Question

A boat heads upstream a distance of 30 miles in the mission river whose current is running at 5 miles per hour. If the trip back takes an hour less, what was the speed of the boat still in the water

Answer 1

DistanceAB = DistanceBA = 30 miles
rateRIV - rate of the river = 5 miles per hour
rateBOAT - rate of the boat in still water
timeAB = time traveled by boat upstream
timeBA = time traveled by boat downstream

(1) timeBA = timeAB - 1
(2) rateBOAT - 5 = 30/timeAB
(3) rateBOAT + 5 = 30/timeBA

Substitute (1) in (3) for timeBA, and solve (2) and (3) simultaneously...

rateBOAT^2 - 325 = 0
rateBOAT = square root (325) = 18.03 miles per hour is the answer....

To check, use (2) and (3)....

timeAB = 2.3 hours
timeBA = 1.3 hours, less 1 hr than 2.3 hours....

:)

Answer 2

let speed of boat =x
speed of boat in upstream=x-5 (boat is going against the current so it will reduce boat speed)
distance=30 miles
time upstream=30/(x-5)
in downstream that is during back...current will speed up the boat
so speed=x+5
distance=30 miles
time in coming back=30/(x+5)=time upstream -1hour
\[30/(x+5)=30/(x-5) -1\]

Answer 3

30/(x+5)=(30-1*(x-5))/(x-5)
30/(x+5)=(30-x+5)/(x-5)
30*(x-5)=(35-x)*(x+5)
30x-150=35x+175-x^2-5x
30x-175=30x+175-x^2
x^2=300
x=300^(1/2)=17.32miles/hour

Answer 4

sorry
30x-150=35x+175-x^2-5x
30x=30x+175+150-x^2
x^2=325
x=18.02miles/hour

Answer 5

... thanks for proving my solution... :)