A and B can do a piece of work in 30 days, while B and C can do the same work in 24 days and C and A in 20 days. They all work together for 10 days when B and C leave. How many days more will A take to finish the work?
2$\left(A + B + C\right)$s 1 days work =$ \left(\dfrac{1}{30} +\dfrac{1}{24} +\dfrac{1}{20} \right) $=$ \dfrac{15}{120} $=$ \dfrac{1}{8} $.
Therefore, $\left(A + B + C\right)$s 1 days work =$ \dfrac{1}{2 \times 8} $=$ \dfrac{1}{16} $.
Work done by A, B, C in 10 days =$ \dfrac{10}{16} $=$ \dfrac{5}{8} $.
Remaining work =$ \left(1 -\dfrac{5}{8} \right) $=$ \dfrac{3}{8} $.
As 1 days work =$ \left(\dfrac{1}{16} -\dfrac{1}{24} \right) $=$ \dfrac{1}{48} $.
Now,$ \dfrac{1}{48} $work is done by A in 1 day.
So,$ \dfrac{3}{8} $work will be done by A in 48$ \times \dfrac{3}{8} $= 18 days.