Question

Suppose that there are two types of tickets to a show: Advanced and same day. Advanced tickets cost $40 and same-day tickets cost $25. For one performance there are 65 tickets sold in all, and the total amount paid for them was $2225. How many tickets of each type were sold?

Answer 1

Solution:

Given:

Two types of tickets; advanced and same-day tickets.

Let a represent advanced tickets

Let s represent same-day tickets.

Developing the word problem (statements) into mathematical expressions, we have;

$\begin{gathered} \text{Advanced tickets cost \ 40. This means;} \\ a=\text{ \ 40} \\ \\ \text{Same}-\text{day tickets cost \ 25. This means;} \\ s=\text{ \ 25} \end{gathered}$$\begin{gathered} \text{Total tickets sold in all is 65. This means;} \\ a+s=65\ldots\ldots\ldots\ldots\ldots\ldots\ldots\text{.}\mathrm{}(1) \\ \\ \text{Total amount paid for advanced tickets = \ 40a} \\ \text{Total amount paid for same-day tickets = \ 25s} \\ \\ \text{Total amount paid for all tickets = \ 2225.} \\ \text{Hence,} \\ 40a+25s=2225\ldots\ldots\ldots\ldots\ldots\ldots\ldots\text{.}(2) \end{gathered}$

Solving equations (1) and (2) simultaneously to get the values of a and s;

$\begin{gathered} \text{From equation (1)},\text{ } \\ a+s=65 \\ a=65-s\ldots\ldots\ldots\ldots\ldots\ldots\ldots\text{.}(3) \\ \\ \text{Substituting equation (3) in equation (2),} \\ 40a+25s=2225 \\ 40(65-s)+25s=2225 \\ 2600-40s+25s=2225 \\ 2600-15s=2225 \\ \text{Collecting the like terms;} \\ 2600-2225=15s \\ 375=15s \\ 15s=375 \\ \text{Dividing both sides by 15 to get s,} \\ s=\frac{375}{15} \\ s=25 \\ \text{Thus, same-day tickets sold were 25 tickets} \end{gathered}$

Substituting the value of s in equation (3) to get the value of a.

$\begin{gathered} a=65-s \\ a=65-25 \\ a=40 \\ \text{Thus, advanced tickets sold were 40 tickets} \end{gathered}$

Therefore, advanced tickets sold were 40 tickets and same-day tickets sold were 25 tickets.