A and B can complete a work in 15 days and 10 days respectively. They started doing the work together but after 4 days B had to leave. Then A working with a new worker C completed the remaining work in 3 days. If C works alone, in how many days he can do 40% of the same work?

9

A = 15 days, B = 10 days,

⇒ A + B worked for 4 days together = (2 + 3) x 4 = 20 units   ..(Efficiency × Days = Total work)

⇒ B left and replaced by C, 

So, A + C completed the remaining work in 3 days.

⇒ Remaining work = 30 -20 = 10 units.

⇒ \(\frac{10}{2 + C}\) = 3               .. (\(\frac{Work}{Efficiency}\) = Time)

⇒ 10 = 6 + 3C 

⇒ 3C = 4 

⇒ C = \(\frac{4}{3}\).

⇒ 40% of 30 = 12 units.

⇒  Time taken by C to complete 40% of the total work = 

$\frac{12}{\frac{4}{3}}$=\(\frac{12 * 3}{4}\)= 9 days.