A certain number of horses and an equal number of men are going somewhere. Half of the owners are on their horses' back while the remaining ones are walking along leading their horses. If the number of legs walking on the ground is 70, how many horses are there?

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 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.

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.

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

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.

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.