A heap of 48 matches are divided into three groups.

If I take as many matches from the first group as there are in the second group and add them to the second, and then take as many from the second group as there are in the third group and add them to the third, and finally take as many from the third group as there are in the first group and add them to the first group, the number of matches in each heap would be equal.

How many matches were in the three groups originally?

Call the original quantities in each pile a, b, and c.
Final number of matches in the 2nd group = 2b - c = 16.
In the 3rd group: 2c - (a - b) = 16
In the 1st group: a - b + a - b = 16.

From the last equation, a - b = 8, so 2c - 8 = 16, so c = 12. 2b - 12 = 16, so b = 14. a - 14 = 8, so a = 22.

Thus, there are 22, 14, and 12 matches in the first, second, and third groups respectively.

Posted by friedlinguini on 2002-06-27 02:22:53