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A and B Are Still At It

A and B, starting at opposite corners of a square whose side is 100 yards long, walk round it in the same direction. A walks 19 yards while B walks 16. At what part of the square will they be together, and how many times will each have passed the corner from whence they started?


Suggested Solution (Adam Miyata & Maggie Jurges, WWU students)

We can see that A walks 19 yards in the same time it takes B to walk 16 yards. In other words, A walks 19 yards per unit of time and B walks 16 yards per unit of time.

Each side of the square is 100 yards, and since A and B start at opposite corners, there are 200 yards between them, regardless of the direction of travel.

However, since both travel in the same dierction, A will gain 3 yards on B each time they walk, or per every unit of time.

So, to find the number of units of time it will take for A to catch B, we divide the 200 yards by the 3 yards that A gains on B each period:

200/3 = 66 2/3

Then to find the distance A has walked, multiply [66 2/3] x 19 = 1266 2/3 yards.

And to find what B has walked, multiply [66 2/3]x 16 = 1066 2/3 yards.

Finally, A and B would have passed their starting corners every 400 yards traveled since there are four 100-yard sides. Therefore, A passed its corner 3 times since it traveled more than 3 x 400 = 1200 yards. B passed its corner twice, since it traveled more than 2 x 400 = 800 yards.

The possible meeting locations on the sides of the square can then be determined, but it will be dependant on whether they are going clockwise or counterclockwise.

Source: Charles Pendlebury's Arithmetic (London: G. Bell & Son, 1918)