Rahul can travel 12 text{ miles} downstream i | vedclass

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(D) Let the speed of the current be $u 	ext{ miles/hour}$ and the speed of rowing in still water be $x 	ext{ miles/hour}$.According to the first con

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$2 \frac{2}{3}$

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(D) Let the speed of the current be $u 	ext{ miles/hour}$ and the speed of rowing in still water be $x 	ext{ miles/hour}$.According to the first condition:$\frac{12}{x-u} - \frac{12}{x+u} = 6$$\frac{12(x+u - (x-u))}{x^2 - u^2} = 6$$\frac{24u}{x^2 - u^2} = 6 \implies x^2 - u^2 = 4u \implies x^2 = u^2 + 4u \quad (1)$According to the second condition,if the rowing rate is doubled $(2x)$:$\frac{12}{2x-u} - \frac{12}{2x+u} = 1$$\frac{12(2x+u - (2x-u))}{4x^2 - u^2} = 1$$\frac{24u}{4x^2 - u^2} = 1 \implies 4x^2 - u^2 = 24u \implies 4x^2 = u^2 + 24u \quad (2)$Substitute $x^2$ from $(1)$ into $(2)$:$4(u^2 + 4u) = u^2 + 24u$$4u^2 + 16u = u^2 + 24u$$3u^2 = 8u$Since $u 
eq 0$,$u = \frac{8}{3} = 2 \frac{2}{3} 	ext{ miles/hour}$.

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