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

Question

A row boat team travelled with the current 36 miles in 2 hours. Coming back against the current, it took 3 hours to cover the same distance. Find the rate of the rowers in calm water and the rate of the current.

Answer 1

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 :)

Answer 2

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

Answer 3

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