Solution:

The password that opens the chamber is 65292.

Solution:

1/3

Explanation
Assuming T1 is the Toy for brother1, T2 is the toy for brother2 and T3 is the toy for brother3.

Following are the possible cases for toys distribution:

Boy1 Boy2 Boy3
T1 T2 T3
T1 T3 T2
T2 T1 T3
T2 T3 T1 .... (A)
T3 T1 T2 .... (B)
T3 T2 T1

In both steps (A) & (B), no one gets the correct toy.
Therefore probability that none brother can get the own toy is 2/6 = 1/3

Solution:

If you are thinking to hold one chocolate from each box in hand and then balancing weight in bare hands, you are thinking all wrong.

Let us begin by labelling boxes as 1, 2, 3 and so on till 10.

Now pick one chocolate from box 1, two chocolates from box 2, three from box 3 and so on. In total, you will have 55 chocolates now. (1 + 2 + 3 + ..... + 10)

The ideal weight of the chocolates should be 55 * 20 = 1100. However, somewhere in there are the defected chocolate/s.

You can judge that clearly by noting down the result of 1100 - total weight of chocolates. If the weight is less than 1 gram, the defected box is box 1, if the weight is less than 2 grams, the defected box is box 2 and so on.

Solution:

The receptionist was pretty clever she played with the mute button of the phone while talking with her mother. She muted everything except the word emergency, call and help. So while talking, she sounded like 'Emergency… Call… Help' to her mother and thus she called the police.

Solution:

There is no mention if you can use sign or not and thus the only possible way to get this is as follows:

(15 - 9) + (13 - 7) + (7 - 1) + (9 - 1) + (13 - 9)

If you solve inside the brackets, you will get the following equation
6 + 6 + 6 + 8 + 4
Adding all these numbers, you will get 30 which is the required sum.

Solution:

7, 4, 6, 5, 8

Solution:

Yes it is possible in Roman Numerals. In Roman Numerals,
9 = IX
10 = X
11 = XI

Thus by removing one from 9, you are getting 10 and by removing one from 11, you are getting 10 again.

Solution:

Solution is below

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:

5 leopard and one sheep

since all animal are rational , once 1st leopard eats the sheep all the rest of leopard would know about this.

Solution:

The man asked the king that he wanted to die of old age.