Solution:

Let us solve this problem with variables to find the general nature. Let us assume that there are x black chocolates and y white chocolates. Three things can happen when you take out two chocolates.
1) You take out two black chocolates.
Then, you will be replacing one black chocolate and so the jar will have x+1 black chocolates and y-2 white chocolates.
2) You take out two white chocolates.
Then, you will be replacing 1 black chocolate and so the jar will have x-1 black chocolates and y white chocolates.
3) You take out 1 black chocolate and 1 white chocolate. Then, you will be replacing a white chocolate and so the jar will have x-1 black chocolates and y white chocolates.
Thus at every step, you are wither removing a black chocolate or replacing two white chocolates by one black chocolate irreversibly till there is just one chocolate left. Therefore each white chocolate is equivalent to two black chocolates.

This, if the number of white chocolates are odd in number, the last one remaining will be white chocolate and if the number of white chocolates are even in number, the last one remaining will be black chocolate.

Solution:

111111111

Explanation:
It's a maths magical square root series as :
Square root of number 121 is 11
Square root of number 12321 is 111
Square root of number 1234321 is 1111
Square root of number 123454321 is 11111
Square root of number 12345654321 is 111111
Square root of number 1234567654321 is 1111111
Square root of number 123456787654321 is 11111111
Square root of number 12345678987654321 is 111111111 (answer)

Solution:

4

No Of circles in the number.

Solution:

I + IX = X

The equation read from the other side becomes X = XI + I which is false.

Solution:

There are 3 fruits (1 grapes, 1 apples, and 1 oranges).

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:

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:

Blue could only be wearing white or red and we know that there is already someone else wearing the white shirt so Mr. Blue could only be wearing the red shirt.

Mr. White could have only been wearing a blue or a red shirt, and red is already taken, so Mr. White is wearing a blue shirt.

Mr. Red now has to be wearing a white shirt.

Solution:

Rhea will open just one present.
Cindy|Y - Y is the 25th alphabet
Duk|E - E is the 5th alphabet
Joh|N - N is the fifteenth alphabet
Rhe|A - A is the first alphabet

Solution:

The equation can be solved by substituting
B = 7
A = 8
S = 8
E = 3
L = 5

Therefore the equation now looks as:
7483 +
7455
-------
14938
-------

Solution:

What must be kept in mind before you start calculating through any method is that the weights can be put on both the pans.

So the weights - 1, 3, 7, 12, 43, 76, 102

The oldest man that can be weighed using these weights is a man aged 122 years