A man can row 40, km upstream and 55, km down | vedclass

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(A) Let the speed of the man in still water be $x\, km/h$ and the speed of the current be $y\, km/h$.The speed downstream is $(x+y)\, km/h$ and the sp

Vedclass · Accepted Answer

$8\, km/h$ and $3\, km/h$

Vedclass · Answer

(A) Let the speed of the man in still water be $x\, km/h$ and the speed of the current be $y\, km/h$.The speed downstream is $(x+y)\, km/h$ and the speed upstream is $(x-y)\, km/h$.According to the problem:$\frac{55}{x+y} + \frac{40}{x-y} = 13 \quad ...(1)$$\frac{44}{x+y} + \frac{30}{x-y} = 10 \quad ...(2)$Let $v = x+y$ and $u = x-y$. The equations become:$\frac{55}{v} + \frac{40}{u} = 13 \quad ...(3)$$\frac{44}{v} + \frac{30}{u} = 10 \quad ...(4)$Multiply equation $(3)$ by $3$ and equation $(4)$ by $4$:$\frac{165}{v} + \frac{120}{u} = 39 \quad ...(5)$$\frac{176}{v} + \frac{120}{u} = 40 \quad ...(6)$Subtracting $(5)$ from $(6)$:$\frac{11}{v} = 1 \Rightarrow v = 11$.Substitute $v=11$ into $(4)$:$\frac{44}{11} + \frac{30}{u} = 10 \Rightarrow 4 + \frac{30}{u} = 10 \Rightarrow \frac{30}{u} = 6 \Rightarrow u = 5$.Now,$x+y = 11$ and $x-y = 5$.Adding these equations: $2x = 16 \Rightarrow x = 8$.Subtracting these equations: $2y = 6 \Rightarrow y = 3$.Thus,the speed of the man in still water is $8\, km/h$ and the speed of the current is $3\, km/h$.

$A$ man can row $40\, km$ upstream and $55\, km$ downstream in $13\, hours$. Also,he can row $30\, km$ upstream and $44\, km$ downstream in $10\, hours$. Find the speed of the man in still water and the speed of the current.

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