Who is the liar in this logic puzzle?

[jammieg]

Here's another one some strange guy said to me once:

So your employers give you more work than you don't want?

What does he mean?
This might seem really simple but for some reason it confused me greatly and I'm wondering if there is more to it.

 

[marqq]

geometry and logic

where is the missing square from the attachement bellow?

 

[NateTG]

The red and green triangles have different slopes, the appearance that the (macroscopic) figures are triangles is misleading.

You can see this by, for example, looking at the height of each of them 5 units from the left point. The lower shape has a height of two, but the upper shape has a height of 15/8 which, even in the illustration, is less than 2.

 

[marqq]

clasique

We have 5 houses, everyone of them painted in a different colour.In every house lives a person with a different nationality, which preferes his own drink, different then other's.
Every man has his own, single, different pet, and smokes his own, type of cigarettes.We know that:
1.The englishman lives in the red house
2.The sweedish has a dog
3.The dutch drinks tea
4.The green house is situated at the left of the white house
5.The person who lives in the green house drinks coffee
6.The person who smokes PallMall has a bird
7.The person who lives in the yellow house smokes Dunhill (like I do )
8.The person who lives in the 3rd house drinks milk
9.The norwegian lives in the 1st house
10.The Blend smoker lives in the house near the house where lives a cat
11.The person who owns a horse lives near the Dunhill smoker
12.The person who smokes Blue Master drinks beer
13.The german smokes Prince
14.The norwegian lives next to the blue house
15.The Blend smoker is the neighbour of the one who drinks only watter

The question is : Which one of them has a fish?

 

[Pergatory]

Originally posted by marqq
We have 5 houses, everyone of them painted in a different colour.In every house lives a person with a different nationality, which preferes his own drink, different then other's.
Every man has his own, single, different pet, and smokes his own, type of cigarettes.

I loved this one! This was created by Einstein originally, claiming that only 2% of the world's population could solve it. It would be interesting to find the percentage of people on this forum who can solve it.

 

[jammieg]

Silly questions

Why doesn't a bird put a roof over it's nest, and yet they will live in birdhouses?

My guess is that anyone can solve it given enough time and effort.

 

[leto]

We have 5 houses, everyone of them painted in a different colour.In every house lives a person with a different nationality, which preferes his own drink, different then other's. Every man has his own, single, different pet, and smokes his own, type of cigarettes.

It is the German, right? That was extremely hard for me; these puzzles are great! This thread has kept me entertained today.

 

[marqq]

Response

Yes,it is the German...I delayed the answer, hoping that i'll find other answers :D guess not...leto I think we r among the 2% (but i doubt that only 2% could solve this problem)

 

[Eyesaw]

I think the solution to the puzzle depended on how
you interpreted the word "near"- the term "next to"
was used elsewhere so the usage of the word "near"
in other descriptions becomes ambiguous.

 

[jammieg]

I mentioned this in another thread and thought it worth repeating, there is a way to solve this type of problem in one's head without pencil and paper, even if you don't have a good imagination although it helps to visualize it, the key is to read through each factual and try to memeorize the logical functions of each factual a bit more, by this I mean to think along the lines of, "well it could go here or here but not here unless that other could go there, ok move on to the next one" and so reading through the list over and over and focusing only on one at a time the functions of those factuals start to engrain in the memory and things may start to snap together seemingly without effort, although it took me way over an hour to do it this way I'm sure most anyone could do it this way with practice, it's the same principle as memorizing a long poem except one is trying to memorize the possibilities and limitations of each factual you would be amazed at how long a poem a person can memorize if they want to.
Let me know if it doesn't work, that means I probably don't know what I'm talking about.

 

[Hurkyl]

It's a trick question; the 5th pet is really a salamander!

 

[marqq]

May I ?

1. Ten years from now Tim will be twice as old as Jane was when Mary was nine times as old as Tim.

2. Eight years ago, Mary was half as old as Jane will be when Jane is one year older than Tim will be at the time when Mary will be five times as old as Tim will be two years from now.

3. When Tim was one year old, Mary was three years older than Tim will be when Jane is three times as old as Mary was six years before the time when Jane was half as old as Tim will be when Mary will be ten years older than Mary was when Jane was one-third as old as Tim will be when Mary will be three times as old as she was when Jane was born.

How old are they now?

 

[Nice coder]

for that robot
"Calculate 1/3 until there is no remainder ,the formula 14700/44100 = the answer is true and using only 2 decimal places (without rounding)"

The robot will never finish that because
1/3 to 2 decimal places (no rounding) is 0.33
0.33 < 0.33333333333 (to infinity)
14700/44100 = 1/3

another one would be,
Find the largest posible numerator in a fraction that would equal pie
That would never work, because pie is a never-ending number, and the highest possible numerator would be infinity!

Those questions would be great to combine and ask the robot, you don't have one by any chance?

 

[marqq]

Nice coder for the 1st question it can answer you 0.(3) :D i think "pi" is the question who may save us :)

 

[pedestrian]

sorry nvm

 

[Eyesaw]

Originally posted by marqq
1. Ten years from now Tim will be twice as old as Jane was when Mary was nine times as old as Tim.

2. Eight years ago, Mary was half as old as Jane will be when Jane is one year older than Tim will be at the time when Mary will be five times as old as Tim will be two years from now.

3. When Tim was one year old, Mary was three years older than Tim will be when Jane is three times as old as Mary was six years before the time when Jane was half as old as Tim will be when Mary will be ten years older than Mary was when Jane was one-third as old as Tim will be when Mary will be three times as old as she was when Jane was born.

How old are they now?

I don't think this problem is solvable. 2 requires Tim's age to be
an odd number while 1 requires Tim's age to be an evem number and 3 requires a lot of tylenol.

 

[gnome]

I don't think this problem is solvable. 2 requires Tim's age to be
an odd number while 1 requires Tim's age to be an evem number and 3 requires a lot of tylenol.

Eyesaw would think that, trapped as he is in his private twilight zone where time has 3 dimensions. But actually, it turns out that it's just algebra. Let N = the current year ("Now"), let T = the year Tim was born, M = the year Mary was born, J = the year Jane was born. And x₁ through x₁₀ can represent the various other unknown years in the clues.

N + 10 - T = 2(x₁ - J)
x₁ - M = 9(x₁ - T)
N - 8 - M = (1/2)(x₂ - J)
x₂ - J = 1 + x₃ - T
x₃ - M = 5(N + 2 - T)
x₄ - T = 1
x₄ - M = 3 + x₅ - T
x₅ - J = 3(x₆ - M)
x₆ = x₇ - 6
x₇ - J = (1/2)(x₈ - T)
x₈ - M = 10 + x₉ - M
x₉ - J = (1/3)(x₁₀ - T)
x₁₀ - M = 3(J - M)

Now, work from the bottom up, substituting out the x's as you go. After eliminating the "x" years, all of part 3 boils down to
-4T -3M + 7J = 1, and the other two equations are
13T - M - 8J = 4N + 40
-6T + 3M = -3N - 27
Solve them in terms of N to get the ages (T-N, M-N and J-N).

[Edit: sorry about that; I meant to write N-T, N-M and N-J]

Tim is 3
Mary is 15
Jane is 8

 

[marqq]

:D

Bravo !
Well Done !

 

[marqq]

Here we go again

we have 10 bags with 10 gold ingots each...Each ingot has 10kg, but we have a bag full of fake ingots (9.900 kg each).we have a coin scale, and a single coin (for a single use of the scale), and we have to find the bag with the fake ingots!
Good Luck!

 

[gnome]

A balance or a scale?

 

[gnome]

Thought so. That makes it too easy. Just number the bags 1 to 10. Take 1 ingot from bag 1, 2 from bag 2, ..., 10 from bag 10 & weigh those 55 ingots together.

Need I say more?

 

[Eyesaw]

Originally posted by gnome
Eyesaw would think that, trapped as he is in his private twilight zone where time has 3 dimensions. But actually, it turns out that it's just algebra. Let N = the current year ("Now"), let T = the year Tim was born, M = the year Mary was born, J = the year Jane was born. And x₁ through x₁₀ can represent the various other unknown years in the clues.

N + 10 - T = 2(x₁ - J)
x₁ - M = 9(x₁ - T)
N - 8 - M = (1/2)(x₂ - J)
x₂ - J = 1 + x₃ - T
x₃ - M = 5(N + 2 - T)
x₄ - T = 1
x₄ - M = 3 + x₅ - T
x₅ - J = 3(x₆ - M)
x₆ = x₇ - 6
x₇ - J = (1/2)(x₈ - T)
x₈ - M = 10 + x₉ - M
x₉ - J = (1/3)(x₁₀ - T)
x₁₀ - M = 3(J - M)

Now, work from the bottom up, substituting out the x's as you go. After eliminating the "x" years, all of part 3 boils down to
-4T -3M + 7J = 1, and the other two equations are
13T - M - 8J = 4N + 40
-6T + 3M = -3N - 27
Solve them in terms of N to get the ages (T-N, M-N and J-N).

[Edit: sorry about that; I meant to write N-T, N-M and N-J]

Tim is 3
Mary is 15
Jane is 8

Goofus, if Tim is 3, 10 years from now, he would be 13. What is 13/ 2? If fractions are allowed in the problem, then your final answer should include the month, day and second they were all born.

 

[gnome]

You're a real piece of work, Eyesore. I guess we were wrong to advise you to re-study relativity.

Based on your comments above, you should first re-study fractions.

I'm in a charitable mood now, so I'll help you through the first few lines.

Ten years from now Tim will be 13. Half of that is 6&frac12;. Now, Jane is 7 years younger than Mary, so when Jane was 6&frac12;, Mary was 13&frac12;. Mary is 12 years older than Tim. So, when Mary was 13&frac12;, Tim was 1&frac12;. 13&frac12; is 9 times 1&frac12;.

The rest is no more difficult than that. Let us know if you need any more help.

No need to worry about months, days, hours, minutes and seconds. Unless you're a masochist. Since you like to do things the hard way, go ahead. Eventually you'll get the same answer if you don't screw it up.

 

[Eyesaw]

Originally posted by gnome
You're a real piece of work, Eyesore. I guess we were wrong to advise you to re-study relativity.

Based on your comments above, you should first re-study fractions.

I'm in a charitable mood now, so I'll help you through the first few lines.

Ten years from now Tim will be 13. Half of that is 6&frac12;. Now, Jane is 7 years younger than Mary, so when Jane was 6&frac12;, Mary was 13&frac12;. Mary is 12 years older than Tim. So, when Mary was 13&frac12;, Tim was 1&frac12;. 13&frac12; is 9 times 1&frac12;.

The rest is no more difficult than that. Let us know if you need any more help.

No need to worry about months, days, hours, minutes and seconds. Unless you're a masochist. Since you like to do things the hard way, go ahead. Eventually you'll get the same answer if you don't screw it up.

Hey, how often do you hear someone say I am 6 years and 6 months older than you? And what are the odds of 3 randomly chosen people being born on the same day, same hour and same second? Based on common practice of excluding all but the year in calculating age differences, this problem is unsolvable.

 

[gnome]

I agree -- if you don't understand fractions, you can't solve it.

 

[Eyesaw]

Originally posted by gnome
I agree -- if you don't understand fractions, you can't solve it.

I knew you could do it! I knew you'd see your mistake
sooner or later. Well done.

 

[leto]

Eyesaw, you're making a fool of yourself.

 

[elibol]

Originally posted by Raven
I have a variation on this idea that is equally as difficult. It goes along these lines:

You have reached a fork in the road. One path leads to the city of truth where everyone always tells the truth and the other path leads to the city of lies where everyone always tells a lie. There is a stranger at the fork of the road who is a native of one of the two cities. Given one question, what can you ask him in order to find out which path leads to the city of truth?

i figured it out the second i noticed the relationship to the first one, but then i got lost again when i tried analyzing the relationships...

i nearly lost my mind trying to figure this one out...

i thought it was impossible... but i realized i just lost track...

it is very complicated, can someone break this down into logical variables? maybe some sort of computer code for me to better understand it?

 

[gnome]

An order of perfectly logical monks lives isolated in their monastery. They are sworn not to communicate with each other in any way. They have no mirrors, and no other means by which a monk can see his own face. They see each other only once each day when they all gather together for afternoon prayers.

There is a demon loose in the land - the relativity demon. If a person becomes possessed by this demon, the demon's sign (e=mc²) appears on his forehead. The monks know that this is a very powerful demon -- one that can never be exorcised -- and so, if a monk would discover that he bore this mark, that evening, in the privacy of his cell, he would commit suicide.

One afternoon, a visitor (one who is not sworn to silence) comes to the prayer meeting. He looks around the room, announces "At least one monk in this room has the demon's mark on his forehead", and immediately leaves. The monks all look around, examining each other's foreheads. Nothing unusual happens that evening.

At the second day's prayer meeting the monks all look at each other, but again that evening nothing unusual occurs.

...and so on for the third, fourth, fifth, and sixth days.

But on the seventh evening, all the monks commit suicide.


What happened? How many monks were there? How many had the demon's mark?



Enjoy.

 

[Eyesaw]

There were 7 monks, all 7 of them had the mark from day 1.