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A-train-which-travels-at-a-uniform-speed-due-to-mechanical-fault-after-traveling-for-an-hour-goes-at-3-5-th-of-the-original-speed-and-reaches-the-destination-2-hours-late-If-the-fault-occured-after

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Question Number 94820 by jdmath last updated on 21/May/20
A train which travels at a uniform speed due to mechanical fault after traveling for an hour goes at 3/5 th of the original speed and reaches the destination 2 hours late. If the fault occured after traveling another 50 miles the train would have reached 40 minutes earlier. What is the distance between the two stations ?
$${A}\:{train}\:{which}\:{travels}\:{at}\:{a}\:{uniform}\:{speed}\:{due}\:{to}\:{mechanical}\:{fault}\:{after}\: \\ $$$${traveling}\:{for}\:{an}\:{hour}\:{goes}\:{at}\:\mathrm{3}/\mathrm{5}\:{th}\:{of}\:{the}\:{original}\:{speed}\:{and}\:{reaches}\:{the}\: \\ $$$${destination}\:\mathrm{2}\:{hours}\:{late}.\:{If}\:{the}\:{fault}\:{occured}\:{after}\:{traveling}\:{another}\:\mathrm{50} \\ $$$${miles}\:{the}\:{train}\:{would}\:{have}\:{reached}\:\mathrm{40}\:{minutes}\:{earlier}.\:{What}\:{is}\:{the}\: \\ $$$${distance}\:{between}\:{the}\:{two}\:{stations}\:? \\ $$
Answered by prakash jain last updated on 21/May/20
Assume distance=s km speed=v (km/hr) expected time=(s/v)=t (I) case when fault occured after 1 hour. distace traveled in 1 hour=v remaining distancd=s−v time taken to cover remaining =((s−v)/((3/5)v)) total time=1+((5(s−v))/(3v))=t+2=(s/v)+2 given that train is late by 2 hrs wrt (I) 1+((5(s−v))/(3v))=t+2=(s/v)+2 (II) case when fault occured after after addition 50 kms distace traveled in 1 hour=v time taken to cover extra 50 km=((50)/v) remaining distancd=s−v−50 time taken to cover remaining =((s−v−50)/((3/5)v)) total time=1+((50)/v)+((5(s−v−50))/(3v)) given that train is late by 40 min wrt (I) 40 minutez=(2/3)hrs 1+((50)/v)+((5(s−v−50))/(3v))=t+(2/3)=(s/v)+(2/3) (III) solve (I1) & (III) to calculate required value.
$$\mathrm{Assume}\:\mathrm{distance}=\mathrm{s}\:\mathrm{km} \\ $$$$\mathrm{speed}=\mathrm{v}\:\left(\mathrm{km}/\mathrm{hr}\right) \\ $$$$\mathrm{expected}\:\mathrm{time}=\frac{\mathrm{s}}{\mathrm{v}}=\mathrm{t}\:\:\left(\mathrm{I}\right) \\ $$$$\boldsymbol{\mathrm{case}}\:\boldsymbol{\mathrm{when}}\:\boldsymbol{\mathrm{fault}}\:\boldsymbol{\mathrm{occured}}\:\boldsymbol{\mathrm{after}}\:\mathrm{1}\:\boldsymbol{\mathrm{hour}}. \\ $$$$\mathrm{distace}\:\mathrm{traveled}\:\mathrm{in}\:\mathrm{1}\:\mathrm{hour}=\mathrm{v} \\ $$$$\mathrm{remaining}\:\mathrm{distancd}=\mathrm{s}−\mathrm{v} \\ $$$$\mathrm{time}\:\mathrm{taken}\:\mathrm{to}\:\mathrm{cover}\:\mathrm{remaining}\:=\frac{\mathrm{s}−\mathrm{v}}{\frac{\mathrm{3}}{\mathrm{5}}\mathrm{v}} \\ $$$$\mathrm{total}\:\mathrm{time}=\mathrm{1}+\frac{\mathrm{5}\left(\mathrm{s}−\mathrm{v}\right)}{\mathrm{3v}}=\mathrm{t}+\mathrm{2}=\frac{\mathrm{s}}{\mathrm{v}}+\mathrm{2} \\ $$$$\mathrm{given}\:\mathrm{that}\:\mathrm{train}\:\mathrm{is}\:\mathrm{late}\:\mathrm{by}\:\mathrm{2}\:\mathrm{hrs}\:\mathrm{wrt}\:\left(\mathrm{I}\right) \\ $$$$\mathrm{1}+\frac{\mathrm{5}\left(\mathrm{s}−\mathrm{v}\right)}{\mathrm{3v}}=\mathrm{t}+\mathrm{2}=\frac{\mathrm{s}}{\mathrm{v}}+\mathrm{2}\:\:\left(\mathrm{II}\right) \\ $$$$\boldsymbol{\mathrm{case}}\:\boldsymbol{\mathrm{when}}\:\boldsymbol{\mathrm{fault}}\:\boldsymbol{\mathrm{occured}}\:\boldsymbol{\mathrm{after}}\: \\ $$$$\boldsymbol{\mathrm{after}}\:\boldsymbol{\mathrm{addition}}\:\mathrm{50}\:\boldsymbol{\mathrm{kms}} \\ $$$$\mathrm{distace}\:\mathrm{traveled}\:\mathrm{in}\:\mathrm{1}\:\mathrm{hour}=\mathrm{v} \\ $$$$\mathrm{time}\:\mathrm{taken}\:\mathrm{to}\:\mathrm{cover}\:\mathrm{extra}\:\mathrm{50}\:\mathrm{km}=\frac{\mathrm{50}}{\mathrm{v}} \\ $$$$\mathrm{remaining}\:\mathrm{distancd}=\mathrm{s}−\mathrm{v}−\mathrm{50} \\ $$$$\mathrm{time}\:\mathrm{taken}\:\mathrm{to}\:\mathrm{cover}\:\mathrm{remaining}\:=\frac{\mathrm{s}−\mathrm{v}−\mathrm{50}}{\frac{\mathrm{3}}{\mathrm{5}}\mathrm{v}} \\ $$$$\mathrm{total}\:\mathrm{time}=\mathrm{1}+\frac{\mathrm{50}}{\mathrm{v}}+\frac{\mathrm{5}\left(\mathrm{s}−\mathrm{v}−\mathrm{50}\right)}{\mathrm{3v}} \\ $$$$\mathrm{given}\:\mathrm{that}\:\mathrm{train}\:\mathrm{is}\:\mathrm{late}\:\mathrm{by}\:\mathrm{40}\:\mathrm{min}\:\mathrm{wrt}\:\left(\mathrm{I}\right) \\ $$$$\mathrm{40}\:\mathrm{minutez}=\frac{\mathrm{2}}{\mathrm{3}}\mathrm{hrs} \\ $$$$\mathrm{1}+\frac{\mathrm{50}}{\mathrm{v}}+\frac{\mathrm{5}\left(\mathrm{s}−\mathrm{v}−\mathrm{50}\right)}{\mathrm{3v}}=\mathrm{t}+\frac{\mathrm{2}}{\mathrm{3}}=\frac{\mathrm{s}}{\mathrm{v}}+\frac{\mathrm{2}}{\mathrm{3}}\:\:\left(\mathrm{III}\right) \\ $$$$\mathrm{solve}\:\left(\mathrm{I1}\right)\:\&\:\left(\mathrm{III}\right)\:\mathrm{to}\:\mathrm{calculate} \\ $$$$\mathrm{required}\:\mathrm{value}. \\ $$
Commented by necxxx last updated on 21/May/20
mr. Prakash Jain welcome back. I sincerely miss you on this forum.
$${mr}.\:{Prakash}\:{Jain}\:{welcome}\:{back}.\:{I} \\ $$$${sincerely}\:{miss}\:{you}\:{on}\:{this}\:{forum}. \\ $$
Commented by jdmath last updated on 21/May/20
thanks rly sir
$${thanks}\:{rly}\:{sir} \\ $$

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