A Broadway theater has 400 seats, divided into orchestra, main, and balcony seating. Orchestra seats sell for $ 80 comma main seats for $ 65 comma and balcony seats for $ 40. If all the seats are sold, the gross revenue to the theater is $ 23 comma 700. If all the main and balcony seats are sold, but only half the orchestra seats are sold, the gross revenue is $ 20 comma 500. How many are there of each kind of seat?

Question

Karter Webb

Mathematics

9 August, 01:10

A Broadway theater has 400 seats, divided into orchestra, main, and balcony seating. Orchestra seats sell for $ 80 comma main seats for $ 65 comma and balcony seats for $ 40. If all the seats are sold, the gross revenue to the theater is $ 23 comma 700. If all the main and balcony seats are sold, but only half the orchestra seats are sold, the gross revenue is $ 20 comma 500. How many are there of each kind of seat?

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Gregory Moran · Answer 1

80 orchestra seats

180 main seats

140 balcony seats

Step-by-step explanation:

Total number of seats = 400

Orchestra seats sell for $80

Main seats sell for $65

Balcony seats sell for $40

If all the seats are sold, gross revenue = $23,700

If all the main seats and balcony seats are sold but only half of the orchestra seats are sold, the gross revenue = $20,500

Let x = Orchestra seats

Let y = main seats

Let z = balcony seats

x+y+z = 400 ... (1)

Orchestra seats sell for $80, Main seats sell for $65 and bslcony seats sell for $40. The gross revenue if all the seats are sold = $23,700

80x + 65y + 40z = 23,700 ... (2)

If half of the orchestra seats are sold while all main and balcony seats are sold, we have

80 (0.5x) + 65y + 40z = 20,500

40x + 65y + 40z = 20,500 ... (3)

we have the simultaneous equations

x+y+z = 400 ... (1)

80x + 65y + 40z = 23,700 ... (2)

40x + 65y + 40z = 20,500 ... (3)

Using elimination method to solve equation 2 and 3, we subtract equation 3 from 2

40x = 3200

x = 3200/40

x = 80 orchestra seats

Put x = 80 into equation 1 and 2

From equation 1,

x + y+z = 400

y + z = 400 - x

y+z = 400 - 80

y+z = 320 ... (4)

From equation 2,

80x + 65y + 40z = 23,700

80 (80) + 65y + 40z = 23,700

6400 + 65y + 40z = 23,700

65y+40z = 23,700 - 6400

65y + 40z = 17,300 ... (5)

Multiply equation (4) by 40 and equation (5) by 1

So we have

40y + 40z = 12,800

65y + 40x = 17,300

subtract both equations from each other. we have

-25y = - 4500

y = - 4500/-25

y = 180 main seats

Put x = 80 and y = 180 into equation (1)

80+180+z = 400

260 + z = 400

z = 400-260

z = 140 balcony seats