An athlete decides to run the same distance in 1/4th less time that she usually took. By how percent will she have to increase her average speed

• let be the original time s1,s2 be the two speeds
  now new time=t-t/4=3t/4
  now distance remaining same
  t*s1=(3t/4)*s2
  s2=4/3s1=(1+1/3)s1
  so the athlete should increase his speed by one-third
• 13 years agoHelpfull: Yes(46) No(2)
• let original speed be s1 and time be t1
  then s1=d/t1 ---eqn 1
  and according to ques
  new speed be s2 and time given is 3t1/4
  therefore s2=d/(3t1/4) -----eqn 2
  dividing eqn 2 by eqn 1
  s2=4s1/3
  increased speed = 4s1/3-s1
  =1s1/3
  percent increase=[(1s1/3)/s1]*100
  =33.33%
• 13 years agoHelpfull: Yes(28) No(1)
• ane is 33.33%
• 13 years agoHelpfull: Yes(5) No(6)
• Let orginal time is 4 min. and now new time is 3 min acc. to question
  
  so increase speed is = (d/4-d/3) / (d/4) = d/3 its means 33.33%
• 9 years agoHelpfull: Yes(1) No(0)