A boat goes 2, km upstream and 3, km downstre | vedclass

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(C) Let the speed of the boat in still water be $x\, km/hr$ and the speed of the stream be $y\, km/hr$.Upstream speed = $(x - y)\, km/hr$ and downstre

Vedclass · Accepted Answer

$\frac{135}{7}$

Vedclass · Answer

(C) Let the speed of the boat in still water be $x\, km/hr$ and the speed of the stream be $y\, km/hr$.Upstream speed = $(x - y)\, km/hr$ and downstream speed = $(x + y)\, km/hr$.Given,time = $	ext{distance} / 	ext{speed}$.For the first case: $\frac{2}{x-y} + \frac{3}{x+y} = \frac{20}{60} = \frac{1}{3} \implies \frac{6}{x-y} + \frac{9}{x+y} = 1$  ---$(1)$For the second case: $\frac{7}{x-y} + \frac{2}{x+y} = \frac{53}{60}$ ---$(2)$Let $u = \frac{1}{x-y}$ and $v = \frac{1}{x+y}$.$6u + 9v = 1$ ---$(3)$$7u + 2v = \frac{53}{60}$ ---$(4)$Multiplying $(3)$ by $2$ and $(4)$ by $9$:$12u + 18v = 2$$63u + 18v = \frac{477}{60} = 7.95$Subtracting the equations: $51u = 5.95 \implies u = \frac{5.95}{51} = \frac{595}{5100} = \frac{7}{60}$.So,$x - y = \frac{60}{7}$.Substituting $u$ in $(3)$: $6(\frac{7}{60}) + 9v = 1 \implies \frac{7}{10} + 9v = 1 \implies 9v = \frac{3}{10} \implies v = \frac{3}{90} = \frac{1}{30}$.So,$x + y = 30$.Adding $x - y = \frac{60}{7}$ and $x + y = 30$:$2x = \frac{60}{7} + 30 = \frac{60 + 210}{7} = \frac{270}{7}$.$x = \frac{135}{7}\, km/hr$.

$A$ boat goes $2\, km$ upstream and $3\, km$ downstream in $20$ minutes. It goes $7\, km$ upstream and $2\, km$ downstream in $53$ minutes. What is the speed (in $km/hr$) of the boat in still water?

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