Hare in the other. The hare starts after the tortoise has covered 1/3 of its distance and that too leisurely3. A hare and a tortoise have a race along a circle of 100 yards diameter. The tortoise goes in one direction and the. The hare and tortoise meet when the hare has covered only 1/5 of the distance. By what factor should the hare increase its speed so as to tie the race?

• Let Vh, Vt be velocity of hare and turtoise
  Now,
  since hare has covered only 1/5 of distance
  therefore, distance covered by turtoise=1-1/5=4/5
  
  but we know that when hare starts running turtoise has already covered 1/3 of distance. Hence the meanwhile distance covered by turtoise when hare covers 1/5 is=4/5-1/3=7/15
  
  Hence, we have
  Vh/Vt=(1/5)/(7/15)
  i,e Vh=(3/7)Vt
  
  again, assume the remaining distance to be travled by hare and turtoise are Dh,Dt
  therefore,
  Dh/Dt=(4/5)/(1/5)
  i,e Dh=4Dt
  
  hence the hare must run 4 times faster than turtoise to tie the match
  so we can say that the,
  velocity of hare to be increased by =4Vt/(3/7)Vt=4*7/3=28/3=9.33....factor
• 12 years agoHelpfull: Yes(23) No(20)
• 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 =Distance/Speed=25/27=175/(5×K)
  Ie., Hare has to increase its speed by a factor K. Solving we get K = 37.8
• 8 years agoHelpfull: Yes(12) No(2)
• Total circumfrance = pie*diameter
  tortoise has covered = 100pie/3
  remaining distance = 200 pie / 3
  1/5th of total distance = 100pie /5 = 20 pie
  let the velocity of tortoise = u and that of hare =v
  on time comparison we get
  140 pie/3u = 20 pie/v
  7v=3u so v=3u/7
  
  now for remaining journey time taken by tortoise = 20 pie/u
  so velocity of hare to make a tie = 4*20 pie u / 20 pie
  =4u
  so 4u/(3u/7) = 28/3 times answer
• 12 years agoHelpfull: Yes(9) No(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/ speed) of tortoise = (dist / speed) of hare
  let speed of tortoise be t and of hare be h
  27/40t = 1/8h
  h = (40/(8*27) )* t = 5/27 * t
  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
  so the factor by which h's speed increases = 7t/ (5/27 * t)
  which we get as 37.8
• 8 years agoHelpfull: Yes(4) No(2)
• let, T= distance covered by Tortoise =1/3
  H=distance covered by hare when meet with Tortoise= 1/5
  Now, The factor by which hare increase its Speed = (1-H)*(1-(H+T)) / H^2
  after putting the value H and T we get 9.33
  N.B: Apply this formula you can solve this type of problems easily.
• 8 years agoHelpfull: Yes(1) No(3)