25 men can reap a field in 20 days. when should 15 men leave the work, if the whole field is to be reaped in 37 ¹/ ₂ days after they leave the work?

Let man's 1 day's work = 1/m
and boy's 1 day's work = 1/n
1 day's work man and boy = 1/24
Man's 6 day's work = 6/m
Now, for 20 days, both man and boy do the work and for last 6 days, only man does the work.
According to the question,
1/m + 1/n = 1/24
? 20(1/m + 1/n) + 6/m = 1
? 20 x (1/24) + 6/m = 1
? 6/m = (1 - 20/24) = 4/24 = 1/6
? 1/n = 1/36
Now from eq. (i)
1/m + 1/n = 1/24
1/36 + 1/n = 1/24
? 1/n = (1/24) - (1/36) = 1/72
Hence, the boy alone can do the work in 72 days.

Let us assume that each soldier eats one unit of food per day. Thus, total units of food at the beginning will be 1000 x 30 = 30000.
After 10 days 1000 soldiers would have eaten 1000 x 10 = 10000 units of food. Thus food left after 10 days equals 20000 units. Now, there are total of 2000 soldiers who eat one unit of food every day. So, the number of days that 20000 units of food will serve 2000 soldiers is 20000/2000 = 10

According to question ,
3 pumps working in 2 days = 8 × 2 = 16 hours
It means that they can empty a tank in 16 hours .
3 pumps 1 hour work = 1/16
1 pumps 1 hour work = 1/( 3 × 16 )
4 pumps 1 hour work =4 × 1/( 3 × 16 ) = 1 / 12
Hence , they will empty the tank in 12 hours .

Suppose the food will last for D days at same rate .
250 men had provisions for 40 days.
( 250 - 50 ) men had provisions for D days.
Less men, More days ( Indirect Proportion )
we can write as :-
? 200 : 250 :: 40 : D ? (200 × D ) =( 250 × 40 )
? ( 250 × 40 ) /200 = 50
Therefore , 50 days is required answer .

12 men ? 18 women
? 1 man ? 18/12 women ? 3/2 women
? 8 men ? (3/2) x 8 = 12 women
? 8 men + 16 women = 12 women + 16 women = 28 women
? 18 women can do the work in 14 days .
? 1 woman can do the same work in (14 x 18) days.
? 28 women will do the same work in (14 x 18)/28 days.
? Required number of days = (14 x 18)/28 = 9 days