Solution:

A lion, king of the jungle! (the head of a Lamb, the middle of a pIg, the hind of a buffalO and the tail of a dragoN)

Solution:

The node can be connected as shown.

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:

mother

We know from (3) that the youngest person was not the victim, from (4) that the youngest person was not the helper and from (6) that the youngest person was not the killer. The youngest person can only have been the witness therefore. If we make up a chart there are now three possible combinations:

Oldest person (father) H H M

Next to oldest (mother) V M H

Next to youngest (son) M V V

Youngest (daughter) W W W

(H = Helper ; V = Victim ; M = Murderer ; W = Witness)

We can work out from (5) that the father was the oldest, from (2) that the youngest person must have been the daughter. Therefore the next to the youngest must have been the son and the next to the oldest, the mother.

Solution:

Kid1, he was left with one rock.

Solution:

NONE, holes are empty.

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 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:

16

Explanation:
9 [1x1] squares
5 [2x2] squares
1 [4x4] square
1 [4x4] square

9 + 5 + 1 + 1 = 16 squares.

Solution:

The password is U472BMT.

Read aloud the line and you will sound something like "U Four Seven 2 B M T".

Solution:

Old password : altered

New Password : distinct