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Rajeshkumar
A man can row at 5 kmph in still water. If the velocity of current is 1 kmph and it takes him 1 hour to row to a place and come back, how far is the place?
The current is 1 so the man can row 4 kmph upstream and 6 kmph downstream distance = rate*time d = 4t1 d = 6t2 t1 + t2 = 1 use substitution t1 = 1 - t2 4(1-t2) = 6t2 solve for t2 4 - 4t2 = 6t2 4 = 10t2 0.4 = t2 then d = 6*0.4 = 2.4 The place is 2.4 km away
t=d/r Assume the place is down stream. Let x = the time it takes to row downstream. Solve the following simultaneous equations for x and d:\[\left\{x\text{=}\frac{d}{6},(1-x)\text{=}\frac{d}{4}\right\}\]\[\left\{x\to \frac{2}{5},d\to \frac{12}{5}\right\} \]If the place is up stream then let x = the time it takes to row upstream.\[\left\{x\text{=}\frac{d}{4},(1-x)\text{=}\frac{d}{6}\right\}\]\[\left\{x\to \frac{3}{5},d\to \frac{12}{5}\right\} \]In either case the place is 12/5 or 2.4 km from the departure point.