A and B can do a piece of work in 12 days; B and C in 15 days; C and A in 20 days.

Solution:

Time taken by $(\mathrm{A}+\mathrm{B})$ to do the work $=12$ days

Time taken by $(\mathrm{B}+\mathrm{C})$ to do the work $=15$ days

Time taken by $(\mathrm{A}+\mathrm{C})$ to do the work $=20$ days

Now,

Work done by $(\mathrm{A}+\mathrm{B})=\frac{1}{12}$

Work done by $(\mathrm{B}+\mathrm{C})=\frac{1}{15}$

Work done by $(\mathrm{A}+\mathrm{C})=\frac{1}{20}$

$\therefore$ Work done together $=(\mathrm{A}+\mathrm{B})+(\mathrm{B}+\mathrm{C})+(\mathrm{A}+\mathrm{C})$

$=\frac{1}{12}+\frac{1}{15}+\frac{1}{20}$

$=\frac{5+4+3}{60}=\frac{12}{60}$

$=\frac{1}{5}$

$\therefore$ Work done together $=2(\mathrm{~A}+\mathrm{B}+\mathrm{C})=\frac{1}{5}$

$\therefore$ Work done by $(\mathrm{A}+\mathrm{B}+\mathrm{C})=\frac{1}{10}$

$\therefore$ Work done by A alone $=(\mathrm{A}+\mathrm{B}+\mathrm{C})-(\mathrm{B}+\mathrm{C})$

$=\frac{1}{10}-\frac{1}{15}=\frac{3-2}{30}=\frac{1}{30}$

Thus, A alone can do the work in 30 days.