Question

A basketball team sells tickets that cost​ $10, $20,​ or, for VIP​ seats,​ $30. The team has sold 487 tickets overall. It has sold 167 more​ $20 tickets than​ $10 tickets. The total sales are ​$8590 . How many tickets of each kind have been​ sold?

Answer 1

Let $10, $20 and $30 tickets sold were x, y and z respectively

The team has sold 487 tickets overall : x + y + z = 487...(1)

It has sold 167 more​ $20 tickets than​ $10 tickets: y-x = 167...(2)

The total sales are ​$8590 : 10x + 20y + 30z = 8590

x + 2y + 3z = 859 ...(3)

get value of y in form from equation (2) : y = 167+x

Put this value of y in eq(3) and eq(1)

(1) x+167+x + z = 487

2x+z = 320...(4)

(2) x + 334+2x + 3z = 859

3x+3z = 859-334 = 525...(5)

Eq (4) multiple by 3 and subtracted from eq(5)

3x = 960-525 = 435

x = 145

z = 30

y = 312

so $10 ticket = 145

$20 tickets = 312

$30 tickets = 30