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

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The discussion revolves around a series of tricky questions and riddles shared among participants. Key questions include the amount of soil in a hole, the safest room for a condemned murderer, and a math challenge involving sequential additions. Participants engage in solving these riddles, offering various answers and reasoning, with some humor and banter included. The thread showcases a mix of logical puzzles and lateral thinking challenges, encouraging creative problem-solving among the members.
  • #361


Here's an alternate solution: we fill the bus with 100 children and buy them ice creams from the change.
 
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  • #362


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
 
  • #363


evra said:
DaveC426913 said:
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.
 
  • #364


DaveC426913 said:
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.
 
  • #365


I like Serena said:
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.
 
  • #366


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?
 
  • #367


davee123 said:
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.
 
  • #368


DaveC426913 said:
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.
 
  • #369


I like Serena said:
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..
 
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  • #370


Soca fo so said:
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).
 
  • #371


DaveC426913 said:
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!
 
  • #372


I PM'd you. Did you get it?
 
  • #373


evra said:
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.
 
  • #374


Char. Limit said:
No bisexuals? That wasn't even mentioned.
He's being preemptive. He means no tricks.
 
  • #375


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.
 
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  • #376


DaveC426913 said:
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.
 
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  • #377


DaveC426913 said:
I PM'd you. Did you get it?

yes.. and i replied.
 
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  • #378


Hmm, I'm confused. :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}}
 
  • #379


evra said:
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?
 
  • #380


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?
 
  • #381


DaveC426913 said:
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.
 
  • #382


Soca fo so said:
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.
 
  • #383


evra said:
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.
 
  • #384


DaveC426913 said:
I think we give up.

Uncle Dave you don't have to. you are stronger than that.
 
  • #385


evra said:
Uncle Dave you don't have to. you are stronger than that.
I think after 40 posts, we're stumped. Time to move on.
 
  • #386


It's not possible. The equations are 10x+1/5 y + 5/4 z = 100, and z+y+z = 100, which gives 7(28x+3z) = 1600 by substituting y from the last equation into the first. 1600 is not divisible by 7, but 28z + 3z is an integer.
 
  • #387


disregardthat said:
It's not possible. The equations are 10x+1/5 y + 5/4 z = 100, and z+y+z = 100, which gives 7(28x+3z) = 1600 by substituting y from the last equation into the first. 1600 is not divisible by 7, but 28z + 3z is an integer.

It's not possible when interpreted literally. Obviously there's a trick to it.

But evra is determined to drag it out as long as possible.

What's black and white and red and can't turn around in hallways?
 
  • #388


DaveC426913 said:
It's not possible when interpreted literally. Obviously there's a trick to it.

But evra is determined to drag it out as long as possible.

What's black and white and red and can't turn around in hallways?

A hallway painted black and white and red?
 
  • #389


daveb said:
A hallway painted black and white and red?
A nun with a javelin through her head! :biggrin:
 
  • #390


Hello my people! i am very very very sorry to all of you. i made a small mistake on the puzzle and it changed everything. forgive my guys, i was a little bit away. and especially to Uncle Dave, I am really sorry. and remeber that to make a mistake is human though not scientific and i hope u will forgive me. anyway, the correct one is;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
1 woman is 4dollars
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??
Honestly i am sorry for the mistake. try this one for me please and let's forget the past.
 
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