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 covered 1/3 of its distance and that too leisurely. The hare and tortoise meet when the hare has covered only 1/4 of the distance. By what factor should hare increase its speed so as the win the race?
(a) 4
(b) 3
(c) 12
(d) 5

• Answer is : (d) 5.
  
  Here is my working:
  
  Let total distance (circumference) = L
  Tortoise(tr) has already covered 1/3 of distance.
  Remainder of distance it needs to cover = 2L/3
  The meeting point for hare (opposite direction from tr) = L/4 (3/4th of circumference for tr)
  Since tr has already covered 1/3 before hare started, the distance covered by it to reach the meeting point: 3L/4 - L/3 = 5L/12
  
  Let T be the time taken for both to reach the meeting point.
  
  speed of tr: (5L/12)/T = 5L/12T
  speed of hare: (L/4)/T = L/4T
  
  Distance remaining for tr= L/4. time it will take to cover= (L/4)/(5L/12T)=12T/(5*4)=3T/5 --- (1)
  Distance for hare = 3L/4. Time to cover = (3L/4)/(L/4T)= 3T ------------------- (2)
  
  For (1) and (2) to be same, hare will have to run 5 time faster. Now, with 5x its initial speed, it will take exactly the same time as tr. So it's a joint winner.
• 12 years agoHelpfull: Yes(16) No(7)
• its simple
  jus(1-1/3-1/4)*3*4=5
  
• 12 years agoHelpfull: Yes(8) No(12)
• x----->first distance
  y------>sec distance
  
  equation (1-y)(1-(x+y))/(1/y^2)
  ans 5.00
• 12 years agoHelpfull: Yes(6) No(7)
• pleaseeeeeeeee xplain it in simple format...
• 12 years agoHelpfull: Yes(5) No(4)
• (d) 5.00 is the answer
  
• 12 years agoHelpfull: Yes(4) No(7)
• pls explain it in a simple way...
  pls pls pls
• 12 years agoHelpfull: Yes(4) No(1)
• Ans is option d) 5
  let speed of hare and tortoise be H and T
  and dist of race is pi*100 simple take 300 (or take 100)
  first tortiose covers 1/3 that means it covers 100
  at the time of meeting hare covered 1/4 ie 75 and at the same time duration tortoise covered 300-75-100=125 hence the time is same 75/H1 = 125/T1
  this is eqn 1.
  after meeting, tortoise need to cover 75 but hare need to 225
  from this 225/H2 = 75/T2 and T2=T1 this is eqn 2
  by solving thesr two eqns we can get H2/H1= 5
• 12 years agoHelpfull: Yes(4) No(4)
• 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)
• 12 years agoHelpfull: Yes(3) No(2)
• can u explain the problem..?
• 12 years agoHelpfull: Yes(2) No(3)
• 3 is the answer...
  ....
  
• 12 years agoHelpfull: Yes(2) No(8)
• when tortoise cover 1/3 distance hare start the race..
  when hare cover 1/4 distance tortoise cover (3/4-1/3)=5/12 distace
  so the ratio of there speed is s(t)/s(h)=1/5
  therefore ans is 5
• 11 years agoHelpfull: Yes(1) No(1)