a hare and tortoise hav a race along a 100 yards diameter .tortoise goes in one n hare in other.hare starts aftr tortoise has covered 1/4 of the distnce.they meet when hare has covered only 1/5 of the distance.by wat factor should hare increase his speed so as to tie the race.?

• 11 times.
  Assume the total distance to be 200.
  distance covered by tortoise before hare start the race - 50.
  so, now remaining distance to be covered by tortoise is 150.
  1/5 of the distance ( 40 ) , which is covered by hare, when they meet, In the mean time distance covered by tortoise is 110.
  Ratio of speed of tortoise and hare- 11:4
  Remaining distance to be covered by tortoise to end the race - 40.
  Time taken by Tortoise to cover remaining distance - 40/11.
  Speed to maintained by hare to tie the race (x)-> x*40/11=160;
  x=4*11;
  x=44;
  so hare has to increase his speed by the factor of 11 to tie the race.
• 11 years agoHelpfull: Yes(24) No(6)
• as given,remaining distance cover for tortoise = 1/5 of total distance.
  remaining distance cover for hare = 1-1/5 = 4/5 of total distance.
  so according to question,
  by wat factor should hare increase his speed so as to tie the race,
  let factor = x
  so, x*1/5 = 4/5
  hence, x=4
  i.e by a factor of 4.
• 11 years agoHelpfull: Yes(22) No(6)
• 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
  
• 11 years agoHelpfull: Yes(15) No(21)
• 4 times
  is it correct??
• 11 years agoHelpfull: Yes(10) No(5)
• 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
• 11 years agoHelpfull: Yes(2) No(2)
• 10 times..
• 11 years agoHelpfull: Yes(1) No(8)
• i think 4 times
• 11 years agoHelpfull: Yes(1) No(0)
• when hare complete its 1/5th=20 yards both are meet that means hare complete its 4/5th =80 yards
  so hare is 20 yard complete
  tortoise is 80 yard complete
  so if hare have to finish as same tortoise so it cover remaining 80 distant in same time as tortoise
  so
  80/x=20/y
  x=4y.
  x= speed of hare
  y=speed of tortoise
• 10 years agoHelpfull: Yes(1) No(0)
• solving gives that tortoise is already 3 times faster than hare.
  moreover, many friends have solved for the factor of 4.
  
  Hence, hare has to increase the speed by 4 times to beat distance and 3 times to beat the speed of tortoise.
  Therefore, answer could be...4X4=12 times the current speed.
• 11 years agoHelpfull: Yes(0) No(6)
• take 100m circum
  hare cover 20m and tortoise cover 55m so we can take tortoise speed 55m/sec and hare speed 20m/sec
  now to tie the match hare have to cover 50m in time of in which tortoise cover 25m
  so time taken by tortoise=25/55
  x*(25/55)=50
  x=110 becoz hare previously speed was 20m/sec
  and know hare have to increase up to 110m/sec
  110/20=5.5
  
• 11 years agoHelpfull: Yes(0) No(0)