A and B can do a work in 30 days. B and C can do the same work in 60 days. C and A can do the same work in 30 days. Then the time taken by A alone is

1 day work of $A+B=30\text{\hspace{0.17em}days}$ 

1 day work of $B+C=60\text{\hspace{0.17em}days}$

1 day work of $C+A=30\text{\hspace{0.17em}days}$

1 day work of $2(A+B+C)=\frac{1}{30}+\frac{1}{60}+\frac{1}{30}$

$=\frac{2+1+2}{60}=\frac{5}{60}$

$A+B+C=\frac{5}{120}$

Mean A,B,C together finish in 24 days

A alone can finish $=(A+B+C)-(B+C)$

$=\frac{5}{120}-\frac{1}{60}=\frac{5-2}{120}=\frac{3}{120}=\frac{1}{40}$

$\therefore$  A alone can do in 40 days.