A man can row 30 text{ km} upstream and 44 te | vedclass

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(A) Let the speed of the man in still water be $u 	ext{ km/h}$ and the rate of the current be $v 	ext{ km/h}$.According to the problem:$\frac{30}{u-

Vedclass · Accepted Answer

$3 	ext{ km/h}, 8 	ext{ km/h}$

Vedclass · Answer

(A) Let the speed of the man in still water be $u 	ext{ km/h}$ and the rate of the current be $v 	ext{ km/h}$.According to the problem:$\frac{30}{u-v} + \frac{44}{u+v} = 10$ $...(1)$$\frac{40}{u-v} + \frac{55}{u+v} = 13$ $...(2)$Let $x = \frac{1}{u-v}$ and $y = \frac{1}{u+v}$.$30x + 44y = 10$ $...(3)$$40x + 55y = 13$ $...(4)$Multiply $(3)$ by $4$ and $(4)$ by $3$:$120x + 176y = 40$ $...(5)$$120x + 165y = 39$ $...(6)$Subtract $(6)$ from $(5)$:$11y = 1 \Rightarrow y = \frac{1}{11}$.So,$u+v = 11$.Substitute $y = \frac{1}{11}$ into $(3)$:$30x + 44(\frac{1}{11}) = 10 \Rightarrow 30x + 4 = 10 \Rightarrow 30x = 6 \Rightarrow x = \frac{1}{5}$.So,$u-v = 5$.Solving $u+v = 11$ and $u-v = 5$:$2u = 16 \Rightarrow u = 8 	ext{ km/h}$.$2v = 6 \Rightarrow v = 3 	ext{ km/h}$.Thus,the speed of the man is $8 	ext{ km/h}$ and the rate of the current is $3 	ext{ km/h}$.

$A$ man can row $30 \text{ km}$ upstream and $44 \text{ km}$ downstream in $10 \text{ hours}$. Also,he can row $40 \text{ km}$ upstream and $55 \text{ km}$ downstream in $13 \text{ hours}$. Find the rate of the current and the speed of the man in still water (in $\text{km/h}$).

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