@Rohit-Rathore
Q3) Let say we had x coins in the beginning
After 1st loot : x - 1 - (x-1)/3 = 2/3(x-1) = (2x-2)/3
After 2nd loot : (2x - 2)/3 - 1 - [(2x - 2)/3 - 1]/3 = 2/3[(2x-2)/3 - 1] = (4x - 10)/9
After 3rd loot : (4x - 10)/9 - 1 - [((4x - 10)/9 - 1]/3 = 2/3 [(4x - 10)/9 - 1] = (8x - 38)/27
Now the final value (after throwing away the fake coin) = (8x - 38)/27 - 1 = (8x - 65)/27 is a multiple of 3.
So we can write, (8x - 65)/27 = 3k
x = (81k + 65)/8
x is an integer, 81K + 65 should be a multiple of 8, means (81K + 65) Mod 8 = 0
We know 81 Mod 8 = 1 and 65 mod 8 = 1. So to get remainder 0, we need to adjust K in a way that 81K mod 8 = 7.
Least value would be K = 7
For K = 7, x = (81 * 7 + 65)/8 = 79, which should (could) be our answer.
Let me know in case of any Logical/Calculation errors.

Q 5) In a family of seven people A, B, C, D, E, F and G there is exactly one pair of twins. B is younger than F but older than E, G and D. C is younger than E but older than A and D. G is younger than F and E. Which of the following can be the pair of twins?
(a) B, C
(b) A, E
(c) C, G
(d) F, G

@shashank_prabhu
1)Option c
2)Option c
3)Option a
4)Option c

@Ashutosh-Singh Hello Ashutosh, can you also share the approach so that others can learn from them

Q26) A person buys 18 local tickets for Rs. 110. Each first class ticket costs Rs. 10 and each second class ticket costs Rs. 3. What will another lot of 18 tickets in which the number of first class and second class tickets are interchanged cost?
a. 112
b. 118
c. 121
d. 124