Question

A stadium has 45,000 seats. Seats sell for $28 in Section A, $24 in Section B, and $20 in Section C. If section C held 300 fewer seats, then it would have half the number of seats as section B. Suppose the stadium takes in $1,139200 from each sold-out event. How many seats does each section hold?

Answer 1

Answer:

Let's define the variables:

A = number of seats in section A.

B = number of seats in section B.

C = number of seats in section C.

We have the equations:

A + B + C = 45,000.

C - 300 = B/2

A*$28 + B*$24 + C*$20 = $1,139,200

This is a system of equations, the first step to solve this is to isolate one variable in one of the equations, and then replace it in the others.

I will isolate C in the second equation:

C = B/2 + 300.

Now let's replace this in the other two equations:

A + B + B/2 + 300 = 45,000

A*$28 + B*$24 + (B/2 + 300)*$20 = $1,139,200

Let's simplify these equations:

A + B*(3/2) = 44,700

A*$28 + B*$34 + $6,000 = $1,139,200

Now let's isolate A in the first equation:

A = 44,700 - B*(3/2)

Let's replace this in the other equation:

(44,700 - B*(3/2))*$28 + B*$34 + $6,000 = $1,139,200

Now let's solve this for B.

-B*$8 + $1,252,200 = $1,139,200

-B*$8 =  $1,139,200 - $1,252,200  = -$113,000

B = $113,000/8 = 14,125

Now we can replace that in the equations:

A =  44,700 - B*(3/2) =  44,700 - 14,125*(3/2) = 23,512.5, that we should round up to 23,513.

And C = B/2 + 300 = 7362.5

As we rounded the previous one up, we should round this one down to 7362.