22. One day, Eesha started 30 minutes late from home and reached her office 50 minutes late while driving 25% slower than her usual speed. How much time in minutes does Eesha usually take to reach her office from home?

• let her usual speed be 100km/h
  D=S*T => D=100*T
  now she travels 25% slower taking 20mins extra for same distance
  therefore
  D=75(T+20)
  equate both eqns
  => 100T=75(T+20)
  => T=60mins
• 10 years agoHelpfull: Yes(45) No(1)
• We know that Speed is inversely proportional to time
  While she drives 25% slower means she drove at 34(S)
  We know that D = S x T
  When speed became 3/4(S) then Time taken should be 4/3(T)
  i.e, She has taken 4/3(T) - T extra to cover the distance.
  Extra Time = T/3 = 20 min (as 20 min late due to slow driving)
  Actual time T = 60 Minutes
• 10 years agoHelpfull: Yes(13) No(1)
• 25% slower means 3/4 of his usual. speed
  now he lates on road by (50-30)= 20 min
  when he drives 3/4 of his usual speed he reached 4/3 of his usual time
  
  4/3 of usual time -usual time= 30
  1/3 usual time= 30
  usual time=(30*3)=90 min
• 10 years agoHelpfull: Yes(1) No(6)
• let the speed be x km per min and the time taken to reach the office be y min.
  distance of office from home=xy
  now,at 25% slower speed distance=.75x(y+20).
  Then,xy=.75x(y+20)
  on solving we get y=60 min
• 10 years agoHelpfull: Yes(1) No(0)
• d/.75x = 20 min
  d/x = .75*20 =15 min
  usual time is 15 min
• 10 years agoHelpfull: Yes(1) No(0)
• we know if speed is reduced by 25% then time increases by 33.3% or by 1/3.
  so 1/3 of usual time=20 min gives us usual time as 60 min...so the ans is 60 min
  
• 10 years agoHelpfull: Yes(1) No(0)
• Let speed be x taking time t1
  Then as she is late speed be3x/4.
  Distance is same let d.
  T1×3X/4=T2×X
  T1 is 20 min.solving we will get 15 min
  
• 9 years agoHelpfull: Yes(1) No(0)
• Here the 30 min data is the redundant data.
  After Eesha started from home then only the speed and time will be calculated .
  And as she reached office 50 min late so let the actual time be t,then she reached at t+50 min.
  Let the speed be x,so 25%of x means 1/4 of it.So the speed on that day is 3x/4.
  As distance is constant so speed is inversely proportional to time.
  x/(3x/4)=(t+50)/t.
  solving t=150min. ANS....
• 9 years agoHelpfull: Yes(1) No(0)
• 60 MINUTES
• 10 years agoHelpfull: Yes(0) No(0)
• santanu equate ur eqn with 20
• 10 years agoHelpfull: Yes(0) No(0)
• 3 hours 20 minutes
  If we take normal speed = x mtr/minute then reduced speed will be 3x/4 mtr/minute .. so let the total distance is 100 metere so 100x = (3x*100/4)+50 so x = 2 total time taken 2x100=200 minutes = 3 hours 20 minutes
• 10 years agoHelpfull: Yes(0) No(0)
• Usual speed=s
  Speed at that day=0.75*s
  total extra time=50 min.
  since she started 30 min.late extra time in traffic=50-30=20
  100*d/(75*s)-d/s=20
  (d/s=t)
  25*t/75=20
  t=60 min.
  0
• 10 years agoHelpfull: Yes(0) No(0)
• ans is 15 min.
  
• 10 years agoHelpfull: Yes(0) No(1)
• let distance be x and speed be k so time taken = x/k
  now new speed = 75%of k = 0.75k
  so new time taken = x/0.75k
  A/Q x/k + 20 = x/0.75k since se started 30 min late and reached 50 min late so delay = 50-30 =20.
  Solving the eqn we get x/k = 60
• 10 years agoHelpfull: Yes(0) No(0)
• .75*x*d=20min
  x*d=20/0.75=40min ans
  
• 10 years agoHelpfull: Yes(0) No(0)
• guys, i think she took 20 minutes to reach her office due to slow driving. so if she was driving at her usual speed the time taken should be less than 20, right??
  anyway i got the answer as 15 minutes.
• 10 years agoHelpfull: Yes(0) No(0)
• 60 min
  assume the usual speed is x kmph and covered distance= d time taken =t hour
  now she takes 20 min more i.e. 1/3 hour more
  t=d/x ------------(1)
  t+ 1/3 =d/(x-(x/4))------------(2)
  comparing two equations for value of t
  x=1 i.e time taken in minutes =1*60= 60 min
• 10 years agoHelpfull: Yes(0) No(0)
• let the usual time taken be x
  x*75%=20 min
  x*(3/4)=20min
  x= 80/3
  x=26.66min
• 9 years agoHelpfull: Yes(0) No(0)