Mr. David usually starts at 9:00AM and reaches his office just in time, driving at his regular speed.Last Wednesday, he started at 9:30AM and drove 25% faster than his regular speed. Did he reach theoffice in time?

(1)Last Monday, he started to his office 20 minutes early, drove 20% slower than his regularspeed, and reached his office just in time.
(2)Last Tuesday, he started to his office 10 minutes early, and reached the office 10 minutesearly driving at his regular speed.

• lets the distance traveled in journey by Mr. David is x and his regular speed is v
  and regular correct time interval he takes will be t=x/v
  
  and if he drive 25% faster then
  (v+v/4) = x/T
  from here time taken T = 4x/5v =4/5 (t) ...eq(1)
  
  and according to statement 1
  (v-v/5)=x/(t+20)
  => 4v/5 = x /(t+20)
  and from statement 2
  v= x/t
  from both relations
  4/5t=1/(t+20)
  5t=4t +80 => t=80 minutes so he takes 80 minutes to reach his office at time now
  from eq(1) above
  T=(4/5)*80
  T=64 minutes but he started after 30 minutes at 9:30 AM so he will be late because he will reach his office at 10:34 AM while he has to reach his office in 80 minutes
  from 9:00 AM to 10:20 AM wo he will be late by 14 minutes :)
• 11 years agoHelpfull: Yes(0) No(0)