A man can swim in still water at 9.6 text{ km | vedclass

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Question

(A) Let the speed of the man in still water be $x = 9.6 	ext{ km/hr}$.Let the speed of the stream be $y 	ext{ km/hr}$.Let the distance to be covered

Vedclass · Accepted Answer

$3.2$

Vedclass · Answer

(A) Let the speed of the man in still water be $x = 9.6 	ext{ km/hr}$.Let the speed of the stream be $y 	ext{ km/hr}$.Let the distance to be covered be $d 	ext{ km}$.The speed downstream is $(9.6 + y) 	ext{ km/hr}$ and the speed upstream is $(9.6 - y) 	ext{ km/hr}$.According to the problem,the time taken to swim upstream is twice the time taken to swim downstream.Time taken upstream = $\frac{d}{9.6 - y}$ and time taken downstream = $\frac{d}{9.6 + y}$.Given: $\frac{d}{9.6 - y} = 2 	imes \frac{d}{9.6 + y}$.Canceling $d$ from both sides: $\frac{1}{9.6 - y} = \frac{2}{9.6 + y}$.Cross-multiplying: $9.6 + y = 2(9.6 - y)$.$9.6 + y = 19.2 - 2y$.$3y = 19.2 - 9.6$.$3y = 9.6$.$y = 3.2 	ext{ km/hr}$.

$A$ man can swim in still water at $9.6 \text{ km/hr}$,but takes twice as long to swim upstream as he does downstream. The speed of the stream is ......... $\text{km/hr}$.

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