Problem 44 - A Graphical Approach to College Algebra

Question

Coin Collecting A coin collection made up of pennies, nickels, and quarters contains a total of 29 coins. The number of quarters is 8 less than the number of pennies. The total face value of the coins is \(\$ 1.77 .\) How many of each denomination are there

Step-by-Step Solution

Answer

12 pennies, 13 nickels, 4 quarters.

Step 1: Define Variables

Let \( p \) represent the number of pennies, \( n \) the number of nickels, and \( q \) the number of quarters.

Step 2: Set Up Equations from the Problem Statement

Based on the given information, we know the following:\1. The total number of coins is 29: \( p + n + q = 29 \).2. The number of quarters is 8 less than the number of pennies: \( q = p - 8 \).3. The total value of the coins is \( \$1.77 \), which is equivalent to 177 cents: \( 1p + 5n + 25q = 177 \).

Step 3: Substitute q in the Equations

Substitute \( q = p - 8 \) into the equations:1. In terms of total coins: \( p + n + (p - 8) = 29 \) simplifies to \( 2p + n = 37 \).2. In terms of value: \( 1p + 5n + 25(p - 8) = 177 \) simplifies to \( 26p + 5n = 377 \).

Step 4: Solve for n using the First Simplified Equation

From the equation \( 2p + n = 37 \), solve for \( n \): \[ n = 37 - 2p \].

Step 5: Substitute n into the Second Simplified Equation

Substitute \( n = 37 - 2p \) into \( 26p + 5n = 377 \): \[ 26p + 5(37 - 2p) = 377 \].This simplifies to: \( 26p + 185 - 10p = 377 \).Further simplification gives: \( 16p = 192 \).

Step 6: Solve for p

Divide both sides of \( 16p = 192 \) by 16: \[ p = \frac{192}{16} = 12 \].Thus, there are 12 pennies.

Step 7: Solve for q

Since \( q = p - 8 \), substitute \( p = 12 \): \[ q = 12 - 8 = 4 \].Thus, there are 4 quarters.

Step 8: Solve for n

Using \( n = 37 - 2p \) and \( p = 12 \): \[ n = 37 - 24 = 13 \].Thus, there are 13 nickels.

Step 9: Verify the Solution

Ensure the number of each type and the total value add up:- Total coins: \( p + n + q = 12 + 13 + 4 = 29 \).- Total value in cents: \( 1 \times 12 + 5 \times 13 + 25 \times 4 = 12 + 65 + 100 = 177 \).Both conditions meet the initial problem statement.

Key Concepts

Coin Collection

When dealing with coin collections in algebra problems, understanding the setup of the problem is key. This problem involves pennies, nickels, and quarters.
Each type of coin has a different value and contributes to the total amount in a distinct way.To begin solving any coin collection problem, it helps to:

• Identify the types of coins involved.
• Determine the number of each type of coin, if possible.
• Calculate the total value made by the coins.

In our specific problem, we had 29 coins in total. We knew that the number of quarters was fewer compared to the number of pennies by a margin of eight coins. Additionally, the combined monetary value of all coins was \(\$1.77\). Using these pieces of information laid the foundation for setting up algebraic equations that mimic these conditions.

Main Keyword

In this context, the main keyword refers to "system of equations." This keyword highlights the central technique used to solve the problem.
A system of equations is a set of equations with multiple variables. These equations are solved together to find the values of the variables that meet all conditions. A classic approach to solve a system of equations involves:

• Formulating the equations based on the information provided in the problem.
• Simplifying and manipulating these equations using substitution or elimination methods.
• Finding solutions for the unknown variables, which in this case are the counts of each coin.

By substituting known relationships, such as the number of quarters being eight less than the number of pennies, the problem becomes manageable, transforming phrases into mathematical expressions. This step is crucial for proceeding with algebraic solving techniques.

Algebra Problem Solving

Algebra problem solving requires a structured process and logical thinking. In coin collection exercises, it's about translating words into formulas, allowing us to handle these problems efficiently with algebra.The following steps are often vital in solving such problems:

• Defining Variables: Represent unknown quantities with algebraic variables. Here, pennies were represented as \( p \), nickels as \( n \), and quarters as \( q \).
• Equation Setup: Use logical relationships to form equations. For example, we had \( p + n + q = 29 \) for the number of coins, and \( 1p + 5n + 25q = 177 \) for their total value.
• Substitutions: Replace one variable in terms of another to simplify the problem. This involves substituting \( q = p - 8 \).
• Simplification: Reduce terms by arithmetic operations. Once any two variables are known, solve for the third using one of the simpler formulations.

These steps make complex problems more approachable and enhance critical thinking skills. The careful structuring and solving of equations turn challenging problems into a step-by-step process.