Mac has £ 3 more than Ken, but then Ken wins on the horses and trebles his money, so that he now has £ 2 more than the original amount of money that the two boys had between them. How much money did Mac and Ken have between them before Ken's win?

Let number of notes of each denomination be x.

Then, x + 5x + 10x = 480 ⟺ 16x = 480 ⟺ x = 30.

Hence, total number of notes = 3x = 90.

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.

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.

Since B and D are twins, so B = D.

Now, A = B + 3 and A = C - 3.

Thus, B + 3 = C - 3 ⟺ D + 3 = C-3 ⟺ C - D = 6.