Three individuals, 'A', 'B', and 'C', are employed to complete a task. How many days will it take for 'A', 'B', and 'C' to finish the work together? Statement I:  'C' requires the same amount of time to complete the work as 'A' would if 'C' operates at 125% efficiency. 'B' can complete the task in 12 days, and 'A' and 'B' together finish it in 48/7 days. Statement II:  'A' can finish the task in 20 days when working at 80% efficiency, while 'B' requires 8 days less than 'C' to complete the task.

ATQ, Statement I: Let, time taken by 'C' to do the work with 100% efficiency = y days So, time taken by 'C' to do the work with 125% efficiency = 4y/5 days Therefore, time taken by 'A'  to do the work = 4y/5 days Work done by 'A' and 'B' in one day = 5/4y + 1/12 = (15 + y)/12y So, (15 + y)/12y = 7/48 60 + 4y = 7y 3y = 60, y = 20 Therefore, 'C' will take 20 days to do the work with 100% efficiency. Work done by all three in one day = 1/20 + 7/48 = (12 + 35)/240 = 47/240 Required number of days = 240/47 days Statement I alone is sufficient to answer the question Statement II: Number of days taken by 'A' to do the work with 100% efficiency = 20 × 4/5= 16 days Let, time taken by 'B' to do the work = x days Then, time taken by 'C' to do the work = (x + 8) days Statement II alone is not sufficient to answer the question.