A and B, working on alternate days with B starting first, complete a piece of work in \(\frac{74}{5}\) days. A alone can complete the work in 20 days. In how many days can B alone do 3 times this work?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 36
Working on alternate days, each person works on every second day. \(\frac{74}{5}=14\frac{4}{5}\) days: B works on the odd days 1, 3, ..., 13 and A on the even days 2, 4, ..., 14, and the last \(\frac{4}{5}\) day is day 15, which is B's turn.
A works 7 full days; B works 7 full days and \(\frac{4}{5}\) of a day, that is \(\frac{39}{5}\) days
Work done by A \(\displaystyle =7\times\frac{1}{20}=\frac{7}{20}\)
Work done by B \(\displaystyle =1-\frac{7}{20}=\frac{13}{20}\), so B does \(\displaystyle \frac{13}{20}\div\frac{39}{5}=\frac{13}{20}\times\frac{5}{39}=\frac{1}{12}\) of the work per day
B alone does the work once in 12 days, so 3 times the work in \(3\times12=36\) days