An Old man and a Young man are working together in an office and staying together in a near by apartment. The Old Man takes 30 minutes and the Young 20 minutes to walk from apartment to office. If one day the old man started at 10:00AM and the young man at 10:05AM from the apartment to office, when will they mee?

• Ratio of old man speed to young man speed = 2:3
  
  The distance covered by old man in 5 min = 10
  
  The 10 unit is covered with relative speed=10/(3-2)=10 min
  
  so, they will meet at 10:15 am.
• 12 years agoHelpfull: Yes(42) No(20)
• they will meet at 10:15 AM
• 12 years agoHelpfull: Yes(8) No(18)
• from trail and error method from the options 10:15minutes the oldman:youngman takes 15:10(ratio) i.e 3:2 which satifies the condition 30 mintes for the oldman and 20 minutes for the youngman
• 8 years agoHelpfull: Yes(4) No(3)
• Let distance = x
  
  speed of young man = x/20
  
  Speed of old man = x/30
  
  relative speed x/20−x/30=x/60
  
  Distance traveled by old man in 5 min =5*x/30=x/6
  
  Time taken by the young man to catch up the old man =Distance travelled /Relative speed=x/6*60/x =10 min
  
  So,young man will catch up man the old man at 10:15 am.
  
  
  
  Hence, 10:15am is the answer.
• 6 years agoHelpfull: Yes(4) No(0)
• the difference of their speed is 10 mins the old man meet office by 10:30 am but the boy by 10:25 am so before 10:25 he should meet the old man but he started after 10:05 so minus that 10 mins from 10:25 we get 10:15 as the answer.
• 6 years agoHelpfull: Yes(0) No(1)
• 60
  /
  3 2 then relative speed is=3-2=1 and old man goes in 5min 5*3=15 then ans=10:15
  /
  20 30
  
  
  
  /
• 6 years agoHelpfull: Yes(0) No(1)
• ans: 0.4270833333333333
  
  verified:
• 5 years agoHelpfull: Yes(0) No(3)
• Ratio of speeds 30units/min : 20 units/min
  relative speed 10 units/minutes
  in 5 min old man travels 100units
  100/(30-20)=10min
  10min+5min =15min
• 4 years agoHelpfull: Yes(0) No(0)