A and B can do a work in 12 days and 18 days, respectively. They worked together for 4 days after which B was replaced by C and the remaining work was completed by A and C in the next 4 days. In how many days will C alone complete 50% of the same work?

18

A = 12 days, B = 18 days, 

Here, A + B worked for 4 days = (6 + 4) x 4 = 40 units .(Efficiency ×Days=work)

⇒ Remaining work = 72 - 40 = 32 units

Now, B is replaced by C and remaining work is completed in 4 days,

⇒ \(\frac{32}{6 + C}\) = 4 days ⇒ 32 = 24 + 4C 

⇒ 4C = 8

⇒ C = 2 (Efficiency)

Hence, Time taken by C to complete 50% of total work = 50% of 72 = 36 units

⇒ \(\frac{36}{2}\) = 18 days