A man has Rs. 480 in the denominations of one-rupee notes, five-rupee notes and ten-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has?

Let Rs. x be the fare of city B from city A and Rs. y be the fare of city C from city A.

Then, 2x + 3y = 77 ...(i) and

3x + 2y = 73 ...(ii)

Multiplying (i) by 3 and (ii) by 2 and subtracting, we get: 5y = 85 or y = 17.

Putting y = 17 in (i), we get: x = 13.

Manick's present age = 12 years, Rahul's present age = 4 years.

Let Manick be twice as old as Rahul after x years from now.

Then, 12 + x = 2 (4 + x) ⟺ 12 + x = 8 + 2x ⟺ x = 4.

Hence, Manick's required age = 12 + x = 16 years.

Let money with Ken = x. Then, money with Mac = x + £ 3.

Now, 3x = (x + x + £ 3) + £ 2 ⟺ x = £ 5.

Therefore Total money with Mac and Ken = 2x + £ 3 = £ 13.

Ayush's present age = 10 years.

His mother's present age = (10 + 20) years = 30 years.

Ayush's father's present age = (30 + 5) years = 35 years.

Ayush's father's age at the time of Ayush's birth = (35 - 10) years = 25 years.

Therefore Ayush's father's age at the time of marriage = (25 - 2) years = 23 years.

Clearly, from 1 to 100, there are ten numbers with 3 as the unit's digit- 3, 13, 23, 33, 43, 53, 63, 73, 83, 93; and ten numbers with 3 as the ten's digit - 30, 31, 32, 33, 34, 35, 36, 37, 38, 39.

So, required number = 10 + 10 = 20.

Let R, G and B represent the number of balls in red, green and blue boxes respectively.

Then, .

R + G + B = 108 ...(i),

G + R = 2B ...(ii)

B = 2R ...(iii)

From (ii) and (iii), we have G + R = 2x 2R = 4R or G = 3R.

Putting G = 3R and B = 2R in (i), we get:

R + 3R + 2R = 108 ⟺ 6R = 108 ⟺ R = 18.

Therefore Number of balls in green box = G = 3R = (3 x 18) = 54.

Total number of routes from Bristol to Carlisle = (4 x 3 x 2) = 24.