A boat goes 32 , km upstream and 36 , km down | vedclass

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(A) Let the speed of the boat in still water be $x \, km/hr$ and the speed of the stream be $y \, km/hr$. It is necessary that $x > y$.The speed of th

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$(10, 2)$

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(A) Let the speed of the boat in still water be $x \, km/hr$ and the speed of the stream be $y \, km/hr$. It is necessary that $x > y$.The speed of the boat downstream is $(x+y) \, km/hr$ and the speed of the boat upstream is $(x-y) \, km/hr$.Using the formula $	ext{time} = \frac{	ext{distance}}{	ext{speed}}$,we have:For the first case: $\frac{32}{x-y} + \frac{36}{x+y} = 7$  --- $(1)$For the second case: $\frac{40}{x-y} + \frac{48}{x+y} = 9$  --- $(2)$Let $\frac{1}{x-y} = a$ and $\frac{1}{x+y} = b$. The equations become:$32a + 36b = 7$ --- $(3)$$40a + 48b = 9$ --- $(4)$Solving these linear equations by elimination or substitution:Multiply $(3)$ by $4$ and $(4)$ by $3$:$128a + 144b = 28$$120a + 144b = 27$Subtracting the two equations: $8a = 1 \implies a = \frac{1}{8}$.Substituting $a = \frac{1}{8}$ in $(3)$:$32(\frac{1}{8}) + 36b = 7 \implies 4 + 36b = 7 \implies 36b = 3 \implies b = \frac{1}{12}$.Now,$x-y = 8$ and $x+y = 12$.Adding these: $2x = 20 \implies x = 10$.Subtracting these: $2y = 4 \implies y = 2$.Thus,the speed of the boat in still water is $10 \, km/hr$ and the speed of the stream is $2 \, km/hr$.

$A$ boat goes $32 \, km$ upstream and $36 \, km$ downstream in $7$ hours. In $9$ hours,it can go $40 \, km$ upstream and $48 \, km$ downstream. Determine the speed of the stream and that of the boat in still water. (Speed of boat in still water is more than the speed of the stream of river.)

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