A man can row at 5 text{ km/hr} in still wate | vedclass

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Question

(A) Let the distance of the place be $x 	ext{ km}$.Speed downstream $= 5 + 1 = 6 	ext{ km/hr}$.Speed upstream $= 5 - 1 = 4 	ext{ km/hr}$.Time taken

Vedclass · Accepted Answer

$2.4$

Vedclass · Answer

(A) Let the distance of the place be $x 	ext{ km}$.Speed downstream $= 5 + 1 = 6 	ext{ km/hr}$.Speed upstream $= 5 - 1 = 4 	ext{ km/hr}$.Time taken to travel downstream is $\frac{x}{6} 	ext{ hours}$ and time taken to travel upstream is $\frac{x}{4} 	ext{ hours}$.Given that the total time taken is $1 	ext{ hour}$,we have:$\frac{x}{6} + \frac{x}{4} = 1$Taking the least common multiple $(LCM)$ of $6$ and $4$,which is $12$:$\frac{2x + 3x}{12} = 1$$\frac{5x}{12} = 1$$x = \frac{12}{5} = 2.4 	ext{ km}$.Therefore,the distance to the place is $2.4 	ext{ km}$.

$A$ man can row at $5 \text{ km/hr}$ in still water. If the velocity of the current is $1 \text{ km/hr}$ and it takes him $1 \text{ hour}$ to row to a place and come back,how far is the place? (in $\text{km}$)

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