algebra precalculus - $16$ men can finish $80$% work in $24$ days..

$16$ men can finish $80$% work in $24$ days. When should $8$ men leave the work so that the whole work is completed in $40$ days? (Answer: $20$days).

MY Attempt:

in $24$ days, $16$ men can do $\dfrac {4}{5}$ work.
In $1$ day, $16$ men can do $\dfrac {1}{30}$ work.
In $1$ day, $1$ man can do $\dfrac {1}{480}$ work.

now, what is the simplest method to complete it further?

Answer

Let $t$ be the number of days when the 8 men leave. Then, the amount of work $w(u)$ finished after working for $u$ days is given by

$$w(u)=\frac{(16-8)\cdot u}{480}+\frac{8t}{480}$$

(assuming that $u>t$). The first term corresponds to 16-8 men working all u days. The second term is the additional 8 men that work only the first $t$ days. Now, solve for $t$ in $w(40)=1$.

Edit: How to get $w(u)$: As another example, assume $p$ men work for $u$ days. As you have already pointed out, one man does $\frac{1}{480}$ of the entire work. So, if $p$ men work for $u$ days, you get done $\frac{pu}{480}$ of the work. You can write $w(u)=\frac{pu}{480}$ to make the number of days worked $u$ a variable.

Now for my expression of $w(u)$: I've split the amount of work done into two parts, depending on the number of men working. You can assume that 16-8 men work for the entire time, i.e. $u$ days. This is the first term. In addition, in the first $t$ days 8 extra men work (which will be removed from the team after $t$ days). This corresponds to the second term.