P. A can complete a piece of work in14 days and B can do the same work in 5 days. If B starts the work and they work on alternative days. Find total no. of days to finish the complete work. Q. pipe M can fill one-third of a tank in 4 hours and pipe N can fill 20% of the same tank in 3 hours. If half the tank is filled already then find time taken to fill the tank completely when both the pipes are opened together. R. 30 men could do a job in 20 days. Due to the inclusion of some more men the work got completed 3/5 times. Find the number of men added.

P. let total work to be done (LCM of 14 & 5) = 70 unit One day work of A = 70/14 = 5 unit One day work of B = 70/5 = 14 unit One day work of A & B = 14+5 = 19 unit In three days they will work = 3×19 = 57 unit Remaining work = 70-57 = 13 unit Since work is started by B Time taken by B to complete rest work = 13/14 = 0.9 days Total time taken to finish the work = 3+0.9 = 3.9 days (approx) Q. Pipe M can fill one-third of the tank in 4 hrs. Pipe M will fill this tank in 4×3 = 12 hrs. Pipe M can fill 20% of the tank in 3 hrs. Pipe M will fill this tank in 5×3 = 15 hrs. Both pipe M & N has to fill (LCM of 12 & 15) = 60 unit Pipe M will fill in 1 hrs. = 60/12 = 5 unit Pipe N will fill in 1 hrs. = 60/15 = 4 unit Both Pipe together will fill = 5+4 = 9 unit Since tank is half filled, hence both tap has to fill only 30 unit Time taken to fill the tank completely = 30/9 = 3.33 hrs. R. let x men included Then 30 men completes the work in 20 days And (30+x) men will complete in 3/5 of 20 days = 12 days 30×20 = (30+x) ×12 X = 20 men Hence we can see that Q < P < R