John covers a certain distance at 60 mph. He reaches the destination 5 mins late . The next day, he covers the same distance at 80 mph. He now reaches 10 minutes early. What speed should he take to reach the destination, such that he reaches at the correct time? (In mph) 1. 720 /11 2. 800/12 3.820/11 4.750/11 5. 650/11 Method 1: Distance =speed x time If speed=60 mph, time = t+5/60 If speed=80 mph, time = t10/60 The distance traveled is constant. Distance = 60 ( t+5/60) = 80( t 10/60) t= >55/60 = 11/12 substitute t in any one of the eq ; D = 60 ( 11/12 + 5/12) = 60 miles. If D=60 m S=60 mph then t= 1 hour If D=60 m, S =80 mph then t = 45 mins So John takes 55 mins to reach D = 60, t = 55,S = ? 60 = s x (55/60) S =720 /11 mph Method 2: If S = 60 mph, Ram is 5 mins late If S = 80 mph, Ram is 10 mins early If D is the distance, Then D/ 60 – D/80 = 15/ 60 D =60 m. If D = 60, and S =60 mph , Ram is 5 mins late, hence t =55 mins 60 = s x (55/60) S =720 /11 mph 
