39 replies to this topic

[ak4su]

three persons on board

What do you think he means

[Smith]

In that case, try this solution:

Spoiler for Yet another attempt

Once again, we see the Smiths on one side and the Jones family on the other. The boat and its owner are on the Smith side of the river.
Mr. Smith, being the great mathematician that he is, realizes that if he plays "by the rules" and can only send 2 passengers on the boat at a time that they'll never get the job done for less than $34.
So, he offers the boat owner an additional $3 if he will allow his son and Mr Jones' son to drive his boat across the river and back. Relieved that Mr. Smith did not ask him to allow the daughters of these men to Captain his fine craft, he agreees and the transfer runs as follows:


SS, SD, SB cross first. JS JD JB return the craft, Now the captain takes his boat across with SF and SM, then returns with JF and JM. Total cost: 2x5 + 2x4 +2x3 +2x2 + 2x1 + 3 = $33

[krisha.patel0110]

First off, does the driver have to be in the boat when it crosses and does the boat hold 3 or 4 ppl?

[perry1843]

Well, I'm sure that there is more to this than is posted, but..

Based on the given information and the way it is stated, I get the impression the Boat Driver has a 3 person minimum for making the trip worth his while. After all, I don't see where it says that 3 is the MAX that he will/can take? That being said, would not each family board the boat on their respective sides, pay the driver the $15 for his services and disembark on the opposite sides of the river, each with $1.50 in hand?

Please let me know what I have missed in the puzzle, because it appears pretty straight forward.
Thanks!

[thoughtfulfellow]

If it is the boat driver PLUS exactly three passengers, the solution is too easy and costs way less than the $66 total available:

Spoiler for Driver Plus 3 Passengers Solution

I name family 1 Smith, so we have SF for Smith Father, etc. Family two is Jones, so we have JM for Jones Mother, etc.
Begin with Smiths plus the boat on the left side, Jones family on the right.
SF SM SS SD SB * JF JM JS JD JB
Smith pays $6 for the Son, Daughter, and Baby to cross.
SF SM * SS SD SB JF JM JS JD JB
Now Jones pays $6 for their Son, Daughter, and Baby to cross.
SF SM JS JD JB * SS SD SB JF JM
Smith pays $13 for the Smith Father and Mother plus the Jones Daughter to cross.
JS JB * SF SM SS SD SB JF JM JD
Now Jones pays $13 for Father, Mother and Daughter to cross.
JF JM JS JD JB * SF SM SS SD SB
$19 each paid, $14 left in pockets for Dinner at McDonalds!

Well, given thoughtfulfellow's response, I now must assume that wolfgang meant that there was a GRAND TOTAL of $33 between the two families' funds. Therefore, my solutions cost $38 and $34, respectively. That also means that thoughtfulfellow spent $4 less on his "Driver + 3 Passengers" solution (I'd like to see it). Oh, well. Back to the drawing board.

(and, post-posting, I see mewminator's comment to the same effect. Thank you, mewminator.)

If you double check your first solution third and forth trips, the steps I changed to red, F M & D only total a cost of $11 so your solution is mine for achieving $34. I did trips in slightly different order but with same results

Edited by thoughtfulfellow, 15 January 2012 - 10:22 AM.

[wolfgang]

The total sum is 33$ for the both two families( i.e.the money ,33$,are devided between them,...somehow...)
The boats driver must be always on his boat.
The boat can take only four persons each time( no more and no less),i.e.the boat driver and exactly three persons.
Each baby must be accompanied with ,at least,one person of his own family.

Edited by wolfgang, 15 January 2012 - 12:13 PM.

[avaidhyan]

Ok guys I could get the solution with a spend of $32
Consider the families as denoted by their first letters as follows : Side1 (f, m, s, d, b) and Side2 (f', m', s', d' b')

Here is the sequence to follow :
1) s,d,b moves from Side1 to Side 2
Cost incurred = $6

2) s', d', b' moves from Side2 to Side1
Cost incurred = $6

3) f, m, b' moves from Side1 to Side 2
Cost incurred = $10

4) f', m' , b' moves from Side2 to Side1
Cost incurred = $10

Total cost incurred = $ 6+6+10+10 = $32

[ak4su]

The problem is only with the third one, baby has to be accompanied with a family member.

[thoughtfulfellow]

Some inconclusive analysis

Spoiler for What I see

1. If we had no restrictions as to number of pasengers, minmum crossing fee would be $5+$4+$3+$2+$1 = $15 for each family. A total of $30.
2. In order to insure 3 passengers on every trip, someone must travel across the river, back to original side and another trip to get to the desired side.
3. With the $33 limit, this person crossing both ways extra must be the baby.

[Morningstar]

Some inconclusive analysis

Spoiler for What I see

In order to insure 3 passengers on every trip, someone must travel across the river, back to original side and another trip to get to the desired side.
With the $33 limit, this person crossing both ways extra must be the baby.

Spoiler for but

If it's the baby, doesn't that present a problem because the baby has to travel with a family member?

Edit: typo.

Edited by Morningstar, 15 January 2012 - 04:31 PM.