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?

R can finish the work in 20 days.

S can finish the work in 15 days.

T can finish the work in 10 days.

R works every day.

S and T work on alternate days, with T starting the work on the first days

Formula Used: 

Total work = efficiency $$\times$$ Time​

LCM of 20, 15, 10 = 60 

the total work is 60 unit. ​

then,  

Efficiency of R = $$\frac{60}{20} = 3$$ unit/day

Efficiency of S = $$\frac{60}{15}$$  = 4 unit/day  

Efficiency of T = $$\frac{60}{10} =$$ 6 unit/day ​

​$$\begin{align*}&\Rightarrow \text{ work done in 2 days} = (R + T) + (R + S) \\&= 9 + 7 = 16 \text{ units} \\[10pt]&\Rightarrow \text{16 units work done in 2 days} \\[10pt]&\Rightarrow \text{48 units work done in } \frac{12}{16} \times 48 = 6 \text{ days} \\[10pt]&\Rightarrow \text{work left} = 60 - 48 = 12 \text{ units} \\[10pt]&\text{Out of this, 9 units done by } (R + T) \text{ in 1 day \& left 3 units done by } (R + S) \text{ in } = \frac{3}{7} \text{ day.} \\[10pt]&\therefore \text{Total time} = 6 + 1 + \frac{3}{7} = \frac{52}{7} \text{ days}\end{align*}$$​​