Three pumps working 8 hours a day, can empty a tank in 2 days. How many hours a day must 4 pumps work to empty the tank in 1 day?

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 .

Let the remaining food will last for D days.
500 men had provisions for (27 ? 3) = 24 days.
( 500 + 300 ) men had provisions for D days.
More men, Less days ( Indirect Proportion )
we can write as :-
? 800 : 500 :: 24 : x ? (800 × D ) =( 500 × 24 )
? ( 500 × 24 ) /800 =15
Therefore , 15 days is correct answer .

Given in the question,
Let y be the persons there originally .
Work done by y persons can finish = 100 days
If there were 10 persons less, it would take more days finish the work in 10 days
i.e. ( y - 10 ) = 110
According to question ,
y × 100 = ( y - 10 ) × 110
? 100y = 110y - 1100 ? 10y = 1100 ? y = 110
Therefore , 110 persons were there originally .

As per given question,
8 persons can build a wall 140 meter long in 42 days.
1 persons can build a wall 140 meter long in 42 x 8 days.
1 persons can build a wall 1 meter long in 42 x 8 / 140 days.
30 persons can build a wall 1 meter long in 42 x 8 / 30 x 140 days.
30 persons can build a wall 100 meter long in 42 x 8 x 100 / 30 x 140 days.
30 persons can build a wall 100 meter long in 42 x 8 / 3 x 14 days.
30 persons can build a wall 100 meter long in 42 x 8 / 42 days.
30 persons can build a wall 100 meter long in 8 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

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 15 men work for m days.
Work done in 1 days = m/20
Remaining work = (1 - m/20)
25 men's 1 days work = 1/20
1 man's 1 days work = (1/20) x (1/25) = 1/500
10 men's 1 days work = 10/500 = 1/50
10 men's 75/2 days work = (1/50) x (75/2) = 75/100 = 3/4
? (1 - m/20) = 3/4
? m/20 = 1/4
? m = (1/4) x 20 = 5
Clearly, 15 men leave after 5 days.

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