What is the product of all the numbers in the dial of a telephone?

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.

Total runs scored = (36 x 5) = 180.

Let the runs scored by E be x.

Then, runs scored by D = x + 5; runs scored by A = x + 8;

runs scored by B = x + x + 5 = 2x + 5;

runs scored by C = (107 - B) = 107 - (2x + 5) = 102 - 2x.

So, total runs = (x + 8) + (2x + 5) + (102 - 2x) + (x + 5) + x = 3x + 120.

Therefore 3x + 120 =180 ⟺ 3X = 60 ⟺ x = 20.

We have : A = 3B ...(i) and

C - 4 = 2 (A - 4) ...(ii)

Also, A + 4 = 31 or A= 31-4 = 27.

Putting A = 27 in (i), we get: B = 9.

Putting A = 27 in (ii), we get C = 50.

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.