X, Y and Z together can do a piece of work in 16 (4/11) days. X and Y together can do the same work in 20 days. In how many days will Z alone finish the 60% of the work?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 54 days
A group that finishes a job in \(n\) days does \(\frac{1}{n}\) of it in one day, and the one-day work of people working together is the sum of their separate one-day works. So Z’s one-day work is the joint figure of X, Y and Z minus that of X and Y.
The one-day work (a group finishing in \(n\) days does \(\frac{1}{n}\) of it a day):
\(16\frac{4}{11}=\frac{180}{11}\) days
So X + Y + Z do \(\frac{11}{180}\) of the work in a day (1)
X + Y do \(\frac{1}{20}=\frac{9}{180}\) of the work in a day (2)
Z alone (subtracting (2) from (1), the work of X and Y cancels):
Z does \(\frac{11}{180}-\frac{9}{180}=\frac{2}{180}=\frac{1}{90}\) of the work in a day
So Z alone finishes the whole work in 90 days
60% of the work (60% of the 90 days):
Z takes \(\frac{60}{100}\times90=54\) days