A man walks 2 km. in first hour and then two thirds of this distance in second hour and then two thirds of the distance covered in the previous hour in the third hour.. and so on. What is the maximum distance walked ?

• total distance covered == 2+ 2*2/3 +2*(2/3)^2+(2/3)^3+....
  =2/(1-2/3)= 2*3/1= 6 kms
  
  sum of GP = a/1-r
  here a=2
  r=2/3
• 12 years agoHelpfull: Yes(5) No(0)
• Nearly 6kms. Exactly, 5.89kms.
  
  Let man walks in first hour, x = 2kms
  And man walks in 2nd hour = 2/3(x)
  .
  .
  .
  Maximum distance = x+ 2/3(x) + 2/3(2/3(x)) + ....
  = 2+1.33+0.88+0.59+0.39+0.26+0.17+0.11+0.07+0.03+0.01
  = 5.8988
  = Roundly 6kms.
• 12 years agoHelpfull: Yes(2) No(3)
• 4 km
  because
  he walked the total distance is
  = 2+ 2/3 +(2/3)^2+(2/3)^3+....
  =2+(2/3)/(1-2/3)
  =2+2*3/3=2+2=4
  4 km..
• 12 years agoHelpfull: Yes(1) No(6)
• 6 kms.
  
  total distance=2+2/3+(2/3)^2+(2/3)³+... g.p with r=2/3
  
  sum=2/1-2/3=6
• 12 years agoHelpfull: Yes(0) No(3)
• 4 kms
  
  2+2/3(1+2/3+(2/3)²+..)=2+2=4
• 12 years agoHelpfull: Yes(0) No(2)