Abasketball team sells tickets that cost $10, $20 or for vip seats $30. the team has sold 3239 tickets overall. it has sold 282 more $20 tickets than $10 tickets. the total sales are $63,720. how many tickets of each kind have been sold?

The number of tickets that cost $ 10 is 923

The number of tickets that cost $ 20 is 1205

The number of tickets that cost $ 30 is 1111

Step-by-step explanation:

Given as :

The total number of three different tickets cost $ 10 , $ 20 and VIP seats $ 30

The total number of all tickets sold = 3239

The sale price for the tickets = $ 63720

Let The type of ticket for $ 10 = x

And The type of ticket for $ 20 = y

And The type of VIP ticket = v

According to question

The number of $ 20 tickets = The number of $ 10 tickets + 282

I.e y = x + 282

And x + y + v = 3239

Or, v = 3239 - ( x + y )

I.e v = 3239 - ( x + x +282 )

Or, v = 2957 - 2 x

Now ,

x × $ 10 + y × $ 20 = v × $ 30

Or, x × $ 10 + ( x + 282 ) × $ 20 = ( 2957 - 2 x ) × $ 30

or, 10 x + 20 x + 5640 = 88710 - 60 x

Or, 30 x + 60 x = 88710 - 5640

or, 90 x = 83070

∴  x =

I.e x = 923

So, The  number of tickets that cost $ 10 = x = 923

Similarly

 y = x + 282

or, y = 923 + 282

I.e y = 1205

So, The  number of tickets that cost $ 10 = y = 1205

And

 v = 2957 - 2 x

∴ v = 2957 - 2 × 923

I.e v = 1111

So, The  number of tickets that cost $ 30 = v = 1111

Hence

The number of tickets that cost $ 10 is 923

The number of tickets that cost $ 20 is 1205

The number of tickets that cost $ 30 is 1111

Answer

1021 of the $10 tickets, 1303 of the $20 tickets and 915 of the $30 tickets.

Step-by-step explanation:

We have 3 types of tickets:

ones that cost $10, others that cost $20 and others that cost $30.

If A is the number of $10 tickets sold, B is the number of $20 tickets sold and C is the number of $30 tickets sold we have that:

A + B + C = 3239

B = A + 282

A*10 + B*20 + C*30 = 63720

We need to solve this system of equations.

From the second equation we have B isolated, we can replace it in the other equations:

A + (A + 282) + C = 3239

A*10 + (A + 282)*20 + C*30 = 63720

Now we can isolate one of the variables in the first equation and replace it in the second, lets isolate C.

C = 3239 - 2*A - 282 = 2957 - 2*A

the third equation is now:

A*10 + (A + 286)*20 + (2957 - 2*A)*30 = 63720

Now we solve it for A:

A*10 + A*20 - A*60 + 282*20 + 2957*30 = 63720

-A*30 + 94350 = 63720

A*30 = - 63720 + 94350 = 30630

A = 30639/30 = 1021

So they sold 1021 tickets of $10.

Now we can replace it in the equation:

C =   2957 - 2*A = 2957 - 2*1021  = 915

They sold 915 $30 tickets  

They sold 1303 tickets of $20.

B = A + 282 = 1021 + 282 = 1303.