Solution:

In the attached picture, you can see that the formation allows 2 castles to be protected by the walls. Congrats!

Solution:

The last marble will be yellow.

Since marbles can only be taken out in pairs and you started off with an odd number of yellows there is always going to be one yellow left over that you'll keep putting back in the box until it's left on it's own.

Solution:

The Bangladeshi seaman is the thief. This is because the Japanese flag cannot be identified as upside down as it is the same both ways. Thus he is just trying to escape the situation.

Solution:

25 Fruits

Remember we told you that Birbal is a witty trader. So his sole motive is to get rid of the sacks as fast as he can.

For the first sack:
He must be able to fill fruits from one sack to other two sacks. Assume that he is able to do that after M check points. Now to find M,

(Space in first sack) M + (Space in second sack) M = (Remaining fruits in Third Sack) 30 – M
M = 10

Thus after 10 checkpoints, Birbal will be left with only 2 sacks containing 30 fruits each.

Now he must get rid of the second sack.
For that, he must fill the fruits from second sack to the first sack. Assume that he manages to do that after N checkpoints.

(Space in First Sack) N = (Remaining fruits in second sack) 30 – N
N = 15

Thus after he has crossed 25 checkpoints, he will be left be one sack with 30 fruits in it. He has to pass five more checkpoints where he will have to give five fruits and he will be left with twenty five fruits once he has crossed all thirty check points.

Solution:

888 + 88 + 8 + 8 + 8 = 1000

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:

One of the cop will move clockwise while the other will move anti-clockwise every day. In such a manner, they will catch the thief in minimum seven days.

Solution:

The young lady was presented with a bill which belonged to the elderly woman. She had asked the waiter to collect the bill from her daughter.

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:

Zouma is 3 and has two footballs.
Ake is 1 and has one football.
Kalas is 4 and has three footballs.
Aina is 2 and has four footballs.

Solution:

13 x 3 + 1 = 40

The number of decks is absolutely not relevant here which means whether we mix 5 or 500 cards still results would be same.
Any card drawn will be a Ace,2,3,4,5,6,7,8,9,10,Jack,Queen or King, so there are 13 possibilities each time a card is drawn.
If you are lucky just 5 cards of the same kind can be obtained in 4 steps

The unluckiest(worst) way is our solution as we need to guarantee a four of a kind.
Draw 3 of each kind =>now we have 39 cards. Next card will guarantee 4 of a kind.