pls assist, tks.
8 tickets cost $50, at least 2 are adult tickets. each adult tickets costs $2 more than each child ticket. how many tickets of each type were bought?
Originally posted by dukedracula:pls assist, tks.
8 tickets cost $50, at least 2 are adult tickets. each adult tickets costs $2 more than each child ticket. how many tickets of each type were bought?
8 tickets = $50
At least 2 are adult tickets.
Each adult ticket cost $2 more than child ticket.
Well, we can have:
2 adult + 6 child
3 adult + 5 child
4 adult + 4 child
5 adult + 3 child
6 adult + 2 child
7 adult + 1 child
8 adult + 0 child
I am not joking, the above are all the possible answers. I believe that you may have accidentally miss out some other requirement.
TS's maths teacher maths fail
lol
Originally posted by Forbiddensinner:8 tickets = $50
At least 2 are adult tickets.
Each adult ticket cost $2 more than child ticket.
Well, we can have:
2 adult + 6 child
3 adult + 5 child
4 adult + 4 child
5 adult + 3 child
6 adult + 2 child
7 adult + 1 child
8 adult + 0 child
I am not joking, the above are all the possible answers. I believe that you may have accidentally miss out some other requirement.
lets go more indepth incase TS dont understand what you mean
8 tickets = $50
At least 2 are adult tickets.
Each adult ticket cost $2 more than child ticket.
to find the price of 1 adult ticket for (where n is the price of 1 adult ticket)
2 adult + 6 child =
2n+6(n+2) =50
3 adult + 5 child =
3n+5(n+2) =50
4 adult + 4 child =
4n+4(n+2) =50
5 adult + 3 child =
5n+3(n+2) =50
6 adult + 2 child =
6n+2(n+2) =50
7 adult + 1 child =
7n+1(n+2) =50
8 adult + 0 child =
8n+0(n+2) =50
hence all the answers above are acceptable
no kidding..i came to same conclusion as to all the possible answers too...don't get what is the real intent of the question...