X can do a piece of work in 40 days. He works at it for 8 days and then Y finished it in 16 days. How long will they together take to complete the work?
Work done by X in 8 days =$ \left(\dfrac{1}{40} \times 8\right) $=$ \dfrac{1}{5} $.
Remaining work =$ \left(1 -\dfrac{1}{5} \right) $=$ \dfrac{4}{5} $.
Now,$ \dfrac{4}{5} $work is done by Y in 16 days.
Whole work will be done by Y in$ \left(16 \times\dfrac{5}{4} \right) $= 20 days.
Xs 1 days work =$ \dfrac{1}{40} $, Ys 1 days work =$ \dfrac{1}{20} $.
$\left(X + Y\right)$s 1 days work =$ \left(\dfrac{1}{40} +\dfrac{1}{20} \right) $=$ \dfrac{3}{40} $.
Hence, X and Y will together complete the work in$ \left(\dfrac{40}{3} \right) $= 13$ \dfrac{1}{3} $days.