Betty and Cathy have some croissants each. If Betty gives Cathy 31 croissants, Betty will have ³₄ as many croissants as Cathy. If Betty gives Cathy 7 croissants, they will have an equal number of croissants. How many croissants does Cathy have at first?

	Case 1		Case 2
	Betty	Cathy	Betty	Cathy
Before	3 u + 31	4 u - 31	3.5 u + 7	3.5 u - 7
Change	- 31	+ 31	- 7	+ 7
After	3 u	4 u	3.5 u	3.5 u

Total number of croissants that Betty and Cathy have
= 3 u + 4 u
= 7 u 

Number of croissants that Betty and Cathy each has in the end is the same.

Number of croissants that Betty and Cathy each has in the end in Case 2
= 7 u ÷ 2
= 3.5 u

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

3.5 u + 7 = 3 u + 31
3.5 u - 3 u = 31 - 7
0.5 u = 24

1 u = 24 ÷ 0.5 = 48

Number of croissants that Cathy has
= 4 u - 31
= 4 x 48 - 31
= 192 - 31
= 161

Answer(s): 161