SOLUTION: Two people are taking a three-day canoe trip. They'll paddle upstream for two days to a campground they have stayed at before, then paddle back downstream the third day in time to

Question 453393: Two people are taking a three-day canoe trip. They'll paddle upstream for two days to a campground they have stayed at before, then paddle back downstream the third day in time to meet their mom at the spot where they dropped them off.
The first day the girls paddle 13 hours. It takes them nine hours of paddling the second day to get to the campground. If they paddle at a rate that's twice that of the current for the whole trip, at what time will they have to leave the campground on the third day in order to meet their mom by 4:00 P.M.?
Your explanation must use algebra to communicate the method you have used to solve the problem. Show all work and fully explain the steps.
PLEASE SOLVE THIS. THANK YOU SO MUCH!!!

Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
let speed of current be x mph
paddling speed = 2x
...
upstream speed = 2x-x = x mph
downstream speed = 2x+x = 3x
...
distance up down is the same
d= rt.
time upstream = 22 hours
d( upstream) = 22x
..
let time downstream be t
d( down stream) = 3x*t
22x=3x*t
t=22x/3x
t= 7.33 hours
They have to leave 7.33 hours before 4.00 pm to meet on time .
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