A hare and tortise hav a race along a circle of 100 yards diameter.the tortise goes in one direction and the hare in the other.the hare starts after the tortise has covered 1/5 of its distance and that too leisurely.the hare and tortise meet when the hare has covered only 1/8 of the distance.by what factor should the hare increase its speed so as to tie the race?

• Total circumfrance = pie*diameter
  tortoise has covered = 20pie
  remaining distance = 80 pie
  1/8th of total distance = 12.5 pie
  let the velocity of tortoise = u and that of hare =v
  on time comparison we get
  67.5 pie/u = 12.5 pie/v
  5.4v=u so v=u/5.4
  
  now for remaining journey time taken by tortoise = 12.5 pie/u
  so velocity of hare to make a tie ,by time comparison
  12.5pie/u=87.5pie/v1, where v1 is the increased factor
  thus v1=7u and u=5.4v
  thus v1=7*5.4=37.8v
  so the hare has to increase his speed by a factor of 37.8 to tie the race.
• 12 years agoHelpfull: Yes(19) No(0)
• (1/8)*(u/v)=1-(1/8)-(1/5)
  =>(u/v)=27/5
  and
  (1/8)*(v'/u)=1-(1/8)
  =>(v'/u)=7
  then,(u/v)*(v'/u)=(27/5)*7
  =>(v'/v)=189/5=37.8
  You can do all this type of apti question by this method.....
• 12 years agoHelpfull: Yes(1) No(0)
• (1/8)*(u/v)=1-(1/8)-(1/5)
  =>(u/v)=27/5
  and
  (1/8)*(v'/u)=1-(1/8)
  =>(v'/u)=7
  then,(u/v)*(v'/u)=(27/5)*7
  =>(v'/v)=189/5=37.8
  You can do all this type of apti question by this method.....
• 12 years agoHelpfull: Yes(1) No(0)