20 men can do a piece of work in 18 days. They worked together for 3 days, then 5 men joined. In how many days is the remaining work completed ?

Answer: Option A

20 men → 18 days
⇒ Work done by 20 men working
Together = 1 work
⇒ Work done by them in 3 days working
Together = 1 × 3 = 3 work
⇒ Remaining work = 18 - 3 = 15 work
⇒ 15 work is to be done by (20 + 5) = 25 men
$$\begin{array}{rl}& \therefore \text{Efficiency of 1 man}=\frac{1}{20}\\ & \Rightarrow \text{Efficincy of 5 men}\\ & =\frac{5}{20}\\ & =\frac{1}{4}\\ & \Rightarrow \text{So, efficiency of }\left(20+5\right)\\ & \Rightarrow 25\text{ men}=1+\frac{1}{4}\\ & =\frac{5}{4}\text{ working days}\\ & \text{Required time}\\ & =\frac{\text{Work}}{\text{Efficiency}}\\ & =\frac{15}{\frac{5}{4}}\\ & =12\text{ days}\end{array}$$
Therefore, 12 more days will be taken to finish the remaining work

Alternate : 20 men can do 18 days
So, total work = 18 × 20 = 360
20 men 3 days work = 20 × 3 = 60
Remaining work = 360 - 60 = 300
After joining 5 men total men = 20 + 5 = 25
So, finish the remaining work in = $$\frac{300}{25}=12\text{ days}$$