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Layover at LaGuardia (Posted on 2008-02-29) Difficulty: 3 of 5
One morning, six travellers had to change planes at LaGuardia airport. Mr. and Mrs. Armstrong landed at 8:00, Mr. and Mrs. Borman at 8:10 and Mr. and Mrs. Collins at 8:15.

One of the men and one of the women departed at 8:45; another man and woman at 8:55 and the remaining two at 9:10.

Each of the six people was at the airport a different length of time, with more than one husband having a shorter wait than his wife.

At what time did each person depart?

  Submitted by Charlie    
Rating: 2.3333 (3 votes)
Solution: (Hide)
No man left with his own wife, as then they would have been on the ground the same length of time.

An Armstrong leaving at 8:45 would be incompatible with a Borman leaving at 8:55, and the same applies to an Armstrong leaving at 8:55 with a Collins leaving at 9:10, as that would also create situations of more than one person being on the ground for the same length of time.

This leaves only four orders of leaving for a given set of A, B and C of a given gender. Listed, with the corresponding number of minutes on the ground, they are:

A(45)  B(35)  C(30)  C(30)
C(40)  C(40)  A(55)  B(45)
B(60)  A(70)  B(60)  A(70)

The only pair of these lists that doesn't contain any duplicate is that consisting of the two middle lists: BCA and CAB.

In the latter list, C has a shorter time than C on the former list, as does A on the latter vs. A on the former. So CAB is the order of the men and BCA is the order of the women:

Mrs. Borman and Mr. Collins left at 8:45
Mrs. Collins and Mr. Armstrong left at 8:55
Mrs. Armstrong and Mr. Borman left at 9:10



Based on Enigma No. 1477, "In transit", by Susan Denham, New Scientist, 19 January 2008.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
AnswerK Sengupta2009-01-07 15:24:54
SolutionSolutionDej Mar2008-02-29 13:14:06
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