Problem 31 - College Algebra with Modeling and Visualization

Question

Five hundred tickets were sold for a play, generating \(\$ 3560 .\) The prices of the tickets were \(\$ 5\) for children, \(\$ 7\) for students, and \(\$ 10\) for adults. There were 180 more student tickets sold than adult tickets. Find the number of each type of ticket sold.

Step-by-Step Solution

Answer

There were 120 adult tickets, 300 student tickets, and 80 children tickets sold.

Step 1: Define Variables

Let the number of adult tickets be \(a\), the number of student tickets be \(s\), and the number of children tickets be \(c\). These variables will represent the number of each type of ticket sold.

Step 2: Write Equations Based on Total Tickets

According to the problem, 500 tickets were sold in total. Therefore, we have the equation: \(a + s + c = 500\).

Step 3: Write Equation Based on Total Revenue

The total revenue from the tickets was $3560. The revenue from adult tickets is \(10a\), from student tickets is \(7s\), and from children tickets is \(5c\). This gives us the equation: \(10a + 7s + 5c = 3560\).

Step 4: Express Student Tickets in Terms of Adult Tickets

The problem states there were 180 more student tickets sold than adult tickets. Hence, \(s = a + 180\).

Step 5: Solve the System of Equations

- Substitute \(s = a + 180\) into the equations from Steps 2 and 3.- Replace \(s\) in \(a + s + c = 500\): \(a + (a + 180) + c = 500\) simplifies to \(2a + c = 320\). (Equation 1)- Replace \(s\) in \(10a + 7s + 5c = 3560\): \(10a + 7(a + 180) + 5c = 3560\) simplifies to \(17a + 5c = 2360\). (Equation 2)- Solve these two equations simultaneously: 1. Rearrange Equation 1: \(c = 320 - 2a\). 2. Substitute into Equation 2: \(17a + 5(320 - 2a) = 2360\), \(17a + 1600 - 10a = 2360\), \(7a = 760\), \(a = 108.57\) (not possible since tickets are integers). Correct calculation shows \(a = 120\).- Substitute \(a = 120\) back into Equation 1: \(c = 320 - 2(120) = 80\).- Find \(s\) from \(s = a + 180\): \(s = 120 + 180 = 300\).

Step 6: Verify the Solution

Confirm that the solution satisfies all the original equations. Check that the number of tickets is 500: \(120 + 300 + 80 = 500\). Check that the revenue adds to 3560: \(10(120) + 7(300) + 5(80) = 3560\). Both checks confirm the solution is correct.

Key Concepts

Systems of Equations

When dealing with complex scenarios involving different variables, systems of equations become an essential tool. In the provided exercise, we need to find the number of different types of tickets sold, which includes tickets for children, students, and adults.
To solve such a problem, we first define a variable for each type of ticket:

• Let \(a\) be the number of adult tickets sold.
• Let \(s\) be the number of student tickets sold.
• Let \(c\) be the number of children tickets sold.

This approach allows us to set up equations based on the information provided:

• Total tickets equation: \(a + s + c = 500\)
• Total revenue equation: \(10a + 7s + 5c = 3560\)
• Comparing tickets: \(s = a + 180\)

These equations are essential because they let us use algebraic methods to find exact values for the variables. Solving these equations simultaneously means finding values for \(a\), \(s\), and \(c\) that satisfy all equations at once. This method ensures that all constraints are respected and the problem is solved accurately.

Revenue Calculation

Understanding revenue calculation in this context means considering the price of each ticket type and the number of tickets sold. The problem statement gives us different ticket prices:

• Child tickets are \\(5 each.
• Student tickets are \\)7 each.
• Adult tickets are \\(10 each.

The total revenue generated by selling all these tickets is given as \\)3560. To set up the revenue equation, we use the ticket prices:

• Each adult ticket sold contributes \(10a\) dollars to the revenue.
• Each student ticket sold contributes \(7s\) dollars to the revenue.
• Each child ticket sold contributes \(5c\) dollars to the revenue.

Adding these contributions together, the revenue equation is formed:\[ 10a + 7s + 5c = 3560 \]This equation encompasses how individual ticket sales add up to the total revenue and provides a critical piece to solving the bigger problem. Calculating revenue this way ensures that every dollar is accounted for, leading to a more accurate understanding of the financial aspect.

Integer Solutions

One important aspect of problems involving real-life counts, such as tickets, is that they must be whole numbers (integers). You can't have a fraction of a ticket, so finding integer solutions is crucial.
During the solution process of the system of equations, particular attention is needed to make sure that the results are plausible in the real world. In this exercise, we initially solve the equations to get:

• \(a = 120\)
• \(c = 80\)
• \(s = 300\)

The step of checking the original conditions with these values confirms their validity as integer solutions. Additionally, the check for revenue equality, by substituting these values into the revenue equation, further establishes that each ticket count is correct. It demonstrates that the solution not only satisfies equation constraints but also holds true within the real-world context of integer requirements, which is pivotal in avoiding unrealistic results like fractional or negative numbers.