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Question

After travelling for 30 minutes a train meets an accident, due to which it has to stop for 45 minutes. Due to the accident its speed is also reduced to 23 of its former value and the train reaches its destination 1 hour 30 minutes late. Had the accident occurred 60 km later, the train would have reached 30 minutes earlier. The length of journey is
  1. 150 km
  2. 180 km
  3. 90 km
  4. 120 km

A
90 km
B
120 km
C
150 km
D
180 km
Solution
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Letthedistancebedandspeedbev.Thereforethenormalrunningtimeisdv.Inthefirstcase,runningtimeis(d+112)hrandinthesecondcase,therunningtimeis(dv+1)hr.Nowineachcasecalculatetherunningtimes,equatethemwiththerespectivetimesasshownabove,solvesimultaneousequationsindandv.

Letthedistancebedkmandspeedofthetrainbevkm/hr.Thennormaltimetoreachthedestinationisdvhr.Inthefirstcaseduetolaterunningby112hrthetotaltimet1=(dv+32)hrt1=2d+3v2vhr(1)Similarlyinthesecondcase,thetrainislatebyonehour.t2=dv+1=d+vvhr(2)FirstcaseNowthetraintravels30kmwithaspeedv,stopsfor45min.=34hrforaccident.Totaltimetakenuptothepointofaccidentis(30v+34)h(a)Restofthedistanceis(d30)kmandthetrainrunswithaspeedof23v.timetakentoreachthedestinationis(d30)×32vhr=3d902vhr(b)Totaltimetakenbythetrainis(a)+(b)=t1=(30v+34+3d90v)hr.takingt1fromequation(1)wehave.30v+34+3d902v=2d+3v2v2d3v=60(3)SecondcaseTheaccidenthappens60kmlater.Thatisitruns(60+30)km=90kmwithaspeedofvkm/hr.Timetakenis90vhr.
Thenitstopsfor34hrTotaltimetakenuptothispointis(90v+34)hr=360+3v4vhr(a)Aftertheaccidentthedistanceleft=(d90)kmandthespeed=2v3km/hrThetotaltimetakentocover(d90)kmis(d90)×32vhr=3d2702vhr(b)Thetotaltimetakentoreachthedestinationis(a)+(b)=t2=[360+3v4v+3d2702v]hr.replacingt2fromequation2.360+3v4v+3d2702v=d+vv2dv=180(4)Multiplyingequation(4)with3andsubtractingfromequation(3)weget4d=480d=120km(Ans)

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