A boat travels 60 km downstream and 20 km ups | vedclass

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(B) Let the speed of the boat in still water be $x 	ext{ km/hr}$ and the speed of the stream be $y 	ext{ km/hr}$.Downstream speed $= (x+y) 	ext{ km

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(B) Let the speed of the boat in still water be $x 	ext{ km/hr}$ and the speed of the stream be $y 	ext{ km/hr}$.Downstream speed $= (x+y) 	ext{ km/hr}$.Upstream speed $= (x-y) 	ext{ km/hr}$.According to the problem:$\frac{60}{x+y} + \frac{20}{x-y} = 4$  --- $(i)$$\frac{40}{x+y} + \frac{40}{x-y} = 6$  --- $(ii)$Let $u = \frac{1}{x+y}$ and $v = \frac{1}{x-y}$.$60u + 20v = 4 \Rightarrow 15u + 5v = 1$ --- $(iii)$$40u + 40v = 6 \Rightarrow 20u + 20v = 3$ --- $(iv)$Multiplying $(iii)$ by $4$: $60u + 20v = 4$.Subtracting $(iv)$ from this: $(60u + 20v) - (40u + 40v) = 4 - 6 \Rightarrow 20u - 20v = -2 \Rightarrow 10u - 10v = -1$.Solving the system,we get $u = \frac{1}{20}$ and $v = \frac{1}{10}$.$x+y = 20$ and $x-y = 10$.Adding the two equations: $2x = 30 \Rightarrow x = 15$.Subtracting: $2y = 10 \Rightarrow y = 5$.Wait,re-evaluating the system: $15u + 5v = 1$ and $20u + 20v = 3$. From $(iii)$,$5v = 1 - 15u \Rightarrow 20v = 4 - 60u$. Substitute into $(iv)$: $20u + 4 - 60u = 3 \Rightarrow -40u = -1 \Rightarrow u = \frac{1}{40}$.Then $5v = 1 - 15(\frac{1}{40}) = 1 - \frac{3}{8} = \frac{5}{8} \Rightarrow v = \frac{1}{8}$.$x+y = 40$ and $x-y = 8$.$2x = 48 \Rightarrow x = 24$.$2y = 32 \Rightarrow y = 16$.The speed of the stream is $16 	ext{ km/hr}$.

$A$ boat travels $60$ km downstream and $20$ km upstream in $4$ hours. The same boat travels $40$ km downstream and $40$ km upstream in $6$ hours. What is the speed (in $km/hr$) of the stream?

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