Question

A basketball team sells tickets that cost​ $10, $20,​ or, for VIP​ seats,​ $30. The team has sold 3168 tickets overall. It has sold 322 more​ $20 tickets than​ $10 tickets. The total sales are ​$61,820. How many tickets of each kind have been​ sold?

How many​ $10 tickets were​ sold?

How many​ $20 tickets were​ sold ?

How many​ $30 tickets were​ sold?

Answer 1

A basketball team sells tickets that cost $10 $20 or VIP seats, $30.
let the no. of three types of tickets = x, y, z
The team has sold 3168 tickets over all.
x + y + z = 3168 ----- eq1
It has sold 322 more $20 tickets than $10 tickets.
y = x + 322 --------2
The total sales are $61820.
10x + 20y + 30z =61820 -----------3

use 2 in 1 and 3
2x + z =3168-322
2x + z =2846--------------------4
and
10x + 20(x+322) + 30z = 61820
10x + 20x + 6440 + 30z =61820

30x + 30z =61820-6440
30x + 30z = 55380
Simplify, divide equation by 30
x + z = 1846-----------5

subtract eqn 5 from eqn 4

a) we get x =1000 i.e no. of   $10 tickets were​ sold

b) use x in eqn 2 y = 1000 +322 = 1322 ie no of ​ $20 tickets were​ sold

c) use x in 4 , z =2846-2000 = 846 ie no of  $30 tickets were​ sold.

hence 1000+1322+846 =3168 confirms our solution