40 men complete one-third of a work in 40 days. How many more men should be employed to finish the rest of the work in 50 more days?

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