Question 763151: a man left his house at 7:00 AM and traveled at an average aped of 60kph. He arrived at his office 10 minutes before the time he was expected to report. Had he left at 7:25 AM and traveled at an average speed of 75kph, he would have arrived 5 minutes after the expected time. How far his office from his house and at what time is was he expected to report?
Answer by josgarithmetic(39613)   (Show Source):
You can put this solution on YOUR website! Time number line using a variable to represent the point for when the man is expected to report. Call this time point p, units of hours.
DEPARTURE__________speed__________time(hours)___________distance(km)
7:00AM_____________60____________ ________(___)
7:25AM_____________75____________ ____(___)
Note very carefully, that p is a point on a number line, for a point in time; NOT a size of a time increment, but a point. The units are in hours, so fractions were used adjusted for time values or expressions in HOURS. We do not yet know the distance. We only need to know (or assume) that the distance from home to office is the same for each departure time. Once p is found, we can find travel times, and then distance.
We can form this equation, equating the distances of each departure time situation:
First, distribute the constants factors over additions on both sides, and continue simplifying.
 hour point on the number line for time.
INTERPRETATION:  hour length was the expected travel time. Actual time spent traveling was 1:00-00:10 = 00:50, meaning 50 minutes, meaning (5/6) hour.
rate*time=distance....  kilometers
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