Solution:

East

Explanation:
- Right Turn => North Direction
- About Turn => south Direction
- Left Turn => East Direction

Solution:

At 01:05

Solution:

Old password : altered

New Password : distinct

Solution:

Let m be initial number of flowers and n be the flowers offered at each temple by the man.
Initial flowers = m
After stepping into the first temple = 2m
Flowers offered at the first temple = n
Remaining flowers = 2m – n

After stepping into the second temple = 4m – 2n
Flowers offered at the second temple = n
Remaining flowers = 2m – n

After stepping into the third temple = 8m – 6n
Flowers offered at the third temple = n
Remaining flowers = 8m – 7n

Now as per the question (8m – 7n) must be equal to 0.
i.e. 8m – 7 n = 0
or 8m = 7y

Now the least numbers required to solve this equation are 7 and 8 for m and n respectively.

Thus m = 7; n = 8
Which means, the man had 7 flowers in the beginning and he offered 8 flowers at each temple.

Solution:

Take 10 number of cards in a new deck and change their face direction. For example- You create a new deck of 10 cards and out of 10, 3 faces up in the new deck. So remaining 7 faces up are in the old deck. But hey! while creating the new deck you reversed the face direction of new cards. So actually the 3 cards which were facing up are actually face down in the new deck while 7 faces up.

Solution:

Ganpat had a black suit and a brown suit.
Harish had a brown suit and a pink suit.
Manoj had a white suit and a black suit.
Tarun had a pink suit and a white suit.

Solution:

Everyone will done one day or other.

Abraham Lincoln era is too old and we expect most of them(if not all) to be dead by now.

Solution:

This puzzle has no answer if you are considering this figure in a 2d domain. However if you follow it on a three-dimensional plane, you can do it like shown in the figure given.

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:

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:

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.