Question

A basketball team sells tickets that cost​ $10, $20,​ or, for VIP​ seats,​ $30. The team has sold

33233323 tickets overall. It has sold 322322 more​ $20 tickets than​ $10 tickets. The total sales are

​$65 comma 87065,870.How many tickets of each kind have been​ sold? A. How many $10 tickets were sold? B. How many $20 tickets were sold? C. How many $30 tickets were sold?

Answer 1

Let $10 ticket be x , $20 ticket be y and $30 ticket be z.

Team sold 3323 tikets overall.

That is

x + y + z = 3323 ----------------------------------------(1)

Team sold 322 more $20 tickets than $10 tickets.

That is

y = 322 + x --------------------------------------------(2)

Total sale is $65870, that is

10x + 20y + 30z = 65870 ---------------------------(3)

Put value of y from equation (2) in Equation (1), we get

x + 322 + x + z = 3323

2x + z = 3323 - 322

2x + z = 3001 ----------------------------------------------------------(4)

Again, Put value of y from equation (2) in equation (3), we get

10x + 20(322 + x) + 30z = 65870

10x + 6440 + 20x + 30z = 65870

30x + 30z = 59430  

Divide this equation by 30

x + z = 1981 ----------------------------------------------(5)

Solve equation (4) and (5)

2x + z = 3001

- x + z = 1981

--------------------

x + 0 = 1020

x = 1020

From equation x + z = 1981,

z = 1981 - 1020

z = 961

From the (1) equation,

x + y + z = 3323

1020 + y + 961 = 3323

y = 1342

So

x = $10 tickets = 1020

y = $20 tickets = 1342

z = $30 tickets = 961

(To confirm solution,

Put values of x, y , and z in Equation (1) OR (3) and see whether both sides same.

Lets put values of x , y , z in equation (1)

1020 + 1342 + 961 = 3323

3323 = 3323

This confirm solution).