Puzzle

[FnG]magnolia;11922771 said:
11 pages of exactly the same discussions can be found there.

The plane DOES take off. The End.

Now back to those puzzles :)
Click to expand...

Of course it does, in that example :)

You are told that the plane is moving forward (with respect to the ground) and so has air passing over its wings. The fact the conveyor belt is moving in the opposite direction, at the same speed, does nothing (or very little) to cancel out the forward propulsion of the engines and so the plane is still moving forwards with respect to the ground. So it still has air passing over the wings. Hence it flys :)

iraiguana was imagining a situation where the plane is not moving (with respect to the ground). Ie the forward propulsion being used by the plane is tiny and only just enough to oppose the backward motion caused by friction between the wheels and the conveyor.
 
thesubstitute said:
Another puzzle since the 2=1 one was immediately solved :<

This one will probably take some explaining along with answer! (yeah, please explain rather than just shouting out colour)

A king wants his daughter to marry the most logical of 3 intelligent princes, and so comes up with a test for them.

The princes are gathered into a room and seated, facing one another, and are shown 2 black hats and 3 white hats. They are blindfolded, and 1 hat is placed on each of their heads, with the remaining hats hidden in a different room.

The king tells them that the first prince to correctly identify the color of his hat will marry his daughter. A wrong guess will mean death.

The blindfolds are then removed. You are one of the princes, you see 2 white hats on the other prince's heads. What color is your hat?
Click to expand...




Since I can see two white hats on the other princes, my hat must be white, assuming the other princes are logical: If my hat were black then one of the other princes could know the colour of their hat, because they'd know that if they had a black hat (and I waswearing a black hat) the other prince wearing a white hat would call out that their hat is white since the 2 black hats are on me and the other prince. Since the other white hat wearing prince hasn't called out they know that their hat would be white.

Since no-one has called out, my hat must be white.
 
Last edited:
I stand corrected on the linked problem. It is similar to the problem I have heard before phased differently depending on who has adapted it.

I agree if the plane is moving at take off speed then it will obv take off as the wings are what give it motion.

Wheels are free spinning so the explanation I used was still correct, just applied to a variant on the problem.

Also White Hat.

String is as long as you want it to be just hand the scissors.

Not sure about the train one yet.
 
you can cut a chunk of ice into one by putting it on a table, metal wire over the top and weights on the end of the wire. It will slice through slowly and the cube will freeze back together

Train wise its probably something to do with the time the trains stay at the station. If it leaves every hour one may arrive at 10 min too the hour, and the other at 10 min past the hour. The earlyer train would wait for 50min and the later one would wait for 10min...

There're two train lines at your local train station, one goes east and the other goes west. Both lines have exactly 1 train leaving at the same time every hour throughout the day.

How come if you arrive at the station at any random time, that the probability of the Eastbound train arriving is 5 times that of the Westbound train?
Click to expand...

yer, so the easterly train only does a 10 min journey (for example) and the westerly train does a 50min journey, the rest of the time they wait at the station :)
 
Last edited:
Siliconslave said:
you can cut a chunk of ice into one by putting it on a table, metal wire over the top and weights on the end of the wire. It will slice through slowly and the cube will freeze back together

Train wise its probably something to do with the time the trains stay at the station. If it leaves every hour one may arrive at 10 min too the hour, and the other at 10 min past the hour. The earlyer train would wait for 50min and the later one would wait for 10min...



yer, so the easterly train only does a 10 min journey (for example) and the westerly train does a 50min journey, the rest of the time they wait at the station :)
Click to expand...


You're onto something with the train puzzle, but it's simplier than that. Just to simplify it a bit - let's assume that both trains stops at at instant and leaves immediately, but you'd be able to catch it if you're present when it arrives.
 
If there are only the two trains then if there is a 10 minute gap between A and B (B being being ten minutes after A) then there is only a 10 minute window where you'd get on B but a 50 minute gap where you would wait until train A so you have 5 times the chance of catching it if you turn up at a random time :)
 
ricky1981 said:
If there are only the two trains then if there is a 10 minute gap between A and B (B being being ten minutes after A) then there is only a 10 minute window where you'd get on B but a 50 minute gap where you would wait until train A so you have 5 times the chance of catching it if you turn up at a random time :)
Click to expand...

Correct :)


Next puzzle:

Imagine if you walk up a hill along the only path up on Monday. You started at 8 in the morning and reach the top at 5pm. Camped out at the top of the hill for the night.

Tuesday you decided to follow the same path down, and you start at 8am. Taking it easy, you stolled down and happened to arrive at the bottom exactly where you started at 5pm again.

Question is, were you at any point of on Tuesday at the same place along the path as you were on Monday at the same time?
 
You are unable to tell, as you do not know if they were walking at the same pace for the entire journey either way, surely?
 
No correct answers yet. But Hxc's right in that you can't tell the pace of which you're walking on either days. That's why I added the fact that you "stoll" down the hill on Tuesday.
 
One I heard on the radio (?) at the bank yesterday:

I am the owner of a pet store that specializes in birds. If I put one canary per cage, I have one canary too many. But if I put two canaries per cage, I have one cage too many. How many canaries and cages do I have?
 
thesubstitute said:
No correct answers yet. But Hxc's right in that you can't tell the pace of which you're walking on either days. That's why I added the fact that you "stoll" down the hill on Tuesday.
Click to expand...

Regardless of the pace you walk you will have to be in the same place at the same time on the previous day.
You can just take the person walking down the slope and move them back by exactly 24 hours.
Then the two people have to cross at a point on the path and this is your place, the exact location depends on how the pace of the journey varies in each case.

Steve2000 said:
Right then: what object can I cut clean through but still be left with one object? :p
Click to expand...

And for this one, a mobius strip?

EDIT, may as well answer this one too...

Phæte said:
One I heard on the radio (?) at the bank yesterday:

I am the owner of a pet store that specializes in birds. If I put one canary per cage, I have one canary too many. But if I put two canaries per cage, I have one cage too many. How many canaries and cages do I have?
Click to expand...

4 canaries, 3 cages.

Let number of canaries be a
Let number of cages be b

b = a - 1 (from the first point)
b = a/2 + 1 (from the second)

Solve for a and b
 
Last edited:
You must log in or register to reply here.
Back
Top Bottom