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?

54 days

The correct answer is Option (2) → 54 days

Total work = LCM of all denominators = 180 units (assumed for calculation)

Work done per day by X, Y, Z together:

$16 \frac{4}{11} = \frac{180}{11}$ days

So (X + Y + Z)'s 1-day work = $\frac{1}{\frac{180}{11}} = \frac{11}{180}$

(X + Y)'s 1-day work = $\frac{1}{20}$

Then Z’s 1-day work:

$\frac{11}{180} - \frac{1}{20} = \frac{11}{180} - \frac{9}{180} = \frac{2}{180} = \frac{1}{90}$

Z does $\frac{1}{90}$ of the work per day

60% of work = $\frac{60}{100} = \frac{3}{5}$

Time Z takes to complete $\frac{3}{5}$ of work:

$\frac{3/5}{1/90} = \frac{3 \times 90}{5} = 54$ days