P, Q, R are three towns on a river which flow | vedclass

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(A) Let the speed of the man in still water be $x \, km/h$ and the speed of the current be $y \, km/h$.Speed downstream $= (x+y) \, km/h$.Speed upstre

Vedclass · Accepted Answer

$5:3$

Vedclass · Answer

(A) Let the speed of the man in still water be $x \, km/h$ and the speed of the current be $y \, km/h$.Speed downstream $= (x+y) \, km/h$.Speed upstream $= (x-y) \, km/h$.Let the river flow from $P$ to $R$ and $PQ = QR = a$. Then $PR = 2a$.Given that the man rows from $P$ to $Q$ and returns in $10 \, h$:$\frac{a}{x+y} + \frac{a}{x-y} = 10$  ---$(1)$Given that he rows from $P$ to $R$ in $4 \, h$ (downstream):$\frac{2a}{x+y} = 4 \implies \frac{a}{x+y} = 2$  ---$(2)$Substituting $(2)$ into $(1)$:$2 + \frac{a}{x-y} = 10 \implies \frac{a}{x-y} = 8$  ---$(3)$Dividing $(2)$ by $(3)$:$\frac{x-y}{x+y} = \frac{2}{8} = \frac{1}{4}$$4x - 4y = x + y$$3x = 5y$$\frac{x}{y} = \frac{5}{3}$Thus,the ratio is $5:3$.

$P, Q, R$ are three towns on a river which flows uniformly. $Q$ is equidistant from $P$ and $R$. $A$ man rows from $P$ to $Q$ and returns in $10 \, h$. He can row from $P$ to $R$ in $4 \, h$. The ratio of the speed of the man in still water to the speed of the current is

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