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# Crossing the River

## 40 posts in this topic

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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 riverĀ®,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.

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Posted · Report post

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.

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

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!

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Posted (edited) · Report post

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:

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
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Posted (edited) · Report post

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
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Posted (edited) · Report post

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
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Posted · Report post

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

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Posted (edited) · Report post

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
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Posted · Report post

Well the riddle dosen't mention that the driver has to be in the boat

... try solving it from there.
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In light of ak4su's comment, I think a clarification is in order...

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:

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

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Posted · Report post

three persons on board

What do you think he means

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In that case, try this solution:

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

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Posted · Report post

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

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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!

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Posted (edited) · Report post

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

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
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Posted (edited) · Report post

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
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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

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Posted · Report post

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

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Some inconclusive analysis

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.

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Posted (edited) · Report post

Some inconclusive analysis

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.

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

Edit: typo.

That is the exact problem I was trying to get around. I also note the implication that the sum should be exactly \$33. To me this suggests that the a daughter makes extra trip back to her original side but somehow a baby makes an extra trip also. I haven't given up looking for a solution, just wonder what I am missing. Wolfgang does offer challenging puzzles!
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Posted · Report post

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

the 2 families are related...say the mothers are sisters, hence they have different names, etc.

That way, the baby could travel back and forth with "a family member!" i.e. Aunt, Uncle, Cousins!

Otherwise, I see no way that the puzzle can be solved with the rules stated for under \$34.00!

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the 2 families are related...say the mothers are sisters, hence they have different names, etc.

That way, the baby could travel back and forth with "a family member!" i.e. Aunt, Uncle, Cousins!

Otherwise, I see no way that the puzzle can be solved with the rules stated for under \$34.00!

...also the boat driver can be a member of a family - a brother or father of any of the parents, which makes him an uncle or grandfather of the baby.

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...also the boat driver can be a member of a family - a brother or father of any of the parents, which makes him an uncle or grandfather of the baby.

And to take the concept one step farther, the Captain of the boat could MARRY the son of one family to the daughter of the other so that a baby could travel with an aunt or uncle.

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let the members be F1,M1,S1,D1,B1 & F2,M2,S2,D2,B2

1) F1,M1,S1 5+4+3=12

2) F2,M2,B2 5+4+1=10

3) D1,B1,B2 2+1+1=4

4) S2,D2,B2 3+2+1=6

total money used 12+10+4+6=32\$ , 1\$ remaaining

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