Tickets for a certain show cost
$15, $24,
or, for VIP seats,
$41.
If
ninenine
times as many
$15
tickets were sold as VIP tickets, and the number of
$15
tickets sold was
55
more than the sum of the number of
$24
tickets and VIP tickets, sales of all three kinds of tickets would total
$46 comma 46,152.
How many of each kind of ticket would have been sold?
If the number of seats for the categories is x,y,z then just list the data:
x = 9z
x = 55+y+z
15x+24y+41z = 46152
Now just solve for x,y,z
Let's solve this problem step-by-step.
Step 1: Assign variables
Let's assume the number of $15 tickets sold as x, the number of $24 tickets sold as y, and the number of VIP tickets sold as z.
Step 2: Set up the equations
From the given information, we can set up the following equations:
Equation 1: x = 99z
Equation 2: x = y + z + 55
Equation 3: 15x + 24y + 41z = 46,152
Step 3: Solve for x in terms of z
From Equation 1, we can substitute x in terms of z in Equation 3:
15(99z) + 24y + 41z = 46,152
1485z + 24y + 41z = 46,152 --> 1526z + 24y = 46,152
Step 4: Solve for y in terms of z
From Equation 2, we can substitute x in terms of y and z:
y + z + 55 = 99z
y = 99z - z - 55
y = 98z - 55
Step 5: Substitute y in terms of z in Equation 4
1526z + 24(98z - 55) = 46,152
1526z + 2352z - 1320 = 46,152
3878z = 47,472
z = 12
Step 6: Solve for y
Using Equation 2:
y = 98z - 55
y = 98(12) - 55
y = 1166
Step 7: Solve for x
Using Equation 1:
x = 99z
x = 99(12)
x = 1188
Step 8: Check the solution
Using Equation 3:
15x + 24y + 41z = 46,152
15(1188) + 24(1166) + 41(12) = 46,152
17,820 + 27,984 + 492 = 46,152
46,152 = 46,152
Step 9: Interpret the solution
Therefore, the number of $15 tickets sold is 1188, the number of $24 tickets sold is 1166, and the number of VIP tickets sold is 12.
To solve this problem, let's represent the number of $15 tickets as "x," the number of $24 tickets as "y," and the number of VIP tickets as "z."
1) We can set up the following equations based on the given information:
Equation 1: x = 99z (ninenine times as many $15 tickets were sold as VIP tickets)
Equation 2: x = y + z + 55 (the number of $15 tickets sold was 55 more than the sum of $24 tickets and VIP tickets)
Equation 3: 15x + 24y + 41z = 46,152 (the total sales of all three kinds of tickets were $46,152)
2) Let's solve this system of equations using substitution:
Substituting Equation 1 into Equation 2, we get:
99z = y + z + 55
98z = y + 55 ----> Equation 4
Now, substituting Equation 1 and Equation 4 into Equation 3, we get:
15(99z) + 24(98z - 55) + 41z = 46,152
1,485z + 2,352z - 1,320 + 41z = 46,152
3,878z = 47,472
z = 12
Using Equation 1, we can find the value of x:
x = 99z
x = 99(12)
x = 1,188
Substituting z = 12 into Equation 4, we can find y:
98z = y + 55
98(12) = y + 55
1,176 = y + 55
y = 1,176 - 55
y = 1,121
So, the solution to the problem is:
Number of $15 tickets (x): 1,188
Number of $24 tickets (y): 1,121
Number of VIP tickets (z): 12