Solution:

60

Explanation:
You can encode MM in the first digit leaving the bottom segment MM (6^2=64>=60).
You will need some order so that they can represent the MM in binary.
Every single element of the display is a binary digit:
"On"=1
"Off"=0

Now, the seconds can be encoded in the similar fashion by using the second digit.

You can now use the remaining four elements of display (two bottom segments and two dots) to encode HH (4^2=16>=12).

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:

C7 => B8
Exchange pawn for a knight

Check-Mate

GameOver

Solution:

You must put labels on the tablets as A and B before using. In that case, if you pour tablets together, you will get 3A, 2A 1B, 1A 2B or 3B. If they are from the same bottles you can take one from another bottle and save the remaining two for another day. If you get two from same and one from other, you can draw one from another bottle and you will have two pairs of which you can eat one and save the other.

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:

It was only a trick

I borrowed my parents (Pound 50 + Pound 50 = Pound 100).
I bought the book for Pound 97 and i had Pound 3 change.
So, the money that i borrowed and bought the book are Pound 100 (borrowing money)
and the change Pound 3 also the borrowing money the book is Pound 97
Total: Pound 97 + Pound 3(the change) = Pound 100
The money that i used to buy the book all i borrowed.
Pound 97 is the borrowing money, change: Pound 3 also the borrowing money. Pound 97 + Pound 3 = Pound 100

Solution:

The stranger said that he thought the room was his. If so, he must have had the keys to the room and thus he must not have knocked.

Solution:

he was genius.

Officials must have thought he was jumping with a hostage, they would never risk giving him a faulty parachute.

Solution:

In the corners of the square, tighten two ropes. Tighten another rope between them by putting rings on each rope. That rope should be tightened on the sheep's neck on the other end.

Solution:

The competition was held on 11 November 2013. The logic is simple. In America, the dates are mentioned as Month/Day/Year and in Britain, dates are written as Day/Month/Year. But on 11 November 2013, the date and month will be the same. And thus both will write in the same manner i.e. 11/11/2013. If it would have been any other date, the evaluators would have been able to identify it.

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.