A does 60% of the work in 18 days. He then calls in B, and they together finish the remaining work in 4 days. How long would B alone take to do the whole work?
15 days
Given: A alone completes 60% of the work in 18 days; then A and B together finish the remaining 40% in 4 days.
First find A's rate. A does 60% (i.e. 0.6 of the work) in 18 days, so A's rate \(= \frac{0.6}{18} = \frac{1}{30}\) work per day.
The remaining work is \(100\% - 60\% = 40\% = 0.4\) of the whole.
In the last 4 days A alone would do \(4 \times \frac{1}{30} = \frac{4}{30} = \frac{2}{15}\) of the work.
So B's share in those 4 days \(= 0.4 - \frac{2}{15} = \frac{6}{15} - \frac{2}{15} = \frac{4}{15}\).
Therefore B's rate \(= \frac{4/15}{4} = \frac{1}{15}\) work per day.
Time for B alone to do the whole work \(= \frac{1}{1/15} = 15\) days.
B alone takes 15 days.
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