A takes 8 more days than B to finish a work. A and B start the work and B leaves the work 9 days before the completion of the work. If A completes 75% of the work. How long would A have taken to finish the work alone?

14

Let the time taken by B working alone = x days

so, time taken by A working alone = (x + 8) days

A completes 75% of work, 

so the day he will work = 75% of (x + 8) = \(\frac{3(x + 8)}{4}\)

⇒ B has to complete 25% of work,

hence, days taken by B = 25% of x = \(\frac{x}{4}\)

B leaves the work 9 days before the work is finished

hence, days for which A worked - days for which B worked = 9 days

\(\frac{3(x + 8)}{4}\) - \(\frac{x}{4}\) = 9

\(\frac{3x + 24 - x}{4}\) = 9

2x = 12

x = 6

Time taken by A to finish the work = x + 8 = 6 + 8 = 14 days