a boat travel at a speed of 20 km/h in still water the current in a river flows at 5 km/h

so that downstream the boat can travel at 25 km/h and upstream it travels at only 15 km/h the boat has only enough fuel for 3 hours.after what time does the boat turn around so that it has enough fuel to return to base

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Let us say boat travel xkm distance from base in t1 hours down stream when it turn around and comes upstream.

`t1 = x/25`

For the return travel let us say it takes t2 hours.

`t2 = x/15`

How ever at the critical time limit;

`t1+t2 = 3`

`x/25+x/15 = 3`

`(x(3+5))/75 = 3`

`x = (75xx3)/8`

`x = 28.125`

`t1 = x/25 = 28.125/25 = 1.125`

*So the boat should turn around in 1.125 hours to return to base without running out of fuel.*

The boat travels at a speed of 20 km/h in still water and water in the river flows at 5 km/h. When the boat travels downstream its speed is 20+5 = 25 km/h and when it travels upstream the speed of the boat is 20 - 5 = 15 km/h.

The fuel in the boat is enough for it to travel for 3 hours. It has to travel along the river, turn around and return to base.

Assume the boat travels downstream first and travels distance D.

The time required for this is D/25.

It then has to travel back upstream in time 3 - D/25.

3 - D/25 is equal to D/15

3 - D/25 = D/15

=> 3 = D(1/15 + 1/25)

D = 3/(1/15 + 1/25) = 225/8

If the boat travels upstream first and has to turn back after distance D, the time spent traveling in this direction is D/15. It has to now complete D traveling at 25 km/h in 3 - D/15 hours.

D/25 = 3 - D/15

D(1/25 + 1/15) = 3

D = 3/(1/25 + 1/15)

D = 225/8

The distance the boat can travel is 28.125 km irrespective of which direction it travels in first.

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