Solution:

As Shyam is helped by Ram and Singhal every third day, Shyam works for 3 days while Ram and Singhal work for 1 day in every 3 days.

Therefore, the amount of work done in 3 days by Shyam, Ram and Singhal:

=3/20+1/30+1/60=9+2+1/60=1260=1/5of the job.

Hence, it will take them 5 times the amount of time=3×5= 15 days.

Solution:

12 men in 36 days can do a work.

1 man in a day can do 1/12×36 work.

8 men in 20 days can do 8×20/12×36=10/27 work.

Similarly, we find that 20 women in 20 days can do 10/27 work.

Remaining work =7/27

Now, because in 60 days a work is done by 20 women.

In 1 day a work done by 20×60 women.

In 4 days 7/27

 work is done by 20×60×7/27×4

= 70 women.

Solution:

The part of the tank filled by A and B in first two hour

⇒ 3/4×(1/5+1/10)+(1/5+1/10)

The part of tank filled by C in first two hours =2×2/3×1/15

Remaining part =139/360

In 1 hour, all the three pipes together will fill =11/30

Hence, the time taken to fill the remaining tank =139/360×30/11=1.0530 hour.

Thus, the total time taken to fill the remaining tank = 3.05 hour.

Solution:

Let 'a' be the number of days in which A can do the job alone. Therefore, working alone, A will complete (1/a)th of the job in a day.

Similarly, let 'b' be the number of days in which B can do the job alone. Hence, B will complete (1/b)th of the job in a day. 

Working together, A and B will complete (1/a+1/b)th of the job in a day. 

The problem states that working together, A and B will complete the job in 7.5 or 15/2 days. i.e they will complete 2/15)th of the job in a day.

Therefore,  1/a+1/b=2/15 ...... (1) 

From the question, we know that if A completes half the job working alone and B takes over and completes the next half, they will take 20 days.

As A can complete the job working alone in 'a' days, he will complete half the job, working alone, in 

a/2 days.

Similarly, B will complete the remaining half of the job in b/2 days.

Therefore, a/2+b/2=20 

⇒ a+b=40 or a=40–b ...... (2) 

From (1) and (2) we get, 

{1/40−b}+1/b=2/15

 ⇒ 600=2b(40−b)

⇒ 600=80b−2b2 

⇒ b2−40b+300=0

⇒ (b−30)(b−10)=0

⇒ b=30 or b=10. 

If b=30, then a=40−30=10 or

If b=10, then a=40−10=30.

As A is more efficient than B, he will take lesser time to do the job alone. Hence A will take only 10 days and B will take 30 days.

Note: Whenever you encounter work time problems, always find out how much of the work can 'A' complete in a unit time (an hour, a day, a month etc). Find out how much of the work can be completed by 'B' in a unit time. Then add the amount of work done by A and B to find the total amount of work that will be completed in a unit time.

If 'A' takes 10 days to do a job, he will do (1/10)th of the job in a day. 

Similarly, if (2/5)th of the job is done in a day, the entire job will be done in 5/2days.

The number of boys-days

1/2[1+(1+2)+(1+2+3)+………..+(1+2+3+....+20)]=1440

⇒ But , each boy =1/2 girls⇒ 770 girl-days.

⇒ 10 girls will take 770/10

= 77 days.