A car travels a total distance of 150 km. After travelling a part of the distance without any trouble ,the car develops an engine problem and proceeds at 2/3rd of its former speed and arrives at the destination 48 mins late. Had the problem developed 24 km further on, the car would have arrived 12 min sooner. Find the original distance it travelled without any problem and the speed over that part of the journey.
a)100km,60 kmph b)48km,36kmph c)72km,50kmph d)54km,60 kmph

• Assume the car travels from A to B (150 km).
  
  In first case, assume it travels till C without any trouble with speed 'v' and covers the remaining distance from C to B with speed '2v/3' once its engine develops a problem and reaches B, 48 minutes late.
  In second case, the car travels an extra 24 km before encountering the engine problem and assume it reaches D (here, D = C + 24 km) with speed 'v' and then travels from D to B with speed '2v/3' and reaches 12 minutes early than in first case i.e. 36 minutes late overall.
  
  In the second case, the 12 minutes it makes is made up over CD.
  Thus,
  24/(2v/3) - 24/v = 12/60
  Or, v = 60 kmph
  
  Over 24 km it makes up 12 minutes of the entire 48 minutes.
  Thus, BC = 24 x 4 = 96 km
  And hence, AC = 150 - 96 = 54 km
  
  Ans: D
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