A can do a piece of work in 18 days and B can do it in 15 days. They work together for 4 days; then B leaves and A continues alone. After 5 more days C joins A, and the work is completed in 3 more days. In how many days can C alone do the work?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 45 days
Take the work as the LCM of 18 and 15, that is 90 units, and add up the work stage by stage. What is left for the last 3 days, minus A's share, is C's work in 3 days.
Daily work: A \(=\frac{90}{18}=5\) units, B \(=\frac{90}{15}=6\) units
A and B for 4 days: \(4\times(5+6)=44\) units
A alone for 5 days: \(5\times5=25\) units
Work left \(=90-44-25=21\) units, done by A and C in 3 days
A's part \(=3\times5=15\) units, so C does \(21-15=6\) units in 3 days, that is 2 units per day
C alone \(\displaystyle =\frac{90}{2}=45\) days