There are two employees in a company. One is younger and the other is older. The older employee leaves his house at 10:00 am and walks to the company with a speed of 20 metres per minute. After 5 minutes the younger employee also leaves his home and walks towards office with a speed of 30 metres per minute. When will he meet the older employee?

a> 11:00
b> 10:10
c> 10:15
d> 10:30

• ans is 10.15
  older run till 10:05=100 younger till 10:05=0
  then at 10:10=200 till 10:10=150
  10:15=300 10:15=300
  they meet when thr distance become equal
  
• 12 years agoHelpfull: Yes(21) No(1)
• A will travel in 100m in 5ins i.e upto 10:05
  then time taken by B to travel 100m with a relative speed of (30-20)=10m
  is 10mins.
  i.e at 10:15am
• 12 years agoHelpfull: Yes(8) No(1)
• let older employee meets younger one at distance x...
  now for both the distance travelled is similar i.e x...
  now let older one takes y min to walk x mtrs
  => younger one takes y-5 min to walk x mtrs
  => speed of old * y = speed of young * (y-5)
  => 20y=30(y-5)
  =>y=15 min
  => at 10:15 both will meet eachother...
• 12 years agoHelpfull: Yes(6) No(1)
• 30(x-5)=20*x
  then x=15
  ie after 15 min later both will meet each other
  
• 12 years agoHelpfull: Yes(6) No(1)
• 10.15-ans
  10.00 -20 meters
  10.05 -30 meters
  thy vil meet at 10.15
• 12 years agoHelpfull: Yes(3) No(1)
• older man:in 5 min----100m
  younger man in 5 min-----150m
  lcm of 100&150 =300m
  younger 300/150=2
  2*5=10min
  ans:-10:15
  
• 12 years agoHelpfull: Yes(3) No(1)
• older : younger
  ---------------------
  10:05=100 : 10:05=0
  10:10=200 : 10:10=150
  10:15=300 : 10:15=300
  
  so ans should be = 10:15
• 12 years agoHelpfull: Yes(1) No(1)