Problem 63 - Algebra 1

Question

Use the following information. A person has quarters, dimes, and nickels with a total value of 500 cents ( 5.00 dollar). The number of nickels is twice the number of quarters. The number of dimes is four less than the number of quarters. Write and solve an equation to find the number of each type of coin.

Step-by-Step Solution

Answer

Therefore, the number of quarters is 9, the number of dimes is 5, and the number of nickels is 18.

Step 1: Set Up Equations

We first need to set up the equations based on the information given. We have the following three equations: \[25q+10d+5n=500\] (from the total value of the coins), \[n=2q\] (from the number of nickels being twice the number of quarters), and \[d=q-4\] (from the number of dimes being four less than the number of quarters)

Step 2: Substitute and Simplify

We substitute \[n\] and \[d\] in the total value equation and simplify it. Substituting \[n=2q\] and \[d=q-4\] into the first equation results in \[25q+10(q-4)+5(2q) = 500\], which simplifies to \[65q-40 = 500\]

Step 3: Solve for q

Next, add 40 to both sides for \[65q = 540\] and divide by 65 to get \[q=540/65=8.31\]. Since we cannot have a fraction of a coin, round up to the nearest whole number to get \[q=9\]

Step 4: Find n and d values

Using the value of \[q\] from the previous step, solve for \[n\] and \[d\] using the equations \[n=2*9=18\] and \[d=9-4=5\]

Key Concepts

Coin Problems

Coin problems are common in algebra and involve finding the number of various types of coins that add up to a specific value. They are often represented with linear equations, where each type of coin has a known value. In this problem, quarters, dimes, and nickels are involved. To approach a coin problem:

• Assign a variable to each type of coin.
• Write equations based on the given relationships among the coins and their total value.
• Solve the equations to find the number of each type of coin.

In the exercise, variables were used to represent quarters ( q ), dimes ( d ), and nickels ( n ). The challenge was to use these coins to reach a total of 500 cents. By understanding the relationships, such as the number of nickels being twice the number of quarters, we form an equation representing this relationship. This method helps simplify complicated problems into manageable steps.

Linear Equations

Linear equations are equations in which each term is either a constant or the product of a constant and a single variable. These equations form straight lines when graphed. In the context of our coin problem, they help create equations based on the value and number of each type of coin.The general form of a linear equation is:\[ ax + by + cz = d \]Where \( a \), \( b \), \( c \) are coefficients, and \( x \), \( y \), \( z \) are the variables. In this problem:

• Quarters have a value of 25 cents, so it's represented as \( 25q \).
• Dimes are 10 cents each, so we have \( 10d \).
• Nickels are 5 cents each, giving us \( 5n \).

These lead to the equation \( 25q + 10d + 5n = 500 \). This is a classic linear equation representing the total value of the coins. Solving this helps us find the specific number of coins of each type that sums to the desired amount.

Substitution Method

The substitution method is a popular technique for solving systems of equations. This method involves solving one equation for one variable and substituting this into another equation. It reduces the system to a simpler equation with one variable.In the original problem, substitution was used as follows:

• First, solve the auxiliary equations \( n = 2q \) and \( d = q - 4 \).
• Next, substitute \( n \) and \( d \) into the primary equation \( 25q + 10d + 5n = 500 \).
• This leads to the equation \( 25q + 10(q-4) + 5(2q) = 500 \), simplifying to \( 65q - 40 = 500 \).

By solving this equation, we find the value of \( q \), the number of quarters. Once the value of one variable is known, it can be substituted back into the equations for \( n \) and \( d \) to find their specific values. This method efficiently breaks down complex problems into simpler, manageable steps.