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.

Let the fixed charge be Rs. x and variable charge be Rs.y per km. Then,

x + 16y = 156 ...(i) and

x + 24y = 204 ...(ii)

Solving (i) and (ii), we get: x = 60, y = 6.

Therefore Cost of travelling 30 km = 60 + 30 y = Rs. (60 + 30 x 6) = Rs. 240.

Clearly, we have to first find two numbers whose difference is 2 and of which the smaller one is a perfect square and the bigger one a perfect cube.

Such numbers are 25 and 27.

Thus, Nitin is now 26 years old. Since the next perfect cube after 27 is 64,

so required time period = (64 - 26) years = 38 years.

Suppose the boy got x sums right and 2x sums wrong.

Then, x + 2x = 48 ⟺ 3x = 48 ⟺ x = 16.

Let the daughter's age be x years.

Then, father's age = (3x) years.

Mother's age = (3x - 9) years; Son's age = (x + 7) years.

So, x + 7 = (3x-9)/2 ⟺ 2x + 14 = 3x - 9 ⟺ x = 23.

Therefore Mother's age = (3X - 9) = (69 - 9) years = 60 years.