Did you solve it? Carry on camping | Mathematics

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Earlier today I set you the following puzzle:

Six friends – Babs, Charles, Hattie, Joan, Kenneth and Sid – are going camping in France. They are travelling across the Channel on a magic carpet that can take only two people at a time. So, in order for everyone to get across there will need to be 9 trips in total from their starting point in England: 5 across carrying two people, and 4 returning carrying one person. (Magic carpets cannot fly empty, which is why one person needs to travel back. All the friends are able to fly alone on the carpet if need be.)

The faster the carpet flies, the bumpier the ride. Babs can fly at a speed that gets her to France in 1 minute. Charles feels queasy at that speed: his maximum speed gets him to France in 5 minutes. The others can get to France without suffering debilitating travel sickness no faster than 6 mins (Hattie), 7 mins (Joan), 8 mins (Kenneth) and 12 mins (Sid).

1. What is the least amount of flying time, in minutes, required to get everyone from England to France without anyone suffering from queasiness? (When two people are on the carpet together, the crossing will take as long as it takes the slowest of the two.)

Solution 42mins

Perhaps the most obvious strategy is for Babs, who can travel fastest, to take everyone over one by one, returning by herself, since she will save everyone time on the return journey.

If she does this, she takes 5 min with Charles, and 6, 7, 8 and 12 mins with the others. Her total time in minutes is thus 5 + 6 + 7 + 8 + 12 + 1 + 1 + 1 + 1 = 42.

This is indeed the best strategy with the given times in the question.

2. Let’s say that the friends’ tolerance for travel-sickness means that their fastest journeys take the following times: Babs (1 minute), Charles (3 mins), Hattie (7 mins), Joan (9 mins), Kenneth (11 mins) and Sid (13 mins).

Solution 39 mins

If you tried the above strategy of Babs taking everyone over one by one, the total time would now be:

3 + 7 + 9 + 11 + 13 + 1 + 1 + 1 + 1 = 47 minutes.

This is not the best strategy any more. You know this because I told you that all the friends can get to France in less than the answer to question 1, which is 42 minutes. So we need to look for a better strategy.

The insight here is that the slowcoaches must go over together. If Sid will take 13 minutes, much better for Kenneth to go with him, which avoids the need for another traverse to take 11 minutes. So let’s put Sid and Kenneth on the same ride.

However, if Sid and Kenneth go on the first traverse, one of them will have to return to pick up someone else, which rather defeats the object. So let the fastest two go over first, leaving someone there to return with the carpet once Sid and Kenneth get there. In other words:

Step 1. Babs and Charles go over. Charles stays in France and Babs returns: 3 + 1 = 4 mins

Step 2: Kenneth and Sid go over, and Charles returns: 13 + 3 = 16 mins.

Step 3: Babs and Charles go over again, and Babs returns: 4 mins.

Step 4: Hattie and Joan go over, and Charles returns: 9 + 3 = 12 mins

Step 5: Babs and Charles go over: 3 mins.

TOTAL: 4 + 16 + 4 + 12 + 3 = 39 mins.

If you used the ‘slowcoaches together’ strategy for Q1, the total time would have been 46 minutes. So that strategy is not always the best one.

I hope you enjoyed today’s puzzle. I’ll be back in two weeks.

I set a puzzle here every two weeks on a Monday. I’m always on the look-out for great puzzles. If you would like to suggest one, email me.

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