SOLUTION: Traveling against the current, the scouts paddled 20 km in 6hrs. The return trip with the current took 1hr. less. Find the scout's rate of paddling in still water, and the rate of

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Question 267333: Traveling against the current, the scouts paddled 20 km in 6hrs. The return trip with the current took 1hr. less. Find the scout's rate of paddling in still water, and the rate of the current.
Please and Thank you.

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The equation for paddling against the current is:
d%5B1%5D+=+%28s+-+c%29%2At%5B1%5D
The equation for paddling with the current is:
d%5B2%5D+=+%28s+%2B+c%29%2At%5B2%5D
where
s = speed paddling in still water
c = speed of the current
-------------------------------
given:
d%5B1%5D+=+20 km
d%5B2%5D+=+20 km
t%5B1%5D+=+6 hrs
t%5B2%5D+=+5 hrs
------------------
Plugging in numbers:
(1) 20+=+%28s+-+c%29%2A6
and
(2) 20+=+%28s+%2B+c%29%2A5
------------------------
(1) 20+=+6s+-+6c
(2) 20+=+5s+%2B+5c
I'll multiply both sides of (2) by 6%2F5
(2) %286%2F5%29%2A20+=+6s+%2B+6c
(2) 24+=+6s+%2B+6c
Now add (1) and (2)
(2) 24+=+6s+%2B+6c
(1) 20+=+6s+-+6c
44+=+12s
s+=+11%2F3 km/hr
From (2):
4+=+s+%2B+c
4+=+11%2F3+%2B+c
c+=+12%2F3+-+11%2F3
c+=+1%2F3 km/hr
In still water, they can paddle 11/3 km/hr
The speed of the current is 1/3 km/hr
check:
(1) 20+=+%28s+-+c%29%2A6
(1) 20+=+%2810%2F3%29%2A6
(1) 20+=+20
and
(2) 20+=+%28s+%2B+c%29%2A5
(2) 20+=+%2812%2F3%29%2A5
(2) 20+=+20
OK