Ryan and Andy have some carnival tickets. If Ryan sells 10 tickets per day and Andy sells 5 tickets per day, Ryan will have 40 tickets left when Andy has sold all his tickets. If Ryan sells 5 tickets per day and Andy sells 10 tickets per day, Ryan will have 70 tickets left when Andy has sold all his tickets.
(a) How many tickets does Ryan have?
(b) Each ticket cost $12. If Ryan and Andy manage to sell all their tickets, how much more did Ryan collect than Andy?
See if this helps :
let No. of tickets Ryan have = x
No. of tickets Andy have = y
No. of days Andy take when he sells 5 tickets each day = y/5.
then No. of tickets Ryan sells in 5 days = (y/5) * 10 = 2y
Ryan will have 40 tickets left when Andy sells all his tickets
So, We can write
No. of tickets Ryan has - No. of tickets Ryan sells = 40
x - 2y = 40 ---- (1)
Now If Andy sells 10 tickets a day then
No. of days Andy take to sell all his tickets y/10
On the same time Ryan sells tickets = y/10 * 5 = y/2
then tickets left to Ryan = x - y/2 = 70 ---- (2)
using eq (1) & eq (2), using elimination method
x - 2y = 40
x - y/2 = 70
-2y + y/2 = -30
y = 20
Tot value of y in eq (2) we get
x - 20/2 = 70
x = 80 (No of tickets Ryan has)
So, Ryan has 80 tickets.
Price of each ticket = $12
Ryan and Andy sold all of their tickets
So, No. of tickets Ryan sells more than Andy
= 80 - 20 = 60
then, money Ryan collects more than Andy = 12 * 60 = $720.
So, Ryan has $720 more than Andy.