A hare and a tortoise have a race along a circle of 100 yards diameter. The tortoise goes in one direction and the hare in the other. The hare starts after the tortoise has coverd 1/4 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 increse its speed so as to tie the race?

• 1/4, 1/8
  4*8=32
  32-4=28
  28-8=20
  28*20/(4^2)=35.00
• 13 years agoHelpfull: Yes(27) No(5)
• speed of hare should increase by a factor of 35..
• 14 years agoHelpfull: Yes(9) No(32)
• Diameter of the circle = 100 yards
  thus perimeter of the circle is 100*3.14 yards
  distance covered by tortoise when both meet =7/8 of total
  speed of hare is 1/8of total per unit
  and speed of tortoise is 5/8 of total per unit of time
  to tie the race hare should complete 7/8 of total distance in the time tortoise completes 1/8 of total distance
  time taken by tortoise to complete 1/8 of total distance =1/5unit of time
  required speed of hare =35/8of total distance per unit of time
  thus hare should increase its speed 35 times to tie the race
• 12 years agoHelpfull: Yes(3) No(4)
• i did not get the solution
• 9 years agoHelpfull: Yes(2) No(0)
• total distance = 50 yards
  tortoise covers=1/3 distance
  hare covers =x+1/3
  hare and tortoise meet=when tortoise covers=1/4th dis(so remaining=3/4)
  tortoise cover=1/3(so remaining to cover 2/3)
  so 3/4-2/3=1/12soit should in crease 12 times ....thats it
• 13 years agoHelpfull: Yes(1) No(18)
• Total circumfrance = pie*diameter
  tortoise has covered = 25pie
  remaining distance = 75 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
  62.5 pie/u = 12.5 pie/v
  5v=u so v=u/5
  
  now for remaining journey time taken by tortoise = 12.5 pie/u
  so velocity of hare to make a tie = 7*12.5 pie u / 12.5 pie
  =7u
  so 7u/u/5 = 35 times answer
  
• 8 years agoHelpfull: Yes(1) No(3)
• Distance covered by tortoise while the hare covers 1/8th of the distance = 1 – (1/4 + 1/8)
  = 5/8
  
  Time = Distance/Speed
  
  Time taken by both is same
  
  Let speed of tortoise be t and of hare be h
  ⇒ 5/(8t) = 1/(8h)
  ⇒ h = (8/(8×5) ) × t = (1/5) × t
  
  Now for the next part hare has to cover 7/8 th distance when tortoise covers 1/8 th distance
  ⇒ 1/(8t) = 7/(8h’)
  h’ = 7t [Assuming speed of tortoise is constant]
  So the factor by which h's speed increases = h’/h = 7t/(t/5)
  = 35
• 6 years agoHelpfull: Yes(1) No(0)
• Assume the circumference of the circle is 200 meters. Hare and tortoise started at the same point but moves in the opposite direction. It is given that by that time tortoise covered 40 m (1/5th of the distance), Hare started and both met after hare has covered 25. This implies, in the time hare has covered 25m, hare has covered 200 - 40 - 25 = 135 meters.
  So Hare : tortoise speeds = 25 : 135 = 5 : 27
  Now Hare and tortoise has to reach the starting point means, Hare has to cover 175 meters and Tortoise has to cover only 25 meters in the same time.
  As time =DistanceSpeed=2527=1755×KDistanceSpeed=2527=1755×K
  Ie., Hare has to increase its speed by a factor K. Solving we get K = 37.8
• 4 years agoHelpfull: Yes(0) No(1)