A man rows to a place 48 , km away and comes | vedclass

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Question

(A) Let the speed of the boat in still water be $x \, km/hr$ and the speed of the stream be $y \, km/hr$.Speed downstream $= (x + y) \, km/hr$ and spe

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$1 \, km/hr$

Vedclass · Answer

(A) Let the speed of the boat in still water be $x \, km/hr$ and the speed of the stream be $y \, km/hr$.Speed downstream $= (x + y) \, km/hr$ and speed upstream $= (x - y) \, km/hr$.Given that he rows $4 \, km$ downstream in the same time as $3 \, km$ upstream:$\frac{4}{x + y} = \frac{3}{x - y} \Rightarrow 4x - 4y = 3x + 3y \Rightarrow x = 7y$.Total time for the round trip of $48 \, km$ is $14 \, hours$:$\frac{48}{x + y} + \frac{48}{x - y} = 14$.Substitute $x = 7y$ into the equation:$\frac{48}{7y + y} + \frac{48}{7y - y} = 14 \Rightarrow \frac{48}{8y} + \frac{48}{6y} = 14$.$\frac{6}{y} + \frac{8}{y} = 14 \Rightarrow \frac{14}{y} = 14 \Rightarrow y = 1 \, km/hr$.The rate of the stream is $1 \, km/hr$.

$A$ man rows to a place $48 \, km$ away and comes back in $14 \, hours$. He finds that he can row $4 \, km$ with the stream in the same time as $3 \, km$ against the stream. The rate of the stream is:

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