Then, number of members owning only 2 cars = 20.

Number of members owning 3 cars = 40% of 80 = 32.

Number of members owning only 1 car = 100 - (20 + 32) = 48.

Thus, 48% of the total members own one car each.

= (No. of digits in 1- digit page nos. + No. of digits in 2-digit page nos. + No. of digits in 3- digit page nos.)

= (1 x 9 + 2 x 90 + 3 x 267) = (9 + 180 + 801) = 990.

So, the bells will toll together after every 420 seconds i.e. 7 minutes.

Now, 7 x 8 = 56 and 7 x 9 = 63.

Thus, in 1-hour (or 60 minutes), the bells will toll together 8 times, excluding the one at the start.

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.

Then, we have :

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

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

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.