Question 1208470
<pre>
I think this type problem should be considered as a problem in VARIATION.

Common sense tells us that:

Time (t) increases as the amount of job-doing (j) increases.
Time (t) decreases as the number of workers (w) increases.

Therefore, time (t) varies directly as the amount of job-doing (j) and 
inversely as the number of workers (w).  That is:

{{{t}}}{{{""=""}}}{{{k*expr(j/w)}}}
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</pre>A certain job [THAT'S 1 JOB] can be done by 72 men in 100 days.<pre> 
{{{t}}}{{{""=""}}}{{{k*expr(j/w)}}}

{{{100}}}{{{""=""}}}{{{k*expr(1/72)}}}

Solve for k

{{{k}}}{{{""=""}}}{{{7200}}}

So

{{{t}}}{{{""=""}}}{{{7200*expr(j/w)}}}
</pre>There were 80 men at the start of the project 
but after 40 days,....<pre>So for those first 40 days,

{{{40}}}{{{""=""}}}{{{7200*expr(j/80)}}}

{{{3200}}}{{{""=""}}}{{{7200*j}}}

{{{3200/7200}}}{{{""=""}}}{{{j}}}

{{{4/9}}}{{{""=""}}}{{{j}}}

So the amount of job-doing during those first 40 days was 4/9ths of a job. 
So that means there is still 5/9 of a job left to be done.</pre>.....30 of them had to be transferred to another project.<pre>So the amount of job-doing left for the remaining 50 men to do was 5/9 of a job.

{{{t}}}{{{""=""}}}{{{7200*expr(j/w)}}}  

{{{t}}}{{{""=""}}}{{{7200*expr("5/9"/50)}}}

{{{t}}}{{{""=""}}}{{{80}}}

Answer: 80 days.

Edwin</pre>