*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?

#### Solution

\[\text{ Time taken by } \left( A + B + C \right) \text{ to do the work } = 15\frac{3}{4} \text{ days } = \frac{63}{4} \text{ days } \]

\[\text{ Time taken by } \left( B + C + D \right) \text{ to do the work = 14 days} \]

\[\text{ Time taken by } \left( C + D + A \right) \text{ to do the work = 18 days } \]

\[\text{ Time taken by } \left( D + A + B \right) \text{ to do the work = 21 days } \]

\[\text{ Now, } \]

\[ \text{ Work done by } \left( A + B + C \right) = \frac{4}{63}\]

\[\text{ Work done by } \left( B + C + D \right) = \frac{1}{14}\]

\[ \text{ Work done by } \left( C + D + A \right) = \frac{1}{18}\]

\[ \text{ Work done by } \left( D + A + B \right) = \frac{1}{21}\]

\[ \therefore \text{ Work done by working together } = \left( A + B + C \right) + \left( B + C + D \right) + \left( C + A + D \right) + \left( D + A + B \right)\]

\[ = \frac{4}{63} + \frac{1}{14} + \frac{1}{18} + \frac{1}{21}\]

\[ = \frac{4}{63} + \left( \frac{9 + 7 + 6}{126} \right) = \frac{4}{63} + \frac{22}{126}\]

\[ = \frac{4}{63} + \frac{11}{63} = \frac{15}{63}\]

\[ \therefore \text{ Work done by working together } = 3\left( A + B + C + D \right) = \frac{15}{63}\]

\[ \therefore \text{ Work done by } \left( A + B + C + D \right) = \frac{15}{63 \times 3} = \frac{5}{63}\]

\[ \text{ Thus, together they can do the work in } \frac{63}{5} \text{ days or } 12\frac{3}{5} \text{ days } .\]