A can do a piece of work in 20 days and B can do it in 12 days. They work together for 3 days; then B leaves and A continues alone. After 4 more days C joins A, and the work is completed in 4 more days. In how many days can C alone do the work?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 20 days
Take the work as the LCM of 20 and 12, that is 60 units, and add up the work stage by stage. What is left for the last 4 days, minus A's share, is C's work in 4 days.
Daily work: A \(=\frac{60}{20}=3\) units, B \(=\frac{60}{12}=5\) units
A and B for 3 days: \(3\times(3+5)=24\) units
A alone for 4 days: \(4\times3=12\) units
Work left \(=60-24-12=24\) units, done by A and C in 4 days
A's part \(=4\times3=12\) units, so C does \(24-12=12\) units in 4 days, that is 3 units per day
C alone \(\displaystyle =\frac{60}{3}=20\) days