# A 6000 seat theater has tickets for sale at $27 and $40....?

My Dad and I were trying to solve these story problems, there’s like 10 in total. He was having a lot of difficulty because he told me that he lacks context/frame of reference for these types of problems. Seeking explanation/breakdown as to how to approach this. Any help is appreciated, thank you in advance.

A 6000 seat theater has tickets for sale at $27 and $40. How many tickets should be sold at each price or sell out performance to generate a total revenue of $185, 400?

the number of tickets for sale at $27 should be ___

### 3 Answers

- charlatanLv 71 month ago
x+y=6000

27x+40y=185400

27x+27y=16200

y=1300

x=4200

if you have learnt /been taught to solve

simultaneous equations the steps shown

will be easy to comprehend.

- oldschoolLv 71 month ago
We have two unknowns so we need two equations. The 6000 seat theater is sold out so let the number of $27 tickets be A and $40 tickets B.

First equation: A+B = 6000

Second equation: 27A+40B = 185,400

Use first equation to express B in terms of A: B = (6000-A)

Substitute into second equation:

27A + 40(6000-A) = 185,400

27A - 40A + 40*6000 = 185,400

-13A = 185,400 - 240,000 = -54,600

A = -54,600/-13 = 4,200 $27 tickets

B = 6000 - 4200 = 1,800 $40 tickets

Check:

4200*27 + 1800*40 = 185,400 Checks!

Hope this helps.

- MichaelLv 71 month ago
x = number of $27 tickets

6000 - x = number of $40 tickets

--------------------------

Multiply the number of tickets by their value to get sales

27x + 40(6000 - x) = 185,400

Distribute

27x + (240,000 - 40x) = 185,400

Combine like terms

-13x + 240,000 = 185,400

Subtract 240,000 from both sides

-13x = -54,600

Divide both sides by -13

x = 4,200 <––––– $27 tickets

6,000 - 4,200 = 1,800 <–––––$ 40 tickets