Xylia and Tammy have some croissants each. If Xylia gives Tammy 31 croissants, Xylia will have ¹₈ as many croissants as Tammy. If Xylia gives Tammy 5 croissants, they will have an equal number of croissants. How many croissants does Tammy have at first?

	Case 1		Case 2
	Xylia	Tammy	Xylia	Tammy
Before	1 u + 31	8 u - 31	4.5 u + 5	4.5 u - 5
Change	- 31	+ 31	- 5	+ 5
After	1 u	8 u	4.5 u	4.5 u

Total number of croissants that Xylia and Tammy have
= 1 u + 8 u
= 9 u 

Number of croissants that Xylia and Tammy each has in the end is the same.

Number of croissants that Xylia and Tammy each has in the end in Case 2
= 9 u ÷ 2
= 4.5 u

Number of croissants that Xylia had at first is the same in Case 1 and Case 2.

4.5 u + 5 = 1 u + 31
4.5 u - 1 u = 31 - 5
0.5 u = 26

1 u = 26 ÷ 0.5 = 52

Number of croissants that Tammy has
= 8 u - 31
= 8 x 52 - 31
= 416 - 31
= 385

Answer(s): 385