Three kinds of tickets were sold at a concert. Child tickets cost $5.00, adult tickets cost $15.00 and student tickets cost $10.00. A total of 110 tickets were sold, bringing in a total of $1250.00. If the number of student tickets sold was twice the number of child tickets sold, how many adult tickets were sold?
2007-09-14 11:29:13 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
refer to the answers in http://answers.yahoo.com/question/index;_ylt=AihtyJKcatCjgtdiRhzTXuLty6IX;_ylv=3?qid=20070914112027AA3BYyN&show=7#profile-info-EqA25hloaa where you will find the general formula for solving these types of problems.
2007-09-14 11:36:24 · answer #1 · answered by Lollipop 5 · 0⤊ 0⤋
let x = child tickets
let y = adult tickets
first equation (money):
5x + 10(2x) + 15y = 1250
second equation (ticket quantity):
x + 2x + y = 110
Solve for y in second equation:
y = 110 - 3x
plug into first equation:
5x + 10(2x) + 15(110 - 3x) = 1250
5x + 20x + 1650 - 45x = 1250
20x = 400
x = 20
y = 110 - 3*20 = 110 - 60 = 50
so there were 20 child tickets sold, 40 student, and 50 adults
2007-09-14 18:33:30 · answer #2 · answered by Anonymous · 0⤊ 0⤋
x = # of child tickets
2x = # of student tickets
(110 - 3x) = # of adult tickets (total sold - child and student)
5x + 10(2x) + 15(110 - 3x) = 1250
5x + 20x + 1650 - 45x = 1250
20x = 400
x = 20 child tickets
40 student tickets
50 adult tickets
2007-09-14 18:36:51 · answer #3 · answered by ChemistryMom 5 · 0⤊ 0⤋
c + s + a = 110
5c + 10s + 15a = 1250
so c + 2s + 3a = 250
and s = 2c
2c + 2s + 2a = 220 = s + 2s + 2a = 3s + 2a
and 2c + 4s + 6a = 500 = s + 4s + 6a = 5s + 6a
3 x the first 9s + 6a = 660
so 4s = 160
s = 40
c = 20
a = 50
2007-09-14 18:36:41 · answer #4 · answered by Beardo 7 · 0⤊ 0⤋