Word Problems On Distributive Property

Mastering Word Problems: A Deep Dive into the Distributive Property

The distributive property, a fundamental concept in algebra, often hides within the seemingly innocuous world of word problems. Understanding how to identify and apply this property is crucial for success in mathematics, paving the way for more advanced concepts. This article provides a comprehensive guide to tackling word problems involving the distributive property, moving from basic understanding to complex applications. We'll explore various scenarios, offer step-by-step solutions, and delve into the underlying mathematical principles. By the end, you'll be confident in your ability to dissect and solve even the trickiest word problems related to the distributive property.

Understanding the Distributive Property

Before diving into word problems, let's solidify our understanding of the distributive property itself. The distributive property states that multiplying a number by a sum is the same as multiplying the number by each term in the sum and then adding the products. Mathematically, this is expressed as:

a(b + c) = ab + ac

or

a(b - c) = ab - ac

This seemingly simple equation unlocks the solution to many complex problems. The key is to recognize when a situation can be represented by this formula.

Identifying Distributive Property in Word Problems

The challenge with word problems lies in translating the written description into a mathematical equation. Look for keywords and phrases that hint at the distributive property. These might include:

• "Total cost" or "Total amount": Often indicating the need to combine several individual costs.
• "Each" or "Every": Suggesting multiplication of a common factor by several quantities.
• "Groups of" or "Sets of": Implying distribution of a quantity over multiple groups.
• "Difference" or "Reduced by": Indicating subtraction within the distributive property.

Let's examine several examples to illustrate these keywords in action.

Example Word Problems & Solutions

Example 1: The Fruit Stand

A fruit stand sells apples for $2 each and oranges for $1.50 each. John bought 3 apples and 4 oranges. What was the total cost?

Solution:

This problem subtly uses the distributive property. We can represent the total cost using the distributive property:

Total cost = Cost of apples + Cost of oranges

Total cost = (3 apples * $2/apple) + (4 oranges * $1.50/orange)

This is equivalent to:

Total cost = $2(3 + 4 * (1.50/2)) = $2(3 + 3) = $12

This can be solved directly as : 32 + 41.5 = 6 + 6 = $12

Notice that we can also use the distributive property to simplify the calculation:

Total cost = (3 + 4) * ($2 + $1.50) which will produce a wrong answer. It is important to note that the distributive property ONLY works when there is a multiplication operation. This is not a direct application of distributive property.

Example 2: The Classroom Supplies

A teacher needs to buy 5 packs of pencils and 5 packs of erasers. Each pack of pencils costs $3, and each pack of erasers costs $2. What is the total cost?

Solution:

This problem is a straightforward application of the distributive property. Let's represent the cost of the pencils as 'p' and the cost of erasers as 'e':

Total cost = 5(p + e) = 5($3 + $2) = 5($5) = $25

We could also calculate it as 5 * 3 + 5 * 2 = 15 + 10 = $25

Example 3: The Rectangular Garden

A rectangular garden has a length of (x + 5) meters and a width of 3 meters. What is the area of the garden?

Solution:

The area of a rectangle is calculated as length * width. In this case:

Area = 3(x + 5) = 3x + 15 square meters.

This is a direct application of the distributive property, expanding the expression to find the area in terms of 'x'.

Example 4: The Discount

A store is offering a 20% discount on all items. If Sarah buys items totaling $100, what is the final price after the discount?

Solution:

The discount amount is 20% of $100, which is 0.20 * $100 = $20. The final price is:

Final price = Original price - Discount

Final price = $100 - $20 = $80

Or, alternatively, and more directly applying distributive property:

Final price = Original price (1 - discount rate) = $100 (1 - 0.20) = $100 * 0.80 = $80

This problem shows a slightly different application where the distributive property helps in efficiently calculating the final price after a discount.

Example 5: More Complex Scenario

A contractor is building a fence around a rectangular area. The length is (2x + 7) feet and the width is (x - 3) feet. If fencing costs $10 per foot, what is the total cost?

Solution:

First, find the perimeter of the rectangle:

Perimeter = 2(length + width) = 2( (2x + 7) + (x - 3) ) = 2(3x + 4) = 6x + 8 feet

Now, calculate the total cost:

Total cost = Perimeter * Cost per foot = (6x + 8) * $10 = $60x + $80

This example demonstrates a more complex application where we combine the distributive property with other geometric formulas to solve the problem.

Step-by-Step Approach to Solving Word Problems

Follow these steps when tackling word problems involving the distributive property:

1. Read carefully: Understand the problem completely. Identify the unknown quantity you need to find.
2. Identify keywords: Look for words and phrases that suggest the distributive property.
3. Define variables: Assign variables to represent the unknown quantities.
4. Write an equation: Translate the word problem into a mathematical equation using the distributive property.
5. Solve the equation: Apply the distributive property and solve for the unknown variable.
6. Check your answer: Make sure your answer is reasonable and makes sense in the context of the problem.

Beyond the Basics: Advanced Applications

The distributive property extends far beyond these basic examples. It's a cornerstone of algebraic manipulation, appearing in more advanced topics like:

• Factoring expressions: The distributive property is used in reverse to factor algebraic expressions.
• Solving equations: It's frequently used to simplify equations before solving for the unknown variable.
• Working with polynomials: The distributive property is essential when multiplying and simplifying polynomials.

Frequently Asked Questions (FAQ)

Q1: Can the distributive property be applied to division?

A1: Not directly. The distributive property applies to multiplication. However, division can be expressed as multiplication by a reciprocal, allowing for indirect application. For instance, (a + b) / c can be rewritten as (1/c)(a + b), enabling the distributive property.

Q2: What if the expression inside the parentheses involves more than two terms?

A2: The distributive property still works. Simply distribute the term outside the parentheses to each term within the parentheses. For example: a(b + c + d) = ab + ac + ad.

Q3: How can I practice more?

A3: Seek out additional word problems in textbooks, online resources, or practice worksheets. Focus on varying the complexity and context of the problems to strengthen your understanding.

Conclusion

Mastering word problems involving the distributive property is a journey, not a destination. By understanding the underlying principle, identifying key phrases, and following a systematic approach, you can confidently tackle increasingly complex problems. Remember, practice is key to solidifying your skills and building confidence. With consistent effort, you'll not only solve these problems but also develop a deeper appreciation for the power and elegance of the distributive property in mathematics.