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

• actual late = 20 min. Distance = speed * time. d=3(s/4) * (t+20) . so, t*s=3(s/4)*(t+20) => t=60min.
• 11 years agoHelpfull: Yes(26) No(3)
• 60 min z crrct..
• 11 years agoHelpfull: Yes(3) No(3)
• she started 30 minutes late and reached the office 50 minute late
  started late(30 min)+travelling time(20 min)=50 min late
  so it takes 20 minutes reach the office when she is 25% slow
  =>(20)*(25/100)=5 min
  so it takes 5 min more when she is 25% slow
  so
  actual time-5=20-5=15
  it takes 15 min to each the office
  
• 11 years agoHelpfull: Yes(3) No(3)
• 60 min is correct
• 11 years agoHelpfull: Yes(2) No(1)
• speed becomes 3/4 of original...thus time will become 4/3...
  increase in time=1/3 of original...
  thus given 1/3 of time=20 minutes
  thus time = 60 minutes......this solution is given in arun sharma
• 11 years agoHelpfull: Yes(2) No(0)
• is it this year question?
• 11 years agoHelpfull: Yes(1) No(4)
• I think 60min
• 11 years agoHelpfull: Yes(1) No(2)
• here we see that it takes her total (50-30)min when she is 25% slow
  i.e 20 min for 25% slow
  total time= 20/25 *100 = 80min
• 11 years agoHelpfull: Yes(1) No(4)