A, B and C can reap a field in

Solution:

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.