a man is living his house at 7:oo AM and traveling at an average speed of 60 km/hr, arrives at his office10 mins. before the expected time. Had he left the house at 7:25 AM and traveled at an average motion of 75km/hr, he should have arrived 5 mins late than the expected time. How far is his office from his house? find also the expected time that he should be at the office

Question

a man is living his house at 7:oo AM and traveling at an average speed of 60 km/hr, arrives at his office10 mins. before the expected time.  Had he left the house at 7:25 AM and traveled at an average motion of 75km/hr,  he should have arrived 5 mins late than the expected time. How far is his office from his house? find also the expected time that he should be at the office

oobleck · Answer

If the distance is x km, then since time = distance/speed, x/60 - 1/6 = x/75 + 1/12 Solve for x. I suspect a typo in one of the specified times.