Two taps $X$ and $Y$ can fill a tank in $15$ hours and $20$ hours respectively. If the two taps are opened at 2 p.m., then at what time (in p.m.) should the tap $X$ be closed to completely fill the tank at exactly 2 a.m.?

Let the total capacity of the tank be the LCM of $15$ and $20$, which is $60$ units.Efficiency of tap $X=\frac{60}{15}=4$ units/hourEfficiency of tap $Y=\frac{60}{20}=3$ units/hourThe total time from 2 p.m. to 2 a.m. is exactly $12$ hours.Since tap $X$ is closed in between, tap $Y$ runs continuously for the entire $12$ hours.Work done by tap $Y$ in $12$ hours:
$${\text{Work}}_Y=12\times 3=36\text{ units}$$
Remaining work to be completed by tap $X$:
$$\text{Remaining Work}=60-36=24\text{ units}$$
Time for which tap $X$ was open:
$${\text{Time}}_X=\frac{24}{4}=6\text{ hours}$$
Since the taps were opened at 2 p.m., tap $X$ should be closed after $6$ hours:
$$2\text{ p.m.}+6\text{ hours}=8\text{ p.m.}$$