/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 45 How much of an alloy that is \(2... [FREE SOLUTION] | 91Ó°ÊÓ

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How much of an alloy that is \(20 \%\) copper should be mixed with 200 ounces of an alloy that is \(50 \%\) copper in order to get an alloy that is \(30 \%\) copper?

Short Answer

Expert verified
Mix 400 ounces of the 20% copper alloy with 200 ounces of the 50% copper alloy.

Step by step solution

01

Define Variables

Let's define the variables necessary to solve the problem. Let \( x \) be the amount (in ounces) of the alloy that is 20% copper. We already know we have 200 ounces of the alloy that is 50% copper.
02

Express Copper Amounts

Calculate the amount of copper in each alloy. The alloy with 20% copper contains \(0.20x\) copper, while the 200-ounce alloy with 50% copper contains \(0.50 \times 200 = 100\) ounces of copper.
03

Set Up the Equation

To find the desired mixture, we need to set the total amount of copper in the final alloy equal to 30% of the total alloy weight. So, we can write the equation: \[0.20x + 100 = 0.30(x + 200)\]
04

Solve the Equation

Solve the equation for \( x \):- Distribute the 0.30 on the right-hand side: \[0.20x + 100 = 0.30x + 60\]- Rearrange the terms: \[100 - 60 = 0.30x - 0.20x\]- Simplify to \[40 = 0.10x\]- Divide each side by 0.10 to find \( x \): \[x = \frac{40}{0.10} = 400\]
05

Interpret the Solution

The solution \( x = 400 \) means that 400 ounces of the 20% copper alloy is needed to be mixed with the 200 ounces of 50% copper alloy to produce a new alloy that is 30% copper.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mixture Problems
Mixture problems are a typical topic in algebra, involving combining different substances to form a new mixture with a specific desired property. In our problem, we are mixing alloys to obtain an alloy with a certain copper percentage.

Whenever dealing with mixture problems, start by clearly defining variables for each part. In this example, we used \( x \) to represent the amount of alloy with 20% copper. This is crucial as it helps lay down the foundation for forming equations.

Next, evaluate the individual components. Compute the exact amount of the key element (copper in this case) in each portion being mixed. For our problem, calculate \(0.20x\) for the alloy with 20% copper, and 100 ounces for the alloy that is 50% copper.

- Problematize: How the elements combine is essential. Look for the sum of the elements and their desired combined percentage.
- Outcome: This approach allows for the identification and arrangement in an equation that represents the situation. This leads us to set up a mathematical model describing the real-world situation.
Percentage Problems
Percentage problems frequently appear in algebra and involve expressing quantities as parts of a whole using percentages. Understanding percentages is crucial for solving problems involving ratios and proportions.

To grasp percentage concepts, remember:
  • 'Percent' means 'per hundred,' which helps visualize ratios.
  • Conversion between percentages and decimal form is essential for algebra calculations, as in this problem: 20% becomes 0.20 and 50% becomes 0.50.
In the exercise, we convert percentage descriptions into equations. By expressing the copper content in alloys as decimals, we establish clearer mathematical expressions.

- Converting percentages to decimals streamlines calculations, ensuring equations remain consistent and manageable. This process enables smooth transitions between pure numbers and percentage-based reasoning.

Mastering percentage problems aids in accurately setting up and solving equations, essential skills not just for mixture problems, but for broader mathematical applications.
Equation Solving
Solving equations is a fundamental algebra skill, particularly useful in mixing and percentage problems. We need to translate problem statements into mathematical expressions, which involves setting up and solving equations.

For instance, starting with a descriptive equation like \[0.20x + 100 = 0.30(x + 200)\], reflects how different aspects of the problem combine to meet criteria. To solve, we must isolate the variable \( x \), which represents the unknown quantity needed.

Steps to solve an algebraic equation include:
  • Expanding and simplifying expressions where possible, as done with 0.30 and \(x + 200\).
  • Rearranging terms to group like terms on one side and constants on the other.
  • Solving by performing inverse operations to isolate and solve for the variable.
In the exercise, simplifying \(40 = 0.10x\) and solving \(x = 400\) involves applying these principles.

Mastering equation solving demystifies complex problems, allowing for logical, step-by-step solutions—a handy tool in various contexts beyond academics.

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