There were some green buttons and brown buttons. The buttons were packed into 2 bags. At first, Box L contained 145 buttons and 40% of them were brown buttons. Box M contained 75 buttons and 80% of them were brown buttons. How many green buttons and brown buttons in total must be moved from Box L to Box M such that 25% of the buttons in Box L are green and 50% of the buttons in Box M are brown?

	Box L		Box M
Total	145		75
	Brown buttons	Green buttons	Brown buttons	Green buttons
Before	58	87	60	15
Change	- ?	- ?	+ ?	+ ?
After	3 u	1 u	1 p	1 p

Number of brown buttons in Box L at first
= 40% x 145
= ⁴⁰₁₀₀ x 145
= 58

Number of green buttons in Box L at first
= 145 - 58
= 87

Number of brown buttons in Box M at first
= 80% x 75
= ⁸⁰₁₀₀ x 75
= 60

Number of green buttons in Box M at first
= 75 - 60
= 15

Box L in the end
25% = ²⁵₁₀₀ =¹₄
Brown buttons : Green buttons = 3 : 1

Box M in the end
50% = ⁵⁰₁₀₀ =¹₂
Brown buttons : Green buttons = 1 : 1

Total number of brown buttons = 3 u + 1 p

3 u + 1 p = 58 + 60
3 u + 1 p = 118 

1 p = 118 - 3 u --- (1)

Total number of green buttons = 1 u + 1 p
1 u + 1 p = 87 + 15
1 u + 1 p = 102

1 p = 102 - 1 u --- (2)

(2) = (1)
102 - 1 u = 118 - 3 u
3 u - 1 u = 118 - 102
2 u = 16

1 u = 16 ÷ 2 = 8

Total number of green buttons and brown buttons that must be moved from Box L to Box M
= 145 - 4 u
= 145 - 4 x 8
= 145 - 32
= 113

Answer(s): 113