Solution:

In the attached picture, you can see that the formation allows 2 castles to be protected by the walls. Congrats!

Solution:

As shown in the picture below.
Cut 1 : Cut the cake across the diameter.
Cut 2 : Cut horizontally and you will get 4 equal pieces of cake
Cut 3 : Cut horizontally from other side and you will get 8 equal pieces of cake

Solution:

Sucre : 1 cupcake
T-Bag : 2 cupcakes
Burrows: 3 cupcakes
Scofield : 5 cupcakes

Explanation :
It's obvious if the prisoner has eaten 5 or more cupcakes than he will not ask this question "did you had more cupcakes than me?" as it will make him the person who had eaten maximum cupcakes.
A) Sucre : Must not have eater 5 cupcakes, therefore he has eaten anywhere between 1-4 cupcakes.

B) T-bag : Since sucre asked him the question T-bag knows sucre have eaten between 1 - 4 cupcakes, and therefore he must have eaten 2 to 4 cupcakes(it can't be 1 else he will reply answer as "No".)

C) Burrows : Replied with the answer as "I don't know and based on logic B) he knows T-bag must have eaten at least 2 and he replied no which implies that he must be eaten 3 or 4.

D) Scofield replied he know all, which indicates he must have eaten 5.
Therefore cupcake division is as below:
Sucre : 1
T-Bag : 2
Burrows: 3
Scofield : 5

Solution:

63.984375

Explanation:
This question is hypothetical because you cannot transact in fractions in bank account. The following is the solution:

Year Amount Interest and Deduction
Start 63.984375
Y1 End 62.968750 63.984375 * 2 - 65
Y2 End 60.937500 62.96875* 2 - 65
Y3 End 56.875000 60.9375 * 2 - 65
Y4 End 48.750000 56.875 * 2 - 65
Y5 End 32.500000 48.75 * 2 - 65
Y6 End 0.000000 32.5 * 2 - 65 = 0

Solution:

The chocolates in the boxes will be as follows:
Box 1 = 1
Box 2 = 2
Box 3 = 4
Box 4 = 8
Box 5 = 16
Box 6 = 32
Box 7 = 64
Box 8 = 128
Box 9 = 256

Here we have taken a geometric progression to facilitate any number of chocolates demanded. In the last box, we will place the number of chocolates left:

Box 10 = 489

If you try now, you will be able to do what is asked.

Solution:

You just need to check $150 bag.

Explanation:
Note: All bags labels are incorrect.
Case 1: If it contains $100 bill, then
A) Current bag contains $200 bills
B) $100 bag contains $150 bills
C) $200 bag contains $100 bills

Case 2: If it contains $50 bill, then
A) Current bag contains $100 bills.
B) $100 bag contains $200 bills
C) $200 bag contains $150 bills

Solution:

73

Explanation:
It's a very famous puzzle known as "Josephus Problem Puzzle". It can be solved in various ways but the trick to solve this problem is illustrated below.

Binay number of 100 is: 1100100
complement pf 100 is 0011011, whose decimal equivalent is 27.
The desired answer will be 100 - 27 = 73

Solution:

The code is 645
Explanation :
A) From Clues 1st and 2nd we know that one 9 is not in the code
B) From Clue 2nd we know that 9 3 and 8 are part of code
C) From Clue 3rd we know that 4 and 6 are part of code
D) From Clue 5th we know that 1 and 2 are not part of the code as we know 6 is part of the code and just 1 of 3 numbers is part of code.
E) From Clue 4th we know that 5 is also in the code as 8 and 1 are not part of code.

So the code is: 645

Solution:

Day 1:
Alibaba Give 1kg gold bar.

Day2:
Now Alibaba gives 2kg gold bar and takes back 1kg gold bar back.

Day3:
Alibaba gives 1kg gold bar.

Day4:
Alibaba gives 4kg gold bar and take back 1kg gold bar back

Day5:
Alibaba gives 1kg gold bar.

Day6:
Alibaba gives 2kg gold bar and takes back 1kg gold bar back.

Day7:
Alibaba gives 1kg gold bar.

Solution:

Rashid's son-1: 1 11 21 31 41 51 61 71 81
Rashid's son-2: 2 12 22 32 42 52 62 72 73
Rashid's son-3: 3 13 23 33 43 53 63 64 74
Rashid's son-4: 4 14 24 34 44 54 55 65 75
Rashid's son-5: 5 15 25 35 45 46 56 66 76
Rashid's son-6: 6 16 26 36 37 47 57 67 77
Rashid's son-7: 7 17 27 28 38 48 58 58 78
Rashid's son-8: 8 18 19 29 39 49 59 69 79
Rashid's son-9: 9 10 20 30 40 50 60 70 80