Question

A basketball team sells tickets that cost​ $10, $20,​ or, for VIP​ seats,​ $30. The team has sold 3334 tickets overall. It has sold 203 more​ $20 tickets than​ $10 tickets. The total sales are ​$63 comma 970. How many tickets of each kind have been​ sold?

Answer 1

Let the number of tickets at $10 be X, the number of tickets at $20 be Y and the number of tickets at $30 be Z. We have 3334 = X + Y + Z

Given that Y = 203 + X and 10X + 20Y + 30Z = 63,970 or X + 2Y + 3Z = 6397

Use  Y = 203 + X in both equations to get

X + 203 + X + Z = 3334 and X + 406 + 2X + 3Z = 6397

2X + Z = 3131 and X + Z = 1997

Subtract them

2X - X + Z - Z = 3131 - 1997

X = 1134, Y = 1134 + 203 = 1337 and Z = 1998 - 1133 = 863

Hence the number of tickets at $10 is X = 1134, the number of tickets at $20 is Y = 1337 and the number of tickets at $30 = 863