Posted by jamia on Thursday, March 13, 2008 at 6:25pm.
HINTS:
1--Since only one person can be permanently taken across at a time, we know that there will be 5 trips made in all; 3 trips bringing a person over and 2 trips where 1 person returns to get another person.
2--By definition, one of the trips across must take 10 minutes dictated by "d's" walking time.
3--Logically, "c and "d", will cross together taking 10 minutes to cross over.
4--This leaves "a" and "b", the two people taking the two shortest times, to arrange their crossings in such a way as to consume the remaining 7 minutes.
5--Four trips consuming 7 minutes can only be achieved by 1 + 2 + 2 + 2 = 7, not necessarily in that order.
6--Person "a's" crossing time can only result from "a" traveling alone since if he was traveling with "b", their crossing time would have to be 2 minutes.
7--Therefore, "a" must make one return trip by himself meaning that he must go across with "b" once and return by himself.
8--We now know that "a" and "b" make at least one trip across together taking 2 minutes, "a" makes one trip returning to the starting side, and "c" and "d" make one trip across taking 10 minutes, consuming in all 2 + 1 + 10 = 13 minutes.
9--With 4 minutes left, it is obvious that "b" accounts for the remaining two trips, one over and oneback.
10--It remains for you to determine the order of these crossings and how and when does person "b" make his trips across and back consuming the remaining 4 minutes.
I think you should now be able to arrange the trips in their proper order.