Asked by vince

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
Answered by oobleck
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.
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