Problem 48 - Algebra for College Students

Question

Solve each equation. Sarah has a collection of nickels, dimes, and quarters worth \(\$ 15.75\). She has 10 more dimes than nickels and twice as many quarters as dimes. How many coins of each kind does she have?

Step-by-Step Solution

Answer

Sarah has 15 nickels, 25 dimes, and 50 quarters.

Step 1: Define Variables

Let \( n \) represent the number of nickels. Since Sarah has 10 more dimes than nickels, let \( d = n + 10 \) represent the number of dimes. She has twice as many quarters as dimes, so let \( q = 2d \) represent the number of quarters.

Step 2: Establish the Value Equation

Nickels are worth \( \\(0.05 \), dimes are worth \( \\)0.10 \), and quarters are worth \( \\(0.25 \). The total value of Sarah's coins is \( \\)15.75 \). Therefore, the equation becomes: \( 0.05n + 0.10d + 0.25q = 15.75 \).

Step 3: Substitute for Quarters

From Step 1, substitute \( q = 2d \) into the value equation: \( 0.05n + 0.10d + 0.25(2d) = 15.75 \). Simplify it to: \( 0.05n + 0.10d + 0.50d = 15.75 \), which further simplifies to \( 0.05n + 0.60d = 15.75 \).

Step 4: Substitute for Dimes

Using \( d = n + 10 \), replace \( d \) in the equation: \( 0.05n + 0.60(n + 10) = 15.75 \). This expands to \( 0.05n + 0.60n + 6 = 15.75 \).

Step 5: Simplify and Solve for Nickels

Combine terms: \( 0.65n + 6 = 15.75 \). Subtract 6 from both sides to get \( 0.65n = 9.75 \). Divide both sides by 0.65 to find \( n \): \( n = \frac{9.75}{0.65} = 15 \).

Step 6: Calculate the Number of Dimes

Substitute \( n = 15 \) back into \( d = n + 10 \) to get \( d = 15 + 10 = 25 \).

Step 7: Calculate the Number of Quarters

Using \( q = 2d \), substitute \( d = 25 \) to find \( q = 2 \times 25 = 50 \).

Step 8: Verify the Solution

Calculate the total value using \( n = 15 \), \( d = 25 \), and \( q = 50 \). \(15 \times 0.05 + 25 \times 0.10 + 50 \times 0.25 = 0.75 + 2.50 + 12.50 = 15.75 \). The values confirm the calculations.

Key Concepts

Systems of Equations

When tackling algebra word problems, especially one involving coin problems, systems of equations become essential. They are like tools that help in connecting multiple relationships and conditions in a problem. A system of equations consists of two or more equations that share variables. In Sarah's coin problem, we need to set up equations based on the condition that involves nickels, dimes, and quarters.

Here's a step-by-step breakdown on how to form these equations:

• First, identify the variables: let them represent unknown quantities such as nickels, dimes, and quarters.
• Next, translate the given conditions into algebraic equations: Sarah has 10 more dimes than nickels, and twice as many quarters as dimes. This can be represented as two different equations.

Solving these equations simultaneously allows us to find specific values for each variable, thus solving the entire problem. Systems of equations offer a structured way to handle multi-variable problems, which is why they're so effective in challenging scenarios like coin problems.

Coin Problems

Coin problems are a typical category of algebraic word problems that revolve around calculating the number of coins of various denominations. These are seen quite often in algebra curriculums because they merge real-world situations with algebraic thinking.

In the context of Sarah's problem, coin problems involve:

• Identifying each type of coin involved: nickels, dimes, and quarters.
• Assigning a value to each type of coin: nickels are worth $0.05, dimes $0.10, and quarters $0.25.
• Setting up equations that not only account for the number of coins but also their total value.

Coin problems beautifully demonstrate the practicality of algebra, illustrating how we can use algebraic concepts to solve real-life scenarios. They also enhance skills in logical thinking and systematic problem-solving.

Problem Solving

Problem solving in algebra involves a methodical approach, where we apply logical and numerical reasoning to arrive at the solution. The key steps include understanding the problem, devising a plan, executing the plan, and finally checking the results.

In Sarah's case, the problem-solving strategy involves:

• Understanding the problem by identifying the variables and the relationships between them.
• Devising a plan by setting up a system of equations based on the given relationships and total values.
• Executing the plan through algebraic manipulations, such as substituting variables and simplifying equations.
• Finally, checking the results by plugging the solution back into the original equations to verify accuracy.

This structured approach not only aids in solving the problem at hand but also equips students with skills that are beneficial in various complex situations outside the classroom.