A and B can do a job in 12 days and B and C can do it in 16 days. A and B started working, without C. A worked for 5 days and quit. B worked for 7 days and quit. C worked for a total of 13 days and finished the work. In how many days can C do the work alone?
A. 12 days
B. 24 days
C. 27 days
D. 29 days
Sol:
The number of days taken by A, B & C is as follows.
(A + B) = 12 days
(B + C) = 16 days
Let the total work be the LCM of the number of days taken by three of them i.e. 48 units
According to the question:
A and B worked together for 5 days and did (4 × 5) 20 units of work.
B and C worked together for 2 days and did (2 × 3) 6 units of work
C finished the remaining work of (48 – 20 – 6) 22 units in days (13 – 2) 11 days.
Hence, Efficiency of C = 22 units / 11 days = 2 units
So, No of days taken by C to do the whole job alone = 48/ 2
∴ 24 days
