Solution:

The player was having food on his plate while he was playing. The food consisted of pork ribs, lips, and chicken thighs. While playing he accidentally knocked down his plate. Therefore, the pig lips and ribs and chicken thighs were busted.

It was not his personal body injuries but the food that is being talked about in the question.

Solution:

Think deeply about the door number 56, people will visit it for every divisor it has. So 56 has 1 & 56, 2 & 28, 4 & 14, 7 & 8. So on pass 1, the 1st person will open the door; pass 2, 2nd one will close it; pass 4, open; pass 7, close; pass 8, open; pass 14, close; pass 28, open; pass 56, close.
Thus we can say that the door will just end up back in its original state for each pair of divisor. But what about the cases in which the pair of divisor has analogous number for example door number 16? 16 has the divisors 1 & 16, 2 & 8, 4&4. But 4 is recurrent because 16 is a perfect square, so you will only visit door number 16, on pass 1, 2, 4, 8 and 16… leaving it open at the end. So only perfect square doors will remain open at the end.

Solution:

If two widows, each having a son married the son of the other widow, and then by the consummation of marriage, both the couples had a daughter, all the aforementioned relationships will turn to be true.

Solution:

Both 3 and 4th

Explanation:
- The water in tank1 will get filled up to the connectivity of tank1 and the tank2 and then it will start flowing to tank2.
- The water in tank2 will get filled up to the connectivity of tank2 and the -tank3 and then it will start flowing to tank3.
- The water in tank3 will get filled up to the connectivity of tank2 and the tank3 and then it will start flowing to tank3.
- As the outlet of tank3 and inlet of tank4 are not at the same level, water will continue to fill tank3 till the inlet of tank4.
- Now both the tank3 and tank4 fill simultaneously until both fill together.

Solution:

Solution can be verified from the picture below.

Solution:

There is a simple logic to solve this question.
The size of the chocolate is 2 x 8. Thus, you need to have 2 x 8 = 16 pieces.
Every time you break the chocolate, you will get one extra piece.
Thus, to get 16 pieces, you must break it (16 - 1) = 15 times.

Solution:

At each draw, the number of strawberry candies are either decreasing by 2 or not decreasing at all. In the case of orange candies, at each draw, they are either increasing by 1 or decreasing by 1.

Thus on an assumed outset with at least one candy in the jar to begin with, if the number of strawberry candies are 0 or are even in numbers, they will finish off leaving an orange candy at the end. If otherwise, the remaining candy will be a strawberry one.

Solution:

It has been clearly mentioned that all the boxes are labeled incorrectly. If he opens the Box3, then he will get either Dark Chocolates or Milk Chocolates as it is labeled incorrectly. Let us suppose he finds Dark Chocolates in there. Now since all are labeled incorrectly, Box B A must contain Milk Chocolates and Box B must contain Milk Chocolates and Dark Chocolates.

Solution:

The teacher will use a simple trick to form a perfect impression. He will ask all his students to raise their hands on each of the question that is asked. But the only catch will be that those who knows the answer correctly will raise their right hand and others will raise their left hand.

In this way, the principal will see all the hands being raised for each question even though all won't be knowing the correct answer. The teacher will ask only those who know the answer and they will always be correct. So the principal will be impressed to full extent.

Solution:

Either your read from start or from ends, it will be same

Solution:

C7 => B8
Exchange pawn for a knight

Check-Mate

GameOver