Latest Logic Riddles

Suppose that there is just one black hat. In such a scenario, the person wearing it will see the other two hats as white and then as the king had announced, he will be easily able to judge that he is wearing a black hat himself. But the other two men will see a black hat and a white hat and will not be able to judge. Thus this will be an unfair situation, so we can rule it out.

Now suppose that there are two black hats. Then, both the person will see a white hat and a black hat. And they have already realized that there must be two black hats, thus they can easily deduce that they are also wearing a black hat. But the one wearing white hat will see two black hats and thus he will not be able to identify on a sure note. Thus the competition will not be fair.

Thus the only situation where the competition can be fair is when all the hats are black. The first one who recognize the fact will stand up and say black.

Clearly as per the data given in the question, the ball will always keep on bouncing half way up and thus it will take infinite bounces before the gravity forces it to stop.

Initially
A – 0, B – 0, c -5, D – 5

Pour C into B
A – 0, B – 4, C – 1, D – 5

Pour C into B
A – 0, B – 4, C – 1, D – 5

Pour half of B into C
A – 0, B – 2, C – 3, D – 5

Pour D into A
A – 3, B – 2, C – 3, D – 2

Pour A into B
A – 1, B – 4, C – 3, D – 2

Pour half of B into D
A – 1, B – 2, C – 3, D – 4

All you have to do is labelling the pills as A and B. Now when you draw out three pills, you will either have 3A or 2A 1B or 1A 2B or 3B. Now if all the pills are from same bottle you can simply take another from the other bottle. If some other combination comes out, you can manage by filling in from the required bottles.

Friend 2 must have taken top 15 cards from the pile and reversed them. Now there are two piles, one with 15 cards and one with 37 cards and both of them will obviously have the same number of inverted cards.

If you want to understand mathematically, let us say that there were x inverted cards in the top 15 cards of the deck. Then the remaining 37 cards will comprise of 15-x number of inverted cards.

If we reverse the 15 cards the number of inverted cards will become 15-x and the number of inverted cards will be same in both the piles.

Just 3.

Refer the picture. All you have to do is cut three links from a single 3-link chain and then you can use them to join three other 3-link chains to make a big 12-link chain

All he need to do is ask one of his son to divide the land in two part and tell him that the other son will have the choice of selecting one of those two halves.

In this manner both of them will be happy.

Suppose that everyone is standing at A corner and they have to reach the B corner

Let the housemaid and Dog go to corner B first and housemaid returns back to corner A.

Next, the housemaid and the first son go to B and she comes back along with the dog.

Now the father and the second son go to B and father comes back.

Mother and Father go to B and the mother comes back.

Housemaid and Dog go to B and the father comes back.

Father and mother goes to B and Mother returns back.

Mother and first daughter go towards B, Housemaid and Daughter returns back.

Housemaid and second Daughter go to B and Housemaid comes back.

Housemaid and Dog goes to B.

So, everyone has reached the other side of the river successfully.

The girl was born on 29 February, thus she put 250 stamps every four years.
In forty years, she put stamps only 10 times which makes the total of 2500 stamps.
Her sister was born on any other day and she took out 50 stamps from the box forty times which makes the total stamp she took out to be 2000.

Thus after forty years, the girl’s box had only 500 stamps.

Anabelle is no dumb trader and she knows the strategy with the help of which she will be able to save 25 artifacts even after passing 30 check points.

All she will focus on is getting rid of the boxes as soon as possible.

To shed the first sack quickest possible, she will start filling artifacts from one box to other two. Assume that she is able to do that after passing through A check points.

(Space in first box) A + (Space in second box) A = (Remaining artifacts in third box) 30 -A
A = 10

Therefore, after 10 check points, she will be left with two boxes with 30 artifacts each. Now to get rid of the second box, she will start filling artifacts from second box to first box. Let us assume that she is able to get rid of the second box after B checkpoints.

(Space in first box) B = (Remaining artifacts in second box) 30 - B
B = 15

Therefore, after fifteen check points, she will be left with one sack with thirty artifacts in it.

Since she has five more check points left, she will have to give five more artifacts and after passing them all, she will still have 25 artifacts.