39 replies to this topic

[wolfgang]

Two families(1 and 2),each consists of 5 persons(Father,Mother, Son,Daughter,and Baby).
Family 1 is at the right side of the river®,and Family 2 at the left side of the river(L).
They want to change places.
There is only one boat with its driver,the boat will not cross the river unless there are three persons on board(baby is one person).
Each father should pay 5$ for each single cross, each mother 4$,each son 3$,each daughter 2$,and each baby 1$.
Both families have a total of 33$.
Any baby should be accompanied with any one of his family.
How can they change their places using that amount of money?
they should use the boat only.

[thoughtfulfellow]

You are going to make me lose sleep on this one! My first try got the total to $34. Will work on it more. I am assuming that you mean the driver plus exactly three passengers and, since the baby must travel with a family member, that it also must have a family member on the side with it.

[Smith]

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!

[Smith]

If the boat driver is included in the three total passengers on each boat trip, the solution still costs way less than the $66 total available:

Spoiler for Driver plus 2 Passengers Solution

Once again, 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 $3 for Daughter and Baby to cross.
SF SM SS * SD SB JF JM JS JD JB
Jones pays $3 for Daughter and Baby to cross.
SF SM SS JD JB * SD SB JF JM JS
Smith pays $7 for Mother and Son to cross.
SF JD JB * SM SS SD SB JF JM JS
Jones pays $7 for Mother and Son to cross.
SF JM JS JD JB * SM SS SD SB JF
Smith pays $7 for himself and the Jones Daughter to cross.
JM JS JB * SF SM SS SD SB JF JD
Jones pays $7 for he and his daughter to cross.
JF JM JS JD JB * SF SM SS SD SB
Each has paid $17 and has $16 remaining (even better than before - now they all get fries!)

Edited by Smith, 14 January 2012 - 06:32 PM.

[mewminator]

Spoiler for maybe

the 2 fathers, mothers, and sons will cross once,one of the family members will need to cross 3 times and the cheapest individual moving alone is the daghter, the baby will cross once, putting that in an equation:
(2 x 5) + (2 x 4) + (2 x 3) + (3 x 2) + (2 x 1) = 32

I've yet to make that solution
@smith

Both families have a total of 33$.

not each family has 33$, 33$ is the maximum both families combined can pay.

Edited by mewminator, 14 January 2012 - 06:40 PM.

[Smith]

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.)

Edited by Smith, 14 January 2012 - 06:46 PM.

[Morningstar]

Wait...is it a combined total of $33 or $66?

[mewminator]

Spoiler for reason of imposibility

I forgot something the first time let me repeat:
the 2 fathers, mothers, and sons will cross once,one of the family members will need to cross 3 times and the cheapest individual moving alone is the daghter, in addition to the daughter of the other family, the babies will cross once, putting that in an equation:
(2 x 5) + (2 x 4) + (2 x 3) + (4 x 2) + (2 x 1) = 34

I missed something
I guess I need more than maths to solve this one
@smith 3 passengers and the money is payed to the driver
@moningstar $33 is the maximum possible you can pay, otherwise ther would be no riddle

Edited by mewminator, 14 January 2012 - 06:54 PM.

[ak4su]

Spoiler for Maybe hint

Well the riddle dosen't mention that the driver has to be in the boat ... try solving it from there.

[Smith]

In light of ak4su's comment, I think a clarification is in order...

Spoiler for Clarifying Question which holds a minor clue, so look at your own risk!

Does the boat hold 3 or 4 total bodies?


AND - IF (and that is a big IF) the owner of the boat is going to allow another person to drive his boat, I think it is obvious that the boat and its owner must end up on the same side of the river in the end.

Now a greater, more philosophical (and dangerous to my peaceful existence here at Brain Den) question becomes:

Spoiler for Dangerous Question for Smith to even ASK

"would the boat owner allow a female to drive his boat?"