A person run from A to B.He took 1/4 of the time less to reach B when compare to run at normal Speed.Then how many percentage he has increased his speed?

• Let the distance be constant = D (A to B)
  consider Speed is x.
  Thus time = T
  Secondly, Distance = D, Speed = x + y , Time= (3/4)T
  Thus, D= xT= (x+y)* (3/4)T
  or x=3y
  implies that previous his speed was 3y and later it becomes 4y.
  thus % increase =(4-3)/3 *100% =33.33 % Ans
  or D(y)
• 13 years agoHelpfull: Yes(14) No(0)
• let A to B time taken T,
  now A to B time taken (T-.25T)
  distance D=S*T
  by taking distance constant speed is (S+x)
  so by taking ratio speed should increase to 33.33%(percentage).
• 13 years agoHelpfull: Yes(1) No(0)