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On a straight road XY, 100 m long, five heavy stones are placed 2 m apart beginning at the end X. A worker, starting at X, has to transport all the stones to Y, by carrying only one stone at a time. The minimum distance he has to travel is
The minimum distance to be walked is 100 (onward)+ 98 (return) + 98 (onward) + 96 (return) + 96( onward) + 94(return) + 94(onward) + 92(return) + 92(onward) = 860 metres
x--------------------100m--------------------------y (Given XY=100 m)
now stones are placed at beginning of X and how many stones ? 5
X---------------------------------------------------------------------------Y
*1------*2--------*3--------*4--------*5---------------------------------Y (Consider * As stone)(*1 =first Stone,*2=Second Stone,*3=Third Stone...so on )
2m 2m 2m 2m 92 m (Stones Are separated by 2m)
1) Truck start and carry *1 stone and move towards Y,
------------------------------------------------------------------------------->
he traveled 100 m
2) He moves from Y and travel upto *2 stone and carry *2 stone and move towards Y
*2<---------------------------------------------------------------------
*2---------------------------------------------------------------------->
he traveled 98+98
3)He moves from Y and travel upto *3 stone and carry *3 stone and move towards Y
*3<---------------------------------------------------------
*3---------------------------------------------------------->
he traveled 96+96
4)Moves from Y and travel upto *4 stone ,carry *4 stone and move towards Y end
*4<-------------------------------------------
*4-------------------------------------------->
he traveled 94+94
5) Last stone *5 to carry,He moves from Y to towards *5 Stone then carry *5 stone and move towards Y
*5<--------------------------------
*5--------------------------------->
he traveled 92+92
Total Distance Covered 100+98+98+96+96+94+94+92+92= 2(50+98+96+94+92)= 2 *430=860 m
