Problem 70 - Precalculus: Functions and Graphs

Question

Gold and silver mixture A ring that weighs 80 grams is made of gold and silver. By measuring the displacement of the ring in water, it has been determined that the ring has a volume of \(5 \mathrm{cm}^{3} .\) Gold weighs \(19.3 \mathrm{g} / \mathrm{cm}^{3},\) and silver weighs \(10.5 \mathrm{g} / \mathrm{cm}^{3} .\) How many grams of gold does the ring contain?

Step-by-Step Solution

Answer

The ring contains approximately 60.31 grams of gold.

Step 1: Define Variables

Let \( x \) be the volume of gold in the ring in cubic centimeters, and \( y \) be the volume of silver in cubic centimeters. Then the total volume is 5 cm³, so we have the equation \( x + y = 5 \).

Step 2: Formulate the Weight Equation

The weight of the gold portion can be expressed as \( 19.3x \) grams and the weight of the silver portion as \( 10.5y \) grams. Since the total weight of the ring is 80 grams, we have \( 19.3x + 10.5y = 80 \).

Step 3: Solve the System of Equations

We have two equations: \( x + y = 5 \) and \( 19.3x + 10.5y = 80 \). Solve the first equation for \( y \), giving \( y = 5 - x \), and substitute into the second equation: \[19.3x + 10.5(5 - x) = 80\]

Step 4: Simplify and Solve for \( x \)

Distribute and simplify the equation from Step 3: \[19.3x + 52.5 - 10.5x = 80\]Combine like terms: \[8.8x + 52.5 = 80\]Subtract 52.5 from both sides: \[8.8x = 27.5\]Divide by 8.8 to solve for \( x \): \[x \approx 3.125\]

Step 5: Calculate the Weight of Gold

Since \( x \) represents the volume of gold in cm³ and knowing the density of gold is 19.3 g/cm³, the weight of the gold is \( 19.3 \times 3.125 \). Calculating this product gives approximately 60.3125 grams.

Key Concepts

Volume Displacement

Volume Displacement is a fundamental concept in physics and is useful in various real-world applications, including determining the volume of an irregular object like our ring. When an object is immersed in a fluid, it displaces an amount of fluid equal to its own volume. In this problem, a ring with a known weight displaces water, revealing its total volume.
Understanding volume displacement helps us determine the volume contributions of multiple materials inside the ring. By knowing the volume displaced, which in this case is 5 cm³, we can start setting up the equations necessary to find the volumes of gold and silver separately in the ring.

System of Equations

A system of equations consists of multiple equations that share common variables. Solving a system of equations allows us to find the values of these variables.
In this exercise, we created two equations based on the given conditions: one for volume and another for weight. They are:

• Volume equation: \( x + y = 5 \)
• Weight equation: \( 19.3x + 10.5y = 80 \)

Here, \( x \) and \( y \) represent the respective volumes of gold and silver in cubic centimeters. By solving these equations together, we determine the exact compositions of gold and silver in the ring.

Solving Linear Equations

Linear equations are equations of the first degree, meaning they can be written in the form \( ax + b = c \). Solving these involves isolating the variable on one side and freeing it from coefficients.
In our case, the steps involve substituting the value of \( y = 5 - x \) from the volume equation into the weight equation. This yields a single equation in terms of \( x \) only:

• \( 19.3x + 10.5(5 - x) = 80 \)

Simplifying this equation step-by-step, we end up with \( 8.8x = 27.5 \). Dividing both sides by 8.8 solves for \( x \), giving us the volume of gold.

Density Calculations

Density is a crucial property that relates the mass of a substance to its volume. It is typically expressed as grams per cubic centimeter (g/cm³). In this problem, knowing the densities of gold and silver allows us to calculate their respective masses.
The density of gold is 19.3 g/cm³, meaning each cubic centimeter of gold weighs 19.3 grams. Once we find the volume of gold, which is approximately 3.125 cm³, we can determine its weight by calculating:

• Weight of gold \( = 19.3 \times 3.125 \approx 60.3125 \text{ grams} \)

Density calculations enable us to convert our results from volume in cubic centimeters to actual weights in grams, combining both physics and algebra for a comprehensive solution.