Answer:
- A: 24,500
- B: 14,300
- C: 10,200
Step-by-step explanation:
Section A seats comprise half the seats in the stadium, so revenue from those seats at a sold-out event will total ...
$28 × 49,000/2 = $686,000
Then the remaining revenue of $1,037,200 -686,000 = $351,200 will come from some mix of B and C seats.
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If all of the seats were C seats, the revenue would be ...
$12 × 24,500 = $294,000
Actual revenue from those seats is $351,200 -294,000 = $57,200 more. Since each B seat contributes $4 more, there must be ...
$57,200 / $4 = 14,300 . . . . B-section seats
Then the number of C-section seats is 24,500 -14,300 = 10,200.
There are 24,500 Section A seats, 14,300 Section B seats, and 10,200 Section C seats.
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You could write 3 equations in 3 unknowns and get the same answer.
a + b + c = 49000 . . . . . where a, b, c are numbers of seats in A, B, and C
a - b - c = 0
28a +16b +12c = 1037200
The solution by your favorite method is (a, b, c) = (24500, 14300, 10200), as above.
