I found this puzzle in the 1950 book Figure for Fun, by Yakov Perelman, where it is problem 29.
It concerns two workers who live in the same apartment block and work in the same establishment. Both walk to work each day and the first worker takes 30 minutes for the journey, whereas their faster colleague takes only 20 minutes.
Suppose the slower worker sets off 10 minutes before the faster and they move at their usual speeds. When will the faster worker catch the slower one up?
Scroll down for my solution.
Solution
(This is one way to solve the problem but I expect other approaches are possible.)
Suppose the distance is km. Then the speed of the slower worker must be km/h and that of the faster km/h. After minutes the slower worker will have travelled km.
Now we calculate how far each will have travelled at time hours (after the faster worker sets off.
Slower worker:
Faster worker:
The faster worker will have caught up when these are equal, so , which reduces to or hours = minutes.