# 25 men are employed to do a work in, which they could finish in 20 days but they drop off by 5 men at the end of every 10 days. In what time will the work be completed?

The work finishes on the 24th day.

#### Explanation:

We start with 25 men. They can do a job in 20 days. This means that the 25 men finish $\frac{1}{20}$ of the job each day. This means that each man (using $m$ to mean the work a man does each day) does:

$25 m = \frac{1}{20} \implies m = \frac{1}{500}$ of the job each day.

At the end of 10 days, the 25 men will have finished:

$\left(\frac{1}{500}\right) \left(25\right) \left(10\right) = \frac{250}{500} = \frac{1}{2}$ of the job.

And now we drop off 5 men, leaving 20 left. We have $\frac{1}{2}$ of the job to do:

$\left(\frac{1}{500}\right) \left(20\right) \left(10\right) = \frac{200}{500} = \frac{2}{5}$

Now the job is $\frac{1}{2} + \frac{2}{5} = \frac{5}{10} + \frac{4}{10} = \frac{9}{10}$ done, leaving $\frac{1}{10}$ to do. We drop off 5 men, leaving 15. They should finish before the 10 days are up:

$\left(\frac{1}{500}\right) \left(15\right) \left(10\right) = \frac{150}{500} = \frac{3}{10}$

And they do. In fact, they can do the remaining work 3 times over. This means they'll finish on the 4th day:

$\left(\frac{1}{500}\right) \left(15\right) \left(3\right) = \frac{45}{500} < \frac{1}{10}$

$\left(\frac{1}{500}\right) \left(15\right) \left(4\right) = \frac{60}{500} > \frac{1}{10}$

Jul 16, 2018

The time taken will be $23 \frac{1}{3}$ days, so the work will be completed by the $24 t h$ day.

#### Explanation:

We are working with inverse proportion where, as the number of people decreases, the time taken to finish a task will increase.
Find the constant first.

$x \times y = k$

The total task requires $25 \times 20 = 500$ 'man-days'

In the first $10$ days, $25 \times 10 = 250$ 'man-days.', which means that the task is half completed.

For the next $10$ days there are only $20$ men.

The amount of work completed is: $20 \times 10 = 200$ 'man-days'.

There are then $15$ men left to complete the remaining amount of work.

$500 - 250 - 200 = 50$ 'man-days'

The work can be completed in:

$\frac{50}{15} = 3 \frac{1}{3}$ days

The total time taken is therefore $10 + 10 + 3 \frac{1}{3}$ days

The work will be completed on the $24 t h$ day.