Chapter 3.5, Problem 32PPSE

Solution

Tocalculate:The number of seats each section can hold if a stadium has 49,000 seats. Seats sell for $25 in Section A, $20 in Section B, and $15 in Section C. The number of seats in Section A equals the total number of seats in Section B and C. Suppose the stadium takes in $1,052,000 from each sold-out event.

Section A holds24,500 seat, Section B holds 14,400 seats and section C holds 10,100 seats.

Given information:

Total number of seats in stadium is 49,000. Seats sell for $25 in Section A, $20 in Section B, and $15 in Section C. The number of seats in Section A equals the total number of seats in Section B and C. Suppose the stadium takes in $1,052,000 from each sold-out event.

Calculation:

Consider the given information.

Let the number of seats in Section A is x , number of seats in Section B is y , and number of seats in Section C is z .

Since the total number of seats are 49,000 so $x+y+z=49,000$

The number of seats in Section A equals the total number of seats in Section B and C $x=y+z$

Put $x=y+z$ in $x+y+z=49,000$

  $\begin{array}{l}x+x=49,000\\ x=24,500\end{array}$

Put $x=24,500$ into $x=y+z$

  $24,500=y+z\mathrm{...}\left(1\right)$

Seats sell for $25 in Section A, $20 in Section B, and $15 in Section C. Also, the stadium takes in $1,052,000 from each sold-out event.

  $\begin{array}{c}25x+20y+15z=1,052,000\\ 5x+4y+3z=210,400\end{array}$

Put $x=y+z$ into above equation.

  $\begin{array}{c}5x+4y+3z=210,400\\ 5\left(y+z\right)+4y+3z=210,400\\ 5y+5z+4y+3z=210,400\\ 9y+8z=210,400\mathrm{...}\left(2\right)\end{array}$

Multiply equation 1 by 9 and subtract from equation 2.

  $\begin{array}{c}9y+8z-9\left(y+z\right)=210,400-9\left(24,500\right)\\ z=10,100\end{array}$

Put $z=10,100$ into equation 1.

  $\begin{array}{c}24,500=y+10,100\\ y=14,400\end{array}$