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?
Answer & explanation
Correct answer: option 2
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