Anonymous
Anonymous asked in Education & ReferenceHomework Help · 1 month ago

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

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  • 1 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.

  • 1 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.

  • 1 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

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