Solution:

30

Explanation:
It's very difficult to count an actual number of balls but it can be counted mathematically as illustrated below.

Balls in lowest level most level, say level 1 : 4 * 4 = 16
Level2 => 3 * 3 = 9
Level3 => 2 * 2 = 4
Leve4 => 1

Summing up 16+9+4+1 = 30.

Solution:

3012

Explanation:
A)
Five Commando alloted 5 room each, Number of rooms = 5 * 5 = 25

Number of huge Tiger = 25 * 5 = 125
Number of small Tigers = 125 * 5 = 625
Total tigers = 625 + 125 = 750

Number of legs of tigers = 750 * 4 = 3000 ..............................(1)

B)
There are 5 soldiers =>
Number of legs of soldiers = 5*2 = 10 .......................(2)

C)
Your leg = 2 ............(3)

Summing all , 3000 + 10 + 2 = 3012

Solution:

301

= Explanation =
Time1 : When there are two pirates, 301/2 => remainder 1
Time2 : When there are three pirates, 301/3 => remainder 1
Time3 : When there are three pirates, 301/4 => remainder 1
Time4 : When there are three pirates, 301/5 => remainder 1
Time5 : When there are three pirates, 301/6 => remainder 1
Time6 : When there are three pirates, 301/7 => remainder 0

Solution:

To add commutative balance in the puzzle above is mistake
You will get this when I show you another way to spend $50.

Spent Balance
5                   45
5                   40
5                   35
5                   30
5                   25
5                   20
5                   15
5                   10
5                   05
5                   00
=========
50                   225

The numbers i choose smartly to confuse readers.

Solution:

Sam : 2
Dean : 5
Castiel : 7

Age diff
7 - 2 = '5' is prime
7 - 5 = '2' is prime
5 - 2 = '3' is prime

Solution:

As shown in the picture below, you should tilt the barrel so the liquid will touch the upper end of the barrel. So if the bottom part of the barrel is visible then it's not half full.

Solution:

111111111

Explanation:
It's a maths magical square root series as :
Square root of number 121 is 11
Square root of number 12321 is 111
Square root of number 1234321 is 1111
Square root of number 123454321 is 11111
Square root of number 12345654321 is 111111
Square root of number 1234567654321 is 1111111
Square root of number 123456787654321 is 11111111
Square root of number 12345678987654321 is 111111111 (answer)

Solution:

They were the same person "Grover Cleveland".

Solution:

Both 3 and 4th

Explanation:
- The water in tank1 will get filled up to the connectivity of tank1 and the tank2 and then it will start flowing to tank2.
- The water in tank2 will get filled up to the connectivity of tank2 and the -tank3 and then it will start flowing to tank3.
- The water in tank3 will get filled up to the connectivity of tank2 and the tank3 and then it will start flowing to tank3.
- As the outlet of tank3 and inlet of tank4 are not at the same level, water will continue to fill tank3 till the inlet of tank4.
- Now both the tank3 and tank4 fill simultaneously until both fill together.

Solution:

The five cards must contain at least one card of the same suit.
Let's assume Rhodes placed one of the two cards of the same suit at the end and the second card was given to Daniel.
In that case, Daniel will know the suit of the card.

Now the real question is how Rhodes made sure that Daniel also knows the number of the card.

Daniel got four cards and one of them will determine the suit, so we are left with 3 cards.
3 cards can be arranged in 3! ways i.e. 6 ways but our card numbers can vary from 1 to 13.
Since we have two cards of the same suit we will make sure that we don't pick the King (13). This leaves us with 12 numbers.

We can also distinguish card from upside down position, therefore, we now have 6 * 2 ways to arrange these card.
"TADA.... We cracked the magic trick".

Let us assume that the smallest card is C1U (upwards position) and C1D (downwards position). Similarly, we have C2U, C2D, C3U and C3D

According to that, the 12 possible arrangements are:
C1U C2U C3U => NUMBER 1
C1U C3U C2U => NUMBER 2
C2U C1U C3U => NUMBER 3
C2U C3U C1U => NUMBER 4
C3U C1U C2U => NUMBER 5
C3U C2U C1U => NUMBER 6
C1D C2D C3D => NDMBER 7
C1D C3D C2D => NDMBER 8
C2D C1D C3D => NDMBER 9
C2D C3D C1D => NDMBER 10
C3D C1D C2D => NDMBER 11
C3D C2D C1D => NDMBER 12

Please note that king (number 13) is not possible as discussed already.