Travelling at 5/6th his normal speed, a man reaches his office 15 minutes later than usual. How long does he normally take to travel to his office? 1. 75 mins 2. 80 mins 3. 85 mins 4. 90 mins 5. 95 mins Method 1: Frame equations: Let the man’s speed be s, and the time taken to travel distance D be t. Distance =speed x time Distance =s x t Distance =(5/6)s x (t+15/60) (5/6)s(t+15/60)=st t = 15/12 hrs or 75 minutes. Method 2: When the speed reduces from s to (5/6)s , the time taken increases from t to (6/5)t. as s α (1/t) Difference in time is 15minutes (6/5)t  t =15/60 => t/5 = 15/60 => t = 5/4 hrs or 75 minutes. 
