Q 4.70 - Intermediate Algebra

Question

The community college fine arts department sold three kinds of tickets to its latest dance presentation. The adult tickets sold for \($20$, the student tickets for \)$12$and the child tickets for \(10.The fine arts department was thrilled to have sold $350$ tickets and brought in \)$4,650$ in one night. The number of child tickets sold is the same as the number of adult tickets sold. How many of each type did the department sell?

Step-by-Step Solution

Answer

The number of adult ticket, student ticket and child ticket sold is $75, 200$ and $75$respectively.

Step 1. Given Information

We are given that the cost of adults, student and child tickets is $\$20, \$12$ and $\$10$. The total number of ticket sold is $350$.

Step 2. Assumptions and formation of system of equation.

Let $x, y, z$ be the number of tickets sold of adults, students and child.

Total number of ticket sold is $350$, so

$x+y+z=350 -\left(1\right)$

The total amount of ticket is $\$4,650$, so

$20x+12y+10z=4650 -\left(2\right)$

It is given that the number of child tickets sold is equal to the number of adult tickets sold, so

$y=z$

Step 3. Solving the equations

Putting $z=x$ in first and second equations, we get

$$\begin{gathered} x+y+x=350 \\ 2x+y=350 -\left(3\right) \\ 20x+12y+10x=4650 \\ 30x+12y=4650 -\left(4\right) \end{gathered}$$

Multiplying by $12$ in third equation, we get

$$\begin{gathered} 12(2x+y)=12\times 350 \\ 24x+12y=4200 -\left(5\right) \end{gathered}$$

Now, subtracting the third and fourth, we get

$$\begin{gathered} 30x-24x+12y-12y=4650-4200 \\ 6x=450 \\ x=\frac{450}{6} \\ x=75 \end{gathered}$$

Now, putting the value of $x$ in third equation, we get

$$\begin{gathered} 2x+y=350 \\ 2\left(75\right)+y=350 \\ 150+y=350 \\ y=350-150 \\ y=200 \end{gathered}$$

Since $x=z$, so

$z=75$

Hence the number of adult ticket, student ticket and child ticket sold is $75, 200$ and $75$ respectively.

Step 4. Checking the solution

Putting the value of $x,y,z$ in the equations, we get

$$\begin{gathered} x+y+z=350 \\ 75+200+75=350 \\ 350=350 \\ 20x+12y+10z=4650 \\ 20\left(75\right)+12\left(200\right)+10\left(75\right)=4650 \\ 1500+2400+750=4650 \\ 4650=4650 \end{gathered}$$

This is true, hence the solution is correct.