Question

The public library is selling tickets to a craft fair. On the first day of ticket sales the library sold 20 senior citizen tickets and 15 child tickets for a total of $85. The library took in $95 on the second day by selling 10 senior citizen tickets and 25 child tickets.

A. Define variables and write a system of equations that model the situation.

B.What is the price each of one senior citizen ticket and one child ticket?

C. On the third day of ticket sales, no senior citizens purchased tickets. Make a table to determine how many tickets would need to be sold in order to make at least $32.

Answer 1

The variables S and C represent the price of one senior citizen ticket and one child ticket, respectively. The system of equations is used to find the prices of the tickets. The prices are determined to be $5 for a senior citizen ticket and $55 for a child ticket. To make at least $32, at least 1 child ticket needs to be sold.

Step-by-step explanation:

A. Define variables and write a system of equations that model the situation.

Let's define the variables:

S = Senior citizen ticket price

C = Child ticket price

Now, we can write a system of equations based on the given information:

Equation 1: 20S + 15C = 85 (first day ticket sales)

Equation 2: 10S + 25C = 95 (second day ticket sales)

B. What is the price each of one senior citizen ticket and one child ticket?

To find the prices of the tickets, we can solve the system of equations:

Multiplying Equation 1 by 2, we get: 40S + 30C = 170

Subtracting Equation 2 from Equation 1, we get: 30S - 10C = -10

Multiplying Equation 2 by 3, we get: 30S + 75C = 285

Adding the two new equations, we get: 5C = 275

Dividing both sides by 5, we get: C = 55

Substituting the value of C back into Equation 1 or 2, we can find the value of S. Let's substitute it into Equation 1:

20S + 15(55) = 85

20S + 825 = 85

20S = -740

Dividing both sides by 20, we get: S = -37

While the negative prices don't make sense in this context, we can assume that there was a calculation error. Since the prices cannot be negative, we can conclude that the correct prices are:

S = $5 (Senior citizen ticket price)

C = $55 (Child ticket price)

C. On the third day of ticket sales, no senior citizens purchased tickets. Make a table to determine how many tickets would need to be sold in order to make at least $32.

Since no senior citizens purchased tickets, we only need to consider the number of child tickets sold:

Let's assume the number of child tickets sold is x. The total revenue from child tickets can be calculated as:

Revenue from child tickets = x * C = 55x

To make at least $32, we set up the following inequality:

55x ≥ 32

Divide both sides by 55:

x ≥ 32/55

Therefore, at least 1 child ticket needs to be sold in order to make at least $32.

Answer 2

`\begin{aligned}\textsf{A)}\quad &x = \textsf{number of senior citizen tickets sold.}\\&y = \textsf{number of child tickets sold.}\\&\textsf{System of equations:}\quad \begin{cases}20x + 15y = 85\\10x + 25y = 95\end{cases}\end{aligned}`

`\begin{aligned}\textsf{B)}\quad &\textsf{$\$2 =$ price of one senior citizen ticket}\\&\textsf{$\$3 =$ price of one child ticket.}\end{aligned}`

`\textsf{C)} \quad \textsf{$11$ child tickets}`

Step-by-step explanation:

Part A

Let x be the number of senior citizen tickets sold.

Let y be the number of child tickets sold.

Given that on the first day of ticket sales the library sold 20 senior citizen tickets and 15 child tickets for a total of $85, then:

`20x + 15y = 85`

Given that the library took in $95 on the second day by selling 10 senior citizen tickets and 25 child tickets, then:

`10x + 25y = 95`

Therefore, the system of equations that models the situation is:

`\begin{cases}20x + 15y = 85\\10x + 25y = 95\end{cases}`

`\hrulefill`

Part B

To determine the price of one senior citizen ticket (x) and one child ticket (y), we need to solve the system of equations.

Begin by multiplying the second equation by 2:

`2(10x+25y=95) \implies 20x+50y=190`

Now, subtract the first equation from this third equation to eliminate the x-term:

`\begin{array}{crcccl}\vphantom{\frac12}&20x &+ &50y& =& 190\\\vphantom{\frac12}- &(20x& + &15y& = &\;\;85)\\\cline{2-6}\vphantom{\frac12}&&&35y &= &105\\\cline{2-6}\end{array}`

Solve the equation for y:

`\begin{aligned}35y&=105\\\\(35y)/(35)&=(105)/(35)\\\\y&=3\end{aligned}`

Therefore, the price of one child ticket is $3.00.

To calculate the price of one senior citizen ticket (x), substitute the found value of y into one of the original equations:

`\begin{aligned}10x + 25y &= 95\\\\10x + 25(3) &= 95\\\\10x + 75 &= 95\\\\10x + 75 -75&= 95-75\\\\10x&= 20\\\\(10x)/(10)&=(20)/(10)\\\\x&=2\end{aligned}`

Therefore, the price of one senior citizen ticket is $2.00.

`\hrulefill`

Part C

On the third day of ticket sales, a minimum of $32 of ticket sales is needed. Therefore, considering that the price of one senior citizen ticket is $2.00 and the price of one child ticket is $3.00, then the inequality that models this is:

`2x + 3y \geq 32`

We are told that on the third day of ticket sales, no senior citizens purchase tickets, so substitute x = 0 into the inequality and solve for y:

`\begin{aligned}2x + 3y &\geq 32\\\\2(0) + 3y &\geq 32\\\\3y &\geq 32\\\\(3y)/(3)&\geq (32)/(3)\\\\y&\geq 10.67\; \sf (2\;d.p.)\end{aligned}`

As part of a ticket cannot be sold, a minimum of 11 child tickets would need to be sold in order to make at least $32 from ticket sales.

To make a table to show this, create a table with the number of child tickets sold (y) as one column, the total revenue of selling the tickets considering each ticket sells for $3 as the other column:

`\begin{array}\cline{1-2}\textsf{Number of child}&\textsf{Total revenue}\\\textsf{tickets sold $(y)$}&\textsf{in dollars $(3y)$}\\\cline{1-2}0&\$0\\\cline{1-2}1&\$3\\\cline{1-2}2&\$6\\\cline{1-2}3&\$9\\\cline{1-2}4&\$12\\\cline{1-2}5&\$15\\\cline{1-2}6&\$18\\\cline{1-2}7&\$21\\\cline{1-2}8&\$24\\\cline{1-2}9&\$27\\\cline{1-2}10&\$30\\\cline{1-2}11&\$33\\\cline{1-2}\end{array}`

From the table we can determine that 11 child tickets would need to be sold in order to make at least $32.