A man can row at the rate of 4 text{ km/hr} i | vedclass

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(D) Let the speed of the man in still water be $x = 4 	ext{ km/hr}$ and the speed of the current be $y 	ext{ km/hr}$.Speed upstream $= (x - y) 	ext

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$2$

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(D) Let the speed of the man in still water be $x = 4 	ext{ km/hr}$ and the speed of the current be $y 	ext{ km/hr}$.Speed upstream $= (x - y) 	ext{ km/hr} = (4 - y) 	ext{ km/hr}$.Speed downstream $= (x + y) 	ext{ km/hr} = (4 + y) 	ext{ km/hr}$.Let the distance be $d 	ext{ km}$.Time taken upstream $T_u = \frac{d}{4 - y}$ and time taken downstream $T_d = \frac{d}{4 + y}$.According to the problem,$T_u = 3 	imes T_d$.$\frac{d}{4 - y} = 3 	imes \frac{d}{4 + y}$.Canceling $d$ from both sides: $\frac{1}{4 - y} = \frac{3}{4 + y}$.$4 + y = 3(4 - y) \Rightarrow 4 + y = 12 - 3y$.$y + 3y = 12 - 4 \Rightarrow 4y = 8$.$y = 2 	ext{ km/hr}$.

$A$ man can row at the rate of $4 \text{ km/hr}$ in still water. If the time taken to row a certain distance upstream is $3$ times as much as to row the same distance downstream,find the speed of the current in $\text{km/hr}$.

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