A hare and tortoise a race along a circle of 100 yardds diameter . The tortoise goes in one direction and the hare in the other . The hare starts after tortoise has covered 1/5 its distance and that leisurely . The hare and tortoise meet when the hare has covered only ¼ of the distance . by what factor should the hare increase its speed so as to tie the race?
a)8 b)37 c)45 d)6.6

• 6.6
  
  After tortoise covers 1/5th distance, Hare starts the race.
  When hare covers 1/4th of distance tortoise meets hare
  So distance covered by tortoise = 1-(1/5 + 1/4)
  = 11/20
  Time taken by both is same
  time = dist/speed
  So (dist/ speed) of tortoise = (dist / speed) of hare
  let speed of tortoise be t and of hare be h
  11/20 t = 1/4 h
  h = (20/(4*11) )* t = (5/11) * t
  Now for the next part hare has to cover 3/4 th distance when tortoise covers 1/4 th distance.
  Let new speed of Hare = H
  so we get
  1/4t = 3/4H
  H= 3t
  so the factor by which h's speed increases = 3t/ (5/11 * t)
  which we get as 33/5 = 6.6
• 13 years agoHelpfull: Yes(10) No(0)
• 1/5,1/4
  5*4=20
  20-5=15
  15-4=11
  15*11/5^2=
  6.6
• 13 years agoHelpfull: Yes(2) No(0)
• After tortoise covers 1/5rd distance hare starts the race
  When hare covers 1/4th of distance tortoise meets hare
  So distance covered by tortoise = 1-(1/5 + 1/4)
  = 11/20
  Time taken by both is same
  time = dist/speed
  So (dist/ speed) of tortoise = (dist / speed) of hare
  let speed of tortoise be t and of hare be h
  11/20t = 1/4h
  h = (20/(11x4) )x t = 5/11 * t
  Now for the next part hare has to cover 3/4 the distance when tortoise covers 1/4 the distance
  so we get
  1/4t = 3/4h
  h = 3t
  so the factor by which h's speed increases = 3t/ (5t/11)
  which we get as 33/5…so ans is 6.6(d)
  
• 13 years agoHelpfull: Yes(1) No(0)