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Sagot :
let s = speed in still water
let c = rate of the current
then
(s-c) = effective speed up-stream
and
(s+c) = effective speed down-stream
Write a distance equation for each way; dist = time * speed
5/6*(s-c) = 5
1/2 *(s+c) = 5
Now we get rid of the fractions
5(s-c)=30
s + c = 10
5s - 5c = 30
s + c = 10
siplifying:
s - c = 6 (1)
s + c = 10 (2)
adding:
2s = 16, then s = 8
8 mph is the boat speed in still water
According with (2):
8 + c = 10, then c = 2
2 mph is the speed of the current
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