A placed three sheets with two carbons to get two extra copies of the original. Then he decided to get more carbon copies and folded the paper in such a way that the upper half of the sheets were on top of the lower half. Then he typed. How many carbon copies did he get?

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.

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.

Clearly, the black cards are either clubs or spades while the red cards are either diamonds or hearts.

Let the number of spades be x. Then, number of clubs = (7 - x).

Number of diamonds = 2 x number of spades = 2x;

Number of hearts = 2 x number of diamonds = 4x.

Total number of cards = x + 2x + 4x + 7 - x = 6x + 7.

Therefore 6x + 7 = 13 ⟺ 6x = 6 ⟺ x - 1.

Hence, number of clubs = (7 - x) = 6.

Numbers from 1 to 60, which are divisible by 6 are : 6,12,18, 24, 30, 36,42, 48, 54, 60.

There are 10 such numbers.

Numbers from 1 to 60, the sum of whose digits is 6 are : 6, 15, 24, 33, 42, 51, 60.

There are 7 such numbers of which 4 are common to the above ones. So, there are 3such uncommon numbers.

Numbers from 1 to 60, which have 6 as one of the digits are 6, 16, 26, 36, 46, 56, 60.

Clearly, there are 4 such uncommon numbers.

So, numbers 'not connected with 6' = 60 - (10 + 3 + 4) = 43.

No. of digits in 1-digit page nos. = 1x9 = 9.

No. of digits in 2-digit page nos. = 2 x 90 = 180.

No. of digits in 3-digit page nos. = 3 x 900 = 2700.

No. of digits in 4-digit page nos. = 3189 - (9 + 180 + 2700) = 3189 - 2889 = 300.

Therefore No. of pages with 4-digit page nos. = (300/4) = 75.

Hence, total number of pages = (999 + 75) = 1074.

Let the number be x. Then, x + 13x = 112 ⟺ 14x = 112 ⟺ x = 8.