X and Y can do a piece of work in 20 days and 12 days respectively. X started the work alone and then after 4 days Y joined him till the completion of the work. How long did the work last?
6 days
10 days
15 days
20 days
Explanation:
Work done by X in 4 days =$ \left(\dfrac{1}{20} \times 4\right) $=$ \dfrac{1}{5} $.
Remaining work =$ \left(1 -\dfrac{1}{5} \right) $=$ \dfrac{4}{5} $.
$\left(X + Y\right)$s 1 days work =$ \left(\dfrac{1}{20} +\dfrac{1}{12} \right) $=$ \dfrac{8}{60} $=$ \dfrac{2}{15} $.
Now,$ \dfrac{2}{15} $work is done by X and Y in 1 day.
So,$ \dfrac{4}{5} $work will be done by X and Y in $\left(\dfrac{15}{2}\times\dfrac{4}{5}\right) $= 6 days.
Hence, total time taken = $\left(6 + 4\right)$ days = 10 days.