P,Q,and R are three towns on a river that flo | vedclass

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(A) Let the distance between $P$ and $Q$ be $d$. Since $Q$ is equidistant from $P$ and $R$,the distance between $Q$ and $R$ is also $d$. The total dis

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$5:3$

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(A) Let the distance between $P$ and $Q$ be $d$. Since $Q$ is equidistant from $P$ and $R$,the distance between $Q$ and $R$ is also $d$. The total distance from $P$ to $R$ is $2d$.Let $x$ be the speed of the boat in still water and $y$ be the speed of the river.Time taken to row from $P$ to $Q$ and back is $10$ $hours$:$\frac{d}{x+y} + \frac{d}{x-y} = 10 \quad \dots (1)$Time taken to row downstream from $P$ to $R$ is $4$ $hours$:$\frac{2d}{x+y} = 4 \Rightarrow d = 2(x+y) \quad \dots (2)$From equation $(1)$:$d \left( \frac{x-y+x+y}{(x+y)(x-y)} ight) = 10 \Rightarrow 2xd = 10(x^2 - y^2) \Rightarrow xd = 5(x^2 - y^2)$Substitute $d = 2(x+y)$ from equation $(2)$ into the expression:x($2$(x+y)) = $5$(x+y)(x-y)Since $x+y 
eq 0$,we can divide both sides by $(x+y)$:$2x = 5(x-y) \Rightarrow 2x = 5x - 5y \Rightarrow 3x = 5y$Therefore,$\frac{x}{y} = \frac{5}{3}$.

$P$,$Q$,and $R$ are three towns on a river that flows uniformly. $Q$ is equidistant from $P$ and $R$. $I$ row from $P$ to $Q$ and back in $10$ $hours$,and $I$ can row downstream from $P$ to $R$ in $4$ $hours$. Compare the speed of my boat in still water with that of the river.

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