Ayush was born two years after his father's marriage. His mother is five years younger than his father but 20 years older than Ayush who is 10 years old. At what age did the father get married?

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.

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.

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.

Let x and y be the number of deer and peacocks in the zoo respectively. Then,

x + y = 80 ...(i) and

4x + 2y = 200 or 2x + y = 100 ...(ii)

Solving (i) and (ii), we get) x = 20, y = 60.