Solving Word Problems Involving Tickets and Stamps

Learning Outcomes

Apply the problem-solving method to solve problems involving tickets with different values
Apply the problem-solving method to solve problems involving stamps with different values

The strategies we used for coin problems can be easily applied to some other kinds of problems too. Problems involving tickets or stamps are very similar to coin problems, for example. Like coins, tickets and stamps have different values, so we can organize the information in tables much like we did for coin problems.

Example

At a school concert, the total value of tickets sold was $\text{\$1,506}$ . Student tickets sold for $\text{\$6}$ each and adult tickets sold for $\text{\$9}$ each. The number of adult tickets sold was $5$ less than three times the number of student tickets sold. How many student tickets and how many adult tickets were sold?

Solution:

Step 1: Read the problem.

Determine the types of tickets involved.

There are student tickets and adult tickets.
Create a table to organize the information.

Type	$\text{Value (\$)}$	$\text{Total Value (\$)}$
Student	$6$
Adult	$9$
		$1,506$

Step 2. Identify what you are looking for.

We are looking for the number of student and adult tickets.

Step 3. Name. Represent the number of each type of ticket using variables.

We know the number of adult tickets sold was $5$ less than three times the number of student tickets sold.

Let $s$ be the number of student tickets.

Then $3s - 5$ is the number of adult tickets.

Multiply the number times the value to get the total value of each type of ticket.

Type	$\text{Number}$	$\text{Value (\$)}$	$\text{Total Value (\$)}$
Student	$s$	$6$	$6s$
Adult	$3s - 5$	$9$	$9\left(3s - 5\right)$
			$1,506$

Step 4. Translate: Write the equation by adding the total values of each type of ticket.

6 s + 9 (3 s - 5) = 1506

$6s+9\left(3s - 5\right)=1506$

Step 5. Solve the equation.

\begin{matrix} 6 s + 27 s - 45 = 1506 \\ 33 s - 45 = 1506 \\ 33 s = 1551 \\ s = 47 students \end{matrix}

$\begin{array}{c}6s+27s - 45=1506\hfill \\ 33s - 45=1506\hfill \\ 33s=1551\hfill \\ s=47\text{ students}\hfill \end{array}$

Substitute to find the number of adults.

$3s-5=$ number of adults

$3(\color{red}{47})-5=136$ adults

Step 6. Check. There were $47$ student tickets at $\text{\$6}$ each and $136$ adult tickets at $\text{\$9}$ each. Is the total value $\text{\$1506}?$ We find the total value of each type of ticket by multiplying the number of tickets times its value; we then add to get the total value of all the tickets sold.

\begin{array}{ccc} 47 \cdot 6 & = & 282 \\ 136 \cdot 9 & = & \underset{_____}{1224} \\ 1506 ✓ \end{array}

$\begin{array}{ccc}\hfill 47\cdot 6& =\hfill & 282\hfill \\ \hfill 136\cdot 9& =\hfill & \underset{\text{_____}}{1224}\hfill \\ & & 1506\quad\checkmark \hfill \end{array}$

Step 7. Answer the question. They sold

47

$47$ student tickets and

136

$136$ adult tickets.

try it

Now we’ll do one where we fill in the table all at once.

Example

Monica paid $\text{\$10.44}$ for stamps she needed to mail the invitations to her sister’s baby shower. The number of $\text{49-cent}$ stamps was four more than twice the number of $\text{8-cent}$ stamps. How many $\text{49-cent}$ stamps and how many $\text{8-cent}$ stamps did Monica buy?

Show Solution

Solution:

The type of stamps are $\text{49-cent}$ stamps and $\text{8-cent}$ stamps. Their names also give the value.

“The number of $\text{49-cent}$ stamps was four more than twice the number of $\text{8-cent}$ stamps.”

$\begin{array}{c}\text{Let}x=\text{number of 8-cent stamps}\\ 2x+4=\text{number of 49-cent stamps}\end{array}$

Type	$\text{Number}$	$\text{Value (\$)}$	$\text{Total Value (\$)}$
$\text{49-cent}$ stamps	$2x+4$	$0.49$	$0.49\left(2x+4\right)$
$\text{8-cent}$ stamps	$x$	$0.08$	$0.08x$
			$10.44$

Write the equation from the total values.	$0.49\left(2x+4\right)+0.08x=10.44$
Solve the equation.	$0.98x+1.96+0.08x=10.44$ $1.06x+1.96=10.44$ $1.06x=8.48$ $x=8$
Monica bought 8 eight-cent stamps.
Find the number of 49-cent stamps she bought by evaluating.	$2x+4\text{ for }x=8$ .
	$2x+4$ $2\cdot 8+4$ $16+4$ $20$
Check. $8\left(0.08\right)+20\left(0.49\right)\stackrel{?}{=}10.44$ $0.64+9.80\stackrel{?}{=}10.44$ $10.44=10.44\quad\checkmark$

Monica bought eight $\text{8-cent}$ stamps and twenty $\text{49-cent}$ stamps.

try it

Watch the following video to see another example of how to find the number of tickets sold given the total sales for two different ticket values.

Module 9: Multi-Step Linear Equations