Worker 'A' takes 16 more days than the combined time taken by workers 'A' and 'B' together to complete a certain job. If worker 'B' alone can finish the job in 60 days, how many days will it take for worker 'A' to complete 125% of the same job working alone?

Let the time taken by 'A' and 'B' to do the work together be 'x' days So, time taken by 'A' alone to do the work = (x + 16) days So, (1/x) = {1/(x + 16)} + (1/60) Or, x(x + 16 + 60) = 60 X (x + 16) Or, x² + 76x = 60x + 960 Or, x² + 16x - 960 = 0 Or, x² + 40x - 24x - 960 = 0 Or, x(x + 40) - 24(x + 40) = 0 Or, (x + 40)(x - 24) = 0 So, x = 24 or x = -40 So, time taken by 'A' to do the work alone = 24 + 16 = 40 days And, time taken to do 125% of the work = 40 X 1.25 = 50 days