# Time and work issue

Question: 45 men can complete a work in 16 days. Six days after they started working, 30 more men joined them. How many days will they now take to complete the remaining work ?

Answer of this question is 6 days but I am not able to understand book's solution.

According to my maths book's solution, it said in first line of solution that = (45*16) men can complete the work in 1 day. (don't know how they calculate this)

please solve this question in simple and easiest way. Thanks for help in advance.

• The work takes in total 45*16 man-days, after six days 45 men have worked 45*6 man-days, so there are (45*16-45*6) man-days remaining that the 45+30 men will work. Also this is site is for questions about the software Mathematica – ssch Sep 25 '13 at 14:33

You've heard, perhaps, of "man-hours"? Let's call a "man-day" the number of days required of one man to complete a project. Or, we the number of men required to complete a project in one day. Then: $$\text{The work requires a total of}\;45\text{ men} \times 16 \;\text{days} = 45\times 16 = 720 \text{ man-days}$$

That is, it would take one man $45\times 16 = 720$ days to complete the project, or alternatively, it would take 720 men 1 day to complete the project.

$45$ men work for $6$ days $\implies$ reducing the number of man days to complete the project by $45\times 6$. We subtract this from the total number of man-days required for the project. $$45\times 16 - 45\times 6 = 45(16 - 6) = 45\times 10 = 450\;\text{man-days of work remain}$$

Now, after those first six days of work, we have $30$ men join the original crew of $45$ men $\implies 75 \;$ men in all. Let $x$ denote the number of days required for the new crew of $75$ men to complete the rest of the work: 450 man-days.

We have that there are $$450 \text{ man-days remaining} = 75 x \iff x = \frac {45\times 10}{75} = 6$$ That is, it would take $75$ men $6$ days to complete the remaining $450$ man-days of work.

## The model...

Before we do anything, we need to assume an underlying model of the problem. We assume there is a function $C(t,w)$ which describes the amount of work completed at time $t$ if we have $w$ workers. Further, we assume $C$ is directly proportional to $wt$. I.e., $$C(t,w)=kwt$$ for some positive constant $k$.

(Note: This is an unrealistic model. This assumes, e.g., (a) each worker works continuously until the task is completed, without breaks for food, sleep, etc., (b) every worker can complete every sub-task, and (c) tasks are infinitely divisible.)

## Finding the parameter...

We know that $45$ workers can complete "a work" in $16$ days. This gives $1=k \times 45 \times 16$, which implies $k=\tfrac{1}{720}$. (The unit for $k$ is (worker day)$^{-1}$, but I'll leave these off for simplicity.) Hence $$C(t,w)=\tfrac{1}{720}wt.$$

Note also that $C(t,w)=1$ is when one unit of work is complete.

## The question...

We now have the situation: $45$ workers work for $6$ days, then $45+30=75$ workers work for $d$ days until one unit of work is complete. We want to find $d$.

Assuming the above model, this situation is described by the equation $$C(6,45)+C(d,75)=1.$$ I.e., $$\frac{46 \times 6}{720}+\frac{75 \times d}{720}=1.$$ If we solve this for $d$, we find $d=6$ days.

45 men completes whole work in 16 days so 45 men's one day's work=1/16 6 day's work =6/16 or 3/8 remaining work=5/8 now 30 more men joining so 45+30=75men by formula m1.d1/w1=m2d2/w2

         45.6/3/8=75.d2/5/8
d2=6days