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Find the rate of the rowers in calm water and the rate of the current

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3 Answers

Hey Donald -- a "verbal" reasoning approach ... 18mph downstream, 12mph upstream ...
 
the midpoint is 15mph for rowers, with +/- 3mph on each side for the current ... Best :)
Hi Donald;
with the current...(36 miles)/(2 hours)=(18 miles)/hour
against the current...(36 miles)/(3 hours)=(12 miles)/hour
x=rate in calm waters
y=current
with the current...x+y=(18 miles)/hour
against the current...x-y=(12 miles)/hour
Let's subtract the two equations from each other...
(x-x)+y-(-y)=(6 miles)/hour
0+y+y=(6 miles)/hour
2y=(6 miles)/hour
y=3 miles/hour
Let's add the two equations together...
(x+x)+(y-y)=(30 miles)/hour
2x+0=(30 miles)/hour
2x=(30 miles)/hour
x=(15 miles)/hour
Let's check our work...
with the current...
[(15 miles)/hour]+[(3 miles)/hour)]=(18 miles)/hour
against the current...
[(15 miles)/hour]-[(3 miles)/hour)]=(12 miles)/hour
One must presume that the 'rate of the rowers' is the same both ways
 
Rr: Rate of rowers
Rs: Rate of stream
Td: Time downstream
Tu: Time upstream
 
1: (Rr+Rs) = D/Td = 18 mi/hr
2: (Rr - Rs) = D/Tu = 12 mi/hr
 
Adding the two equations (1 & 2)yields
 
2Rr = 30 mi/hr
Rr = 15 mi/hr
 
Substituting Rr back into 1)
 
Rs = 3 mi/hr