Question

A stadium has 49,000 seats. Seats sell for ​$28 in Section​ A, ​$16 in Section​ B, and ​$12 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 ​$1,037,200 from each​ sold-out event. How many seats does each section​ hold?

Answer 1

• A: 24,500
• B: 14,300
• C: 10,200

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.