Q. 76 - Precalculus Enhanced with Graphing Utilities

Question

Theater Revenues 

A movie theater charges $\backslash (8.00$ for adults, $\backslash )4.50$ for children, and $\backslash (6.00$ for senior citizens. One day the theater sold $405$ tickets and collected $\backslash )2320$ in receipts. Twice as many children’s tickets were sold as adult tickets. How many adults, children, and senior citizens went to the theater that day?

Step-by-Step Solution

Answer

There were $110$ adults, $220$ children and $75$ senior citizens.

Step 1. Given Information

Cost of adult ticket $=\$8.00$

Cost of children ticket $=\$4.50$

Cost of senior citizens ticket $=\$6.00$

Total tickets sold $=405$

Total money earned $=\$2320$

Twice as many children’s tickets were sold as adult tickets 

Step 2. Using elimination and substitution effect

Let $x$ be the number of adults ticket sold, $y$ be the number of children tickets sold and $z$ be the number of senior citizens ticket sold.

The required equations are

$$\begin{gathered} x+y+z=405.......\left(i\right) \\ 8x+4.5y+6z=2320........\left(ii\right) \\ y=2x........\left(iii\right) \end{gathered}$$

Putting the value of equation $\left(iii\right)$ in equations $\left(i\right)$ and $\left(ii\right)$ we get,

$$\begin{gathered} x+2x+z=405 \\ \Rightarrow 3x+z=405........\left(iv\right) \\ 8x+4.5y+6z=2320 \\ \Rightarrow 8x+4.5\left(2x\right)+6z=2320 \\ \Rightarrow 8x+9x+6z=2320 \\ \Rightarrow 17x+6z=2320.........\left(v\right) \end{gathered}$$

Multiplying equation $\left(iv\right)$ by $6$ we get,

$$\begin{gathered} 6\times \left(3x+z\right)=6\times 405 \\ \Rightarrow 18x+6z=2430.......\left(vi\right) \end{gathered}$$

Subtracting equation $\left(v\right)$ from $\left(vi\right)$ we get,

$x=110$

From equation $\left(iii\right)$ we get,

$$\begin{gathered} y=2x \\ \Rightarrow y=2\times 110 \\ \Rightarrow y=220 \end{gathered}$$

From equation $\left(i\right)$ we get,

$$\begin{gathered} z=405-110-220 \\ \Rightarrow z=405-330 \\ \Rightarrow z=75 \end{gathered}$$