SOLUTION: Boudreaux rowed his pirogue from his camp on the bayou to his crab traps. Going down the bayou , he caught a falling tide that increased his normal speed by 2 mph but coming back

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Question 123077: Boudreaux rowed his pirogue from his camp on the bayou to his crab traps. Going down the bayou , he caught a falling tide that increased his normal speed by 2 mph but coming back he decreased his speed by 2mph. GOing with the tide, the trip only took 10 min; going against the tide the trip took 30 minutes. How far is it from Boudreaux's camp to his crab traps.
HINT: With the tide his rate is x+2 mph and against the tide his rate is x-2.
Answer by rajagopalan(174) (Show Source):
You can put this solution on YOUR website!
Onward journey along the stream:
speed=x+2
time=10min=10/60=1/6 hour
distance=speed*time=(1/6)*(x+2)
return Journey against stream:
speed=x-2
time=30min=30/60=1/2 hour
distance=speed*time=(1/2)*(x+2)
Equating both distances
we get
(1/6)*(x+2)=(1/2)*(x+2)
multiply throughout by 6,
x+2=3(x-2)
x+2=3x-6
interchange sides
3x-6=x+2
3x-x=2+6
2x=8
x=4
***
Speed DISABLED_event_onward= x+2=6mph
time=1/6
Distance=6*1/6=1 mile
The answer = 1 mile
***
Verification
Return speed=x-2=4-2=2 mph
Time=1/2 hour
Distance=2*(1/2)=1 mile
okay