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Question Number 12699 by chux last updated on 29/Apr/17

A man can row a boat at 4km/hr  in still water.He rows the boat 2km  upstream and 2km back to his  starting place in 2hours.How fast  is the stream moving?      please help

$$\mathrm{A}\:\mathrm{man}\:\mathrm{can}\:\mathrm{row}\:\mathrm{a}\:\mathrm{boat}\:\mathrm{at}\:\mathrm{4km}/\mathrm{hr} \\ $$$$\mathrm{in}\:\mathrm{still}\:\mathrm{water}.\mathrm{He}\:\mathrm{rows}\:\mathrm{the}\:\mathrm{boat}\:\mathrm{2km} \\ $$$$\mathrm{upstream}\:\mathrm{and}\:\mathrm{2km}\:\mathrm{back}\:\mathrm{to}\:\mathrm{his} \\ $$$$\mathrm{starting}\:\mathrm{place}\:\mathrm{in}\:\mathrm{2hours}.\mathrm{How}\:\mathrm{fast} \\ $$$$\mathrm{is}\:\mathrm{the}\:\mathrm{stream}\:\mathrm{moving}? \\ $$$$ \\ $$$$ \\ $$$$\mathrm{please}\:\mathrm{help} \\ $$

Answered by mrW1 last updated on 29/Apr/17

x=speed of water  (2/(4−x))+(2/(4+x))=2  (8/(16−x^2 ))=1  x^2 =8  x=2(√2)≈2.83 (km/h)

$${x}={speed}\:{of}\:{water} \\ $$$$\frac{\mathrm{2}}{\mathrm{4}−{x}}+\frac{\mathrm{2}}{\mathrm{4}+{x}}=\mathrm{2} \\ $$$$\frac{\mathrm{8}}{\mathrm{16}−{x}^{\mathrm{2}} }=\mathrm{1} \\ $$$${x}^{\mathrm{2}} =\mathrm{8} \\ $$$${x}=\mathrm{2}\sqrt{\mathrm{2}}\approx\mathrm{2}.\mathrm{83}\:\left({km}/{h}\right) \\ $$

Commented by chux last updated on 29/Apr/17

thanks sir.      pls i dont understand the concept  of the speed of the upper and lower  stream.

$$\mathrm{thanks}\:\mathrm{sir}. \\ $$$$ \\ $$$$ \\ $$$$\mathrm{pls}\:\mathrm{i}\:\mathrm{dont}\:\mathrm{understand}\:\mathrm{the}\:\mathrm{concept} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{the}\:\mathrm{upper}\:\mathrm{and}\:\mathrm{lower} \\ $$$$\mathrm{stream}. \\ $$

Commented by mrW1 last updated on 30/Apr/17

when the man rows against the stream,  if the water were still, the could row  4 km in one hour. in this one hour the  water brings ihm x km back, i.e. his  speed relatively to the coast is (4−x) km/hr.  when he rows in the direction of the  steam, his speed relatively to the  coast is (4+x) km/hr. since he rows  2 km upper steam and 2 km back in  totally 2 hours, we have  (2/(4−x))+(2/(4+x))=2

$${when}\:{the}\:{man}\:{rows}\:{against}\:{the}\:{stream}, \\ $$$${if}\:{the}\:{water}\:{were}\:{still},\:{the}\:{could}\:{row} \\ $$$$\mathrm{4}\:{km}\:{in}\:{one}\:{hour}.\:{in}\:{this}\:{one}\:{hour}\:{the} \\ $$$${water}\:{brings}\:{ihm}\:{x}\:{km}\:{back},\:{i}.{e}.\:{his} \\ $$$${speed}\:{relatively}\:{to}\:{the}\:{coast}\:{is}\:\left(\mathrm{4}−{x}\right)\:{km}/{hr}. \\ $$$${when}\:{he}\:{rows}\:{in}\:{the}\:{direction}\:{of}\:{the} \\ $$$${steam},\:{his}\:{speed}\:{relatively}\:{to}\:{the} \\ $$$${coast}\:{is}\:\left(\mathrm{4}+{x}\right)\:{km}/{hr}.\:{since}\:{he}\:{rows} \\ $$$$\mathrm{2}\:{km}\:{upper}\:{steam}\:{and}\:\mathrm{2}\:{km}\:{back}\:{in} \\ $$$${totally}\:\mathrm{2}\:{hours},\:{we}\:{have} \\ $$$$\frac{\mathrm{2}}{\mathrm{4}−{x}}+\frac{\mathrm{2}}{\mathrm{4}+{x}}=\mathrm{2} \\ $$

Commented by chux last updated on 30/Apr/17

thanks ....i really appreciate this.

$$\mathrm{thanks}\:....\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{this}. \\ $$

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