Tickets for an event cost $4 for children, $12 for adults, and $7 for senior citizens. The total ticket sales were $1920. There were 50 more adult tickets sold than child tickets, and the number of senior citizens tickets were 4 times the number of child tickets. How many of each ticket were sold?
2007-01-07 17:56:18 · 4 answers · asked by Rene 5 in Education & Reference ➔ Homework Help
x = # of child tickets
therefore x + 50 = # of adult tickets
and 4x = senoir tickets
$4x + $12(x + 50) + $7(4x) = 1920
4x + 12x + 600 + 28x = 1920
44x + 600 - 600 = 1920-600
44x = 1320
x=30
therefore
30 Children $120
30 + 50 = 80 adults $960
4 X 30 = 120 seniors $840
_____
$1920
2007-01-07 18:12:38 · answer #1 · answered by ccrstitch2003 2 · 1⤊ 0⤋
let 1 child ticket cost $x,1 adult ticket cost $y ,1sen cit ticket cost $z
since no of adult tickets is x+50,y=x+50 & z=4*x
hence we get the equation,
4x+12y+7z=1920
^ 4x+12(x+50)+7(4*x)=1920
4x+12x+600+28x=1920
44x+600=1920
44x=1920-600
44*x=1320
x= 30
therefore , number of
child tickets = 30
adult " = 80
senior citizen " = 120
2007-01-07 18:25:07 · answer #2 · answered by sreeva06 1 · 0⤊ 0⤋
$4x + $12(x+50) + $7(4x) = $1920, where x equals the number of child tickets. So $4x + $12x + $600 + $28x = $1920, which equals $44x + $600 = $1920, which equals $44x = $1320, and x = 30. So the number of child tickets equals 30, the number of adult tickets equals (x + 50) 80, and the number of senior tickets equals (4x) 120.
2007-01-07 18:11:46 · answer #3 · answered by kkline03 2 · 0⤊ 0⤋
ITS BEEN A WHILE SINCE IVE DID ALGEBRA , IM PRETTY SURE THE EQUI. IS SET UP LIKE THIS
$4C+$12A+$7S=1920
(50+C)=A
(4XC)=S
$4C+$12(50+C)+$7(4C)=$1920 I WILL KEEP LOOKING AT IT UNTILL I CAN FIGURE IT OUT
2007-01-07 18:33:37 · answer #4 · answered by comacati 3 · 0⤊ 0⤋