Question 726648
6 people worked for 9 days to complete  two fifths of a project. If three
joined the original 6 people, how many days were needed to complete the project?
<pre>

You can do it in your head.  I'll show you that way and then I'll show you
how to do it by algebra:

IN YOUR HEAD.
When the 3 joined them, there were one and a half times as many people,
but the remaining {{{3/5}}}ths of the job is one and a half times the {{{2/5}}}ths they had
already completed, so it would take them the same time, or 9 days.

BY ALGEBRA:
Let x = the number of days needed to complete the remaining {{{3/5}}} of the project.

The 6 people's combined rate is {{{2/5}}}ths of project per 9 days or {{{"2/5_project"/"9_days"}}} or 

{{{(2/5)/9}}}{{{(project)/(day)}}} or {{{(2/5)*(1/9)}}} or {{{2/45}}}{{{(project)/(day)}}}

Since 3 people is half of 6 people, the three who joined them's combined rate
is half that or {{{1/45}}}{{{(project)/(day)}}}, since half as many people work at only half the rate 
of 6 people.

Therefore the combined rate of all 9 people is {{{2/45}}}{{{""+""}}}{{{1/45}}} or {{{3/45}}} or {{{1/15}}}{{{(project)/(day)}}}

After the three joined them after {{{2/5}}}ths of the project was completed, there was 
only {{{3/5}}}ths of the project left to do, so we set up this proportion:

1 project is to 15 days as {{{3/5}}}ths of a project is to x days.

          {{{1/15}}}{{{""=""}}}{{{(3/5)/x}}}

Cross multiply:

          x = {{{3/5}}}·15
          x = 9

So 9 more days were needed to complete the project.

Edwin</pre>