A man can row 7 frac{1}{2} , km/hr in still w | vedclass

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(B) Let the distance to the place be $x \, km$.Speed of the man in still water $= 7.5 \, km/hr$.Speed of the river stream $= 1.5 \, km/hr$.Upstream sp

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$3$

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(B) Let the distance to the place be $x \, km$.Speed of the man in still water $= 7.5 \, km/hr$.Speed of the river stream $= 1.5 \, km/hr$.Upstream speed $= 7.5 - 1.5 = 6 \, km/hr$.Downstream speed $= 7.5 + 1.5 = 9 \, km/hr$.The total time taken is $50 \, minutes = \frac{50}{60} \, hours = \frac{5}{6} \, hours$.Using the formula $	ext{Time} = \frac{	ext{Distance}}{	ext{Speed}}$,we have:$\frac{x}{6} + \frac{x}{9} = \frac{5}{6}$Taking the least common multiple $(LCM)$ of $6$ and $9$,which is $18$:$\frac{3x + 2x}{18} = \frac{5}{6}$$\frac{5x}{18} = \frac{5}{6}$Multiplying both sides by $18$:$5x = \frac{5 	imes 18}{6}$$5x = 15$$x = 3 \, km$.Therefore,the place is $3 \, km$ away.

$A$ man can row $7 \frac{1}{2} \, km/hr$ in still water. If in a river running at $1.5 \, km/hr$,it takes him $50 \, minutes$ to row to a place and back,how far off is the place? (in $km$)

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