If A, B and C work together for a day then they will finish (1/5+1/10+1/15)th work =(11/30)th work.

Therefore working together for two days they will finish 2×11/30=(11/15)th work.

C alone does remain (11/15–1)=(4/15)th work.But C finishes (1/15)th work in one day. 

Therefore C will finish (4/15)th work in 4 days.

Solution:

Since Jay does (1/30)th of the work in one day, total work done by Jay in 18 days =18/30=3/5

Remaining (2/5)th work was done by Anup.

Since Anup does (1/15)th of the work =(2/5)/(1/15)=6 days.

So, Anup joined Jay after (18−6)= 12 days.

Solution:

Pipe A fills the tank in 'a' hours. 

Therefore, 1/a of the tank gets filled in an hour. 

On account of the leak it takes 3a hours to fill the tank. 

Therefore, 1/3a of the tank gets filled in an hour. 

Let the leak at the bottom of the tank take 'x' hours to empty the tank. 

Hence, 1/xof the tank gets emptied every hour.

1/a–1/x=1/3a

⇒ 1/x=1/a=1/3a=2/3a

Hence, x = 3a/2

Solution:

Here we can find after 6 hours, how much part was filled by the three pipes.

⇒ 16/12+4/18−6/36=5/9.

Remaining part =1−5/9=4/9.

Since, this part is to be filled by A and B together. 

Further, A and B together can fill in one hour =1/12+1/10=5/36 part

4/9 part will be filled by A and B together in time =16/5 hour

Total time required =6h+3h+12 min = 9 hours 12 minutes

Solution:

Ram takes 24 days to complete the work, if he works alone.

As Krish is twice as efficient as Ram is, Krish will take half the time to complete the work when Krish works alone, i.e., in 12 days.

Ram completes  (1/24)th of the work in a day.

Krish completes (1/12)th of the work in a day.

When they work together, they will complete  1/24+1/12=1+2/24=1/8 work in a day.

Therefore, when they work together they will complete the work in 8