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?
a) 40 b) 44.4 c) 33.3 d) 22.2

• Let the distance be D
  time for normal run = t
  new time = t - t/4 = 3t/4
  normal speed, s = D/t or D = s*t ...(1)
  changed speed, s' = D/(3t/4) or D = s' *(3t/4) or D = 3t * s'/4 ...(2)
  from 1 and 2 ,we get
  s * t = 3t s'/4
  or s' = 4s/3
  or s' = 1.33333*s
  hence he have increase his speed by
  = [(changed speed - normal speed)/normal speed]*100
  = [(1.3333s - s)/s]*100
  = .3333 * 100
  = 33.33
  hence ANSWER is 33.3
  
  
• 13 years agoHelpfull: Yes(19) No(0)
• i have simple logic,1-(1/4)=(3/4),then reverse it so it is 4/3=1.333,
  1.33-1=33.33
• 13 years agoHelpfull: Yes(13) No(0)
• initialy---
  speed-s1 ,distance -d ,time -t1
  s1=d1/t1
  now speed-s2 ,distance - d, time- t2=(1-1/4)=3/4s1
  so, s1/s2=3/4.
  s2= 4/3s1 = 1.33s1
  % increase = (1.33s1 - s1)/s1 *100 =33.3%
• 13 years agoHelpfull: Yes(2) No(0)