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(D) Let the speed of the motorboat in still water be $u \, km/h$ and the speed of the stream be $v \, km/h$.Then,the speed of the motorboat downstream

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$10, 4$

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(D) Let the speed of the motorboat in still water be $u \, km/h$ and the speed of the stream be $v \, km/h$.Then,the speed of the motorboat downstream is $(u+v) \, km/h$ and the speed of the motorboat upstream is $(u-v) \, km/h$.According to the first condition,the time taken to travel $30 \, km$ upstream and $28 \, km$ downstream is $7 \, hours$:$\frac{30}{u-v} + \frac{28}{u+v} = 7 \quad ...(i)$According to the second condition,the time taken to travel $21 \, km$ upstream and return $21 \, km$ downstream is $5 \, hours$:$\frac{21}{u-v} + \frac{21}{u+v} = 5 \quad ...(ii)$Let $x = \frac{1}{u+v}$ and $y = \frac{1}{u-v}$.Substituting these into equations $(i)$ and $(ii)$:$30y + 28x = 7 \quad ...(iii)$$21y + 21x = 5 \Rightarrow x + y = \frac{5}{21} \quad ...(iv)$From $(iv)$,$x = \frac{5}{21} - y$. Substituting this into $(iii)$:$30y + 28(\frac{5}{21} - y) = 7$$30y + \frac{20}{3} - 28y = 7$$2y = 7 - \frac{20}{3} = \frac{1}{3} \Rightarrow y = \frac{1}{6}$Now,$x = \frac{5}{21} - \frac{1}{6} = \frac{10-7}{42} = \frac{3}{42} = \frac{1}{14}$.Since $y = \frac{1}{u-v} = \frac{1}{6}$,we have $u-v = 6 \quad ...(v)$Since $x = \frac{1}{u+v} = \frac{1}{14}$,we have $u+v = 14 \quad ...(vi)$Adding $(v)$ and $(vi)$:$2u = 20 \Rightarrow u = 10 \, km/h$.Substituting $u=10$ in $(vi)$:$10 + v = 14 \Rightarrow v = 4 \, km/h$.Thus,the speed of the boat in still water is $10 \, km/h$ and the speed of the stream is $4 \, km/h$.

$A$ motor boat can travel $30 \, km$ upstream and $28 \, km$ downstream in $7 \, hours$. It can travel $21 \, km$ upstream and return in $5 \, hours$. Find the speed of the boat in still water and the speed of the stream (in $km/h$).

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