Puzzle: Crossing the English Channel

[TheOriginalHappyGoat]

Haven't posted one of these in a while, but Alex Bellos had a fun little puzzle today that wasn't too mentally taxing:

Six friends – Babs, Charles, Hattie, Joan, Kenneth and Sid – are going camping in France.

They are travelling across the Channel on a magic carpet (of course) 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 with no one on board, 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.)

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).

Now what is the least time, in minutes, required to fly all friends to France?

Here’s a tip: the answer to question 1 is less than 45 minutes. And the answer to question 2 is less than the answer to 1.​

As per usual, answers in spoilers, please.

 

[sglowrider]

During Covid times wouldn't it be better if they supported local businesses by taking a ferry or the train?

 

[Marvin the Martian]

Spoiler: Answer I hope

I believe the secret is to get the two slowest over once at the same time. I used the 1st initial for each

B 1
C 5
H 6
J 7
K 8
S 12

B and C over 5
B back 1
S and K over 12
C back 5
H and B over 6
B back 1
J and B over 7
B back 1
B and C over 5

Total time 43

 

[sdhoosier]

 

[76-1]

Unless they've hacked the French Air and Beach Defense systems they'll be downed by a Roland AA missile (SAM) within seconds of entering the French Air I D zone...

So the answer is: at least two individuals and one carpet will be destroyed in a flaming spiral and no one completes their flight...

 

[Marvin the Martian]

Unless they've hacked the French Air and Beach Defense systems they'll be downed by a Roland AA missile (SAM) within seconds of entering the French Air I D zone...

So the answer is: at least two individuals and one carpet will be destroyed in a flaming spiral and no one completes their flight...

Or the more likely possibility, France surrenders to the 2 people on a carpet.

 

[TheOriginalHappyGoat]

Spoiler: Answer I hope

I believe the secret is to get the two slowest over once at the same time. I used the 1st initial for each

B 1
C 5
H 6
J 7
K 8
S 12

B and C over 5
B back 1
S and K over 12
C back 5
H and B over 6
B back 1
J and B over 7
B back 1
B and C over 5

Total time 43

Nope.

Spoiler: Half way

You have the right strategy for Q2, but not Q1.

 

[Lucy01]

This is a really dumb and juvenile thread!

 

[outside shooter]

Spoiler: quick guess

B 1
C 5
H 6
J 7
K 8
S 12

B and C over 5
B back 1
B and H over 6
B back 1
B and J over 7
B back 1
B and K over 8
B back 1
B and S over 12

Total time 42

 

[Noodle]

Spoiler: Spoiler

First one:
B and S over 12
B back 1
B and K over 8
B back 1
B and J over 7
B back 1
B and H over 6
B back 1
B and C over 5
Total = 42 minutes (doesn't matter the order, as long as B goes on every trip)

Second one:
B and C over 3
C back 3
K and S over 13
B back 1
B and J over 9
B back 1
B and H over 7
B back 1
B and C over 3
Total = 41 minutes (you could do this in a variety of other ways as long as K and S make a trip together after B and C go over, with C making two trips over and one back, and B making four trips over and three back)

 

[TheOriginalHappyGoat]

Spoiler: Spoiler

First one:
B and S over 12
B back 1
B and K over 8
B back 1
B and J over 7
B back 1
B and H over 6
B back 1
B and C over 5
Total = 42 minutes (doesn't matter the order, as long as B goes on every trip)

Second one:
B and C over 3
C back 3
K and S over 13
B back 1
B and J over 9
B back 1
B and H over 7
B back 1
B and C over 3
Total = 41 minutes (you could do this in a variety of other ways as long as K and S make a trip together after B and C go over, with C making two trips over and one back, and B making four trips over and three back)

Good but you can actually make the second one even shorter.

 

[Noodle]

Good but you can actually make the second one even shorter.

Yep. I see that now.

Spoiler

It’s 39 minutes for the second one if B and C make 3 trips together, and each makes 2 return journeys. Other two trips are H+J and K+S