A basketball team sells tickets that cost $10, $20, or, for VIP seats, $30. The team has sold 33003300 tickets overall. It has sold 136136 more $20 tickets than $10 tickets. The total sales are $64 comma 19064,190. How many tickets of each kind have been sold?

Number of each ticket is;

$10 tickets = 1115

$20 tickets = 1251

$30 tickets = 934

Step-by-step explanation:

Let x,y and z represent the number of $10,$20 and $30 tickets sold.

Given;

Total number of tickets n = 3300

x+y+z = 3300 1

Total sales = $64,190

10x + 20y + 30z = 64,190 2

It has sold 136 more $20 tickets than $10 tickets

y = x +136 3

Substituting equation 3 into equation 1 and 2;

For 1;

x+y+z = 3300

x+(x+136)+z = 3300

2x + z = 330-136

2x + z = 3164 4

For 2;

10x + 20y + 30z = 64,190

10x + 20(x+136) + 30z = 64,190

10x + 20x + 2720 + 30z = 64190

30x + 30z = 64190-2720

30x+30z = 61470

divide through by 30

x+z = 2049 5

Subtract equation 5 from 4

2x-x +z-z = 3164-2049

x = 1115

From equation 3

y = x + 136 = 1115+136

y = 1251

From equation 1;

z = 3300 - (x+y)

z = 3300- (1115 + 1251)

z = 934

Number of each ticket is;

$10 tickets = 1115

$20 tickets = 1251

$30 tickets = 934

the answer is 1/4

step-by-step explanation:

step-by-step explanation: