# Some puzzles

bobbob1056th
Posers and Puzzles 25 Jun '05 03:52
1. 25 Jun '05 03:52
In this thread I will post some puzzles:

Three people are standing in a line, one behind the other, such that the last person can see both of the two people in front of him, the middle person can see only the one person in front of him, and the first person can see no one else. From a pile of five hats - - three of them white and two of them black - - which each person knows about, you place a hat on each person's head. None of them can see their own hats, but the last person can see each hat on the other two, and the middle person can see the one hat of the person in front of him. The last person in the line is asked what color hat he is wearing and he says, truthfully, he does not know. The middle person is then asked, after hearing all this, what color hat he has on, and he also truthfully says he does not know. The woman in the front then is asked what color hat she has on, and even though she cannot see any of the hats, she says she knows what color hat she has on. How does she know, and what color hat is it?
2. 25 Jun '05 08:23
Originally posted by bobbob1056th
In this thread I will post some puzzles:

Three people are standing in a line, one behind the other, such that the last person can see both of the two people in front of him, the middle person can see only the one person in front of him, and the first person can see no one else. From a pile of five hats - - three of them white and two of them black - ...[text shortened]... hats, she says she knows what color hat she has on. How does she know, and what color hat is it?
If the last person doesn't know his own hat, then he doesn't see two black ones in front of him (otherwise he would wear a white one). So, one white and one black, or two whites. If the middle person doesn't know his own hat colour, then he must be seeing a white one in front of him (otherwise he would know his colour is white). So the lady in front knows for sure she is wearing a white hat.
3. 26 Jun '05 01:33
Originally posted by Mephisto2
If the last person doesn't know his own hat, then he doesn't see two black ones in front of him (otherwise he would wear a white one). So, one white and one black, or two whites. If the middle person doesn't know his own hat colour, then he must be seeing a white one in front of him (otherwise he would know his colour is white). So the lady in front knows for sure she is wearing a white hat.
Well done. here's a few more (easy ones, I'll post some hard ones later...):

1. Imagine there is a smooth ball the size of the earth, 24,000 miles around, and you tied a ribbon tight around the equator of the ball, then spliced in one additional yard of ribbon (36 inches) of ribbon, so there was a small loop at one point on the surface of the ball. Then imagine that you smooth out the loop so that the slack in the ribbon is spread evenly over the whole 24,000 mile surface. Will the ribbon be very far off the ground? About how high off the ground do you think it will be?

2. To qualify for an automobile race on a particular one-mile oval track, drivers must do two laps, averaging 60 mph. On his first lap one driver has an engine problem that holds his speed down to 30 mph for that lap. How fast must he do the second lap in order to average 60 mph for the both of them?
4. XanthosNZ
Cancerous Bus Crash
26 Jun '05 05:34
1. If the radius of a circle is increased by r then it's circumfrence increases by 2*pi*r.
2*pi*r = 0.9144
r = 0.1455 m
14.6 cm or just 5.73 inches (in olden day measurements).

2. For a driver to average two laps at 60mph the driver must finish the two laps in 1/30th of an hour. Driving at 30mph for a mile takes 1/30th of an hour. Therefore the driver must take zero time for the second lap to average 60mph.
Obviously this can't be done.

Try some harder problems next time.
5. 26 Jun '05 05:53
Originally posted by XanthosNZ
1. If the radius of a circle is increased by r then it's circumfrence increases by 2*pi*r.
2*pi*r = 0.9144
r = 0.1455 m
14.6 cm or just 5.73 inches (in olden day measurements).

2. For a driver to average two laps at 60mph the driver must finish the two laps in 1/30th of an hour. Driving at 30mph for a mile takes 1/30th of an hour. Therefore the dri ...[text shortened]... nd lap to average 60mph.
Obviously this can't be done.

Try some harder problems next time.
Good job.

This next one is more of a fun maze thingy. You have to work your way through the "doors" to get to the next door and sadly, the inevitable end. The signs on the doors are what I will be posting, they are not nessisarily true or false. Which door should be opened and why? In order for the answer to be correct you must explain why the correct door is the correct one.

The first choice:
door A. "Only one of these signs is false"
door B. "This is the door you should go through"

The seconds choice:
door A. "These signs are both false"
door B. "This is the way to go"

Please note this is not the whole "maze", there are six more choices! After this is finished, I will post progressively more difficult puzzles.
6. 26 Jun '05 08:11
Originally posted by bobbob1056th
Good job.

This next one is more of a fun maze thingy. You have to work your way through the "doors" to get to the next door and sadly, the inevitable end. The signs on the doors are what I will be posting, they are not nessisarily true or false. Which door should be opened and why? In order for the answer to be correct you must explain why the ...[text shortened]... are six more choices! After this is finished, I will post progressively more difficult puzzles.
First choice: door A
The term 'only one' is ambiguous. I read it as 'exactly one', or 'one'. Otherwise it would be better to state 'at most one', which would include 'none'. And in that case, the outcome remains uncertain (because the two statements could be either both false or both true).
Given that, B must be false. If b were true, then A is a contradictio in terminis (if true then false and vice versa). With B false, it doesn't matter wether A is true (meaning only B is false) or false (meaning both A and B are false).

Second choice: door B
Statement A is false. If it were true, then it contradicts itself. Hence B is true.
7. 26 Jun '05 08:161 edit
Originally posted by bobbob1056th

The first choice:
door A. "Only one of these signs is false"
door B. "This is the door you should go through"

The seconds choice:
door A. "These signs are both false"
door B. "This is the way to go"

Please note this ...[text shortened]... his is finished, I will post progressively more difficult puzzles.
For the first one:
Through door A
It doesn't matter if "A" is true or false, B is always false. (Okay I will elaborate: "A" true implies "B" false, "A" false implies "B" false)

For the second:
Through door B.

Let "A" be true, then "A" is false?! and "B" is false. Let "A" be false then "B" is true. Since "A" cannot be true and false at the same time, this means "A" is false and "B" is true.
EDIT: Mephisto was quicker ðŸ™‚
8. Bowmann
Non-Subscriber
27 Jun '05 00:17
Originally posted by XanthosNZ
2. For a driver to average two laps at 60mph the driver must finish the two laps in 1/30th of an hour. Driving at 30mph for a mile takes 1/30th of an hour. Therefore the driver must take zero time for the second lap to average 60mph.
Obviously this can't be done.
No? What if he travelled at c?
9. XanthosNZ
Cancerous Bus Crash
27 Jun '05 01:05
Originally posted by Bowmann
No? What if he travelled at c?
It would still take 1/670 616 629th of an hour to travel the second lap.
10. Bowmann
Non-Subscriber
27 Jun '05 02:20
Originally posted by XanthosNZ
It would still take 1/670 616 629th of an hour to travel the second lap.
By whose clock?
11. 27 Jun '05 02:25
Originally posted by Bowmann
No? What if he travelled at c?
Better, what if he was omnipresent?
12. 27 Jun '05 02:38
3rd choice
A. "This is not the door to go through unless the sign on the adjacent door is true"
B. "Exactly two of these signs are false"
C. "This is the door to open unless the sign on the adjacent door is false"

4th choice
A. "No fewer than 2 of these signs are false"
B. "None of these signs are false"
C. "Go through either this door or the door with the sign that is true"

5th choice
A. "Do not go through door C"
B. "At least one of these signs is false"
C. "If door A is not the one to go through, then door B is"
13. XanthosNZ
Cancerous Bus Crash
27 Jun '05 02:51
Originally posted by Bowmann
By whose clock?
The person in a stationary inertial frame.
14. 27 Jun '05 03:32
Originally posted by XanthosNZ
It would still take 1/670 616 629th of an hour to travel the second lap.
In a real life situation, a race car probably cannot travel at c.
Also I presume the "average mph lap counter thingy" thingy does not measure decimal places smaller than the hundredths place. So if the guy did the two laps averaging 60.0000000000000000000000000000000000000000001 mph, based on the rules he does not qualify. However, it would be difficult to measure this, so I doubt he would not pass the qualifications.
15. 27 Jun '05 08:38
Originally posted by bobbob1056th
3rd choice
A. "This is not the door to go through unless the sign on the adjacent door is true"
B. "Exactly two of these signs are false"
C. "This is the door to open unless the sign on the adjacent door is false"

4th choice
A. "No fewer than 2 of these signs are false"
B. "None of these signs are false"
C. "Go through either this do ...[text shortened]... ast one of these signs is false"
C. "If door A is not the one to go through, then door B is"
3rd door: B
A is saying: B false => not door A, B true => door A
C is saying: B true => door C, B false => not door C
If B were true, then A and C are false and with the above we have from A (being false): not door A, and from C (being false) : not door C, leaving door B as the solution
If on the other hand, B were false then there are two options: A and B are both true, or both false. If both true, then with B false the two are stating: not door A (says A) and not door C (says C) leading again to door B. If both are false, however, we have a contradiction: A says not A, but it is false, hence door A. But C says not C, but it is false , hence door C. But it cannot be both A and C, hence A and C are not false at the same time.

4th door: B
B is false. If it were true then it would contradict itself. Given that, A must be true, because if it were false, it would be true together with the false B.
So, with A true, C must be false. C states that it is either door C or door A (the true one). Hence it is the third one: door B.

5th door: C although with a degree of uncertainty (if A true and B false)
B is true. If not it would be true which is a contradiction. Three options:
- A false and B false cannot happen: from A (false) follows it has to be door C. But from C (false) follows that if it were door C (i.e. not A) then it is door B, which contradicts
- A true and B false: A says it is not door C, hence A or B; if A then from C (false) no conclusion can be drawn (no statement about what happens if A is the door). Assuming it is a clear solution, though, then A is not the door then C leads us to B, but it is false, hence it must be C
-A false and C true: A false leads to door C. But from C follows that if it is door C (not door A) then it is door B which is a contradiction.