Problem 55 - Prealgebra

Question

Sewing. If \(\frac{6}{7}\) yard of material is needed to make a blanket, how many blankets can be made from 12 yards of material?

Step-by-Step Solution

Answer

14 blankets can be made from 12 yards of material.

Step 1: Understanding the Problem

To solve this problem, we need to find out how many times \( \frac{6}{7} \) yards can fit into 12 yards. This will tell us how many blankets can be made.

Step 2: Setting Up the Division

Translate the word problem into a division problem: divide 12 yards by \( \frac{6}{7} \) yard per blanket. This setup is \( 12 \div \frac{6}{7} \).

Step 3: Reciprocal of a Fraction

To divide by a fraction, multiply by its reciprocal. The reciprocal of \( \frac{6}{7} \) is \( \frac{7}{6} \). So, we rewrite the problem as \( 12 \times \frac{7}{6} \).

Step 4: Multiplication of a Fraction and Whole Number

Multiply 12 by \( \frac{7}{6} \). Start with multiplying the whole number by the numerator: \( 12 \times 7 = 84 \).

Step 5: Final Division

Divide the result by the denominator of the fraction: \( 84 \div 6 = 14 \).

Step 6: Conclusion

Thus, 14 blankets can be made from 12 yards of material.

Key Concepts

Fraction Division

In prealgebra, fraction division is a crucial concept that facilitates solving everyday problems like the one about making blankets. When you encounter a division problem involving fractions, you often aim to determine how many times a fraction fits into another number. Here's the step-by-step understanding:

• Identify the division: The given problem requires finding out how many times \( \frac{6}{7} \) fits into 12. This is expressed as \( 12 \div \frac{6}{7} \).
• Recontextualize using division: You can reframe the problem of how many \( \frac{6}{7} \) yards fit into 12 yards, treating it as a division problem.

This context helps you grasp why it transforms into a specific math operation and connects the real-world application of fractions to mathematical operations.

Reciprocal of a Fraction

Understanding reciprocals is key to dividing fractions. A reciprocal of a fraction inverts the numerator and the denominator. When dividing by a fraction, you multiply by its reciprocal. Here's how this applies:

• Define the reciprocal: The reciprocal of \( \frac{6}{7} \) is \( \frac{7}{6} \).
• Application in division: Instead of dividing, you multiply 12 (the whole number) by the reciprocal of \( \frac{6}{7} \).
• Simplification: This conversion helps simplify calculations as multiplication is generally more straightforward than division with fractions.

Grasping how reciprocals function is vital since it underpins the process of converting division into a manageable multiplication problem.

Multiplication of Fractions

Once the division is converted into a multiplication problem using the reciprocal, the process becomes straightforward. Multiplying fractions or a fraction by a whole number involves few steps:

• Multiply numerators and denominators: Multiply 12 by the new numerator (7), which gives 84.
• Complete the calculation: The result is an intermediate product that needs to be divided by the original denominator (6).

These steps highlight the ease with which multiplication can solve fraction division problems, transforming a complex division into simple arithmetic.

Word Problem in Mathematics

Word problems in mathematics can sometimes seem daunting, but they are essential for applying mathematical concepts to real-life scenarios. Solving word problems involves a few strategic steps:

• Understand the scenario: Determine what the problem is asking. Here, it's how many blankets can be made from 12 yards of material.
• Translate into math: Converting the words into a math problem allows us to apply arithmetic or algebra to find a solution.
• Work through steps: Apply the proper operations—here, division by a fraction, converted to multiplication—to solve the problem.
• Conclude with interpretation: Finally, interpret the math result back into the context of the problem to provide a meaningful answer.

By breaking down word problems into these manageable steps, you can effectively solve them using prealgebra techniques.