A basketball 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?

Answer: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 1111Step-by-step explanation:Given as :The total number of three different tickets cost $ 10 , $ 20 and VIP seats $ 30The total number of all tickets sold = 3239The sale price for the tickets = $ 63720Let The type of ticket for $ 10 = xAnd The type of ticket for $ 20 = yAnd The type of VIP ticket = vAccording to questionThe number of $ 20 tickets = The number of $ 10 tickets + 282I.e y = x + 282And x + y + v = 3239Or, v = 3239 - ( x + y )I.e v = 3239 - ( x + x +282 )Or, v = 2957 - 2 xNow , x × $ 10 + y × $ 20 = v × $ 30Or, x × $ 10 + ( x + 282 ) × $ 20 = ( 2957 - 2 x ) × $ 30or, 10 x + 20 x + 5640 = 88710 - 60 xOr, 30 x + 60 x = 88710 - 5640or, 90 x = 83070∴  x = [tex]\frac{83070}{90}[/tex]I.e x = 923So, The  number of tickets that cost $ 10 = x = 923Similarly  y = x + 282or, y = 923 + 282 I.e y = 1205So, The  number of tickets that cost $ 10 = y = 1205And  v = 2957 - 2 x∴ v = 2957 - 2 × 923I.e v = 1111So, The  number of tickets that cost $ 30 = v = 1111Hence 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 1111Answer