Solution:

No Excuse

Explanation:
Since there is no X listed =>No Excuse

Solution:

The answer is 2. First, divide the coins into 3 equal piles. Place a pile on each side of the scale, leaving the remaining pile of 3 coins off the scale. If the scale does not tip, you know that the 6 coins on the scale are legitimate, and the counterfeit is in the pile in front of you. If the scale does tip, you know the counterfeit is in the pile on the side of the scale that raised up. Either way, put the 6 legitimate coins aside. Having only 3 coins left, put a coin on each side of the scale, leaving the third in front of you. The same process of elimination will find the counterfeit coin.

Solution:

It can surely be true because it is not mentioned that it was raining.

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:

The brain teaser can be solved as shown in the image below.

Solution:

The first step that they will take will be a leap of logic. What it means is that they will deduce that every color must appear twice at least. Why? Because the master logician has assured them that the puzzle will not be impossible for anyone of them. And if a color appears only once in the circle, the person wearing it will have no clue about that color which will not be fair for him.

Then the logicians will follow the same and look for all the colors of hats in the circle. If one of them sees a color just once, he can safely assume that he is also wearing the hat of the same color as by leap of logic, no color can appear just once. Thus when the bell is rung, he will leave.

In the similar fashion, if anyone sees another color just once, he can determine that he is wearing the hat of the same color and will leave when the bell rings or will be disqualified and sent home. Unvaryingly, if a color is seen twice, they will be eliminated after the first bell. Hence, there must be at least three hats of any of the remaining color.

Assume that you are sitting in the circle and you don’t see a color once but see it twice. Then if they were the only two hats of the same color, the logicians must have left at the first bell already. But they did not. Which means that there are three hats of that color and you are wearing one. Thus you will leave after the second bell.

Solution:

In the picture you can find the green matchsticks. We have moved them from the bottom and placed them in the middle. It was a bit tricky but now you have your desired result.

Solution:

1/2

one of two is the possibility
1. If any of the first 99 people sit in Abhishek seat, Abhishek will not get to sit in his own seat.
2. If any of the first 99 people sit in Aishwarya's seat, Abhishek will get to sit in his seat.

Solution:

C7 => B8
Exchange pawn for a knight

Check-Mate

GameOver

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:

2100 miles.

Each vowel equivalent to 500 and each consonant equivalent to -200