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.

Clearly, we have :

A = B - 3 ...(i)

D + 5 = E ...(ii)

A+C = 2E ...(iii)

B + D = A+C = 2E ...(iv)

A+B + C + D + E=150 ...(v)

From (iii), (iv) and (v), we get: 5E = 150 or E = 30.

Putting E = 30 in (ii), we get: D = 25.

Putting E = 30 and D = 25 in (iv), we get: B = 35.

Putting B = 35 in (i), we get: A = 32.

Putting A = 32 and E = 30 in (iii), we get: C = 28.

Let number of horses = number of men = x.

Then, number of legs = 4x + 2 x (x/2) = 5x.

So, 5X = 70 or x = 14.

Clearly, while counting, the numbers associated to the thumb will be : 1, 9,17, 25,.....

i.e. numbers of the form (8n + 1).

Since 1994 = 249 x 8 + 2, so 1993 shall correspond to the thumb and 1994 to the index finger.

There were all sparrows but six' means that six birds were not sparrows but only pigeons and ducks.

Similarly, number of sparrows + number of ducks = 6 and number of sparrows + number of pigeons = 6.

This is possible when there are 3 sparrows, 3 pigeons and 3 ducks i.e. 9 birds in all.

When Rahul was born, his brother's age = 6 years; his father's age = (6 + 32) years = 38 years, his mother's age = (38 - 3) years = 35 years; his sister's age = (35 - 25) years = 10 years.

Clearly, the smallest such number is 3.

Three ducks can be arranged as shown above to satisfy all the three given conditions.

Clearly, every member except one (i.e. the winner) must lose one game to decide the winner. Thus, minimum number of matches to be played = 30 - 1 = 29.

Let the total number of sweets be (25x + 8).

Then, (25x + 8) - 22 is divisible by 28

⟺ (25x - 14) is divisible by 28 ⟺ 28x - (3x + 14) is divisible by 28

⟺ (3x + 14) is divisible by 28 ⟺ x = 14.

Therefore Total number of sweets = (25 x 14 + 8) = 358.