Question

A boat man rowed to a place along the current and returned against the current in 15hrs. If the speed of the boat in still water is 6km/h and that of the current be 2km/h then find the distance of the place.​

Answer 1

$\large{\red{ \bf{ \underline {\underline{Answer}}}}} \\ \\ \purple{ \sf{\mapsto Distance=40 \: km}} \\ \\ \mathrm{\green{ \underline{Given}}} \\ \\ \sf{ \rightsquigarrow Total\:time \: taken = 15 \: hr} \\ \\ \sf{ \rightsquigarrow Speed \: of \: boat=6\:km/hr} \\ \\ \sf{ \rightsquigarrow Speed \: of \: current=2\:km/hr}\\ \\ \mathrm {\blue{ \underline{ To \: Find}}} \\ \\ \sf{ \rightsquigarrow Distance \: of \: the \: place\:=\:?}$

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• According to given question

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$\sf{Let \: the \: distance \: be \:x} \\ \\ \sf{The \: time \: taken \: while \: going \: to \: place \: be \: t} \\ \\ \sf{ \therefore \: \: Speed \: against \: current \: = (6 - 2 )\: Km/hr} \\ \\ \sf{ = \: 4 \: Km/hr} \\ \\ \sf{Speed \: along \: current = (6 + 2)Km/hr} \\ \\ \sf{= 8 \: km/hr } \\ \\ \sf{\dashrightarrow Sailed \: along \: with \: current \: while \: going } \\ \\ \sf{ \therefore \: \: \boxed{ \sf \orange{Distance = Speed \times Time}}}$

$\sf{\implies x = 8 \times t} \\ \\ \sf{\implies x = 8t \: - - - - (1)} \\ \\ \sf{\dashrightarrow Sailed \: against \: current \: while \: returning \: } \\ \\ \sf{ \therefore \: \: x = 4 \times (15 - t)} \: \: \: \: \: [Total \: time \: taken = 15] \\ \\ \sf{ \implies x = 60 - 4t \: - - - - (2)} \\ \\ \sf {\underline{ \bf{Solving \: equation \: (1) \: and \: equation \: (2)}}} \\ \\ \sf{ \implies 8t = 60 - 4t} \\ \\ \sf{ \implies 12t = 60} \\ \\ \sf{ \implies t = 5} \\ \\ \sf{Therefore, \: Distance = Speed \times Time} \\ \\ \sf{\implies Distance = 8 \times 5} \\ \\ \sf {\purple{ \implies}} \: \purple{ \underline{ \boxed{ \sf{Distance = 40 \: Km}}}} \: \orange{ \Large{\dag}}$

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

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$\green{\underline \bold{Given :}}$

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• Time taken by the boat = 15hours

• Speed of boat in still water = 6 km/hr

• Speed of current is = 2 km/hr

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$\red{\underline \bold{To \: Find:}}$

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• Distance of the place.

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$\large{\orange{\underline{\tt{Solution :-}}}}$

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$\huge{\bf Case 1 }$

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Let the distance covered by the boat be 'd'

Let the time taken while going be 't'

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$\footnotesize{ \underline{\bold{\texttt{Boat is rowed along the current while going}}}}$

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That means this is a condition of downstream.

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$\sf \footnotesize{ Speed \: in\: downstream \: = \: Speed \: of \: current \: + \: Speed \: of \: boat \: in \: still \: water}$

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Speed while going = (2 + 6) km/hr

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$\purple{\sf Speed \: while \: going \: = \: 8 \: km/hr}$

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$\sf \longmapsto Speed = \dfrac { Distance } { Time }$

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$\sf \longmapsto 8 = \dfrac { d } { t }$

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$\sf \longmapsto d = 8t$ -------(1)

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$\huge{\bf Case 2 }$

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Again,

Distance covered is 'd'

Time taken while returning is '15 - t'

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$\footnotesize{ \underline{\bold{\texttt{Boat is rowed against the current while returning }}}}$

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That means this is a condition of upstream

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$\sf \footnotesize{ Speed \: in\: upstream \: = \: Speed \: of \: boat \: in \: still \: water \: - \: Speed \: of \: current}$

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Speed while returning = (6 - 2) km/hr

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$\purple{\sf Speed \: while \: returning \: = \: 4 \: km/hr}$

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$\sf \longmapsto Speed = \dfrac { Distance } { Time }$

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$\sf \longmapsto 4 = \dfrac { d } { 15 - t}$

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$\sf \longmapsto d = 60 - 4t$ -----(2)

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$\footnotesize{ \underline{\bold{\texttt{As distance will remain same hence (1) = (2)}}}}$

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$\sf \longmapsto 8t = 60 - 4t$

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$\sf \longmapsto 12t = 60$

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$\sf \longmapsto t = \dfrac { 60 } { 12 }$

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$\sf \longmapsto t = 5$

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Hence time taken while going is 5hrs.

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So,

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Time taken while returning is '15 - 5'

i.e 10hrs

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$\underline{\bold{\texttt{Putting t = 5 in (1)}}}$

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$\sf \longmapsto d = 8(5)$

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$\sf \longmapsto d = 40$

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Hence distance of the place will be 40 km

$\rule{200}5$