Problem 6 - Prealgebra

Question

Solve each of the following word problems by translating the statement into a proportion. Be sure to show the proportion used in each case. [Examples \(1-4]\) Nutrition If 10 ounces of a certain breakfast cereal contain 3 ounces of sugar, how many ounces of sugar do 25 ounces of the same cereal contain?

Step-by-Step Solution

Answer

25 ounces of cereal contain 7.5 ounces of sugar.

Step 1: Identify Known Values

Read the problem carefully and identify the known quantities. We know that 10 ounces of cereal contain 3 ounces of sugar.

Step 2: Identify Unknown Value

Determine the unknown value that we need to find. In this case, we need to find out how many ounces of sugar are in 25 ounces of cereal.

Step 3: Set Up the Proportion

A proportion compares two ratios. We can set up the proportion using the known and unknown values: \( \frac{3}{10} = \frac{x}{25} \), where \(x\) represents the unknown quantity of sugar.

Step 4: Cross-Multiply to Solve

To solve for \(x\), cross-multiply: \(3 \times 25 = 10 \times x\). This gives us the equation \(75 = 10x\).

Step 5: Solve for the Unknown

Solve the equation \(75 = 10x\) by dividing both sides by 10. \(x = \frac{75}{10} = 7.5\).

Step 6: Interpret the Result

The result \(x = 7.5\) means that 25 ounces of cereal contain 7.5 ounces of sugar.

Key Concepts

Word Problems

Word problems in mathematics are scenarios presented in textual form that require interpretation and translation into a mathematical context before they can be solved. The key to successfully solving them is to thoroughly understand the given situation and identify the relevant quantities and relationships involved.

In the provided problem, we are told that 10 ounces of a cereal contain 3 ounces of sugar. We need to determine how many ounces of sugar are in 25 ounces of the same cereal. This problem is inherently about finding a relationship between the two quantities, represented here as a proportion.

To solve a word problem effectively:

• Read the problem carefully to identify known and unknown values.
• Formulate a question from the text that clearly defines what you need to find.
• Translate the verbose sentence into a mathematical statement or equation.

By translating the word problem into a proportion, we make a comparison between two ratios, which paves the way for further solving techniques such as cross-multiplication.

Cross-Multiplication

Cross-multiplication is a useful mathematical technique often employed to solve proportions. It involves multiplying across the equal sign in a way that eliminates the denominators, allowing you to solve for the unknown variable.

To illustrate using the previous example: Given the proportion \( \frac{3}{10} = \frac{x}{25} \), cross-multiply the terms to simplify the solving process. You multiply the numerator of the first fraction by the denominator of the second (3 × 25) and the numerator of the second fraction by the denominator of the first (10 × x). This yields the equation:

• \( 3 \times 25 = 10 \times x \)

Cross-multiplication effectively converts the proportion into a quicker-to-solve equation format that doesn't rely on maintaining the fractions. This technique is especially useful for dealing with algebraic equations involving fractions and can simplify the solving process significantly.

Solving Equations

Once a proportion has been translated into a simple equation through cross-multiplication, solving for the unknown becomes a matter of performing arithmetic operations to isolate the variable.

Continuing with our cereal problem, we have the equation \( 75 = 10x \) after cross-multiplication. The goal now is to solve for \( x \). This is achieved by isolating \( x \) on one side of the equation, which typically involves dividing both sides by the numerical coefficient of \( x \).

Perform the following step:

• Divide both sides by 10: \( x = \frac{75}{10} \)

This results in:

• \( x = 7.5 \)

Thus, we find that 25 ounces of cereal contain 7.5 ounces of sugar. Solving equations is about simplifying and systematically breaking down the equation using algebraic principles until the unknown is isolated. This approach helps in understanding and solving not just proportions, but all types of algebraic problems.