I was going to take a trip to my friend's house in San Jose by airplane. If my airplane had arrived on time as scheduled, my friend would drive out to pick me up from San Jose's airport. She would get to the ariport just as I would be leaving it, and drive me to her house. This way, I would come to her house at 12:00.
But as it happened, my airplane arrived at the airport 1 hour earlier. So I walked from the airport directly to my friend's house. On the way to her house, I met her driving to pick me up. I got into the car, and we drove me to the house. This way, I arrived at the house at 11:40.
How long did I spend walking?
(The only thing known about my walking speed and my friend's driving speed is that they are different and they are constant.)
(In reply to
answer by K Sengupta)
Let the time taken by the friend to drive to the airport be t hours. Since, her driving speed is a constant, it must take her precisely the same time (t hours) to drive back to her house. Thus, the total time taken by the friend to drive to the airport and back is 2t hours.
Now, we know that it took the friend 12:00 - 11:40 = 20 minutes or one-third hour less to pick up the individual and drive back to her house, so that the total time taken for the onward truncated journey is (2t - 1/3)/2 = (t - 1/6) hours.
Thus, the friend's onward journey by car to the airport was shorter by one-sixth hour.
Therefore, if the individual had arrived at the airport precisely m hours before the designated time, then it follows that he would have to walk for (m - 1/6) hours, whenever m >= 1/6
By the problem, it is known that m=1, and consequently, the individual walked for (1 - 1/6) = 5/6 hours, or 50 minutes.
Edited on May 17, 2008, 7:48 am
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