Question : 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?
Option 1: $\frac{61}{7}$
Option 2: $\frac{50}{7}$
Option 3: $\frac{52}{7}$
Option 4: $\frac{57}{8}$
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Correct Answer: $\frac{52}{7}$
Solution :
Given: R alone can complete a task in 20 days.
T alone can complete the same task in 15 days.
S alone can complete the same task in 10 days.
Total work is LCM of 20,15 and 10 = 60 units
So, the efficiency of R $=\frac{60}{20} = 3$ units per day
Efficiency of S $=\frac{60}{15} = 4$ units per day
Efficiency of T $=\frac{60}{10} = 6$ units per day
R works on all days, and S and T work on alternate days, starting with T.
So, work done in 2 days = ((efficiency of (R + T)) × 1days) + ((efficiency of (R + S)) × 1 days).
= ((3 + 6) × 1days) + ((3 + 4) × 1 days) = 16 units.
Work done in 2 days = 16 units.
So, work done in 6 days = 48 units.
Work left = 12 units.
Now (R + T) start again and do 9 units of work on the 7th day.
So, their 7 days of work is 57 units.
Work left now is 3 units.
Now, (R + S) start and do 7 units of work in 1 day.
So (R + S) do 3 units of work in $\frac{3}{7}$ days.
So, the total time taken is 7 days + $\frac{3}{7}$ days = $\frac{52}{7}$ days
Hence, the correct answer is $\frac{52}{7}$.
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