Problem 74 - Elementary Algebra

Question

Set up an algebraic equation and then solve. Billy has a pile of quarters, dimes, and nickels that values \(\$ 3.75 .\) He has 3 more dimes than quarters and 5 more nickels than quarters. How many of each coin does Billy have?

Step-by-Step Solution

Answer

Billy has 8 quarters, 11 dimes, and 13 nickels.

Step 1: Define Variables

Let's define variables for each type of coin Billy has. Let \( q \) be the number of quarters, \( d \) be the number of dimes, and \( n \) be the number of nickels.

Step 2: Establish Relationships

According to the problem, Billy has 3 more dimes than quarters and 5 more nickels than quarters. We can express this as:\[ d = q + 3 \]\[ n = q + 5 \]

Step 3: Set Up the Value Equation

The total value of the coins is \$3.75, which is 375 cents. The value equation considering each coin type is:\[ 25q + 10d + 5n = 375 \]

Step 4: Substitute Relationships into the Value Equation

Substitute \( d = q + 3 \) and \( n = q + 5 \) in the value equation:\[ 25q + 10(q+3) + 5(q+5) = 375 \]

Step 5: Simplify and Solve for \( q \)

Expand and simplify the equation:\[ 25q + 10q + 30 + 5q + 25 = 375 \]Combine like terms:\[ 40q + 55 = 375 \]Subtract 55 from both sides:\[ 40q = 320 \]Divide by 40:\[ q = 8 \]

Step 6: Find \( d \) and \( n \)

Using \( q = 8 \), calculate \( d \) and \( n \):\[ d = q + 3 = 8 + 3 = 11 \] \[ n = q + 5 = 8 + 5 = 13 \]

Step 7: Verify the Solution

Check the calculated values to ensure they add up to \$3.75:- Value from quarters: \( 8 \times 25 = 200 \) cents- Value from dimes: \( 11 \times 10 = 110 \) cents- Value from nickels: \( 13 \times 5 = 65 \) centsTotal: \( 200 + 110 + 65 = 375 \) cents. The solution checks out.

Key Concepts

Coin problem

Coin problems are a popular type of algebraic problem where you need to determine the number of coins of different denominations given certain conditions. These problems typically involve the values of coins adding up to a particular amount of money and specific relationships between the coins. Let's use Billy's coin problem as an example. Billy has quarters, dimes, and nickels that together make up $3.75. However, the problem introduces an additional twist: there are fixed relationships between the amounts of different coins. For example, Billy has 3 more dimes than quarters and 5 more nickels than quarters. In essence, coin problems test your ability to translate word problems into mathematical equations and solve them efficiently using basic algebraic principles.

Variable substitution

Variable substitution is a key strategy used in solving algebraic equations, especially in coin problems. This technique involves replacing one variable with another variable or expression to simplify the equation. In our coin problem, we have three variables: the number of quarters (\(q\)), the number of dimes (\(d\)), and the number of nickels (\(n\)).

By substituting the expressions \(d = q + 3\) and \(n = q + 5\) into the main equation, we reduce the problem to a simpler form. This helps eliminate some of the variables, making it easier to solve for one variable first. Once you have one variable figured out, solving for the others becomes a straightforward task. Variable substitution lightens the complexity of dealing with multiple variables at once.

Value equation

The value equation is at the heart of any coin problem. It represents the total monetary value of all the coins combined. In our problem, the total value of Billy's quarters, dimes, and nickels is $3.75 or 375 cents.

Each coin type contributes its own particular value to the equation:

• Quarters contribute \(25q\) cents.
• Dimes contribute \(10d\) cents.
• Nickels contribute \(5n\) cents.

Putting them together, we formed the equation \(25q + 10d + 5n = 375\). This equation needs to be solved alongside the substitution equations to find the exact number of each type of coin Billy has. By solving these equations step by step, one can determine the quantity of each coin that sums up to the given total value.

Solving equations

Solving equations is the final step in tackling coin problems. Once all variables and equations are accurately set up, you need to solve them to find precise numeric answers. Here, the process began by simplifying the value equation after the substitution:

- Substitute \(d = q + 3\) and \(n = q + 5\) into \(25q + 10d + 5n = 375\).
- Expand and simplify the equation to form \(40q + 55 = 375\).
- Isolate the variable \(q\) by subtracting 55 and then dividing by 40, giving \(q = 8\).

With \(q\) known, finding \(d\) and \(n\) becomes easy:

• \(d = q + 3 = 11\)
• \(n = q + 5 = 13\)

Finally, verify the solution by ensuring that the calculated coin values equate to the total of 375 cents. Solving these steps systematically reflects the relationships and values dictated by the problem, leading to a satisfying solution.