Question

A can do a piece of work in 16 days and the same work can be done by B in 24 days, If they work individually on alternate days (i.e., on the first day A does the work and on the second day A was on leave and B done the second-day work and so on.) and A starts the work, then find the total days required to complete the work.

• A. 19 days
• B. 18 days
• C. 20 days
• D. 10 days

Answer 1

Answer:

A. 19 days

Solution: Given: Time is taken by A to complete the work = 16 days Time is taken by B to complete the work = 24 days Formula used: Time is taken by A to complete the work = a days Time is taken by B to complete the work = b days Total unit of work = L.C.M of (a and b) units Calculation: Total unit of work = L.C.M of (16 and 24) = 48 units Efficiency of A = 4816 = 3 units a day Efficiency of B = 4824 = 2 units a day According to the question, Amount of work done by A on the first day = 3 units a day Amount of work done by B on the second day = 2 units a day In 2 days, the amount of work done = 5 units On multiplying 9 on both sides, In (2 × 9) days, the amount of work done = 5 × 9 units In 18 days, the amount of work done = 45 units Remaining amount of work = 48 - 45 = 3 units Amount of work done by A on the first day = 3 units a day Since the amount of work done by A on one day = 3 units a day Therefore, A will complete the 19th-day work. Therefore, The total time required to complete the work = 19 days Therefore, '19 days' is the required answer.