A, B, C and D, working separately, can do a piece of work in 5 days, 6 days, 8 days and 10 days, respectively. In how many days can A and B complete the work together, if they are assisted by C and D on alternate days, starting with C?

Correct option is C

Given:
A can complete the work in 5 days
B can complete the work in 6 days
C can complete the work in 8 days
D can complete the work in 10 days
A and B work daily, assisted by C and D on alternate days, starting with C.
Solution:
Total work = LCM(5, 6, 8, 10) = 120 units
Efficiency of A = 120 ÷ 5 = 24 units/day
Efficiency of B = 120 ÷ 6 = 20 units/day
Efficiency of C = 120 ÷ 8 = 15 units/day
Efficiency of D = 120 ÷ 10 = 12 units/day
Work done by (A+B+C) on Day₁ = 24+20+15 = 59 units
Work done by (A+B+D) on Day₂ = 24+20+12 = 56 units
2-day cycle = 59+56 = 115 units
After 2 days, Remaining work = 120−115 = 5 units
Time taken to complete by  (A+B+C) on Day₃ = $$\frac 5{59} \text{days}$$​
Total time = 2 + $$\frac 5{59} \text{days}$$​ = $$2 \frac 5{59} \text{days}$$​