Chetan is returning home from his friend's home and his dog is with him. It takes chetan 45 min to reach home .His dog runs twice as fast as chetan walks.The moment chetan and his dog leave his friend 's house the dog runs ahead and reaches home before chetan ,the dog again runs towards home.The moment the dog reaches home it returns again to meet chetan .
After how many minutes from the start of the trip does chetan meet his dog for the second time ?

• 40 mins.
  As Chetan's dog runs twice as fast as Chetan,
  When Chetan reaches half distance(50%), his dog reaches home.
  When Chetan reaches still more 16.67%, his dog first time meets him.
  When Chetan reaches still more 16.67%, his dog reaches home again.
  When Chetan reaches still more 5.56%, his dog second time meets him.
  So totally Chetan reaches 88.90% of total distance when he meets his dog for second time. So time taken = (45/100)*88.90 = 40.005 mins
• 13 years agoHelpfull: Yes(2) No(3)
• If distance is x, speed of chetan is c and speed of dog is d,and d=2c (given ) , x/c= 45 ( given ),
  then
  First time they will meet after time t ( in mins) when combined distance travelled by them is 2x
  t= 2x/c+d = 2x/3c= 45*2/3 = 30 mins
  Distance travelled by chetan till first meeting = x*30/45 = 2x/3
  Distance left = x/3
  Second time they will meet after time T after first meeting ( in mins) when combined distance travelled by them is 2x/3.
  T= (2x/3)/(c+d) = (2x/3)/3c = 2x/9c = 45*2/9 = 10 mins.
  so total time taken till second meeting from start = 30+10 = 40 mins.
• 13 years agoHelpfull: Yes(2) No(1)