Question

A movie theater charges $15 for standard viewing, $20 for 3D viewing, and $35 for Dinner and a

Answer 1

The quantity of each type of seats sold are as follows:

• Standard viewing = 2000

• 3D seats = 800

• Movie and a dinner seat = 200

According to the question,there are four times as many 3D, x seats as Dinner and a Movie, y seats.

That is, x = 4y

Also, total seats

= (x) + (y) + (z) = 3000...….............eqn(1)

Also, If the theater brings in $53,000 when tickets to all 3000 seats are sold.

• 20x + 35y + 15z = 53000...........eqn(2)

By substituting 4y for x in equations 1 and 2; we have;

5y + z = 3000..…........eqn(3) and

115y + 15z = 53000.........eqn(4)

By solving equations 3 and 4 simultaneously; we have;

y = 200 and z = 2000

and since x = 4y

x = 800

The quantity of each type of seats sold is as follows:

• Standard viewing = 2000

• 3D seats = 800

• Movie and a dinner seat = 200

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