Question
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. Information about three values have been given. Calculate the value of all three quantities and answer accordingly.Solution
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
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