Solution:

5*(X+6) = (X+6)+21
4*(X+6) = 21
X+6 = 5.25
X = -0.75

So Brad is at top of Kukki.

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:

13 x 3 + 1 = 40

The number of decks is absolutely not relevant here which means whether we mix 5 or 500 cards still results would be same.
Any card drawn will be a Ace,2,3,4,5,6,7,8,9,10,Jack,Queen or King, so there are 13 possibilities each time a card is drawn.
If you are lucky just 5 cards of the same kind can be obtained in 4 steps

The unluckiest(worst) way is our solution as we need to guarantee a four of a kind.
Draw 3 of each kind =>now we have 39 cards. Next card will guarantee 4 of a kind.

Solution:

Although it looks complicated, the answer is simple. The password to his computer is JIGSAW.
1st letter of Jug = J
2nd letter of Birthdays = I
3rd letter of Fights = G
4th letter of Cars =S
2nd letter of Laptops = A
1st letter of Watch = W

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:

First fill the 8 liters jug complete - 4, 8, 0
Fill the 5 liters jug with 8 liters jug - 4, 3, 5
Pour back the beer from 5 liters jug to 12 liters jug - 9, 3, 0
Pour the 3 liters from 8 liters jug to 5 liters jug - 9, 0, 3
Fill the 8 liters jug completely from 12 liters jug - 1, 8, 3
Fill the 5 liters jug from the 8 liters jug - 1, 6, 5
Pour the entire 5 liters jug back in 12 liters jug - 6, 6, 0

You have successfully split the beer into two equal parts.

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

Solution:

We now have a square of 3x3 size, 2 squares of 2x2 size and 5 squares of 1x1 size.

Solution:

At first, there were 2 apples on the tree. After the wind blew, one apple fell on the ground. So there where no apples on the tree and there were no apples on the ground.
*apples => more than one apple

Another solution(non tricky one)
The wind blew so hard that the apples fell of the tree and blew along the ground.

Solution:

Think deeply about the door number 56, people will visit it for every divisor it has. So 56 has 1 & 56, 2 & 28, 4 & 14, 7 & 8. So on pass 1, the 1st person will open the door; pass 2, 2nd one will close it; pass 4, open; pass 7, close; pass 8, open; pass 14, close; pass 28, open; pass 56, close.
Thus we can say that the door will just end up back in its original state for each pair of divisor. But what about the cases in which the pair of divisor has analogous number for example door number 16? 16 has the divisors 1 & 16, 2 & 8, 4&4. But 4 is recurrent because 16 is a perfect square, so you will only visit door number 16, on pass 1, 2, 4, 8 and 16… leaving it open at the end. So only perfect square doors will remain open at the end.

Solution:

Both will rotate at Same speed.

Explanation:
Gears of the same size will always rotate at constant speeds despite any smaller or larger gears in between them.