4/7 of a pole is in the mud. When 1/3 of it is pulled out, 250 cm of the pole is still in the mud. Find the full length of the pole .

L.C.M of 6, 7, 8, 9, 12 is 504
So, the bells will toll together after 504 sec.
In hour, they will toll together
= (60 x 60) / 504 times
= 7 times

Let the average for last 4 matches be y.
Sum of score for 10 matches = sum of score for first 6 matches + sum of score of last 4 matches.
So 53 x 6 + 4y = 10 x 43.9
? 4x= 439-318
?4x= 121
? x= 30.25

1 men's one day's work = 1/96
12 men's 3 day's work = 3 x (1/8) = 3/8
Remaining work = (1 - 3/8) = 5/8
15 men's 1 day's work = 15/96
Now, 15/96 work is done by them in 1 day
? 5/8 work will be done by them in = (96/15) x (5/ 8) i.e., 4 days

3 yr before, total age of 5 members = 17 x 5 = 85 yr
? The present age of all members of family = 85 + 3 x 5 = 100 yr

Let the age of child be N yr.
? Present average age of family including the child
= (100 + N)/6
? 17 = (100 + N)/6
? N = 102 - 100 = 2 yr

Let the number be N.
Therefore N = 5k + 3

On squaring both sides, we get
N² = ( 5k + 3)²
= 5( 5k² + 6k +1 ) + 4
? On dividing x² by 5, the remainder is 4

Let Tarun and Varun's ages are 3k yr and 7k yr, respectively.
? 7k + 4 = 39
? k = 5

Hence, Tarun's age 4 yr ago = 3 x 5 - 4 = 11 yr

Let k yr before, the ratio of their ages was 3 : 5.
According to the question
(40 - k)/(60 - k) = 3/5
? 200 - 5k = 180 - 3k
? 2k = 20
? k = 10

Correct mean = ( sum of total marks - 57 + 75 ) / 30
= ( 30 x 58.5 - 57 + 75 ) / 30
= (1755 + 18 ) / 30
= 59.1

Total number of books = a + 2b + 3c + d . Since there are 'b' copies of each of two books, 'c' copies of each of three books and single copy of 'd' book.
Therefore, the total number of arrangements is = (a + 2b + 3c + d )! / {a! (b!)² (c!)³}

Total number of ways of selection of 3 marbles out of 12
= n(S) = ¹²C₃ = 220
Total number of favorable events = n(E)
= ³C₃ + ⁴C₃ = 1 + 4 = 5
? Required probability
= 5/220 = 1/44