Question

A stadium has 50,000 seats. Seats sell for $42 in section A, $36 in section B, and $30 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 $1,888,800 from each sold out event. How many seats does each section hold?

Answer 1

A=B+C
2A=54000
A=27000 seats $30/seat=$810000
B+C=27000
24B+18C=571200
24B+18(27000-B)=571200
24B+486000-18B=571200
6B=85200
B=14200
C=12800
A=27000 seats, B=14200 seats, C=12800

Answer 2

Section A has 25,000 seats.

Section B has 14,800 seats.

Section C has 10,200 seats.

Explanation:

Let the seats in section A be = x

Let the seats in section B be = y

Let the seats in section C be = z

The equations forms as follows:

`x+y+z=50000` .....(1)

`42x+36y+30z=1888800` .......(2)

`x=y+z` ......(3)

Substituting the value of x in (1) to get equation in two terms.

`y+z+y+z=50000`

=>
`2y+2z=50000`

taking out 2 common, we get;

`y+z=25000` .........(4)

And substituting the value of x in (2), we get

`42(y+z)+36y+30z=1888800`

=>
`42y+42z+36y+30z=1888800`

=>
`78y+72z=1888800`

Taking out 2 common, we get;

`39y+36z=944400` ........(5)

Multiplying (4) by 39 and subtracting (5) from (4), we get

`3z=30600`

We get z = 10200

And
`y+z=25000`

`z=25000-10200=14800`

We get y = 14800

Also
`x=y+z`

`x=14800+10200=25000`

We get x = 25000

Therefore,

Section A has 25,000 seats.

Section B has 14,800 seats.

Section C has 10,200 seats.