With-a-constant-throttle-setting-a-motor-boat-travels-36-miles-downstream-and-then-returns-The-downstream-takes-6-hours-less-than-the-return-trip-when-the-speed-of-the-motor-boat-is-doubled-the-

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Question Number 53188 by Otchere Abdullai last updated on 18/Jan/19

$$Withaconstantthrottlesetting$$$$amotorboattravels36miles$$$$downstreamandthenreturns.The$$$$downstreamtakes6hourslessthan$$$$thereturntrip.whenthespeedof$$$$themotorboatisdoubled,thetrip$$$$downstreamis1hourlessthanthe$$$$returntrip.whatistherateofthe$$$$streamscurrent?$$$$helppleasesir$$

Answered by tanmay.chaudhury50@gmail.com last updated on 18/Jan/19

$$speedofmotorboat=xmile/hour$$$$speedofstream=ymiles/hour$$$$down[stresmspeed=(x+y)miles/hour$$$$uostreamspeed=(x-y)miles/hour$$$$$$$$\frac{d}{x+y}+6=\frac{d}{x-y}\dots .eqn1$$$$whensppedofmotordouble$$$$\frac{d}{2x+y}+1=\frac{d}{2x-y}\dots eqn2$$$$2d=36$$$$so$$$$\frac{18}{x+y}+6=\frac{18}{x-y}$$$$\frac{18}{2x+y}+1=\frac{18}{2x-y}$$$$\frac{18}{x-y}-\frac{18}{x+y}=6$$$$\frac{18x+18y-18x+18y}{}=6x^2-6y^2$$$$36y=6x^2-6y^2$$$$6y=x^2-y^2\dots .\left(3\right)[x^2=6y+y^2]$$$$\frac{18}{2x-y}-\frac{18}{2x+y}=1$$$$\frac{36x+18y-36x+18y}{}=4x^2-y^2$$$$36y=4x^2-y^2$$$$now$$$$36y=4(6y+y^2)-y^2$$$$36y-24y-4y^2+y^2=0$$$$12y-3y^2=0$$$$3y(4-y)=0$$$$y=4miles/hour$$$$x^2=6y+y^2$$$$x^2=24+16=40$$$$x=\sqrt{40}miles/hour$$$$speedofstream\left(y\right)=4miles/hour$$$$$$

Commented by Otchere Abdullai last updated on 18/Jan/19

$$powerful!solutionsir!thanksalot$$

Commented by tanmay.chaudhury50@gmail.com last updated on 19/Jan/19

$$mostwelcome..$$

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