The cashier in a movie box office sold 200 more adult admission tickets at $11.00 eachthan children’s admission tickets at $8.00 each. What is the minimum number of each typeof ticket that the cashier had to sell for the total receipts to be at least $5000?
Question
The cashier in a movie box office sold 200 more adult admission tickets at 8.00 each. What is the minimum number of each typeof ticket that the cashier had to sell for the total receipts to be at least $5000?
Solution
Let's denote the number of children's tickets sold as x. Then the number of adult tickets sold is x + 200.
The total revenue from the sale of tickets is the sum of the revenue from the sale of children's tickets and the revenue from the sale of adult tickets. This can be represented by the equation:
8x + 11(x + 200) >= 5000
Solving this equation will give us the minimum number of each type of ticket that the cashier had to sell.
First, distribute the 11 to the terms inside the parentheses:
8x + 11x + 2200 >= 5000
Combine like terms:
19x + 2200 >= 5000
Subtract 2200 from both sides of the equation:
19x >= 2800
Finally, divide both sides by 19 to solve for x:
x >= 147.37
Since the number of tickets sold must be a whole number, we round up to the nearest whole number. So, the cashier had to sell at least 148 children's tickets.
Substitute x = 148 back into the equation for the number of adult tickets:
x + 200 = 148 + 200 = 348
So, the cashier had to sell at least 348 adult tickets.
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