A person runs 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?

• 33.3333% he has increased his speed.
  
  
  Let, Distance(d) = Speed(s)*Time(t)
  
  d =s*t
  When t1 =(3/4)t
  Then,
  d = s1*t1
  
  Apply t1 value,
  s*t = s1*(3/4)t
  Then
  s1/s =4/3
  
  So, increase in speed = (s1/s - 1)
  =4/3 - 1
  =1/3
  =33.33%
• 12 years agoHelpfull: Yes(11) No(2)