Theater Revenues

Problem:

Solution:

This is not an easy question - it involves three variables, now let's set them up

Let $x$ be the number of orchestra seating.
Let $y$ be the main floor seating.
Let $z$ be the balcony seating.

We know there are 600 seats, therefore we have

$x + y + z = 600$ (Equation 1)

Next, we know if they are all sold, the revenue would be 33,500, that means

$80x + 60y + 25z = 33500$ (Equation 2)

Last but not least, we know on that evening, this equation holds.

$80x + \frac{3}{5}60y + \frac{4}{5}25z = 24640$.

The simplify to

$80x + 36y + 20z = 24640$.  (Equation 3)

These give us three equations with 3 unknowns, now we are all set to solve them.

Subtract equation 2 by the equation 3, we get

$24y + 5z = 8860$.  (Equation 4)

Subtract 80 times equation 1 by the equation 2, we get

$20y + 55z  = 14500$  (Equation 5)

Finally subtract the 5 times equation 4 by 6 times equation 5, we get

$-305z = -42700$, or simply $z = 140.$

Substitute this result to equation 4, solving get $y = 340$.

Finally, substitute both values back to equation 1, get $x = 120$.