A, B and C together can complete a work in 12 days and A and B together can complete the same work in 24 days and the ratio of the efficiency of B and C is 2:5. A, B and C started the work and worked for 6 days. After 6 days, A and B left the work and C completed the remaining work. Find the time taken by C to complete the remaining work.
a) 24 days
b) 12 days
c) 8 days
d) 20 days
e) 15 days
Sol:
1/A + 1/B + 1/C = 1/12
1/A + 1/B = 1/24
1/C = 1/12-1/24 = 1/24
Time taken by C alone to complete the work =
24 days
Time taken by B alone to complete the work = 24
* 5/2 = 60 days
1/A + 1/60 = 1/24
1/A = 1/40
Time taken by A alone to complete the work = 40
days
LCM of 40, 60, 24 = 120
Total work = 120 units
A = 120/40 = 3 units per day
B = 120/60 = 2 units per day
C = 120/24 = 5 units per day
A, B and C work for 6 days
Work completed by A, B and C together =
(3+2+5)*6 = 60 units
Remaining work = 120 – 60 = 60 units
Time taken by C to complete the remaining work
= 60/ 5 = 12 days
