Solution:

Man1 is carrying 5 bags and the Man2 is carrying 7 bags.

Let's assume that the Man1 was carrying M1 bags and the Man2 was carrying M2 bags.
M2 + 1 = 2 * (X-1)
M2 + 1 = 2M1 - 2
M2 = 2M1 - 3

Also,
M2 - 1 = M1 + 1
M2 = M1 + 2

=>
2M1 - 3 = M1 + 2
M1 = 5

=>
M2 = 7

Solution:

Flyball should pay Lavy $11 instead of accepting his counter offer as in that case the club will end up paying almost double the amount.

Solution:

Let us assume that all of them are standing on the X shore and they have to travel to the Y shore.

X = mother, first daughter, second daughter, father, first son, second son, maid and dog
Y = None

Maid and dog go first and maid returns back
X = mother, first daughter, second daughter, father, first son, second son and maid
Y = dog

Maid and first son go and maid returns with dog
X = mother, first daughter, second daughter, father, second son, maid and dog
Y = first son

Father and second son go and father returns back
X = mother, first daughter, second daughter, father, maid and dog
Y = first son and second son

Mother and father go and mother comes back
X = mother, first daughter, second daughter, maid and dog
Y = father, first son and second son

Maid and Dog go and Father returns back
X = mother, first daughter, second daughter and father
Y = first son, second son, maid and dog

Father and Mother go and mother comes back
X = mother, first daughter and second daughter
Y = father, first son, second son, maid and dog

Mother and first daughter go and maid comes back with dog
X = second daughter, maid and dog
Y = mother, first daughter, father, first son and second son

Maid and second daughter go and maid comes back
X = Maid and dog
Y = mother, first daughter, second daughter, father, first son and second son

Maid and dog go
X= none
Y = mother, first daughter, second daughter, father, first son, second son, maid and dog

Thus everyone has reached the other end.

Solution:

All Digits [0-9] follow the following pattern: There is one occurrence of all digits in every 10 numbers Example. Digit-9 occurred one time between 10 and 19 in 9, between 20 and 29 in 29
However, for tens(91,92,93...99) and hundreds (900,901.... 999), the occurrence is much more. Since we have included 1000, Therefor 1 is the most occurred digit.
Note: Despite three '0s' in 1000, 0 is not most occurred digit as they don't have tens and hundreds like other digits.

Solution:

I filled a glass full of water and held it over my head for 20 minutes.

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:

13 + 11 + 6 = 30

Explanation
The trick is that you just need to rotate '9' to form '6'.

Solution:

Michelangelo juggles the diamonds across while crossing the bridge.

Solution:

If you notice the animated gif, you know that D, C and B are going to die. D dies directly after coming in contact with the object, C dies after being thrust by the spikes of the see-saw and B dies with the blue ball.

But consider the situation that the person at E stands for Ebola. Eventually A dies as well.

We thank 9gags for this awesome gif.

Solution:

This can be done if you draw a five pointed star and place the balls on the corners and intersection points. See the figure to understand better.

Solution:

19

Explanation:
The children can purchase 19 chocolates in return.

Out of 77 wrappers, 75 will be used to buy 15 chocolates and two will be left spare.
The 15 chocolates will create 15 empty wrappers that can be exchanged to get three chocolates.
Three chocolates will return three wrappers which will help them buy another chocolate.
Now the wrapper from this chocolate and the two spare that were left earlier will get them another chocolate.
15 + 3 + 1 = 19