A country baker sent off his boy with a message to the butcher in the next village, and at the same time the butcher sent his boy to the baker.
One ran faster than the other, and they were seen to pass at a spot
720 yards from the baker’s shop.
Each stopped ten minutes at his destination and then started on the return journey, when it was found that they passed each other at a spot 400 yards from the butcher’s.
How far apart are the two tradesmen’s shops?
Of course each boy went at a uniform pace throughout.
Source: Henry Dudeney’s Canterbury Puzzles.
Let D be the distance between the two villages.
At the first meeting the baker's boy has traveled 720 yards and the butcher's boy traveled D-720 yards.
Similarly, at the second meeting the baker's boy has traveled D+400 yards and the butcher's boy traveled 2D-400 yards.
The ten minute stop is immaterial as both boys stopped for the same time length. Each boys' speed was constant otherwise. Then the ratios of distance covered must be equal at both meetings. This implies the equation:
720/(D-720) = (D+400)/(2D-400)
Cross multiplying and simplifying yields:
720*(2D-400) = (D+400)*(D-720)
1440D - 288000 = D^2 - 320D - 288000
D^2 - 1760D = 0
Discard the D=0 root to leave the answer D=1760 yards. In general if the first meeting was A yards from the baker and B yards from the butcher then the distance D=3*A-B