Two taps A and B can fill an overhead tank in 10 hours and 15 hours respectively.

Pipe A can fill the tank in 10 hours, and pipe B can fill the tank in 15 hours.

$\therefore$ In 1 hour, A can fill $\frac{1}{10}$ th part of the tank.

In 1 hour, $B$ can fill $\frac{1}{15}$ th part of the tank.

$\therefore$ In 1 hour, $\mathrm{A}$ and $\mathrm{B}$ can fill $\left(\frac{1}{10}+\frac{1}{15}\right)$

$=\frac{3+2}{30}=\frac{5}{30}=\frac{1}{6}$ th part of the tank

$\therefore$ In 4 hours, $\mathrm{A}$ and $\mathrm{B}$ can fill $\left(\frac{1}{6} \times 4\right)=\frac{2}{3} \mathrm{rd}$ part of the tank

Remaining part of the $\operatorname{tank}=1-\frac{2}{3}=\frac{1}{3}$

Now, A can fill the tank in 10 hours.

$\therefore \frac{1}{3} \mathrm{rd}$ part of the tank can be filled by $\mathrm{A}$ in $\left(\frac{1}{3} \times 10\right)$ hours or $\frac{10}{3}$ hours or $3 \frac{1}{3}$ hours.