A man can row at 10 text{ km/h} in still wate | vedclass

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(A) Let the speed of the boat in still water be $u = 10 	ext{ km/h}$ and the speed of the current be $v 	ext{ km/h}$.The speed downstream is $(10 +

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$2$

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(A) Let the speed of the boat in still water be $u = 10 	ext{ km/h}$ and the speed of the current be $v 	ext{ km/h}$.The speed downstream is $(10 + v) 	ext{ km/h}$ and the speed upstream is $(10 - v) 	ext{ km/h}$.The total time taken for the round trip is given by the sum of time taken to go downstream and upstream:$\frac{24}{10 + v} + \frac{24}{10 - v} = 5$Taking $24$ as a common factor:$24 \left( \frac{10 - v + 10 + v}{(10 + v)(10 - v)} ight) = 5$$24 \left( \frac{20}{100 - v^2} ight) = 5$Simplifying the equation:$480 = 5(100 - v^2)$$96 = 100 - v^2$$v^2 = 4$$v = 2 	ext{ km/h}$ (since speed cannot be negative).Thus,the speed of the water current is $2 	ext{ km/h}$.

$A$ man can row at $10 \text{ km/h}$ in still water. If he takes a total of $5 \text{ h}$ to go to a place $24 \text{ km}$ away and return,then the speed of the water current is $... \text{ km/h}$.

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