A and B complete a piece of work in 24 days. B and C do the same work in 36 days, and A, B and C together finish it in 18 days. In how many days will:

(i) A alone,

(ii) C alone,

(iii) A and C together, complete the work?

(i) (A + B)'s 1 day work = $\dfrac{1}{24}$

(B + C)'s 1 day work = $\dfrac{1}{36}$

(A + B + C)'s 1 day work = $\dfrac{1}{18}$

A's 1 day work = (A + B + C)'s 1 day work - (B + C)'s 1 day work

= $\dfrac{1}{18} - \dfrac{1}{36}$

= $\dfrac{2 - 1}{36}$

= $\dfrac{1}{36}$

Number of days required by A alone = 36 days

Hence, A requires 36 days to complete the work alone.

(ii) C's 1 day work = (A + B + C)'s 1 day work - (A + B)'s 1 day work

= $\dfrac{1}{18} - \dfrac{1}{24}$

= $\dfrac{4 - 3}{72}$

= $\dfrac{1}{72}$

Number of days required by C alone = 72 days

Hence, C requires 72 days to complete the work alone.

(iii) (A + C)'s 1 day work = $\dfrac{1}{36} + \dfrac{1}{72}$

= $\dfrac{2 + 1}{72}$

= $\dfrac{3}{72}$

No. of days required to complete the work with A and C working together = $\dfrac{72}{3} = 24$ days

Hence, A and C together will complete the work in 24 days.