A, B and C can do a piece of work in 30, 20 and 10 days respectively. A works every day; he is assisted by B on the first day, by C on the second day, by B on the third day, and so on alternately. In how many days will the work be finished?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(9\frac{3}{8}\)
Take the work as the LCM of 30, 20 and 10, that is 60 units. The days repeat in a 2-day cycle (A with B, then A with C), so count whole cycles first.
Daily work: A \(=2\) units, B \(=3\) units, C \(=6\) units
Day with B: \(2+3=5\) units; day with C: \(2+6=8\) units; one 2-day cycle \(=13\) units
4 cycles \(=8\) days, \(4\times13=52\) units
Day 9 (A with B): \(52+5=57\) units, so 3 units are left
Day 10 is A with C at 8 units a day: 3 units take \(\frac{3}{8}\) day
The work is finished in \(9\frac{3}{8}\) days.