Question

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A stadium has 55,000 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 ​$1,448,500 from each​ sold-out event. How many seats does each section​ hold?

Section A holds _________ seats.
Section B holds _________ seats.
Section C holds _________ seats.

Answer 1

Section A: 27,500 seats.

Section B: 14,700 seats.

Section C: 12,800 seats.

Explanation:

We set up a system of 3 equations where a, b and c are the number of seats in Section A B and C,

a = b + c (1)

a + b + c = 55000 (2)

35a + 20b + 15c = 1448500 (3)

From equation 1;

a - b - c = 0

Now if we add this to equation 2:

2a = 55,000

a = 27,500 So there are 27,500 seats in Section A.

Substitute for a in equation 3:

35* 27,500 + 20b + 15c = 1448500

20b + 15c = 486000 (4)

b + c = 27,500

b = 27,500 - c

Substitute for b in equation 4:

20*(27500 - c) + 15c = 486000

-20c + 15c = 486000 - 550000

-5c = -64000

c = 12800

So b = 27500-12800

= 14700

Answer 2

• A: 27,500
• B: 14,700
• C: 12,800

Explanation:

Section A holds half the seats in the stadium, so 27,500 seats.

Revenue from a sold-out section A will be ...

$35×$27500 = $962,500

so the combined revenue from B and C seats is ...

$1,448,500 -962,500 = $486,000

If we let "b" represent the number of section B seats, then the number of section C seats is (27,500 -b) and the revenue from sold-out sections B and C will be ...

20b +15(27500 -b) = 486000

5b = 73,500 . . . . . . . . subtract 412,500

b = 14,700 . . . . . . . . . . divide by 5; number of Section B seats

27,500-b = 12,800 . . . number of Section C seats

There are 27,500 Section A seats, 14,700 Section B seats, and 12,800 Section C seats.