Tickets to a football match cost $8,$15 or $20 each. The number of $15 tickets sold was double the number of $8 tickets sold. 6000 more $20 tickets were sold than $15 tickets. If the gate receipts totalled $783 000, how many of each type of ticket were sold?
f = 15 , e = 8 , t = twenty
f = 2 e ... e = f/2
t = f + 6000
8 e + 15 f + 20 t = 783000
substituting ... 4 f + 15 f + 20 f + 120000 = 783000
solve for f , then substitute back to find e and t
To solve this problem, let's use a system of equations to represent the given information.
Let's denote the number of $8 tickets as "x," the number of $15 tickets as "y," and the number of $20 tickets as "z."
From the given information, we can form three equations:
1) The number of $15 tickets sold was double the number of $8 tickets sold:
y = 2x
2) 6000 more $20 tickets were sold than $15 tickets:
z = y + 6000
3) The gate receipts totalled $783,000:
8x + 15y + 20z = 783,000
Now, let's substitute the values from equations 1) and 2) into equation 3) to solve for x, y, and z:
8x + 15y + 20z = 783,000
8x + 15(2x) + 20(y + 6000) = 783,000
8x + 30x + 20y + 120,000 = 783,000
38x + 20y + 120,000 = 783,000
Rearranging the equation:
38x + 20y = 783,000 - 120,000
38x + 20y = 663,000
Now, we need to further simplify the equation by dividing through by 2:
19x + 10y = 331,500
To make it easier to solve, let's multiply both sides of the equation by 2:
38x + 20y = 663,000
We now have a system of two equations:
1) y = 2x
2) 38x + 20y = 663,000
Using substitution method, you can substitute equation 1) into equation 2) to solve for x:
38x + 20(2x) = 663,000
38x + 40x = 663,000
78x = 663,000
x = 663,000 / 78
x ≈ 8500
Now, substitute the value of x back into equation 1) to find the value of y:
y = 2(8500)
y = 17,000
Finally, substitute the values of x and y into equation 2) to find the value of z:
38(8500) + 20(17,000) = 663,000
323,000 + 340,000 = 663,000
663,000 = 663,000
Therefore, there were approximately 8,500 $8 tickets, 17,000 $15 tickets, and 34,000 $20 tickets sold.