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. At what time can A, B, C, and D together reap it?

Time taken by A, B, and C together to reap the field $=15\frac{3}{4}$ days $= \frac{63}{4}$ days

Time taken by B, C, and D together to reap the field $= 14$ days

Time taken by C, D, and A together to reap the field $= 18$ days

Time taken by D, A, and B together to reap the field $= 21$ days

To do :

We have to find the time to be taken A, B, C, and D together can reap it.

Solution :

Work done by A, B and C together in a day $=15\frac{3}{4}$ days $= \frac{63}{4}$ days ----(1)

Work done by B, C and D together in a day $= \frac{1}{14}$ -----(2)

Work done by C, D, and A together in a day $= \frac{1}{18}$ ----(3)

Work done by D, A, and B together in a day $= \frac{1}{21}$ -----(4)

Adding (1), (2), (3), and (4), we get,

Work done by A, B and C together in a day $+$ Work done by B, C and D together in a day $+$ Work done by C, D and A together in a day $+$ Work done by D, A and B together in a day $= \frac{4}{63} + \frac{1}{14} + \frac{1}{18} + \frac{1}{21}$

3(Work done by A, B, C and D in a day) $= \frac{4}{63} + \frac{(1\times9+1\times7+1\times6)}{126} = \frac{4}{63} + \frac{(9+7+6)}{126} = \frac{4}{63} + \frac{22}{126} = \frac{4}{63} + \frac{11}{63} = \frac{(4+11)}{63} = \frac{15}{63}$

Work done by A, B, C and D in a day $= \frac{15}{63} \times \frac{1}{3} = \frac{5}{63}$

Time required for reaping the field if A, B, C and D work together is $\frac{1}{\frac{5}{63}} days = \frac{63}{5} days = 12\frac{3}{5}$ days.

Therefore, A, B, C, and D together can reap in $ 12\frac{3}{5}$ days.