Rohit, Harsh and Sanjeev are three typists, who working simultaneously, can type 216 pages in four hours . In one hour, Sanjeev can type as many pages more than Harsh as Harsh can type more than Rohit. During a period of five hours, Sanjeev can type as many passages as Rohit can, during seven hours. How many pages does each of them type per hour?

The weight of baggage deliver in 1 min by 1st belt = 3/5 tonne
The weight of baggage deliver in 1 min by 2nd belt = 1/2 tonne
The weight of baggage deliver in 1 min by both belt = 3/5 + 1/2 = 11/10 tonne
So, the two belts delivers 1 tonne in 10/11 min.
Hence, required time to deliver 33 tonne = (10/11) x 33 = 30 min

Work done by C in one min = (1/40 + 1/60) - 1/30 = 1/120
Hence, C can empty the tank in 120 min or 2 h.

Let the number of men originally employed = x
So, (x - 6) men could finish the work in 15 days and x men could finish the work in 9 days.
? 9x = 15(x - 6)
? x = 15

(A + B +)'s 2 h work = 1/8 + 1/12 = 5/24
(A + B)'s 8 h work = (5 x 8)/(24 x 2) = 5/6
Work done by A in 9th h = 1/8
Total work done upto 9th h = 5/6 + 1/8 = 23/24
Remaining wok = 1/24
B's 1 h work = 1/12
B can do 1/24 of the work = (1/24) x 12 = 1/2 h
So, both can finish the job in 9¹/₂ h.

Let us assume that each soldiers eats one unit of food per day. Thus, total units of food at the beginning will be 1000 x 30 = 30000. After 10 days the 100 soldiers would have eaten 1000 x 10 = 10000 units of food. Thus, food left after 10 days equals 20000 units. Now, there are a total of 2000 soldiers who eat one unit of food everyday.

Now, 30 x 1000 = 1000 x 10 + 2000 x D (here, D is number of days food will last)
? 30000 - 10000 = 2000 x D
? D = 20000/2000 = 10 days

(A + B)'s one day work = 1/12
(B + C)'s one day work = 1/16
Now, from the question,
A's 5 days work + B's 7 days work + C's 13 days work = 1
? A's 5 days work + B's 5 days work + B's 2 days work + C's 2 days work + C's 11 days work = 1
(A + B)'s 5 days work + (B + C)'s 2 days work + C's 11 days work = 1
? 5/12 + 2/16 + C's 11 days work = 1
? C's 11 days work = 1 - (5/12 + 2/16) = 11/24
? C's 1 day work = (11/24) x 11 = 1/24
Hence, C can do this work in 24 days.

Fraction of work complete by (A and B) + Fraction of work complete by (B and C) + Fraction of work completed by C = 1
As B did the work for 8 days we can say B did the work for 3 days with A and B did the work for 5 days with C. So, A and B together did the work for 3 days, B and C together did the work for 5 days, C alone did the work for 5 days.
? 3/12 + 5/15 + 5/x = 1 or 5/x = 5/12
i.e., x = 12 or C alone can complete the work in 12 days

Total man hours work done by men supplied by A, B and C
= (20 x 8 x 6), (15 x 9 x 7), (10 x 6 x 8) and the wages must be in the ratio of the work done.
So, ? 636 has to divided among A, B and C in the ratio (20 x 8 x 6) : (15 x 9 x 7) : (10 x 6 x 8) = 64 : 63 : 32
? C's share = (32/159) x 636 = ? 128

By the fundamental of work and time,
x/45 + (x + 23)/40 = 1
? x = 9 days

Since, 50 men can do a job in 50 days.
So, work done by 1 man in a day = 1/(50 x 50)

Also, 80 women can do the job in 50 days.
So, work done by 1 women in 1 day = 1/(50 x 80)

Now, work done by 40 men and 48 women in first 10 days
= (40 x 10)/(50 x 50) + (48 x 10)/(50 x 80)
= 4/25 + 3/25 = 7/25

Now, 5 men and 8 women are removed after 10 days,
So work done by 35 men and 40 women in 10 days = (35 x 10)/(50 x 50) + (40 x 10)/(50 x 80)
= 7/50 + 1/10 = (7 + 5)/50
= 6/25

Again, 5 men and 8 women are removed after 10 days,
So work done by 30 men and 32 women in 10 days = (30 x 10)/(50 x 50) + (32 x 10)/(50 x 80) = 5/25

Now, after every 10 days as the number of men and
women decrease, work done also decreased by 1/25th past.
So, work done after every 10 days upto 50 days = 7/25 + 6/25 + 5/25 + 4/25 + 3/25
= 25/25 = 1

So, it will take 50 days for them to complete the work.