Suppose the stadium takes in ​$1,448,500 from each​ sold-out event. How many seats does each section​ hold?

Question

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?

Answer 1

a+b+c=55k, k=1000

a=b+c

b+c+ b+c= 55K=55

b +c = 27,5k=a= 27,500 seats for section A

there's a problem with this problem it leas to a negative solution for b which doesn't make sense

maybe a typo, misprint or mis copied

35a +25b +20c = 1448.5k=1448.5

35(27.5) + 20b +20c + 5b = 1448.5

35(27.5) + 20(27.5)+5b = 1448.5

5b = 1448.5 - 55(27.5)

divide by 5

b= 289.7 -11(27.5) = 289.7- 275-27.5=14.7-27.5=-13.2= negative 13,200 in Section B

Answer 2

Hi there!

So let's start by writing down everything we know about the situation in terms of equations. The stadium has three sections, A, B, and C, which in total is 55,000 seats. We can write that as:

A + B + C = 55,000 (Eq. 1)

The number of seats in Section A equals the total number of seats in Sections B and C. Therefore:

A = B + C (Eq. 2)

Lastly, in a sold-out show (where all seats are purchased), the total revenue is $1,448,500. Seats sell for ​$35 in Section​ A, ​$20 in Section​ B, and ​$15 in Section C. We can write this as:

35A + 20B + 15C = 1,448,500 (Eq. 3)

We have three equations written, with three unknowns. This is known as a system of equations, and one way of working through this is by the substitution method. To substitute, we can arrange the equations in terms of one variable, and plug it into the other equations.

Let's start with Eq. 2:

A = B + C

It already defines A in terms of B and C, so this is a good place to begin. We'll move to Eq. 1 and replace every A with its equivalent, B + C:

A + B + C = 55,000

(B + C) + B + C = 55,000

2B + 2C = 55,000

Now we'll arrange this equation to define B in terms of C:

2B + 2C = 55,000

2B = 55,000 - 2C

B = 27,500 - C (Eq. 4)

Our last step of substitution is to plug Eq. 4 into Eq. 3, the last equation of our system. So for every A, we'll replace it with B + C. Then, for every B, we'll substitute it with 27,500 - C:

35A + 20B + 15C = 1,448,500

35(B + C) + 20B + 15C = 1,448,500

35[(27,500 - C) + C] + 20(27,500 - C) + 15C = 1,448,500

962,500 + 550,000 - 20C + 15C = 1,448,500

-5C = -64,000

C = 12,800 seats

Now we have the number of seats in Section C. We can plug this into Eq. 4 to solve for B:

B = 27,500 - C

B = 27,500 - 12,800

B = 14,700 seats

Lastly, we can plug our values for B and C into either Eq. 1, 2, or 3 to find A. Let's do Eq. 2:

A = B + C

A = 12,800 + 14,700

A = 27,500 seats

Hope that helps!