A basketball team sells tickets that costs 10 20 or for VIP

A basketball team sells tickets that costs $10, $20, or, for VIP seats, $30. The team has sold 3236 tickets overall. It has sold 195 more $20 tickets than $10 tickets. The total sales are $62,190. How many of each kind have been sold?

How many $10 tickets were sold?

How many $20 tickets were sold?

How many $30 tickets were sold?

Solution

Let the number of $ 10, 20 and 30 tickets be x , y and z respectively. Then x + y +z = 3236...(1). Also, y = x + 195...(2) and 10x + 20y + 30z = 62190 or, on dividing both the sides by 10, x +2y +3z = 6219...(3).

On substituting y = x +195 in the 1st and the 3rd equations, we get x +x +195 = 3236 or, 2x +z = 3041...(4) and x + 2(x+195)+3z = 6219 or, 3x +3z = 6219 - 585 or, 3x +3z = 5634. On dividing both the sides by 3, we get x +z = 1878...(5)

Now, on subtracting the 5th equation from the 4th equation, we get 2x+z -x -z = 3041- 1878 or, x = 1163. Then y = x +195 = 1163+195 = 1358. Also, on substituting x = 1163 in the 5th equation, we get 1163 +z = 1878 so that z = 1878-1163 = 715. We can verify the result by substituting the values of x,y,z in the 1st or the 3rd equation. The answer is x = 1163, y = 1358 and z = 715 i.e. the numbers of $ 10,20 and 30 tickets sold are 1163,1358 and 715 respectively.