# Andy, Bob, Charlie & Doug

talzamir
Posers and Puzzles 17 Oct '11 09:39
1. talzamir
Art, not a Toil
17 Oct '11 09:39
Four guys have gathered for a get-together at point (0,0). They all live on small, separate islands, and call a boat to take them home. They agree to divide the costs of the boat fairly - that is, everyone pays costs that he causes. The cost is \$10 for the boat to show up at all plus \$1 per mile.

Andy lives at (0,3).
Bob lives at (4,0).
Charlie lives at (7,0).
Doug lives at (7,4).

The boat travels (0,0) -> (0,3) -> (4,0) -> (7,0) -> (7,4)

The route length is 3 + 5 + 3 + 4 = 15 miles and the total costs therefore \$25.

How much should each person pay?
2. 17 Oct '11 10:08
Think this is right:
Andy:
\$4.22

Bob:
\$5.85

Charlie:
\$6.82

Doug:
\$8.11

Like the problems, keep 'em coming.
3. 17 Oct '11 10:251 edit
I get the following:
A :
\$4.54

B :
\$5.22

C :
\$7.26

D :
\$7.98

This was based on :
splitting the fixed cost 4 ways, and allocating the \$15 to each person using their theoretical mileage as a proportion of a theoretical total.
4. 17 Oct '11 10:45
Originally posted by andrew93
I get the following:
A : [hidden][/hidden]
B : [hidden][/hidden]
C : [hidden][/hidden]
D : [hidden][/hidden]
This was based on : [hidden][/hidden]
My explanation of how I got to my answers.

I judged splitting the fixed cost 4 ways wasn\&#039;t fair as some travelled further than others, mine is based on the relative cost each one incurred.

E.g. The cost while A is on the boat is \$13, summing the cost for everyone like this I got 77. Hence I had A pay 13/77 of the total cost.
5. 17 Oct '11 12:21
A :
\$5,24

B :
\$5,64

C :
\$6,85

D :
\$7,28

They fight for the \$0,01 they paid too much.
6. 17 Oct '11 17:487 edits
Originally posted by VelvetEars
My explanation of how I got to my answers.

[hidden] I judged splitting the fixed cost 4 ways wasn\'t fair as some travelled further than others, mine is based on the relative cost each one incurred. [/hidden]

[hidden] E.g. The cost while A is on the boat is \$13, summing the cost for everyone like this I got 77. Hence I had A pay 13/77 of the total cost. [/hidden]
A causes more than 13/77 though.

From 0,0 to 0,3 is three miles to 4,0 is five more miles. A causes an extra mile to get to B's house.
If you cut out A that is four dollars less that needs to be paid.
After cutting out A then B doesn't cause any mileage because they have to pass by B's house to get to C's house.
C causes 2 more dollars because if C were cut out it would only be 5 miles from B to D.

This is harder than you guys make it seem. I think you have to start at the end and work backwards. Eg the cost to go straight to 7,4 is 8.06, the cost to go straight to 7,0 is 7, the cost to go straight to 4,0 is 4 and the cost to go straight to 0,3 is 3; then work forwards and find how much more each added. Eg A added 1 B added 0 C added 2.

This would be A = 4 B= 4 C=9 D=8

I like this answer (though the method can be perfected) as nobody has to use any change. ðŸ™‚

Edit: using a different and probably more accepted method I get

A=5.24
B=5.65
C=6.85
D=7.26

a = the cost of A if he were alone
b= the cost of B if he were alone
c= the cost of C if he were alone
d= the cost of D if he were alone

a = 13
b = 14
c = 17
d = 18

I rounded to the nearest dollar.

a+b+c+d = 62

[a/(a+b+c+d)]*25 = A
[b/(a+b+c+d)]*25 = B
[c/(a+b+c+d)]*25 = C
[d/(a+b+c+d)]*25 = D

I rounded to the nearest hundredth.
7. 17 Oct '11 17:53
Originally posted by talzamir
Four guys have gathered for a get-together at point (0,0). They all live on small, separate islands, and call a boat to take them home. They agree to divide the costs of the boat fairly - that is, everyone pays costs that he causes. The cost is \$10 for the boat to show up at all plus \$1 per mile.

Andy lives at (0,3).
Bob lives at (4,0).
Charlie lives a ...[text shortened]... 3 + 5 + 3 + 4 = 15 miles and the total costs therefore \$25.

How much should each person pay?
Here's way that could be just a beginning of the argument.

A causes B to go an extra mile. He pays that for B. So he pays for 4 miles or \$6.67.

B causes no one to go more miles so he pays the 4 miles he caused or \$6.67.

C causes no one to go more miles so he pays the 3 miles he caused or \$5.00.

D pays for his 4 miles so pays 6.67.

This is just a beginning if we get into causing someone to pay for fewer miles because it can also be said that B saves both C and D 4 miles!
8. 17 Oct '11 18:05
Originally posted by JS357
Here's way that could be just a beginning of the argument.

[hidden] A causes B to go an extra mile. He pays that for B. So he pays for 4 miles or \$6.67.[/hidden]
[hidden] B causes no one to go more miles so he pays the 4 miles he caused or \$6.67.[/hidden]
[hidden] C causes no one to go more miles so he pays the 3 miles he caused or \$5.00.[/hidden]
[hidde ...[text shortened]... to pay for fewer miles because it can also be said that B saves both C and D 4 miles![/hidden]
Correct but C does cause D to go 2 more miles from B's house.
9. talzamir
Art, not a Toil
17 Oct '11 18:271 edit
My version - no more nor less true than any of the above.

All four contribute equally to the need to call a boat in the first place so that's \$2.50 apiece.

O->B direct route 4 miles vs O->A->B 3+5 = 8 miles so A causes +4 miles
A->C direct route sqrt 58 miles vs A->B->C = 8 miles so B causes about +0.4 miles
B->D direct route 5 miles vs B->C->D = 7 miles so C causes +2 miles as tomtom noted
the boat would others stop so the last +4 miles are due to D.

Dividing the distance-based \$15 relative to these miles and adding the \$2.50 would give

A pays \$8.28, B pays \$3.06, C pays \$5.39, D pays \$8.28
10. 17 Oct '11 18:511 edit
Originally posted by tomtom232
Correct but C does cause D to go 2 more miles from B's house.
Without extra cost to D.
11. 17 Oct '11 18:521 edit
Originally posted by talzamir
My version - no more nor less true than any of the above.

All four contribute equally to the need to call a boat in the first place so that's \$2.50 apiece.

O->B direct route 4 miles vs O->A->B 3+5 = 8 miles so A causes +4 miles
A->C direct route sqrt 58 miles vs A->B->C = 8 miles so B causes about +0.4 miles
B->D direct route 5 miles vs B->C->D = 7 ...[text shortened]... the \$2.50 would give

[hidden]A pays \$8.28, B pays \$3.06, C pays \$5.39, D pays \$8.28[/hidden]
That is a neat way of solving it but if I were A I wouldn't accept simply because all the others are saving >10 dollars on what they would have to pay if they were alone (except D who is close to 10) while A is saving <5.

If I were A I would say, "Party at my house! We just have to pay equal parts cost of the boat ride"

Then it is a simple 3.25 each. ðŸ™‚
12. 17 Oct '11 19:056 edits
Originally posted by JS357
Without extra cost to D.
It is a very convoluted problem because if you eliminate everybody's route and go straight to D it is about 8.06 miles; If you only eliminate B and C then from A it is about 7.07 + 3 = 10.07 miles; if you only eliminate C then it is about 3+5+5 = 13 miles; and if you eliminate nobody then it is 15 miles.
Add 2.50 to each of these numbers and you get:

A= 4.51
B= 5.43
C= 4.50
D= 10.56

I am sure there are many many ways to do this as you could argue any version put forward.

Edit: There is probably a "fair" but different version for each where Version A would be the version where A saves the most money, Version B would be the version where B saves the most etc. A version AB; A version BC; A version CD etc.
13. talzamir
Art, not a Toil
17 Oct '11 20:43
Without a doubt there is. Andy would certainly not like my version much, and "party at my house" becomes very appealing, unless of course Bob, Charlie, and Doug would be likely to cause so much damage in the process that a boat cab fare pales in comparison.. making it "I'll pay the whole \$25, just don't step off on my isle until you're sober" one possibility. As Bob, Charlie, and Doug would get home for free and Andy save a fortune, perhaps that too would in a way to 'fair'.

I've had this problem as a conversation starter in some math classes, and it serves very well at that. Incidentally, I showed this to a puzzle-loving friend of mine who came up with costs for A..D that are perfectly plausible and different from any of the above.
14. 17 Oct '11 21:02
Originally posted by talzamir
Without a doubt there is. Andy would certainly not like my version much, and "party at my house" becomes very appealing, unless of course Bob, Charlie, and Doug would be likely to cause so much damage in the process that a boat cab fare pales in comparison.. making it "I'll pay the whole \$25, just don't step off on my isle until you're sober" one possibilit ...[text shortened]... with costs for A..D that are perfectly plausible and different from any of the above.
Yes, this isn't a math puzzle; it's a puzzle in distributive justice.
15. forkedknight
Defend the Universe
17 Oct '11 21:10
I would just divide what it would cost each of them to rent their own boat by the total of what renting 4 different boats would and have each pay that percentage of the fare.