Can a monkey outrun a bullet and still save her litter?

[Soca fo so]

There seems to be no solution when you treat the three groups, men, women, children as distinct but if you consider that each child is also male or female then consider a female child as being half a woman so should be charged half the fare for a woman as well as the fare for a child so 5/8 + 1/5 dollars in total and similarly for a male child with the fare in this case being 5 + 1/5 dollars, then 4 male children plus 96 female children would give you 100 passengers and 100 dollars.

 

[DaveC426913]

there is! just do your best and see. if you can't, that does not mean that there is no solution.

Clearly there is s trick to it, as

10x + .2x + 1.25x = 100 does not have an integer solution.

 

[Soca fo so]

If you were to charge pregnant women the fare for a woman plus the charge for a child i.e. 5/4 + 1/5 dollars that would give a solution of 15 pregnant women, 21 non pregnant women, 60 children, and 4 men. The issue would be whether to count a pregnant woman as two passengers, but since they would be taking up one seat I would think they would be counted as one passenger.

Another thing that comes to mind is that perhaps there is a difference in the fare for a male child and a female child in the same way as for adults i.e. an 8 to 1 ratio male to female then let b be the number of boys and g be the number of girls, both natural numbers or zero, and let 1/f be the fare for girls, then the following two equations could be used to fix the two different fares for girls and boys:

b + g = 5

b*(8/f) + g*(1/f) = 1

these can be solved for f by noting that b must be a natural number or zero and must be less than or equal to 5 the first constraint means f must be equivalent to 12 mod 7 and the second constraint in addition to the first means that f must in fact be 12.

Then that leads to the following two simultaneous equations:

m + w + b + g = 100

10*m + (5/4)*w + (2/3)*b + (1/12)*g = 100

where m is the number of men and w is the number of women both natural numbers or zero.

If a solution exists it can be found by brute force I expect, I look forward to seeing the solution! (if it exists of course).

 

[DaveC426913]

If you were to charge pregnant women the fare

I would consider that a cheating answer. Charging a pregnant woman for her unborn child is unheard of.

 

[DaveC426913]

i have a bus that takes 100 passengers and i need only 100dollars id the bus is full. the fare for;
1 man is 10dollars
5 children is 1dollar
4 women is 5dollars
i need the bus to be full and i need 100dollars only out of it..

how many children, men and women should I've in the bus??

Perhaps we are overthinking the answer.

I can answer the question quite simply, really:

Q: how many children, men and women should I've in the bus??
A: 100.

 

[Soca fo so]

Another thing that comes to mind is that perhaps there is a difference in the fare for a male child and a female child in the same way as for adults i.e. an 8 to 1 ratio male to female then let b be the number of boys and g be the number of girls, both natural numbers or zero, and let 1/f be the fare for girls, then the following two equations could be used to fix the two different fares for girls and boys:

b + g = 5

b*(8/f) + g*(1/f) = 1

these can be solved for f by noting that b must be a natural number or zero and must be less than or equal to 5 the first constraint means f must be equivalent to 12 mod 7 and the second constraint in addition to the first means that f must in fact be 12.

I've made a bit of a mistake in the above; there are 6 possible values of f from those equations of which 12 is one.

Perhaps we are overthinking the answer.

I can answer the question quite simply, really:

Q: how many children, men and women should I've in the bus??
A: 100.

Yes of course!

 

[evra]

This might be a stupid question, but is the driver included in the 100 passengers?

no he's not!

 

[evra]

Clearly there is s trick to it, as

10x + .2x + 1.25x = 100 does not have an integer solution.

there is! very correct infact! try!

 

[evra]

Perhaps we are overthinking the answer.

I can answer the question quite simply, really:

Q: how many children, men and women should I've in the bus??
A: 100.

lol! nooo that's not it.

 

[I like Serena]

there is! very correct infact! try!

You skipped over the pregnant-woman-solution.
So is that it?

 

[I like Serena]

Here's an alternate solution: we fill the bus with 100 children and buy them ice creams from the change.

 

[davee123]

If the bus driver is one of the 100 passengers, I still don't think there's a solution, although if he has to pay a fee just like the other passengers (IE, there are 101 seats), then I believe there are a few solutions (I get 3 of them).

Otherwise, there's obviously something wrong-- either the problem isn't stated correctly, or there's a trick (like having some people not pay, or paying full fare for 4 women when in fact , say, only 2 of them actually show up). If evra posts the "intended" solution, I'm sure we can determine which is the case.

DaveE

 

[DaveC426913]

10x + .2x + 1.25x = 100 does not have an integer solution.

there is! very correct infact! try!

No, there may be an integer solution to the puzzle, but there is not an integer solution to the equation I posted. The solution to the equation I posted is x=8.733. So clearly there's some trick to the puzzle.I surmise that the trick is that, through fancy footwork with the wording, at least one of the criteria need not be met. Either the solution does not actually require 100 people, or it does not require 100 dollars.

 

[I like Serena]

No, there may be an integer solution to the puzzle, but there is not an integer solution to the equation I posted. The solution to the equation I posted is x=8.733. So clearly there's some trick to the puzzle.

LOL. I didn't even notice.
But there's also no integer solution to 10x + .2y + 1.25z = 100 with x + y + z = 100.

 

[evra]

You skipped over the pregnant-woman-solution.
So is that it?

no the pregnant woman stuff is stupid. i don't want to talk about it.. all the seats should be occupied. and we've 100 for the passengers.

 

[DaveC426913]

You know, I don't believe he said it all has to be on the same trip.

Is there an integer solution to 10x + .2y + 1.25z = 100 with x + y + z = 200?

 

[evra]

If the bus driver is one of the 100 passengers, I still don't think there's a solution, although if he has to pay a fee just like the other passengers (IE, there are 101 seats), then I believe there are a few solutions (I get 3 of them).

Otherwise, there's obviously something wrong-- either the problem isn't stated correctly, or there's a trick (like having some people not pay, or paying full fare for 4 women when in fact , say, only 2 of them actually show up). If evra posts the "intended" solution, I'm sure we can determine which is the case.

DaveE

the driver is not paying. and we have 100 seats for 1OOpeople and all of them will pay.

 

[I like Serena]

You know, I don't believe he said it all has to be on the same trip.

Is there an integer solution to 10x + .2y + 1.25z = 100 with x + y + z = 200?

No there isn't.

 

[evra]

Here's an alternate solution: we fill the bus with 100 children and buy them ice creams from the change.

hahahahahaaha.. that's not correct.. you are sooo funny..

 

[DaveC426913]

There seems to be no solution when you treat the three groups, men, women, children as distinct but if you consider that each child is also male or female then consider a female child as being half a woman so should be charged half the fare for a woman as well as the fare for a child so 5/8 + 1/5 dollars in total and similarly for a male child with the fare in this case being 5 + 1/5 dollars, then 4 male children plus 96 female children would give you 100 passengers and 100 dollars.

Yes, there could be a valid solution if you say that the children are men or women, but I'd also consider it a cheating one. One does not pay two fees for two different criteria (age and gender).

 

[evra]

Yes, there could be a valid solution if you say that the children are men or women, but I'd also consider it a cheating one. One does not pay two fees for two different criteria (age and gender).

no manipulation. children are children, men are men and women are women. no prenancy, no bisexual no other thing!

 

[DaveC426913]

I PM'd you. Did you get it?

 

[Char. Limit]

no manipulation. children are children, men are men and women are women. no prenancy, no bisexual no other thing!

No bisexuals? That wasn't even mentioned.

 

[DaveC426913]

No bisexuals? That wasn't even mentioned.

He's being preemptive. He means no tricks.

 

[DaveC426913]

Some of the children are young enough that they would be in their parents' arms. Thus, they are charged a fee totalling to $100, but they do not take up any of the 100 seats.

Unfortunately, I think there are many solutions to this one.

Here is one:
0 men.
28 children (28 seats, $5.60)
72 mothers (72 seats, $90)
babes-in-arms (0 seats, $4.40)
The bus is full, all 100 seats are occupied, driver has received $100.

 

[evra]

Some of the children are young enough that they would be in their parents' arms. Thus, they are charged a fee totalling to $100, but they do not take up any of the 100 seats.

Unfortunately, I think there are many solutions to this one.

unfortunately you are wrong uncle Dave. there is only one answer to it. lol! I will send the person who answers a present.

 

[evra]

I PM'd you. Did you get it?

yes.. and i replied.

 

[I like Serena]

Hmm, I'm confused.
Is Dave's statement correct or not?

If it is, WolframAlpha gives me the following solutions:
{{x == 0, y == 25, z == 76}, {x == 0, y == 50, z == 72}, {x == 1, y == 50, z == 64}, {x == 2, y == 50, z == 56}, {x == 3, y == 50, z == 48}}

 

[DaveC426913]

unfortunately u are wrong uncle Dave. there is only one answer to it. lol! I will send the person who answers a present.

Are you saying there are no children being carried by their parents?

 

[Soca fo so]

Do all the passengers board the bus at the same stop? Because thinking about it, where I'm from at each stop generally some passengers get off a bus and new ones get on so in theory you could have a full bus at every stage of the journey while at the end of the bus route more than 100 people would have traveled on the bus. Is this bus such a bus?

 

[evra]

Are you saying there are no children being carried by their parents?

no child is carried by his or her mum. all the children are capable of taking seats.

 

[evra]

Do all the passengers board the bus at the same stop? Because thinking about it, where I'm from at each stop generally some passengers get off a bus and new ones get on so in theory you could have a full bus at every stage of the journey while at the end of the bus route more than 100 people would have traveled on the bus. Is this bus such a bus?

no its not! it is a bus that goes from a destination to the other without a stop and it won't move without being full and can't collect less or more than 100Dollars.

 

[DaveC426913]

no its not! it is a bus that goes from a destination to the other without a stop and it won't move without being full and can't collect less or more than 100Dollars.

I think we give up.

 

[evra]

I think we give up.

Uncle Dave you don't have to. you are stronger than that.

 

[DaveC426913]

Uncle Dave you don't have to. you are stronger than that.

I think after 40 posts, we're stumped. Time to move on.