In a cricket match, five batsmen A, B, C, D and E scored an average of 36 runs. D Scored 5 more than E; E scored 8 fewer than A; B scored as many as D and E combined; and B and C scored 107 between them. How many runs did E score?

L.C.M. of 6, 5, 7, 10 and 12 is 420.

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.

Total number of digits

= (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.

Let total number of members be 100,

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.

Let number of girls = x and number of boys = 3x.

Then, 3x + x = 4x = total number of students.

Thus, to find exact value of x, the total number of students must be divisible by 4.

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.

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.

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

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

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