Solution:

1. Start both the 7 minute hour glass & 11 minute hour glass.
2. Wait till the 7 minute hour glass times out. Time is 7 minute!
3. Restart the 7 minute hour glass. At this time 11 minute hour glass will have 4 minutes left to time out.
4. As soon as 11 minute glass times out invert the 7 minute hour glass. Total time now is 11 minutes.
5. After inverting 7 minute hour glass, it will now have 4 minutes left for time out.
6. After these 4 minutes times out, the total time is 15 minutes.

Solution:

charles put the ring into the box, secure it with one of your locks, and send the box to the girl firend.
She should then attach one of his own locks and return it. When you receive it again, remove your lock and send it back with her lock. Now she can unlock his own lock and retrieve the object.

Solution:

The husband had booked two tickets for departure and only one for the return. The travel agent told this information to the police and they immediately came to know that he had planned the murder right from the beginning.

Solution:

To do this, he must be pushing the cork inside the bottle and then taking out the coin.

Solution:

10 prisoners must sample the wine. Bonus points if you worked out a way to ensure than no more than 8 prisoners die.

Number all bottles using binary digits. Assign each prisoner to one of the binary flags. Prisoners must take a sip from each bottle where their binary flag is set.

Here is how you would find one poisoned bottle out of eight total bottles of wine.

Bottle 1 Bottle 2 Bottle 3 Bottle 4 Bottle 5 Bottle 6 Bottle 7 Bottle 8
Prisoner A X X X X
Prisoner B X X X X
Prisoner C X X X X
In the above example, if all prisoners die, bottle 8 is bad. If none die, bottle 1 is bad. If A & B dies, bottle 4 is bad.

With ten people there are 1024 unique combination so you could test up to 1024 bottles of wine.

Each of the ten prisoners will take a small sip from about 500 bottles. Each sip should take no longer than 30 seconds and should be a very small amount. Small sips not only leave more wine for guests. Small sips also avoid death by alcohol poisoning. As long as each prisoner is administered about a milliliter from each bottle, they will only consume the equivalent of about one bottle of wine each.

Each prisoner will have at least a fifty percent chance of living. There is only one binary combination where all prisoners must sip from the wine. If there are ten prisoners then there are ten more combination where all but one prisoner must sip from the wine. By avoiding these two types of combination you can ensure no more than 8 prisoners die.

One viewer felt that this solution was in flagrant contempt of restaurant etiquette. The emperor paid for this wine, so there should be no need to prove to the guests that wine is the same as the label. I am not even sure if ancient wine even came with labels affixed. However, it is true that after leaving the wine open for a day, that this medieval wine will taste more like vinegar than it ever did. C'est la vie.

Solution:

Take 10 number of cards in a new deck and change their face direction. For example- You create a new deck of 10 cards and out of 10, 3 faces up in the new deck. So remaining 7 faces up are in the old deck. But hey! while creating the new deck you reversed the face direction of new cards. So actually the 3 cards which were facing up are actually face down in the new deck while 7 faces up.

Solution:

We now have a square of 3x3 size, 2 squares of 2x2 size and 5 squares of 1x1 size.

Solution:

I spoke with Joe.

Explanation:
There can be only two cases. Either Zoe is a liar or Joe is a liar.

First Case
Let us assume that Zoe is a liar and Joe is a truthful person.
If I asked the question from Joe, the answer will be yes. If I asked the question from Zoe, the answer will be no. Thus in this case, I must have asked from Joe.

Second Case
Let us assume that Zoe is a truthful and Joe is a liar.
If I asked the question from Joe, the answer will be yes. If I asked the question from Zoe, the answer will be no. Thus in this case as well, I must have asked from Joe only.

Solution:

It is best if the person sits on the back of the train. After the stop, the train will be accelerating as it goes into the tunnel. Thus the train will be much master when the back of the train enters the tunnel. In this way, he will have to spend much less time in the tunnel.

Solution:

The women is a of short stature. She can't reach the upper elevator buttons, but can push is with his umbrella.

Solution:

There are two things to keep in mind:
Firstly there is at least one red hat. (There can be two or three as well).
Secondly the competition is fair for everyone.

Thus if there is only one red hat, that person will see two green hats on other heads and will be able to deduce his own color as red. However the other students will see one red and one green hat and can never be sure. In such manner, the competition will prove to be partial for one student.

Suppose if there are two red hats. Then the students who are wearing red hats will see one red and one green hat on others. Now they must have deduced that there can’t be just one red hat. Thus they will know that they are also wearing a red hat. But the one who is wearing a green hat will see two red hats and can never be sure of his own color. In this case as well, the competition will not be fair.

Thus the only possible and fair means is if all of them are wearing a red hat. The one who is able to deduce the situation first, will raise his hand and will tell the correct answer.