A stadium has 49,000 seats. Seats sell for 25 dollars in Section A, 20 dollars in Section B, and 15 dollars 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 1,052,000 dollars from each sold out event. How many seats does each section hold?

Let section B has x seat , section C has y seat.

section A has (x+y)seat
(x+y)+x+y =49000
x+y = 24500 equation1
25(x+y)+20x +15y = 1052000 equation2
9x+8y= 210400 equation2
solving eqn.1 & eqn. 2 by substitution method you will get
x=14400
y=10100
section A =24500
section B =14400
section C= 10100

Well, it seems like Section A is the most expensive and has the fewest seats, which makes sense because life is all about balance. Let's call the number of seats in Section A as "x".

Now, according to the information, the number of seats in Section B and Section C combined is equal to x. So, the total number of seats in both Section B and C would be 2x.

Adding up the number of seats in all three sections, we get the equation: x + 2x + 2x = 49,000 (since we know there are 49,000 seats in total).

Simplifying this equation, we have 5x = 49,000.

Dividing both sides by 5, we find x = 9,800.

So, Section A has 9,800 seats, and since Section B and C combined is also equal to x, we get that each of them has 9,800 seats as well.

Now, let's calculate the revenue from each sold-out section:

Section A: 9,800 seats x $25 per seat = $245,000
Section B: 9,800 seats x $20 per seat = $196,000
Section C: 9,800 seats x $15 per seat = $147,000

Adding up these values, we find that the total revenue from a sold-out event is $588,000. This is quite far from the given revenue of $1,052,000.

It seems like someone isn't good at math! Maybe the stadium should hire a better accountant or maybe I should enroll in a clown university!

Let's assume the number of seats in Section A is x.

Since the number of seats in Section A equals the total number of seats in Sections B and C, we have:

x = (number of seats in Section B) + (number of seats in Section C) ...(equation 1)

According to the problem, the total number of seats in the stadium is 49,000, so we can write:

x + (number of seats in Section B) + (number of seats in Section C) = 49,000 ...(equation 2)

Since we know the price of seats in each section, we can calculate the total revenue from each section:

Revenue from Section A = 25 * x
Revenue from Section B = 20 * (number of seats in Section B)
Revenue from Section C = 15 * (number of seats in Section C)

Given that the total revenue from a sold-out event is 1,052,000 dollars, we have:

25 * x + 20 * (number of seats in Section B) + 15 * (number of seats in Section C) = 1,052,000 ...(equation 3)

Now, we have a system of equations (equations 1, 2, and 3) that we can solve to find the values of x (number of seats in Section A), (number of seats in Section B), and (number of seats in Section C).

To solve this problem, we can set up a system of equations. Let's denote the number of seats in Section A as "x," the number of seats in Section B as "y," and the number of seats in Section C as "z."

Based on the given information, we know the following:
1) x + y + z = 49,000 (The total number of seats in all sections is 49,000)
2) The price of a seat in Section A is $25, in Section B is $20, and in Section C is $15.
3) The total revenue from a sold-out event is $1,052,000.

Using equation 1, we can rewrite it as x = 49,000 - (y + z).

Next, we can calculate the revenue from each section based on the number of seats and their respective prices:
Section A revenue: 25x
Section B revenue: 20y
Section C revenue: 15z

According to the problem, the total revenue is $1,052,000. So we can write the equation:
25x + 20y + 15z = 1,052,000

Now, substitute the value of x from equation 1 into the revenue equation:
25(49,000 - (y + z)) + 20y + 15z = 1,052,000

Simplify the equation:
1,225,000 - 25y - 25z + 20y + 15z = 1,052,000
(20y - 25y) + (15z - 25z) = 1,052,000 - 1,225,000
-5y - 10z = -173,000

Divide both sides by -5:
y + 2z = 34,600 ----- (Equation 2)

Now, we have two equations:
x = 49,000 - (y + z) ----- (Equation 1)
y + 2z = 34,600 ----- (Equation 2)

To find the solutions, we can use substitution or elimination method. Let's solve this system of equations.

From Equation 2, we have y = 34,600 - 2z.

Substituting the value of y in Equation 1, we get:
x = 49,000 - (34,600 - 2z + z)
x = 49,000 - 34,600 + 3z
x = 14,400 + 3z ----- (Equation 3)

Now, we have three equations:
x = 14,400 + 3z ----- (Equation 3)
y = 34,600 - 2z ----- (Equation 4)
x + y + z = 49,000 ----- (Equation 5)

Substituting the values of x and y from Equations 3 and 4 into Equation 5, we get:
(14,400 + 3z) + (34,600 - 2z) + z = 49,000

Simplifying the equation:
14,400 + 3z + 34,600 - 2z + z = 49,000
48,000 + 2z = 49,000
2z = 49,000 - 48,000
2z = 1,000
z = 1,000 / 2
z = 500

Now, substitute the value of z into Equation 4:
y = 34,600 - 2(500)
y = 34,600 - 1,000
y = 33,600

Finally, substitute the values of y and z into Equation 1:
x = 14,400 + 3(500)
x = 14,400 + 1,500
x = 15,900

Therefore, Section A has 15,900 seats, Section B has 33,600 seats, and Section C has 500 seats.