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

a. 80min
b. 50min
c. 60min
d. 70 min

• distance remains constant.hence s*t=3*s*(t+20)/4
  hence t=60
• 9 years agoHelpfull: Yes(27) No(3)
• home---------------office(time taken =t,speed=u)
  one day
  time taken=t+20
  speed=3u/4
  t*u=(t+20)*(3u/4)
  solving t=60min
• 9 years agoHelpfull: Yes(19) No(0)
• 25% decrease in the speed results in 33.33% increase in time due to inverse relationship of them.
  so here 33.33%= 50-30 i.e 20 mins
  then 100% = 20*3 =60mins
• 9 years agoHelpfull: Yes(5) No(1)
• Let the distance be D and the usual speed S. Then the time taken to travel is D/S=T(1).The new speed is (S- 1/4 S) = ¾ S. The new time taken is D/3/4 S = T + 20 (50 – 30)(20) we get 60 as answer.
• 9 years agoHelpfull: Yes(4) No(6)
• c.60 min
  100/75=t+20/t
  t=60
• 9 years agoHelpfull: Yes(3) No(3)
• Speed is inversely proportional to time
  While she drives 25% ,i.e 1/4S ..slower means she drove at 3/4(S)
  
  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 (20 min late due to slow driving)
  Actual time T = 60 Minutes
• 9 years agoHelpfull: Yes(3) No(0)
• d=st s1*t1=s2*t2 100t=75(t+20) t=60min
  
• 9 years agoHelpfull: Yes(1) No(3)
• speed is 3/4 then time is 4/3 which is 1+1/3
  so 1/3 of actual speed =20
  speed = 60
  
• 9 years agoHelpfull: Yes(1) No(0)
• ans-b because
• 9 years agoHelpfull: Yes(0) No(3)
• 25%=20
  thn 100=80
• 9 years agoHelpfull: Yes(0) No(1)