If a person walks at 4/5th of his usual spee he reaches 40min late. If he walks at his usual speed how much time does he travels.

• If he walks at 4/5 of his usual speed he will take 5/4 of the usual time as speed and time are inversely proportional
  
  Hence he is taking (1/4) of the time extra coz of which he is late
  
  hence 1/4 of usual time = 40 min
  usual time taken = 160 min
• 13 years agoHelpfull: Yes(29) No(7)
• Ans. 160 minutes.
  
  Let his usual speed is x and time taken is t minutes,
  so distance travelled is xt.
  If his speed is (4/5)*x then time taken is (t+40) minutes to cover the same distance.
  therefore (4/5)*x (t+40)= xt => t/5 = 160/5
  => t = 160 minutes.
• 13 years agoHelpfull: Yes(22) No(5)
• 160 min or 2hr 40 min
  
  (5/4) x - x = 40 min
  
  1/4 x = 40 min
  
  so x=160 min.
• 13 years agoHelpfull: Yes(14) No(2)
• let d=distance
  s=speed
  t=time
  so,(4*s)*(t+40)/5=st
  4t+160=5t
  t=160min
• 13 years agoHelpfull: Yes(4) No(1)
• 160 minutes
  distance d=speed*time
  speed: y
  time:x
  normally d=xy;-------1
  in particular time increases 40 min ie x+40
  when speed=y(4/50
  but d remains same
  d=(x+40)(y4/5);------2
  from 1&2
  xy=(x+40)(y4/5)
  =((x+40)*y4)/5
  5xy=4xy+160y
  5xy-4xy=160y
  xy=160y
  cancelling y both sides
  x=160 minutes;
• 12 years agoHelpfull: Yes(4) No(2)
• Let d =dis. travelled (km)
  u=usual speed(kmph)
  t=time taken(hrs.) ....
  So. d/u =t...............(1)
  Also , d/(4u/5)=t+(40/60).......(2)
  
  So from 1 & 2 :
  d/u =8/3 or t=8/3 hrs = 8/3 * 60 min = 160 min
• 13 years agoHelpfull: Yes(3) No(1)
• Lets Say S is the Usual speed in KM per hour, T is the time in Minutes
  
  person walks at 4/5th of his usual spee he reaches 40min late means Ditance = (4S/(5*60))*(T+40) ------ 1
  
  he walks at his usual speed Distance becomes (S/60)*T -------2
  Equating 1 and 2 time does he travels becomes 160 mins
  
  
• 13 years agoHelpfull: Yes(1) No(4)
• speed(decrease) = 1- (4/5) = 1/5
  time(increase) = 1/(5-1)= 1/4
  1/4 = 40 mins
  1 = 160 mins
• 6 years agoHelpfull: Yes(0) No(0)