Let A be the number of adult tickets sold.
As 325 tickets in total were sold, the number sold to students was 325 - A.
The income from the sale of adult tickets = A x 5 = 5A
The income from the sale of student tickets = (325 - A) x 2 = 650 - 2A
Total Income = 995 = 5A + 650 - 2A = 3A + 650
3A = 995 - 650 = 345
A = 115, therefore S = 325 - 115 = 210.
115 Adult tickets were sold and 210 Student tickets.
Sounds like a system of equations. X = adult ticket Y = student ticket Change all to pennies for convenience. X + Y = 105 300X + 150Y = 25000 -150(X + Y = 105) - 150X - 150Y = - 15750 300X + 150Y = 25000 -----------------------------------add them 150X = 9250 X = 62 adult tickets ------------------------ Y = 43 student tickets -------------------------- Those two prices added together = $250.50 So, we are a little off. I rounded the adult ticket number, but pieces of tickets can not be sold, so check work.
answer is 325 adult tickets were sold ( fmtickets.com )
a=# of adult tickets, s=# of student tickets 8a+5s=3475 a+s=500 Solve for s in this equation: s=500-a Now substitute 500-a in for s in the equation 8a+5s=3475: 8a+5(500-a)=3475 8a+2500-5a=3475 3a=975 a=325 ans. There were 325 adult tickets sold.
x=adult y=student x+y=810 4x+3y=2853 3240-y=2853 y=387 x=423 check: 4*423=1692 3*387=1161 add together 2853 423 adult tickets 387 student tickets
s- student, price of ticket is $2 a- adult, price for ticket i s$ 3 2*s money collected from students 3*s money collected from adults if a+s=400 and 3*a+2*s=1050 we have system of linear equations so a=400-s substitute in second equation a with 400-s then 3*(400-s) +2*s=1050 1200-3*s +2s=1050 1200 - s = 1050 -1200 -1200 -s = - 150 s=150 if s=150 then a=400-s so a=400-150, thus a=250 At the game sold 250 adult tickets and 150 student tickets
Sounds like a system of equations. X = adult ticket Y = student ticket Change all to pennies for convenience. X + Y = 105 300X + 150Y = 25000 -150(X + Y = 105) - 150X - 150Y = - 15750 300X + 150Y = 25000 -----------------------------------add them 150X = 9250 X = 62 adult tickets ------------------------ Y = 43 student tickets -------------------------- Those two prices added together = $250.50 So, we are a little off. I rounded the adult ticket number, but pieces of tickets can not be sold, so check work.
Your question lacks the info necessary to answer it.How many tickets were sold in total?What percentage of either student or adult tickets sales were sold?Your question cannot be answered as written.
325 Tickets sold to adults 400 tickets sold to students
answer is 325 adult tickets were sold ( fmtickets.com )
700/2 = 350350 - 25 = 325325 adult tickets700 - 325 = 375375 kids tickets
Adults: 325 Student: 175
A + S = 50 ===> S = 50 - A5A + 3S = 1805A + 3(50 - A) = 1805A + 150 - 3A = 1802A = 30[ A = 15 ][ S = 35 ]=====================Check:5(15) + 3(35) = 75 + 105 = 180 yay!
If A is number of adult tickets, and B is number of student tickets, we get a system of two equations for two unknowns: 4A+2B=2446 A+B=920 By solving this system, we obtain B=617
220 Tickets at 3 = 660 241 Tickets at 4 = 964 ---------------------------- 461 1624
a=# of adult tickets, s=# of student tickets 8a+5s=3475 a+s=500 Solve for s in this equation: s=500-a Now substitute 500-a in for s in the equation 8a+5s=3475: 8a+5(500-a)=3475 8a+2500-5a=3475 3a=975 a=325 ans. There were 325 adult tickets sold.
x=adult y=student x+y=810 4x+3y=2853 3240-y=2853 y=387 x=423 check: 4*423=1692 3*387=1161 add together 2853 423 adult tickets 387 student tickets
s- student, price of ticket is $2 a- adult, price for ticket i s$ 3 2*s money collected from students 3*s money collected from adults if a+s=400 and 3*a+2*s=1050 we have system of linear equations so a=400-s substitute in second equation a with 400-s then 3*(400-s) +2*s=1050 1200-3*s +2s=1050 1200 - s = 1050 -1200 -1200 -s = - 150 s=150 if s=150 then a=400-s so a=400-150, thus a=250 At the game sold 250 adult tickets and 150 student tickets