A and B can complete a task in 20 days and 25 days, respectively. They started working together, but B left after 5 days. How many more days will A take to finish the remaining work?
11 days
A completes the task in 20 days, so A's one-day work is \(\frac{1}{20}\); B completes it in 25 days, so B's one-day work is \(\frac{1}{25}\).
Together in one day they do \(\frac{1}{20} + \frac{1}{25} = \frac{5 + 4}{100} = \frac{9}{100}\).
In 5 days together they finish \(5 \times \frac{9}{100} = \frac{45}{100} = \frac{9}{20}\) of the work.
The remaining work is \(1 - \frac{9}{20} = \frac{11}{20}\).
A alone finishes this in \(\frac{11}{20} \div \frac{1}{20} = \frac{11}{20} \times 20 = 11\) days.
Hence, A takes 11 more days to finish the remaining work.
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