SOLUTION: A stadium has 50,000 seats. Seats sell for ​$35 in Section​ A, ​$20 in Section​ B, and ​$15 in Section C. The number of seats in Section A equals the total number of se

Algebra ->  College  -> Linear Algebra -> SOLUTION: A stadium has 50,000 seats. Seats sell for ​$35 in Section​ A, ​$20 in Section​ B, and ​$15 in Section C. The number of seats in Section A equals the total number of se      Log On


   



Question 1208221: A stadium has 50,000 seats. Seats sell for ​$35 in Section​ A, ​$20 in Section​ B, and ​$15 in Section C.
The number of seats in Section A equals the total number of seats in Sections B and C.
Suppose the stadium takes in ​$1,323,000 from each​ sold-out event. How many seats does each section​ hold?

Answer by ikleyn(52776) About Me  (Show Source):
You can put this solution on YOUR website!
.
A stadium has 50,000 seats. Seats sell for ​$35 in Section​ A, ​$20 in Section​ B, and ​$15 in Section C.
The number of seats in Section A equals the total number of seats in Sections B and C.
Suppose the stadium takes in ​$1,323,000 from each​ sold-out event. How many seats does each section​ hold?
~~~~~~~~~~~~~~~~~~~~

We assume that all tickets are sold and the amount $1,323,000  relates to 50,000 sold tickets.


Since "the number of seats in Section A equals the total number of seats in Sections B and C<
it means that the number of seats in section A is half of the total 50,000 seats, i.e. 25,000 seats,
while the total number of seats in sections B and C is another half of the total 50,000 seats,
i.e. 25.000 seats.


Thus, part of the problem is just solved: we derived that section A has 25,000 seats.


From it, we conclude that the cost of tickets sold in sections B and C is

    1323000 - 25000*35 = 448000  dollars.


Thus we know that total number of seats in sections B and C is 25,000  and the total cost of tickets
sold in these section B and C is 448000 dollars.


Let x be the number of seats in section B.  Then the number of seats in section C is 25000-x.


Having it, write an equation for the total cost of tickets in sections B and C

    20x + 15(25000-x) = 448000.


Simplify and find x

    20x + 15*25000 - 15x = 448000

    20x - 15x = 448000 - 15*25000

        5x    =     73000

         x    =     73000/5 = 14600.


ANSWER.  There are  25000 seats in section A;  14600 seats in section B  and  25000-14600 = 10400 seats in section C.


CHECK.   The total cost of all tickets is  35*25000 + 20*14600 + 15*10400 = 1,323000  dollars.

         It coincides with the given value.  Hence, the solution is correct.

Solved.