Question 1100563: A stadium has 50,000 seats. Seats sell for $25 in section A, $20 in section B, $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 1,074,500 from each sold-out event. How many seats does each section holds?
Answer by ikleyn(52831)   (Show Source):
You can put this solution on YOUR website! .A stadium has 50,000 seats. Seats sell for $25 in section A, $20 in section B, $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 1,074,500 from each sold-out event. How many seats does each section holds?
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Let x be the number of seats in section B, and y be the number of seats in section C.
Then the number of seats in section A is x+y, the sum of seats in sections B and C, according to the condition.
The condition says
seatsA + seatsB + seatsC = 50000, or
(x+y) + x + y = 50000, which is the same as
2(x+y) = 50000, which implies
x + y =  = 25000.
Thus we found that the sum of seats in sections B and C is 25000,
and, therefore, the number of seats in section A is 25000.
Thus we have now the problem FOR TWO UNKNOWNS only, since we just excluded A.
Now our system of two equations is
x + y = 25000, (1)
20x + 15y = 1074500 - 25*25000. (2)
Equation (2) express the money payed for seats in sections B and C together.
Simplify (1),(2) step by step:
x + y = 25000, (3)
20x + 15y = 449500. (4)
Multiply eq(1) by 20 (both sides). Then subtract eq(2) from it:
20x + 20y = 500000, (3)
20x + 15y = 449500. (4)
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====> 5y = 500000-449500 = 50500 ====> y =  = 10100.
Then x = 25000 - y = 25000 - 10100 = 14900.
Answer. There are 25000 sears in section A, 14900 in section B and 10100 in section C.
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