$12$ men can finish a job in $16$ days. $5$ men work at the start; and after 8 days, 3 men were added. How many days needed to finish the whole job? 
Twelve men can finish a job in 16 days. 5 men were working at the start and after 8 days, 3 men were added. How many days will it take to finish the whole job?

Solution:
So, the job will take $12 \times 16 = 192$ man days to finish.
In the first $8$ days we have done $8 \times 5 = 40$ man days.
Now we are doing $8 \times 8 = 64$ man days per day and need to do the remaining $192 - 40 = 152 $man days.


*

*Day $1 = 152 - 64 = 88$ man days left

*Day $2 = 88 - 64 = 24$
So, we’ll finish on day $3$ after the extra $3$ men are added.
So based on my solution, I came up with $11$ days but I feel like it's wrong. Can someone point out my mistakes if any? 
 A: $\dfrac{12 \times 16 - 5 \times 8}{8}=19$ days of the $8$-man team 
or $8+19=27$ days in total
A: $12\times16=5\times8+8x$ 
(12 men @ 16 days = 5 men @ 8 days + 8 men @ x days)
So, $192=40+8x\implies x=19$.
Therefore $8+19=27$ days.
A: We know that $\frac{S}{12\cdot v_M}=16d$ where $S$ is the lenght of the work, $v_M$ is the velocity of one men and $d$ stands for days. Also, we know that:
$$5v_M\cdot (8d+t)+3v_M\cdot t=S$$
Substituing, $S=16d\cdot12v_M$, we obtain:
$$5v_M\cdot (8d+t)+3v_M\cdot t=16d\cdot12v_M\leftrightarrow 8v_M\cdot t=152d \cdot v_M\leftrightarrow t=19d$$
In total, we have: $19d+8d=27d$.
A: One man does $\frac1{192}$ of the job in one day – call this one unit. When the extra men are added, $40$ units have been completed, and the new task force completes $8$ units per day. So it takes $\frac{192-40}8=19$ days for them to complete the job, and $27$ days in all.
A: $12$ men can finish the job in $16$ days.
In $8$ days, $24$ men can finish the job.
So $5$ men would've finished $\frac 5{24}$ of the job within those $8$ days, leaving $\frac{19}{24}$ of the job remaining.
Add $3$ men, you get $8$ men now. $8$ men would take $24$ days to finish the job.
So the reinforced workforce would take an additional $\frac{19}{24} \times 24 = 19$ days to finish up.
The total time from the start is $8 + 19 = 27$.
A: In steps:
$1$ man can finish the job in $12×16$ days.
$1$ man works  $1$ day to finish $1/(12×16)$ of the job.
$5$ man work $8$ days to finish $(5×8)/(12×16)=5/24$ of the job.
$19/24$ of the job left to complete by $8$ people.
Let $X$ be the number of days.
$(8/(12×16))×X=19/24$;
$X= 19$;
Altogether:
$8+19=27$ days.
A: It would take one man 12 × 16 = 192 days to finish by himself
In the first 8 days, with 5 men, they finish the same amount of work one person could finish in 8×5 = 40 days.
192 - 40 = 152, 
meaning there is still the amount of work one person could finish in 152 days left.
With the crew of 5 + 3 = 8, it would only take 152/8 = 19 days
However, remember to add the 8 days from before:
19 + 8 = 27 
Thus, it would take 27 days
