A and B, working on alternate days with B starting first, complete a piece of work in \(\frac{34}{3}\) days. A alone can complete the work in 48 days. In how many days can B alone do 4 times this work?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 27
Working on alternate days, each person works on every second day. \(\frac{34}{3}=11\frac{1}{3}\) days: B works on the odd days 1, 3, ..., 11 and A on the even days 2, 4, ..., 10, and the last \(\frac{1}{3}\) day is day 12, which is A's turn.
B works 6 full days; A works 5 full days and \(\frac{1}{3}\) of a day, that is \(\frac{16}{3}\) days
Work done by A \(\displaystyle =\frac{16}{3}\times\frac{1}{48}=\frac{1}{9}\)
Work done by B in 6 days \(\displaystyle =1-\frac{1}{9}=\frac{8}{9}\), so B does \(\displaystyle \frac{8}{9}\div6=\frac{4}{27}\) of the work per day
B alone does the work once in \(\frac{27}{4}\) days, so 4 times the work in \(\displaystyle 4\times\frac{27}{4}=27\) days