Question 91078: Tickets for and event cost $4 for children, $12 for adults, and $10 for senior citizens. The total ticket sales were $2160. There were 60 more adult tickets than child tickets, and the number of senior citizens tickets were 2 times the number of child tickets. How many of each tickets were sold? Thanks in advance
Answer by Earlsdon(6294)   (Show Source):
You can put this solution on YOUR website! Start by assigning variables to the unknown quantities.
Let C = # of child tickets sold at $4.00 each.
Let A = # of adult tickets sold at $12.00 each.
Let S = # of senior tickets sold at $10.00 each.
The amount collected from child tickets was: C($4.00)
The amount collected from adult tickets was: A($12.00)
The amount collected from senior tickets was: S($10.00)
The sum of these is given as: $2,160.00, so you can write:
1) C($4.00) + A($12.00) + S($10.00) = $2,160
Now, you are told that:
There were 60 more (+60) child tickets than adult tickets, so C = A+60
The number of senior tickets was two times the number of child tickets, so S = 2C
Now we have three unknowns and only one equation, so we need to express everthing in terms of one unknown (A, C, or S).
We already have C = A+60 and S = 2C so let's substitute the C here in terms of A.
S = 2(A+60) = 2A+120
Now we have everything in terms of A so we go back to the equation 1) and substitute.
C = A+60
A = A
S = 2A+120
1a) (A+60)($4.00) + A($12.00) + (2A+120)($10.00) = $2,160.00 Simplify and solve for A.
(4A+240) + (12A) + (20A+1,200) = 2,160 Collect the A's together and the constants together.
(4A+12A+20A) + (240+1200) = 2,160
36A+1,440 = 2,160 Subtract 1,440 from both sides.
36A = 720 Finally, divide both sides by 36.
A = 20 This is the number of adult tickets sold.
C = A+60
C = 20+60
C = 80 This is the number of child tickets sold.
S = 2C
S = 2(80)
S = 160 This is the number of senior tickets sold.
Check:
80($4.00)+20($12.00)+160($10.00) = $320+$240.00+$1,600 = $2,160
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