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(A) Let the rate of the stream be $x \, km/h$.Speed downstream = $(10 + x) \, km/h$.Speed upstream = $(10 - x) \, km/h$.According to the problem,the t

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(A) Let the rate of the stream be $x \, km/h$.Speed downstream = $(10 + x) \, km/h$.Speed upstream = $(10 - x) \, km/h$.According to the problem,the total time taken for the round trip is $20 \, hours$:$\frac{91}{10 + x} + \frac{91}{10 - x} = 20$Factor out $91$:$91 \left( \frac{10 - x + 10 + x}{(10 + x)(10 - x)} \right) = 20$Simplify the numerator:$91 \left( \frac{20}{100 - x^2} \right) = 20$Divide both sides by $20$:$\frac{91}{100 - x^2} = 1$$100 - x^2 = 91$$x^2 = 100 - 91 = 9$$x = \sqrt{9} = 3 \, km/h$.Thus,the rate of flow of the river is $3 \, km/h$.

$A$ motor-boat can travel at $10 \, km/h$ in still water. It travelled $91 \, km$ downstream in a river and then returned to the same place,taking altogether $20 \, hours$. Find the rate of flow of the river (in $km/h$).

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