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
• 9 years agoHelpfull: Yes(27) No(4)
• Easha need total 30+50min=80 min to reach her office for that day.She got 50-30 min =20 min late to reach her office,so we just simply substract 80min-20 min=60 min
  and thus she takes 60 min usaully to reach her office.
• 9 years agoHelpfull: Yes(10) No(3)
• Let distance=y and his usual speed=x
  new speed=x*75/100=3x/4.
  Nw distance is same but time diff=20min
  time=distance/speed
  t2-t1=20
  y/(3x/4)-y/x=20=>y=60x
  now usual time=y/x=>60x/x=60min
  
• 9 years agoHelpfull: Yes(4) No(1)
• Eesha usually takes time to go her house to office takes 20min in 25% speed. Eesha maintains as usual speed she can reach office 5min only.
• 9 years agoHelpfull: Yes(1) No(11)
• 25% --> 20 then
  100% --> ??
  ==> (20*100)/25
  ==> 80 min
  
• 9 years agoHelpfull: Yes(1) No(3)
• actual delay by Esha = 50-30=20 min
  since v = k*1/t
  v->75/100v
  =>v1/v2=t2/t1
  =>4/3=t/t+20
  =>t=60 min answer
• 9 years agoHelpfull: Yes(1) No(1)
• 25%slow=20min late;as he start 30 min late n rich 50min let
  so 100%=1hr 20 min
• 9 years agoHelpfull: Yes(0) No(6)
• as speed is getting reduced by 1/4.so time will increase by 1/3.so T/3 =20 and T = 60 mins.
  using % eqivalence method
• 9 years agoHelpfull: Yes(0) No(1)
• D=S*T
  S*T=(S*0.75)*(T+20)
  .25*S*T=15*S
  T=15/.25
  60
• 9 years agoHelpfull: Yes(0) No(1)
• Here the speed is 75/100s=3s/4
  now time = 4t/now
  4t/3-1=20
  so t=60 min
  Hope u got it
• 9 years agoHelpfull: Yes(0) No(1)
• distance is constant:
  so
  s*t=3/4(s)(t+20)
  st=(3st+60s)/4
  4st=3st+60s
  : . st=60s
  ---> t=60 min
• 9 years agoHelpfull: Yes(0) No(0)