A man leaves office everyday at 5 p.m. His driver leaves home everyday at such a time that he can pick up his master from the office at 5 p.m. The driver drives at a constant speed. Now one day, the man left office at 4.40 p.m. and started walking towards home. On that day too the driver had left home on such a time that he could pick up his master at 5 p.m. But since the man had started walking at 4.40 p.m. they met somewhere in between from where the driver took his master home. That day, the man reached home 10 minutes earlier. One week later, a similar thing happened, the man left office at 4.50 p.m. and met the driver on his way home from where the driver took his master. How much earlier did the man reach home the second day?
Solvers, Please show your methods, because answer is 5 minutes, as expected , but how you get it is of interest.
The main points are the bullets, a reasoning for each bulleted proposition is right below it.
The reason for this is that the driver's route was ten minutes shorter, which means he must have turned around 5 minutes earlier. Picking up the man five minutes earlier, plus a five minutes shorter trip back home totals 10 minutes.
What the man walked in 15 minutes (from 4.40 to 4.55), the driver was expected to drive in 5.
This means he has walked for 7.5 minutes, and the driver meets him two and a half minutes earlier than expected. This is in line with the argument above, that the driver drives three times the man's walking speed.
The driver picked him up two and a half minutes earlier than planned, and had two and a half minutes shorter drive home, which adds up to five minutes.