# $12$ 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?

• 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 – Henry Apr 8 at 9:31
• @Henry Exactly... That is the mistake in the solution presented. – peter.petrov Apr 8 at 9:33
• 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? – Mike Scott Apr 8 at 18:23

$$\dfrac{12 \times 16 - 5 \times 8}{8}=19$$ days of the $$8$$-man team

or $$8+19=27$$ days in total

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

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

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

Therefore $$8+19=27$$ days.

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

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.

$$12$$ 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 $$8 + 19 = 27$$.

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.

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