The worlds first riddle!

Tryagain wrote:

172.012
Two trains, A and B, leave their respective starting points at the same time and travel in opposite directions. They travel at constant speeds, and pass at point M. One travels at twice the speed of the other. If one of the trains leaves five minutes late they pass at a point 2 miles from point M. What is the speed of the slow train, in miles per hour

is it possible that the slow train travels with 36 mph Question

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MyOwnUsername wrote:

are you sure you provided enough information? I don't think so, honestly.
The problem is that it seems there are more possible solutions, or, two of them should know colour of their hats, so maybe one of them is just smarter? But, that's very shaky then because in further developments other solutions are available as well...

First of all, let's eliminate things that are not possible:
a) 3R and B) 2R and 1G - those are not possible because in those cases not everyone would raise hand

c) if they all have green hats then every single one of them thinks that at least two of them have to have green hat (because if only one would have it, then he wouldn't raise hand, seeing only red on heads of two others)...and it still can't help them, because they cannot know which one is on their head...

so, it has to be:

d) 2 G and 1 R
and then...

1) the one with red hat sees two greens, so he can still have both colours on his head
b) those that have green see one red and one green - therefore, they know that they cannot possibly have red - because if they would have red too, then one they see with the green hat would see two reds and therefore he wouldn't raised hand at first place

BUT, in this case two of them can come to a clue that they have green hat...so we should probably assume that one of them is smarter?

it is because that one is faster then the other i guess the riddle doesn;t naswer that question nut your answer is right:lol:

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Tryagain wrote:

The start of my decline.

A merchant was hiring an assistant, and two young men showed up for the job. The trader said: "Let's see which of you two is the smarter."
He took the two boys into a room, took out a box, and said: "There are five hats in this box, two red, three black. And this is what we'll do: I turn off the light, and each of us three pick a hat from the box and put on one's own head. I then close the box, turn on the light, and let's see which of you two can be the first to tell me the color of the hat that is on your own head."

Actually the merchant did not tell the whole truth. Yes there were five hats in the box, but only one of them is red and the merchant knew how to get that red hat for himself. The two young men, however, did not know this and they believed what they were told: there were five hats, two red and three black.

The light was turned out, and each of the three picked a hat to put on. When the light was back on, the two young men saw a red hat on the merchant's head. They then looked at each other, but for a second neither of the two could tell what hat he himself was wearing. Finally, one of them cried out: "I know! I know!"

He got it right. Now, how did he figure it out

it got to be black
they assumed there were 2 red hats
so they saw a red one on the merchants head and a black on each others. so the one who knew answer knew it had to be black cause if it was red the other one would know that he had a black one cause there were only 2 red hats :wink: but he didn't he his color Surprised

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there, Magnum was sooner here Smile

But i completely agree with his answer Very Happy

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Magnum wrote, "Is it possible that the slow train travels with 36 mph"
Answer; It sure is. Very well done. Very Happy

In fact, that goes for both of you. What took me days, you managed in minutes. Excellent. Cool

Merchant's Hats

The young man who got the correct answer first reasoned like this:
There are two red hats and three black hats, and the merchant is wearing a red one. This leaves one red and three black. If the other felllow is wearing the red hat, then I must be wearing a black and I know the answer right away. But I look at the other fellow and see that he is not wearing the red hat--he has a black on. This makes it a bit more difficult. I cannot tell if I'm wearing a red or black hat. What to do, what to do... But wait! The other fellow is puzzled too! He has not said anything yet! This must mean that I am not wearing the red hat, otherwise he would have said so. I am wearing a black hat!

Can you save a life, or two?

10 straight-jacketed prisoners are on death row. Tomorrow they will be arranged in single file, all facing one direction. The guy in the front of the line (he can't see anything in front of him) will be called the 1st guy, and the guy in the back of the line (he can see the heads of the other nine people) will be called the 10th guy. An executioner will then put a hat on everyone's head; the hat will either be black or white, totally random. Prisoners cannot see the color of their own hat.

The executioner then goes to the 10th guy and asks him what color hat he is wearing; the prisoner can respond with either "black" or "white". If what he says matches the color of the hat he's wearing, he will live. Else, he dies. The executioner then proceeds to the 9th guy, and asks the same question, then asks the 8th guy ... this continues until all of the prisoners have been queried.

This is the night before the execution. The prisoners are allowed to get together to discuss a plan for maximizing the number of lives saved tomorrow. What is the best plan

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well I have to go to sleep now, since it's 1.35am in Croatia and I work tomorrow morning Smile

And I will not be executed tomorrow, so I'll skip discussion...
So, in very very quick response I can say that five can be saved easily (by 10th saying the colour of first, 9th saying the colour of second, etc... - of course, still some of them can actually guess his own colour, but five is minimum)
I'll have to think tomorow if they can save some more of them

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of course, this works if they all hear responses of other prisoners, and I suppose they do

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Quote:

16.012
Five pumpkins are weighed two at a time in all ten sets of two. The weights are recorded as 16, 18,19, 20, 21, 22, 23, 24, 26, and 27 pounds. All individual weights are also integers. How much does each pumpkin weigh

Total weight of the pumpkins is;

(16+18+19+20+21+22+23+24+26+27) / 4 = 54lbs

The 2 heaviest weigh 27lbs together, the 2 lightest 16lbs so the middle one weighs;

54 - 27 - 16 = 11lbs

The second heaviest weighing is 26lbs. This has to contain the 11lb pumpkin so the heaviest one must weigh 15lbs and the second heaviest 12lbs.

The second lightest weighing is 18lbs. This has to contain the 11lb pumpkin so the lightest one must weigh 7lbs and the second lightest 9lbs.

So, the pumpkins weigh 7lbs, 9lbs, 11lbs, 12lbs and 15lbs.

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Quote:

164.014
A courier pigeon departs Las Vegas for Reno at the same time as another courier pigeon departs Reno for Las Vegas. Both pigeons fly at constant speeds, although different from each other. They cross paths 2x miles from Las Vegas. After each arrives at their destination, they immediately turn around, going back and forth without breaks. They cross paths the second time x miles from Reno. Where will they cross paths the third time

Taking a dump on the MGM lion.

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Quote:

173.013
Three playing cards, removed from an ordinary deck, lie face down in a horizontal row. Immediately to the right of the King there's a Queen or two. Immediately to the left of a Queen there's a Queen or two. Immediately to the left of a Heart there's a Spade or two. Immediately to the right of a Spade there's a Spade or two. Name the three cards in order.

Oops! just noticed the answer to this one has already been posted.

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MyOwnUsername wrote:

well I have to go to sleep now, since it's 1.35am in Croatia and I work tomorrow morning And I will not be executed tomorrow, so I'll skip discussion...
So, in very very quick response I can say that five can be saved easily (by 10th saying the colour of first, 9th saying the colour of second, etc... - of course, still some of them can actually guess his own colour, but five is minimum)
I'll have to think tomorow if they can save some more of them

i was thinking in that direction either but if you look that way you cannot maximise the number of people who can be saved

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well I think that maximize here means "maximum possible with given options" and I still think that this is kinda it - five for sure, and other five still have 50/50 chance that maybe they will guess too...so far I have no idea how could they save six FOR SURE...but I am also still at my job so I haven't used my all capacities Very Happy

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Adrian, newly returned from the annual Dingo roundup really knows his melons the weights are correct. Very Happy

Having just completed the higher standards of Art and Literature, he writes, "Taking a dump on the MGM lion." May I respectfully suggest you apply for a refund. You have clearly wasted the $2.99 for the six week residential course. Shocked

However, from the ?'fan mail' received your absence was clearly missed. Well my mother asked after you! :wink:

As for you other guys, I am counting on you doing a bit better, as I am number 6 in the row and there the others to think of. Please save the children. Crying or Very sad

Tension increased here today as protesters, some armed with placards, denouncing the inclusion of a question that seemed to encourage child slave labour. When asked how much the child received, no one could give an answer, so they all went home. A spokesperson for the WMC confirmed all child labour laws were being observed.
What do you think?

177. 017 Three men and a boy are gathered in your orchard to pick apples. The boy can shake the apples off the trees as fast as the men can pick them up. Any one of the men can shake apples off the trees 25% faster than the other two men and the boy can pick them up. The same number of apples are gathered regardless of who is the shaker, as long as everybody is always doing something and no excess apples are left on the ground.

(Note: anybody can divide their time between shaking and gathering.) How should you divide $500 between the 3 men and the boy, according to their contribution to the team effort. Question

178.018 You pick three different numbers from 0 to 9. What is the probability that a dollar bill chosen at random will contain all 3 of your numbers in the serial number? Note: the serial number has 8 digits, you may assume that an equal number of each combination is in circulation. Question

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but of course....code...they talked last night, damn it!! Smile

they have code and the best one I found out so far (I think best at all) is that 10th says this colour that pair number of persons in front of him have - because he sees nine hats there must be pair of either black or red hats and only black OR red (not both)
So, if there are let's say 6 reds, he says "red"
9th - if he sees 6 reds (or 2,4,8), he has black - if he sees 5 (or 1,3,7), he has red
8th - he only have to think about what 9th said and calculate...
and so on
this way 9 are saved, and possibly even 10 if No.10 is lucky enough that code turns out to be colour of his hat.

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brillant mr username Very Happy

but are they allowed to say more than just the color?

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" The prisoner can respond with either "black" or "white". Evil or Very Mad

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well of course, but they DO say only one colour
they agree before lining that the colour 10th sees in pair number is colour he will say as a code
So, let's imagine that he sees pair number of reds and unpair number of blacks in front of him
So:
10th: RED (if he is lucky maybe he lives too)
9th: he knows that there is pair number of reds in front of tenth, because of code they all together made - so, if he sees, let's say one red in front of him (or 3, or 5, or 7) he knows that he has red and he says: RED (because 10th saw pair number of reds, and he sees unpair number, so he has to have red) - IF he sees 0 or 2 or 4 or 6 or 8 reds, then he knows that he has black: so he says BLACK
8th: he knows everything that 9th knows, but he also has to calculate the colour 9th said - so if he sees one red in front of him, and 10th used code that there is pair of reds, he also must think about what 9th said - e.g. if he sees one, and 9th also said red, then 8th has black - if he sees one, and 9th said black, then he says "RED"
And so on...

So, with that code everybody can know for sure what colour of hat he has. Of course, everybody except first one - 10th in a row. But in this option they save 9 of 10 lives, and I am sure that they cannot possibly save all ten (except with pure luck) because first one (10th in a row) simply can't have any help or code or knowledge about hat on his head.

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If I understand correctly you have the answer. Cool

It is agreed upon that the person ANSWERING first (last in line) will answer BLACK for an ODD, and WHITE for an EVEN. Odd/Even applying to only ONE color hat(must be memorized by all...we'll say they picked BLACK to be the counted color)

Regardless of whether the 1st person answered correctly, the rest of them can now deduce what color hat they have on by counting the same color hats in front of themselves.

For instance:
1st person answers = BLACK (odd)...
now 2nd person counts the black hats in front of himself and finds an odd number, therefore he knows his hat is white (for those not following: if he counts and finds odd he CAN'T have black on or the count would have come up even when the 1st person counted).
The 3rd person now knows his hat is BLACK, because its even in front of him. The 4th person, knowing that the 3rd must have counted even, starts the whole process over again, bearing in mind that the person behind him saw an EVEN number of black hats... etc etc etc
FINALLY note that the 1st person to answer did so based solely on his count, therefore he had 50% of dying...

To summarize:
The prisoners are only allowed to say "Black" or "White. The prisoners agree in their meeting that the first person will call "black" if he sees an even number of black hats. Otherwise, he says "white".
Each remaining prisoner counts the number of "blacks" called out behind him, and the number of black hats he can see. If on his turn the total is odd, he must have on a white hat. If the total is even his hat must be black. He calls out the appropriate color.

George Brown, Tom Green and Bill White were talking. ?'Funny thing, said Brown. ?'I've just noticed we're wearing different colored hats the same as our names.' ?'Yes', said one of the others. ?'But none of us is wearing a hat the same color as our name. For instance, I'm wearing a green hat.' Who's wearing what color hat Question

Train A leaves from Boston to New York. An hour later, Train B leaves from New York to Boston. The two trains are going at exactly the same speed. Which train will be nearer to Boston when they meet Question

(a) Train A
(b) Train B
(c) None of the above

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easy one

that's good for refreshing Smile

Brown has white hat (he is not one talking that he has green), therefore Bill is one with the green (cause Tom's surname is Green), and Tom has brown.

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Train A leaves from Boston to New York. An hour later, Train B leaves from New York to Boston. The two trains are going at exactly the same speed. Which train will be nearer to Boston when they meet
(a) Train A
(b) Train B
(c) None of the above

it's answer C when they meet they will be at the same point at the same time so one train cannot be closer than the other train

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