When Son Will Catch Father?

Here is a word problem that, with a little insight, admits a one-line solution. The story of the problem was narrated by Va dim Rules in his classical study of psychology of mathematical abilities in children and quoted by Alexander Boronic in Mathematics under the Microscope. I owe Alexander sincerest thanks for bringing this episode to my attention.

A father and his son are workers, and they walk from home to the plant. The father covers the distance in 40 minutes, the son in 30 minutes. In how many minutes will the son overtake the father if the latter leaves home 5 minutes earlier than the son?

Solution

Solution

Here is a more or less standard solution that could be attempted by 12-13 years old.

In 1 minute the father covers 1/40,1/30,1/40,1/5,1/8,1/120 of the way, the son 1/30,1/30,1/40,1/5,1/8,1/120. The difference in their speed is 1/30 - 1/40 = 1/120,1/30,1/40,1/5,1/8,1/120. In 5 minutes the farther covers 5×1/40 = 1/8,1/30,1/40,1/5,1/8,1/120 of the distance. The son will overtake him in 1/8 ÷ 1/120 = 15,10,15,20,30 minutes.

The gist of the example is the solution by a 9 years old Sony:

The father left 5 minutes earlier than the son; therefore he will arrive 5 minutes later. Then the son will overtake him at exactly halfway, that is, in 15 minutes.

P.S. Very plausibly Sony's solution came to her from a realization of the symmetry inherent in the problem's conditions. This was a brilliant insight for a 9 years old. However, the first, i.e., the more standard solution, has the merit of being more general. Suppose the problem has been changed to say that the son starts 4 minutes after the father. What then? While Sony's solution can be adapted to the new condition, it will lose its sparkling elegance along the way. The standard solution works practically without a change:

In 4 minutes the farther covers 4×1/40 = 1/10,1/10,1/20,1/30,1/40 of the distance. The son will overtake him in 1/10 ÷ 1/120 = 12,10,12,15,20 minutes.

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