SOLUTION: Tickets to the school dance cost $4, and the projected attendance is 300 people. It is further projected that for every 10 cent increase in ticket price, the average attendance wil

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Tickets to the school dance cost $4, and the projected attendance is 300 people. It is further projected that for every 10 cent increase in ticket price, the average attendance wil      Log On


   



Question 307177: Tickets to the school dance cost $4, and the projected attendance is 300 people. It is further projected that for every 10 cent increase in ticket price, the average attendance will decrease by 5. At what ticket price will the receipts from the dance be $1,248? Show the quadratic equation used to solve this problem and show all work in solving it.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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Tickets to the school dance cost $4, and the projected attendance is 300 people.
It is further projected that for every 10 cent increase in ticket price, the
average attendance will decrease by 5.
At what ticket price will the receipts from the dance be $1,248?
Show the quadratic equation used to solve this problem and show all work in solving it.
:
Let x = no. .10 increases and no. of 5 people decreases
:
Cost * Attendance = $1248
(4 + .10x)(300 - 5x) = 1248
FOIL
1200 - 20x + 30x - .5x^2 = 1248
Arrange as a quadratic equation
-.5x^2 + 10x + 1200 - 1248 = 0
:
-.5x^2 + 10x - 48 = 0
Multiply equation by -2
x^2 - 20x + 96 = 0
Factors to
(x-12)(x-8) = 0
two solutions
x = 12
and
x = 8
:
:
Find the revenue when x=12
(4 + .10*12))(300 - 5(12)) =
5.20 * 240 = 1248
:
Revenue when x = 8
(4 + .10*8))(300 - 5(8)) =
4.80 * 260 = 1248 also