X can do a piece of work in 24 days. When he had worked for 4 days, Y joined him. If complete work was finished in 16 days, Y can alone finish that work in:

Solution (By Examveda Team)

According to the question,
X → 24 days
⇒ Work done by X in 4 days alone = 4 × 1 = 4 units
⇒ Remaining work = 24 - 4 = 20 units
⇒ 20 units done by both together in (16 - 4 days) = 12 days
⇒ Then efficiencies of (X + Y) $$ = \frac{{{\text{work}}}}{{{\text{days}}}} = \frac{{20}}{{12}} = \frac{5}{3} = 1 + \frac{2}{3}$$
⇒ Efficiency of Y = $$\frac{2}{3}$$
⇒ Time taken by Y alone to complete the total work $$ = \frac{{24}}{{\frac{2}{3}}} = 36{\text{ days}}$$

Alternate solution:
$$\eqalign{ & X \times 20 = \left( {X + Y} \right) \times 12 \cr & \frac{X}{{X + Y}} = \frac{{12}}{{20}} \cr & = \frac{{3 \to {\text{Efficiency of }}X}}{{5 \to {\text{Efficiency of }}\left( {X + Y} \right)}} \cr} $$
Efficiency of Y = 5 - 3 = 2 units/day
Total work = 24 × 3 = 72 units
Total time taken by Y = $$\frac{{72}}{2}$$ = 36 days