A hare and a tortoise race along a circle of 100 yards diameter. The tortoise goes in one direction and the hare in the other. The hare starts after tortoise has covered 1/5 of its distance and that too leisurely. The hare and tortoise 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?
(a)8
(b) 37.80
(c) 40
(d) 5

• After tortoise covers 1/5th distance hare starts the race
  When hare covers 1/8th of distance tortoise meets hare
  So distance covered by tortoise = 1-(1/5 + 1/8)
  = 27/40
  Time taken by both is same
  time = dist/speed
  So (dist covered by tortoise/ speed of tortoise) = (dist covered by hare / speed of hare)
  let speed of tortoise be t and speed of hare be h
  27/40t = 1/8h
  h = (40/(8*27) )* t = 5/27 * t which is initial speed of hare
  Now for the next part hare has to cover 7/8 th distance when tortoise covers 1/8 th distance
  so we get
  1/8t = 7/8h
  h = 7t which is final speed of hare
  so the factor by which hare's speed should increase = 7t/ (5/27 * t)= 189/5
  = 37.8
• 12 years agoHelpfull: Yes(28) No(1)
• hare increase its speed 37.80 times so as to tie the race.
• 12 years agoHelpfull: Yes(3) No(1)
• garima pls explain the 2nd part. 1/8t = 7/8h
  h = 7t which is final speed of hare.
  does tie up means vice versa of distance travelled by them. becoz for hare it was 1/8 now you took 1/8 for tortoise.
  
• 12 years agoHelpfull: Yes(1) No(0)
• Garima how can u say the distance covered by tortoise is 27/40
  .
• 12 years agoHelpfull: Yes(1) No(1)
• hare has to increase his speed by 37.80 times
• 12 years agoHelpfull: Yes(0) No(0)