Q 4.72 - Intermediate Algebra

Question

The community college soccer team sold three kinds of tickets to its latest game. The adult tickets sold for \($10$, the student tickets for \)$8$ and the child tickets for \($5$. The soccer team was thrilled to have sold $600$ tickets and brought in \)$4,900$for one game. The number of adult tickets is twice the number of child tickets. How many of each type did the soccer team sell?

Step-by-Step Solution

Answer

The number of adult ticket, student ticket and child ticket sold is $200, 300$ and $100$respectively. 

Step 1. Given Information

We are given that the cost of adults, student and child tickets is $\$10, \$8$ and $\$5$. The total number of ticket sold is $600$. 

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 $600$, so

$x+y+z=600 -\left(1\right)$

The total amount of ticket is $\$4,900$, so

$10x+8y+5z=4900 -\left(2\right)$

Also, it is given that the number of adult tickets is twice the number of child tickets, so

$x=2z -\left(3\right)$

Step 3. Solving the equations

Putting $x=2z$ in first and second equations, we get

$$\begin{gathered} 2z+y+z=600 \\ y+3z=600 -\left(4\right) \\ 10\left(2z\right)+8y+5z=4900 \\ 8y+20z+5z=4900 \\ 8y+25z=4900 -\left(5\right) \end{gathered}$$

Multiplying by $8$ in fourth equation, we get

$8y+24z=4800 -\left(6\right)$

Now, subtracting fifth and sixth equation, we get

$$\begin{gathered} 8y-8y+25z-24z=4900-4800 \\ z=100 \end{gathered}$$

Putting the value of $z$ in third equation, we get

$$\begin{gathered} x=2z \\ x=2\times 100 \\ x=200 \end{gathered}$$

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

$$\begin{gathered} x+y+z=600 \\ 200+y+100=600 \\ y=600-300 \\ y=300 \end{gathered}$$

Step 4. Checking the solution

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

$$\begin{gathered} x+y+z=600 \\ 200+300+100=600 \\ 600=600 \\ 10x+8y+5z=4900 \\ 10\left(200\right)+8\left(300\right)+5\left(100\right)=4900 \\ 2000+2400+500=4900 \\ 4900=4900 \end{gathered}$$

This is true, hence the solution is correct.