A man rows at 9 frac{1}{3} text{ km/h} in sti | vedclass

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(C) Let the speed of the man in still water be $x = 9 \frac{1}{3} = \frac{28}{3} 	ext{ km/h}$ and the speed of the current be $y 	ext{ km/h}$.The sp

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$4 \frac{2}{3}$

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(C) Let the speed of the man in still water be $x = 9 \frac{1}{3} = \frac{28}{3} 	ext{ km/h}$ and the speed of the current be $y 	ext{ km/h}$.The speed upstream is $(x - y) 	ext{ km/h}$ and the speed downstream is $(x + y) 	ext{ km/h}$.Let the distance be $d$. According to the problem,the time taken to row upstream is thrice the time taken to row downstream:$\frac{d}{x - y} = 3 	imes \frac{d}{x + y}$Canceling $d$ from both sides:$\frac{1}{x - y} = \frac{3}{x + y}$$x + y = 3(x - y)$$x + y = 3x - 3y$$4y = 2x$$y = \frac{x}{2}$Substituting $x = \frac{28}{3}$:$y = \frac{28/3}{2} = \frac{14}{3} = 4 \frac{2}{3} 	ext{ km/h}$.

$A$ man rows at $9 \frac{1}{3} \text{ km/h}$ in still water and finds that it takes him thrice as much time to row up than to row down the same distance in the river. The speed of the current is.......$\text{km/h}$

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