2 workers ,one old and one young,live together and work at the same office.the old man takes 30 mins where as the young man takes only 20 mins to reach the office.when will the young man catch up the old man ,if the old man starts at 10.00am and the young man starts at 10.05am?

• Distance = Speed * Time;
  Let the distance will be 30m between Home and Office,
  Hence Old Man's speed is 1m/min
  and Young Man's speed is 1.5m/min
  
  Old man starts at 10:00, and Young man starts at 10:05,
  
  Old man travels 5m, Young man travels 0m at 10:05;
  Old man travels 10m, Young man travels 7.5m at 10:10;
  Old man travels 15m, Young man travels 15m at 10:15;
  
  So both will meet at 15m apart from home at 10:15 A.M
• 10 years agoHelpfull: Yes(56) No(4)
• 30 mins of old man=20 min of young man=same dist
  15 min of old man=10 min of young man=same dist
  Old man starts at 10.00 whereas young man starts at 10.05.
  Thus at 10.15 young will catch up the old
• 10 years agoHelpfull: Yes(36) No(0)
• suppose the distance b/w the starting point to the office = 30 m
  so,in 15 min the old man covers 15 m bcoz speed= 1m/min
  and in 10 min the young man covers 15 m bcoz speed = 1.5m/min
  thus 10:15 is the right answer.
• 10 years agoHelpfull: Yes(22) No(5)
• young man catches old man at 10:25 am..
• 10 years agoHelpfull: Yes(4) No(44)
• both will reach at same time...
  suppose old man travels x m in 1 min..so 30x m in 30 min.
  young man travels 1.5 x in 1 min..so 30x in 20 min.graph of "TRAVELLED DISTANCE" is given
  old:young==5:0,7:3,9:6,11:9,13:12,15:15(so at 10:15 0 clock old man will be caught up.)
• 10 years agoHelpfull: Yes(4) No(8)
• AT 10:15
  suppose old man travels x m in 1 min..so 30x m in 30 min.
  young man travels 1.5 x in 1 min..so 30x in 20 min.graph of "TRAVELLED DISTANCE" is given
  old:young==5:0,7:3,9:6,11:9,13:12,15:15(so at 10:15 0 clock old man will be caught up.)
• 10 years agoHelpfull: Yes(4) No(2)
• as old man start at 10:00 am n young man start at 10:05 am .young man reach 10 min.before old man. so he catch up the old man at 10:10 am.
• 10 years agoHelpfull: Yes(3) No(18)
• @Anjali: young man reaches 5 min before old man not 10 min.
  The old man will reach at 10:30 and young man @ 10:25
  I think answer should be 10:20 am
• 10 years agoHelpfull: Yes(2) No(8)
• at 10:15
  let the distance be 30 km and the speed of old man be 60 km/hr and the speed of young man be 90 km/hr..so they will meet at 10:15 as the distance travelled by both will be =15 km
• 10 years agoHelpfull: Yes(2) No(1)
• in 5mins oldman trvls d/30*5=d/6mtrs
  so,
  (d/6)/(d/20 - d/30)=10 mins..
• 10 years agoHelpfull: Yes(2) No(1)
• the ratio between the speed of the two is 3:2 i.e 1:1.5 which means if the man covers 1.5 km then the old man would cover 1km in 1min.
  So, initially the old man would have covered a distance of 5km wen the man just made a start.
  The man would cover 7.5kms n the old man will b at 10kms. The time is 10:10
  in the next 5 mins the man would cover 7.5kms more i.e. will be at 15km n same will be the old man as he covers 5kms more in the 5mins.
  hence the meeting time is 10:15
  
• 8 years agoHelpfull: Yes(1) No(0)
• @SIVARANJANI.S Plzzzzzz tell the what the correct ans is???
• 10 years agoHelpfull: Yes(0) No(0)
• let the distance b/w office and home be x km.
  speed of old man=x/(30/60)=2x
  speed of young man=x/(20/60)=3x
  now,dist travelled by old man in 5 min.=2x*(5/60)
  now time taken by young man=(dist travelled by old man in 5 min)/(rel. speed b/w two)
  =(2x*5)/(3x-2x)
  =10 min
  so,young man wil catch old man at 10:10am
• 10 years agoHelpfull: Yes(0) No(3)
• suppose distance is 100m, young travels 5m/min n old travels 3.33m/min
  at 10:05 old would have already walk 16.65
  now adding distance to young and old min by min at 10:15 young will catch old
• 10 years agoHelpfull: Yes(0) No(0)
• time ratio is 3:2, so
  3 min for old = 2 min fro young
  1 min for old = 2/3 min fro young
  Now we will take multiple of 3 both side,
  3 min for old = 2 min fro young
  6 min for old = 4 min fro young
  9 min for old = 6 min fro young
  12 min for old = 8 min fro young
  15 min for old = 10 min fro young and so on
  Now the difference of time to start for old and young is 5 min. So when they meet then old will take 5 minute more than young ( or their time difference should be 5 min) which is given by only
  15 min for old = 10 min fro young
  So at 10:15 for old and for young it is 10:05 + 10 min = 10:15
  Answer is 10:15
• 8 years agoHelpfull: Yes(0) No(0)
• time taken by old=x
  time taken by young=(x-5)
  and the difference between their distance =0
  therefore,
  (x-5)/20-x/30=0
  x=15
• 4 years agoHelpfull: Yes(0) No(0)
• let the total distance between the home and the office is 30 km.
  in 1min the old man covers 1km
  and the young man covers 3/2km
  
  Now in next 5 min the old man covered 5km whereas the young man 0km ( at 10:05 am).
  again at next 5 min the old man covered (5+5)km whereas the young man {(3/2)*5}km=15/2km (at 10:10 am).
  again for next 5 min old man cover (5+5+5)km=15km and young man covers (15/2+15/2)=15 km (at 10:15 am).
  
  both meet at 10:15 am
• 4 years agoHelpfull: Yes(0) No(0)