Solution:

February(just 28 days)

Solution:

Fifteen Minutes.

After frying for five minutes, you can take out one steak and put the third one inside while turning the other one inside. After five more minutes, one steak is fried from both sides and you can take it out. Now put in the steak that we removed and turn the other one inside. After five more minutes, they both will also be fried from both sides.

Solution:

192

Explanation:
I started with the number 192.
Adding the reversed 291 to 192 gave me 483.
Similarly,
483 + 384 = 867
867 + 768 = 1635
1635 + 5361 = 6996

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:

Anybody will prefer the shop in the eastern market for sure. There are two reasons for it:
Firstly, since there are only two barbers in the town, the one running the shop in eastern market must have gotten his hair done by the one running the shop in the northern market and vice versa. Thus, if the barber in the eastern market has a shabby haircut, it must be due to the incompatibility.
Secondly, the neat and clean shop can be due to the scarcity of customers as well. While the untidy place refers to the fact that so many customers get their hair done in that shop.

Thus any one will opt for the barber shop in the eastern market.

Solution:

This statement is true but only in a specific case i.e. if both of the siblings continue to live till the day the younger is as old as the elder one was during the time of their birth.

Solution:

31

Explanation:
Ramanujan lighting the candles in binary.
11110 =30
11111 =31

Solution:

zero

I had 3 green socks and 1 Pink sock.

=> chances of pair of green socks = 1/2
chances of pair of mixed socks = 1/2
chances of pair of green socks = NULL

Solution:

The women is a of short stature. She can't reach the upper elevator buttons, but can push is with his umbrella.

Solution:

2 minutes (duration between 9:59 and 10:01)

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.