Question 1197109: vip seating reserved seating and general admission tickets were sold for the school play at $15 $10 and 5$ each respectively the drama department sold 360 tickets for a total of $2800 if there were 40 more general admission tickets than the total number of vip and reserved tickets how many of each type of ticket were sold
Found 3 solutions by math_tutor2020, MathTherapy, greenestamps:
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
x = number of VIP tickets
y = number of reserved seating tickets
z = number of general admission tickets
x,y,z are nonnegative integers
| Ticket Type | Cost per ticket | # of tickets | Subtotal cost | | VIP | 15 | x | 15x | | Reserved Seating | 10 | y | 10y | | General Admission | 5 | z | 5z | | Total | | 360 | 2800 |
The second to last column forms this equation
x+y+z = 360
this is to represent the idea adding the ticket counts gets to 360 tickets total
The last column forms the equation 15x+10y+5z = 2800
15x = cost of just the VIP tickets (selling x of them for $15 each)
10y = cost of just the reserved seating tickets
5z = cost of just the general admission tickets
We're told that "there were 40 more general admission tickets than the total number of vip and reserved tickets"
So we know that z = x+y+40
Think of it like saying
(# of general admission) = ( (# of VIP) + (# of reserved seating) ) + 40
The system of equations is
x+y+z = 360
15x+10y+5z = 2800
z = x+y+40
Because z = x+y+40, we can replace each z with x+y+40 in the first two equations
Let's do so for the first equation
x+y+z = 360
x+y+x+y+40 = 360
2x+2y+40 = 360
2x+2y = 360-40
2(x+y) = 320
x+y = 320/2
x+y = 160
and the second equation as well
15x+10y+5z = 2800
15x+10y+5(x+y+40) = 2800
15x+10y+5x+5y+200 = 2800
20x+15y+200 = 2800
20x+15y = 2800-200
5(4x+3y) = 2600
4x+3y = 2600/5
4x+3y = 520
We have this reduced system of equations
x+y = 160
4x+3y = 520
Let's solve the first equation for y
x+y = 160
y = 160-x
Then plug it into the second equation and solve for x.
4x+3y = 520
4x+3(160-x) = 520
4x+480-3x = 520
x+480 = 520
x = 520-480
x = 40
There were 40 VIP tickets sold
Then we can determine y based on this x value.
y = 160-x
y = 160-40
y = 120
There were 120 reserved seat tickets sold.
Lastly let's compute z
z = x+y+40
z = 40+120+40
z = 200
There were 200 general admission tickets sold.
Here's what the table looks like after replacing x,y,z with 40,120, and 200 in that order.
| Ticket Type | Cost per ticket | # of tickets | Subtotal cost | | VIP | 15 | 40 | 600 | | Reserved | 10 | 120 | 1200 | | General Admission | 5 | 200 | 1000 | | Total | | 360 | 2800 |
Check:
40+120+200 = 360
600+1200+1000 = 2800
200 general admission = (40 VIP + 120 reserved) + 40 = 40+120+40 = 200
The answers are confirmed.
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Answers:
40 VIP tickets
120 reserved seating tickets
200 general admission tickets
Answer by MathTherapy(10555)  (Show Source):
You can put this solution on YOUR website! vip seating reserved seating and general admission tickets were sold for the school play at $15 $10 and 5$ each respectively the drama department sold 360 tickets for a total of $2800 if there were 40 more general admission tickets than the total number of vip and reserved tickets how many of each type of ticket were sold
Keep it SIMPLE as possible.
Let number of VIP and reserved tickets sold, be V and R, respectively
Then number of general-admission tickets sold = V + R + 40
We then get: V + R + (V + R + 40) = 360____2V + 2R = 320_____2(V + R) = 2(160)____V + R = 160 ----- eq (i)
Also, 15V + 10R + 5(V + R + 40) = 2,800_____20V + 15R = 2,600____5(4V + 3R) = 5(520)____4V + 3R = 520 ----- eq (ii)
3V + 3R = 480 ----- Multiplying eq (i) by 3 ----- eq (iii)
V = 520 - 480 ----- Subtracting eq (iii) from eq (ii)
Number of VIP tickets sold, or V = 40
40 + R = 160 ----- Substituting 40 for V in eq (i)
R = 160 - 40
Number of reserved tickets sold, or R = 120
Number of general-admission tickets sold: V + R + 40 = 40 + 120 + 40, or 360 - (40 + 120) = 200
Answer by greenestamps(13203)  (Show Source):
You can put this solution on YOUR website!
Start with this: the total number of tickets is 360, and the number of general admission tickets was 40 more than the rest.
Use formal algebra if you want; or use logical reasoning and simple arithmetic to determine that 200 of the 360 tickets were general admission.
Those 200 general admission tickets cost a total of 200($5) = $1000, meaning the other 160 tickets cost a total of $2800-$1000 = $1800.
You can then use formal algebra to set up a pair of equations to find the numbers of VIP seating tickets and reserved seating tickets -- something like
x+y = 160 (total number of tickets)
15x+10y=1800 (total cost of the tickets)
That system of equations is easily solved by any of several standard algebraic techniques.
But this remaining part of the problem can also be solved using logical reasoning and simple arithmetic, as follows.
(1) If all 160 remaining tickets were reserved seating tickets, their total cost would have been $1600
(2) But the actual cost is $1800, which is $200 more
(3) Each VIP seating ticket costs $5 more than each reserved seating ticket
(4) To make the additional $200, the number of VIP seating tickets had to be 200/5 = 40
So of the remaining 160 tickets, 40 were VIP seating tickets, meaning 120 were reserved seating tickets.
ANSWER:
VIP seating: 40
reserved seating: 120
general admission: 200
CHECK: 40(15)+120(10)+200(5) = 600+1200+1000 = 2800
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