If a 1 mm thick paper is folded so that the area is halved at every fold, then what would be the thickness of the pile after 50 folds?

Clearly, the required number would be such that it leaves a remainder of 1 when divided by 2, 3 or 4 and no remainder when divided by 5. Such a number is 25.

Let number of keepers be x. Then,

Total number of feet = 2 x 50 + 4 x 45 + 4 x 8 + 2x = 2x + 312.

Total number of heads = 50 + 45 + 8 + x= 103 + x.

Therefore (2x + 312) = (103 + x) + 224 or x = 15.

Income on the first day = Re. 1.

Income on the 2nd day = Rs. (1x2) = Rs. 2¹.

Income on the 3rd day = Rs. (2¹ x 2) = Rs. 2² and so on. Thus, Income on the rath day = Rs. 2^n-1.

Therefore Income on the 10th day = Rs. 2⁹.

The seven pieces consist of 6 smaller equal pieces and one half cake piece.

Weight of each small piece = 20 g.

So, total weight of the cake = [2 x (20 x6)]g= 240 g.

Let the number of 20-paise coins be x. Then, number of 25-paise coins = (324 - x).

Therefore 0.20 x x + 0.25 (324 - x) = 71 ⟺ 20x + 25 (324 - x) = 7100

⟺ 5x= 1000 ⟺ x = 200. Hence, number of 25-paise coins = (324 - x) - 124.

Let the father's age be x and the son's age be y.

Then, x - y = y or x = 2y,

Now, x = 36. So, 2y = 36 or y = 18.

Therefore Son's present age = 18 years.

So, son's age 5 years ago = 13 years.

Since the number of carbons is 2, only two copies can be obtained.