A bus started from bustand at 8.00am, and after staying for 30 minutes at a destination, it returned back to the busstand. The destination is 27 miles from the busstand. The speed of the bus is 18mph. During the return journey bus travels with 50% faster speed.At what time does it return to the busstand?

• Time = Distance/Speed
  Reaching Time(destination) = 27/18 = 1.5hrs
  50% faster the speed = (18*50)/100=9mph
  Return speed = 18+9 = 27mph
  Return time(Bus stand) = Distance/new speed = 27/27 = 1hr
  Total = Reaching time + Staying time + return time
  = 1.5hrs + 30min + 1hr = 3hrs
  Therefore answer is 11am
  
• 10 years agoHelpfull: Yes(37) No(0)
• 
  Time taken to Reach destination is 27/18 hrs. = 3/2 hrs
  50% faster speed means speed is 3/2 times the original speed, so it will take 2/3 times time to cover the same distance.
  So time taken during return journey = (3/2)*(2/3) = 1hr
  Stay time = 30 minutes = 1/2 hrs
  Total time = Time for onward journey + Staying time + Time for return journey
  = 3/2 hrs + 1/2 hrs. + 1 hr = 3 hrs
  It will return to the bus stand after 3 hrs.
  Hence it will return to the bus stand at 11am
  
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• 10 years agoHelpfull: Yes(5) No(0)
• can any of u explain speed unit ???
  is it meters/hour or miles/hour???
  
• 10 years agoHelpfull: Yes(2) No(0)
• bus will return in 3 hours i.e. at 11 am
• 10 years agoHelpfull: Yes(1) No(0)
• t=d/s
  (d1/s1)+(d2/s2)=time taken +stoppage time
  (27/18)+(27/27)=x+(30/60)
  x=3hrs
  hence 8am +3hrs=11am
• 10 years agoHelpfull: Yes(1) No(0)
• 2hrs 42 min
• 10 years agoHelpfull: Yes(0) No(0)