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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?

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    $\begingroup$ You are not doing $8\times 8=64$ man-days per day, you are doing $8$ man-days per day as you have $8$ men each day $\endgroup$ – Henry Apr 8 at 9:31
  • $\begingroup$ @Henry Exactly... That is the mistake in the solution presented. $\endgroup$ – peter.petrov Apr 8 at 9:33
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    $\begingroup$ What is the job? If a 50-person orchestra can play Beethoven’s Fifth Symphony in 40 minutes, does that mean it will take a 25-person orchestra 80 minutes? $\endgroup$ – Mike Scott Apr 8 at 18:23
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$\dfrac{12 \times 16 - 5 \times 8}{8}=19$ days of the $8$-man team

or $8+19=27$ days in total

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$12\times16=5\times8+8x$

(12 men @ 16 days = 5 men @ 8 days + 8 men @ x days)

So, $192=40+8x\implies x=19$.

Therefore $8+19=27$ days.

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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: $19d+8d=27d$.

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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.

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$12$ 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 $8 + 19 = 27$.

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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.

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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

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