R, S and T can finish a work in 20, 15 and 10 days, respectively. R works on all days and S and T work on alternate days with T starting the work on the first day. In how many days is the work finished?
Show answer & explanation
Correct answer: C
Explanation
The correct answer is option 3 i.e. 52/7.
Let the total work to be done (LCM of 20, 15 and 10) be 60 units.
Efficiency of R = 60/20 = 3 units/day
Efficiency of S = 60/15 = 4 units/day
Efficiency of T = 60/10 = 6 units/day
Now, work done by (R + T) in 1 day = 9 units
And work done by (R + S) in 1 days = 7 units
Work done in 2 days = 16 units
Work done in (2 × 3 =) 6 days = 48 units
Now, work done in 7 days = (48 + 9 =) 57 units
Remaining work = 60 - 57 = 3 units
3 units of work completed by (R + S) in 3/7 days.
Total time = 7 + (3/7) = 52/7 days