Solution:

7 Trips

Explanation:
Send 2+5+9 = 16
Bring back 9
Send 3+4+9 = 16
Bring back 9
Send 1+7+8 = 16
Bring back 1
Send 1+6+9 = 16

Solution:

Pour 3 lumps on cup-1 and cup-3 and 4lumps on cup4 and place them as shown in the picture.

Explanation:
Now,
Cup-1 got three sugar lumps
Cup-3 got three sugar lumps
Cup-2 got three seven lumps (3 of cup-1 and 4 of its own)

Solution:

Federer serves first

Explanation:
The one who serves first can serve 5 and other will serve four.

== Option A ==
If Federer serve 4 and Nadal serve 5
Then five service breaks and Federer winning 6:3 not possible.

== Option B ==
Federer serve 5 and Nadal serve 4
There are five service breaks and score is Federer 6 - Nadal 3
i.e
On Federer serve, Federer wins 3 and lost 2
On Nadal serve, Federer wins 3 and lost 1

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

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:

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:

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:

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:

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:

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