$A, B$ and $C$ can reap a field in $15 \frac{3}{4}$ days; $B, C$ and $D$ in 14 days; $C, D$ and $A$ in 18 days; $D, A$ and $B$ in 21 days. In what time can $A, B, C$ and $D$ together reap it?
Time taken by $(\mathrm{A}+\mathrm{B}+\mathrm{C})$ to do the work $=15 \frac{3}{4}$ days $=\frac{63}{4}$ days
Time taken by $(\mathrm{B}+\mathrm{C}+\mathrm{D})$ to do the work $=14$ days
Time taken by $(\mathrm{C}+\mathrm{D}+\mathrm{A})$ to do the work $=18$ days
Time taken by $(\mathrm{D}+\mathrm{A}+\mathrm{B})$ to do the work $=21$ days
Now,
Work done by $(\mathrm{A}+\mathrm{B}+\mathrm{C})=\frac{4}{63}$
Work done by $(\mathrm{B}+\mathrm{C}+\mathrm{D})=\frac{1}{14}$
Work done by $(\mathrm{C}+\mathrm{D}+\mathrm{A})=\frac{1}{18}$
Work done by $(\mathrm{D}+\mathrm{A}+\mathrm{B})=\frac{1}{21}$
$\therefore$ Work done by working together $=(\mathrm{A}+\mathrm{B}+\mathrm{C})+(\mathrm{B}+\mathrm{C}+\mathrm{D})+(\mathrm{C}+\mathrm{A}+\mathrm{D})+(\mathrm{D}+\mathrm{A}+\mathrm{B})$
$=\frac{4}{63}+\frac{1}{14}+\frac{1}{18}+\frac{1}{21}$
$=\frac{4}{63}+\left(\frac{9+7+6}{126}\right)=\frac{4}{63}+\frac{22}{126}$
$=\frac{4}{63}+\frac{11}{63}=\frac{15}{63}$
$\therefore$ Work done by working together $=3(\mathrm{~A}+\mathrm{B}+\mathrm{C}+\mathrm{D})=\frac{15}{63}$
$\therefore$ Work done by $(\mathrm{A}+\mathrm{B}+\mathrm{C}+\mathrm{D})=\frac{15}{63 \times 3}=\frac{5}{63}$
Thus, together they can do the work in $\frac{63}{5}$ days or $12 \frac{3}{5}$ days.