A is three times as old as B. C was twice-as old as A four years ago. In four years' time, A will be 31. What are the present ages of B and C?

Since one of the numbers on the dial of a telephone is zero, so the product of all the numbers on it is 0.

Since there are socks of only two colours, so two out of any three socks must always be of the same colour.

Required sum = (80 - 3 x 3) years = (80 - 9) years = 71 years.

Let x and y be the ten's and unit's digits respectively of the numeral denoting the woman's age.

Then, woman's age = (10X + y) years; husband's age = (10y + x) years.

Therefore (10y + x)- (10X + y) = (1/11) (10y + x + 10x + y)

⟺ (9y-9x) = (1/11)(11y + 11x) = (x + y) ⟺ 10x = 8y ⟺ x = (4/5)y

Clearly, y should be a single-digit multiple of 5, which is 5.

So, x = 4, y = 5.

Hence, woman's age = 10x + y = 45 years.

Let d and s represent the number of daughters and sons respectively.

Then, we have :

d - 1 = s and 2 (s - 1) = d.

Solving these two equations, we get: d = 4, s = 3.

When Ravi's brother was born, let Ravi's father's age = x years and mother's age = y years.

Then, sister's age = (x - 28) years. So, x - 28 = 4 or x = 32.

Ravi's age = (y - 26) years. Age of Ravi's brother = (y - 26 + 3) years = (y - 23) years.

Now, when Ravi's brother was born, his age = 0 i.e. y - 23 = 0 or y = 23.

Let son's age be x years. Then, father's age = (3x) years.

Five years ago, father's age = (3x - 5) years and son's age = (x - 5) years.

So, 3x - 5 = 4 (x - 5) ⟺ 3x - 5 = 4x - 20 ⟺ x = 15.

Originally, let number of women = x. Then, number of men = 2x.

So, in city Y, we have : (2x - 10) = (x + 5) or x - 15.

Therefore Total number of passengers in the beginning = (x + 2x) = 3x = 45.