Work Problem that deals with Number of Men, Days, Leaving

Question

A project can be done by 70 men in 100 days. There were 80 men at the start of the project but after 50 days, 20 of them had to be transferred to another project. How long will it take the remaining workforce to complete the job?

The correct answer is 50.

Any hints on how to go about this? I have encountered work problems before with the general formula $$\frac1A + \frac1B + \dots = \frac1T.$$

There's also problems with time involved:

$$t_A\left(\frac1A + \frac1B\right) + t_B\left(\frac1C + \frac1D\right) \dots = 1.$$

This problem incorporates people leaving, remaining days. But I am not sure how to combine them concepts.

Answer 1

Think about the required amount of work in man-days. The project requires $70*100=7000$ man-days of total work. After $80*50=4000$ man-days of work, there are $7000-4000=3000$ man-days of work remaining, and there are $60$ remaining workers, so the project will take another $3000/60=50$ days.

Answer 2

One man does one-seventieth of the project in 100 days. Therefore, one man does one-seventhousandth of the project per day.

For 50 days, 80 men each did $\frac{1}{7000}$ of the project, leading to a total progress of $\frac{50\cdot 80}{7000} = \frac{4000}{7000} = \frac47$ of the project being done.

Now, 60 men remain. Solve $\frac{60x}{7000} = 1-\frac47$. Step by step, we have

$$\frac{60x}{7000} = 1-\frac47 = \frac37 \\ \frac{60x}{7000} = \frac{3000}{7000} \\ \frac{60x}{7000} \times 7000 = \frac{3000}{7000} \times 7000 \\ 60x = 3000 x = \frac{3000}{60} = \frac{300}{6} = \frac{100}{2} = 50.$$