Problem 3 - A History of Mathematics: An Introduction

Question

I have 25 pounds of silver alloy that contain 8 ounces of pure silver per pound and 16 pounds that have \(9 \frac{1}{2}\) ounces of silver per pound. How much copper must be added to the total so that I can make coins containing \(7 \frac{1}{2}\) ounces of silver per pound?

Step-by-Step Solution

Answer

Answer: \(\frac{89}{15}\) pounds of copper should be added.

Step 1: Calculate the total amount of silver in the first alloy

To find the total amount of silver in the first alloy, multiply the weight of the alloy (25 pounds) by the ounces of silver per pound (8 ounces): Total silver in first alloy = 25 * 8 = 200 ounces.

Step 2: Calculate the total amount of silver in the second alloy

To find the total amount of silver in the second alloy, multiply the weight of the alloy (16 pounds) by the ounces of silver per pound (\(9 \frac{1}{2}\) ounces): Total silver in second alloy = 16 * 9.5 = 152 ounces.

Step 3: Calculate the total amount of silver and the total weight of the alloy

Add the total silver from both alloys: Total silver = 200 + 152 = 352 ounces. Add the weight of both alloys: Total weight of the alloy = 25 + 16 = 41 pounds.

Step 4: Determine the weight of the new alloy

Let x be the weight of copper added (in pounds). The new alloy will have a total weight of 41 + x pounds, and the proportion of silver per pound should be \(7 \frac{1}{2}\) ounces. Write an equation for this proportion: \(\frac{352}{41+x}=\frac{15}{2}\)

Step 5: Solve the equation for x

To solve the equation for x, first multiply both sides by (2 * (41 + x)): \(2(352)=15(41+x)\) Next, simplify the equation: \(704=615+15x\) Subtract 615 from both sides: \(89=15x\) Finally, divide by 15 to find the value of x: \(x=\frac{89}{15}\).

Step 6: Calculate the amount of copper to add

The value of x represents the weight of copper to be added to the alloy, which is \(\frac{89}{15}\) pounds. Therefore, we need to add \(\frac{89}{15}\) pounds of copper to reach the desired proportion of \(7 \frac{1}{2}\) ounces of silver per pound in the new alloy.

Key Concepts

Alloy Mixtures

When creating alloy mixtures, blending different metals can change the properties of the final product. In this exercise, an alloy made of silver is being adjusted for its silver content.
There are two alloys in consideration:

• The first alloy weighs 25 pounds and contains 8 ounces of silver per pound. This equates to a total silver content of 200 ounces (25 pounds x 8 ounces per pound).
• The second alloy weighs 16 pounds, with each pound containing\(9\frac{1}{2}\) ounces of silver. The total silver from this alloy is 152 ounces (16 pounds x \(9\frac{1}{2}\) ounces per pound).

Understanding these basic calculations helps us determine how to adjust them to achieve a desired mixture. The goal is to mix these alloys and possibly add copper so that the final mixture contains \(7\frac{1}{2}\) ounces of silver per pound. This requires understanding mutually the properties of silver and copper in the context of their weights and purity.

Algebraic Equations

In this scenario, we use algebraic equations to adjust the mix of metals, particularly balancing the total silver content with the weight of the alloy.
Here's how:

• We know the combined silver content from the two alloys is 352 ounces (200 + 152 ounces).
• The total weight of these two is 41 pounds (25 + 16 pounds).
• To find the correct mixture, we set up an equation: \( \frac{352}{41+x} = \frac{15}{2} \)

Here, \(x\) represents the weight of copper added. This equation equates the present silver proportion with the target of \(7\frac{1}{2}\) silver ounces per pound. Through algebraic manipulation, we solve for \(x\) demonstrating the pivotal role of algebraic equations in achieving the desired alloy mixture.

Weight Calculations

Accurate weight calculations are fundamental when adjusting the composition of an alloy like in this task.
In the process:

• We needed to calculate the total weight of the silver within each existing alloy before mixing.
• After determining the total silver and the initial overall weight of the mixtures, it was necessary to measure and adjust by solving for \(x\), which stands for the weight of additional copper required.
• The solution concluded that the amount of copper required is \( \frac{89}{15} \) pounds (approximately 5.93 pounds).

These weight calculations verify the precision needed to achieve the target alloy composition and ensure uniformity in the final product.