Question Number 53188 by Otchere Abdullai last updated on 18/Jan/19
$${With}\:{a}\:{constant}\:{throttle}\:{setting}\: \\ $$$${a}\:{motor}\:{boat}\:{travels}\:\mathrm{36}\:{miles} \\ $$$${downstream}\:{and}\:{then}\:{returns}.\:{The} \\ $$$${downstream}\:{takes}\:\mathrm{6}\:{hours}\:{less}\:{than}\: \\ $$$${the}\:{return}\:{trip}.\:{when}\:{the}\:{speed}\:{of}\: \\ $$$${the}\:{motor}\:{boat}\:{is}\:{doubled},\:{the}\:{trip} \\ $$$${downstream}\:{is}\:\mathrm{1}\:{hour}\:{less}\:{than}\:{the}\: \\ $$$${return}\:{trip}.\:{what}\:{is}\:{the}\:{rate}\:{of}\:{the} \\ $$$${streams}\:{current}? \\ $$$${help}\:{please}\:{sir} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 18/Jan/19
![speed of motor boat=x mile/hour speed of stream=y miles/hour down[stresm speed=(x+y)miles/hour uo stream speed=(x−y)miles/hour (d/(x+y))+6=(d/(x−y))....eqn 1 when spped of motor double (d/(2x+y))+1=(d/(2x−y))...eqn 2 2d=36 so ((18)/(x+y))+6=((18)/(x−y)) ((18)/(2x+y))+1=((18)/(2x−y)) ((18)/(x−y))−((18)/(x+y))=6 ((18x+18y−18x+18y)/)=6x^2 −6y^2 36y=6x^2 −6y^2 6y=x^2 −y^2 ....(3) [x^2 =6y+y^2 ] ((18)/(2x−y))−((18)/(2x+y))=1 ((36x+18y−36x+18y)/)=4x^2 −y^2 36y=4x^2 −y^2 now 36y=4(6y+y^2 )−y^2 36y−24y−4y^2 +y^2 =0 12y−3y^2 =0 3y(4−y)=0 y=4miles/hour x^2 =6y+y^2 x^2 =24+16=40 x=(√(40 )) miles/hour speed of stream(y)=4miles/hour](https://www.tinkutara.com/question/Q53191.png)
$${speed}\:{of}\:{motor}\:{boat}={x}\:{mile}/{hour} \\ $$$${speed}\:{of}\:{stream}={y}\:{miles}/{hour} \\ $$$${down}\left[{stresm}\:{speed}=\left({x}+{y}\right){miles}/{hour}\right. \\ $$$${uo}\:{stream}\:{speed}=\left({x}−{y}\right){miles}/{hour} \\ $$$$ \\ $$$$\frac{{d}}{{x}+{y}}+\mathrm{6}=\frac{{d}}{{x}−{y}}….{eqn}\:\mathrm{1} \\ $$$${when}\:{spped}\:{of}\:{motor}\:{double} \\ $$$$\frac{{d}}{\mathrm{2}{x}+{y}}+\mathrm{1}=\frac{{d}}{\mathrm{2}{x}−{y}}…{eqn}\:\mathrm{2} \\ $$$$\mathrm{2}{d}=\mathrm{36} \\ $$$${so} \\ $$$$\frac{\mathrm{18}}{{x}+{y}}+\mathrm{6}=\frac{\mathrm{18}}{{x}−{y}} \\ $$$$\frac{\mathrm{18}}{\mathrm{2}{x}+{y}}+\mathrm{1}=\frac{\mathrm{18}}{\mathrm{2}{x}−{y}} \\ $$$$\frac{\mathrm{18}}{{x}−{y}}−\frac{\mathrm{18}}{{x}+{y}}=\mathrm{6} \\ $$$$\frac{\mathrm{18}{x}+\mathrm{18}{y}−\mathrm{18}{x}+\mathrm{18}{y}}{}=\mathrm{6}{x}^{\mathrm{2}} −\mathrm{6}{y}^{\mathrm{2}} \\ $$$$\mathrm{36}{y}=\mathrm{6}{x}^{\mathrm{2}} −\mathrm{6}{y}^{\mathrm{2}} \\ $$$$\mathrm{6}{y}={x}^{\mathrm{2}} −{y}^{\mathrm{2}} ….\left(\mathrm{3}\right)\:\:\left[{x}^{\mathrm{2}} =\mathrm{6}{y}+{y}^{\mathrm{2}} \right] \\ $$$$\frac{\mathrm{18}}{\mathrm{2}{x}−{y}}−\frac{\mathrm{18}}{\mathrm{2}{x}+{y}}=\mathrm{1} \\ $$$$\frac{\mathrm{36}{x}+\mathrm{18}{y}−\mathrm{36}{x}+\mathrm{18}{y}}{}=\mathrm{4}{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \\ $$$$\mathrm{36}{y}=\mathrm{4}{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \\ $$$${now} \\ $$$$\mathrm{36}{y}=\mathrm{4}\left(\mathrm{6}{y}+{y}^{\mathrm{2}} \right)−{y}^{\mathrm{2}} \\ $$$$\mathrm{36}{y}−\mathrm{24}{y}−\mathrm{4}{y}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{12}{y}−\mathrm{3}{y}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{3}{y}\left(\mathrm{4}−{y}\right)=\mathrm{0} \\ $$$${y}=\mathrm{4}{miles}/{hour} \\ $$$${x}^{\mathrm{2}} =\mathrm{6}{y}+{y}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} =\mathrm{24}+\mathrm{16}=\mathrm{40} \\ $$$${x}=\sqrt{\mathrm{40}\:}\:{miles}/{hour} \\ $$$${speed}\:{of}\:{stream}\left({y}\right)=\mathrm{4}{miles}/{hour} \\ $$$$ \\ $$
Commented by Otchere Abdullai last updated on 18/Jan/19
$${powerful}!\:{solution}\:{sir}!\:{thanks}\:{a}\:{lot} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 19/Jan/19
$${most}\:{welcome}.. \\ $$