A river is flowing with a steady speed of 4, | vedclass

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Question

(B) Let the speed of the boat in still water be $x\, km/h$ and the speed of the river current be $v = 4\, km/h$.The total distance covered is $42\, km

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(B) Let the speed of the boat in still water be $x\, km/h$ and the speed of the river current be $v = 4\, km/h$.The total distance covered is $42\, km$,so the distance for one way is $d = 21\, km$.The speed downstream is $(x + 4)\, km/h$ and the speed upstream is $(x - 4)\, km/h$.The time taken for the outward (downstream) journey is $t_1 = \frac{21}{x + 4}$.The time taken for the return (upstream) journey is $t_2 = \frac{21}{x - 4}$.According to the problem,the return journey takes $2\, h$ more than the outward journey,so $t_2 - t_1 = 2$.Substituting the values: $\frac{21}{x - 4} - \frac{21}{x + 4} = 2$.$21 \left( \frac{(x + 4) - (x - 4)}{(x - 4)(x + 4)} \right) = 2$.$21 \left( \frac{8}{x^2 - 16} \right) = 2$.$168 = 2(x^2 - 16) \Rightarrow 84 = x^2 - 16$.$x^2 = 100 \Rightarrow x = 10\, km/h$.

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