. A person run from A to B.He took ¼ of the time less to reach B when compare to run at normal Speed.Then how many percentage he has increased his speed?

• new speed be Y and old speed X
  
  .75Y=X
  
  Y=1.33X
  
  hence he increased his speed by 33.33%
• 13 years agoHelpfull: Yes(20) No(5)
• let normal speed be s1 and time be t1
  then s1=d/t1 ---eqn 1
  and according to ques
  new speed [increased one] 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%
• 12 years agoHelpfull: Yes(6) No(1)
• walking speed = x
  time = t
  
  running speed = y
  time = 3t/4
  
  distance is constant, dist = speed * time
  so, x*t = y*3t/4
  
  t gets cancelled
  so x=3y/4
  
  y=4/3x
  
  increase in speed = 4*100/3 = 133.33
  
  therefore the person has increased his speed by 33.33%
• 10 years agoHelpfull: Yes(3) No(0)
• as distance is constant, speed * time = constant. When time becomes 3/4th, speed becomes 4/3rd of original speed.
  increment = 1/3rd
  % = 33.33%
• 6 years agoHelpfull: Yes(0) No(0)