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mastergmat1
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For rate problems, it helps to write: Distance = rate * time
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sch
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let's say:
clam speed=X (m/h)
resistance (or propeller) speed of the water= Y (m/h)

if you take 3 hours at a speed 3mph while take 4 hours to get back, the first trip should be downstream:

X+Y=3 (mph)

in this trip, the distance you travel is 3 (hours)*3(mph)=9 miles

the sencond trip is upstream, the speed should be:
X-Y

the same distance takes you 4 hours, that is

9 (miles) /(X-Y)=4 (hours)


OK, lets see what we have now:

X+Y=3

9/(X-Y)=4

=> X=2.625 mile/hour


************
there is another quicker way to figure out this problem

first trip distance= 3*3=9 miles

second trip speed= 9 miles/ 4 hours=2.25 m/h

first trip speed =3 m/h

(first speed+second speed)/2=5.25/2=2.625 m/h------->because the MEAN of the two speeds (to and back) will counteract the infulence of water.


open to discuss



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