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

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?

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.

• One frequent cause of difficulties is that people often look for a formula that will do the job, instead of thinking directly about the problem. Note how simple and natural Fred's approach is. – André Nicolas Jun 3 '15 at 2:23
• My grandchildren routinely solve such problems in a high school ("Gymnasium") acceptance test. Sometimes it's about how many days longer a supply of hay lasts when three of the 25 cows suddenly died. – Christian Blatter Jun 3 '15 at 10:03

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