Question

A basketball team sells tickets that cost​ $10, $20,​ or, for VIP​ seats,​ $30. The team has sold 3061 tickets overall. It has sold 271 more​ $20 tickets than​ $10 tickets. The total sales are ​$58,670. How many tickets of each kind have been​ sold?

Answer 1

Let x, y and z represents the number of tickets that cost​ $10, $20,​ or, for VIP​ seats,​ $30 respectively.

Team has sold 3061 tickets overall. Thus,

x + y + z = 3061 ----(1)

Team has sold 271 more​ $20 tickets than​ $10 tickets. Thus,

y = 271 + x ----(2)

The total sales are ​$58,670. So,

10x + 20y + 30z = 58670

=> x + 2y + 3z = 5867 ---(3)

Multiplying both sides of equation (1) by 3, we get

3 * (x + y + z) = 3 * 3061

3x + 3y + 3z = 9183 ----(4)

Subtract (4) from (3), we get

(3x + 3y + 3z) - (x + 2y + 3z) = 9183 - 5867

2x + y = 3316

2x + 271 + x  = 3316 (from (2))

3x = 3316 - 271

x = 3045 / 3 = 1015

y = 271 + x = 271 + 1015 = 1286

From (1), x + y + z = 3061

=> 1015 + 1286 + z = 3061

=> z = 3061 - 2301

=> z = 760

So, number of  $10, $20,​ and for VIP​ seats,​ $30 sold are 1015, 1286 and 760 respectively.