraic methods to isolate the variable and find its value.

• Answer the question: Substitute back to find all required quantities.

Step 4: Check Your Solution

• Verify your answer: Substitute your solution back into the original context to ensure it makes sense and satisfies all conditions.

Examples of Comparison Word Problems

Example 1: Basketball Game Score

Problem: In a basketball game, the Vikings scored 12 points more than the Mustangs. A total of 78 points were scored. How many points were scored by each team?

• Let $m$ = number of points scored by the Mustangs

• Vikings scored: $m + 12$

• Total points: $m + (m + 12) = 78$

Solving:

• $m + m + 12 = 78$

• $2m + 12 = 78$

• $2m = 66$

• $m = 33$

• Mustangs: 33 points; Vikings: $33 + 12 = 45$ points

Example 2: Rectangle Perimeter

Problem: A small plot of land is in the shape of a rectangle. The length is 7 meters longer than the width. The perimeter of the rectangle is 86 meters. Find the dimensions of the rectangle.

• Let $w$ = width

• Length: $w + 7$

• Perimeter equation: $w + (w + 7) + w + (w + 7) = 86$

Solving:

• $4w + 14 = 86$

• $4w = 72$

• $w = 18$

• Width: 18 meters; Length: $18 + 7 = 25$ meters

General Steps for Solving Comparison Word Problems

1. Read the problem carefully and identify what is being compared.

2. Assign variables to unknown quantities.

3. Express all other quantities in terms of the variable(s).

4. Write an equation that models the situation.

5. Solve the equation using algebraic techniques.

6. Answer the question and check your solution in the context of the problem.

Practice Problems

Practice 1: Rope Cutting

A deck hand on a fishing boat is working with a rope that measures 89 feet. He needs to cut it into two pieces. The long piece must be 17 feet longer than the short piece. Find the length of each piece of rope.

Practice 2: Heating Bill Comparison

A social services worker was comparing the cost incurred by three families in heating their homes for the year. The first family had an annual heating bill that was $360 more than that of the second family. The third family had a heating bill that was $200 less than double the heating bill of the second family. The total annual heating bill for the three families was $3960. What was the annual heating bill for each family?

Practice 3: Triangle Fencing

A farmer needs 720 meters of wire fencing to enclose a pasture. The pasture is in the shape of a triangle. The first side of the triangle is 30 meters less than the second side. The third side is one-half as long as the second side. Find the dimensions of the triangle.

Worked Example: Animal Hospital

Problem: The Center City Animal Hospital treated a total of 18,360 dogs and cats last year. The hospital treated 1376 more dogs than cats. How many dogs were treated last year? How many cats were treated last year?

• Let $c$ = number of cats

• Dogs: $c + 1376$

• Total: $c + (c + 1376) = 18,360$

Solving:

• $2c + 1376 = 18,360$

• $2c = 16,984$

• $c = 8,492$

• Cats: 8,492; Dogs: $8,492 + 1,376 = 9,868$

Worked Example: Airport Departures

Problem: An airport filed a report showing the number of plane departures from the airport during each month last year. The number of departures in March was 50 more than the number of departures in January. In July, the number of departures was 150 less than triple the number of departures in January. In those three months, the airport had 2,250 departures. How many departures were recorded for each month?

• Let $j$ = number of departures in January

• March: $j + 50$

• July: $3j - 150$

• Total: $j + (j + 50) + (3j - 150) = 2,250$

Solving:

• $5j - 100 = 2,250$

• $5j = 2,350$

• $j = 470$

• January: 470; March: 520; July: 1,260

Worked Example: Rectangle Perimeter (with Diagram)

Problem: A small plot of land is in the shape of a rectangle. The length is 7 meters longer than the width. The perimeter of the rectangle is 86 meters. Find the dimensions of the rectangle.

• Let $w$ = width

• Length: $w + 7$

• Perimeter equation: $w + (w + 7) + w + (w + 7) = 86$

Solving:

• $4w + 14 = 86$

• $4w = 72$

• $w = 18$

• Width: 18 meters; Length: 25 meters

Key Concepts and Strategies

• Comparison statements are translated into algebraic expressions (e.g., "7 more than x" becomes $x + 7$).

• Perimeter problems require adding all sides and setting the sum equal to the given perimeter.

• Always check your answer by substituting back into the original context.

Summary Table: Steps for Solving Comparison Word Problems

Step	Description
1. Understand	Identify what is being compared and define variables.
2. Write Equation	Translate relationships into an algebraic equation.
3. Solve	Use algebra to find the value of the variable(s).
4. State Answer	Answer the question in the context of the problem.
5. Check	Verify the solution by substituting back into the problem.