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
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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