There were some red buttons and black buttons. The buttons were packed into 2 bags. At first, Packet L contained 240 buttons and 10% of them were black buttons. Packet M contained 1000 buttons and 60% of them were black buttons. How many red buttons and black buttons in total must be moved from Packet L to Packet M such that 25% of the buttons in Packet L are red and 50% of the buttons in Packet M are black?

	Packet L		Packet M
Total	240		1000
	Black buttons	Red buttons	Black buttons	Red buttons
Before	24	216	600	400
Change	- ?	- ?	+ ?	+ ?
After	3 u	1 u	1 p	1 p

Number of black buttons in Packet L at first
= 10% x 240
= ¹⁰₁₀₀ x 240
= 24

Number of red buttons in Packet L at first
= 240 - 24
= 216

Number of black buttons in Packet M at first
= 60% x 1000
= ⁶⁰₁₀₀ x 1000
= 600

Number of red buttons in Packet M at first
= 1000 - 600
= 400

Packet L in the end
25% = ²⁵₁₀₀ =¹₄
Black buttons : Red buttons = 3 : 1

Packet M in the end
50% = ⁵⁰₁₀₀ =¹₂
Black buttons : Red buttons = 1 : 1

Total number of black buttons = 3 u + 1 p

3 u + 1 p = 24 + 600
3 u + 1 p = 624 

1 p = 624 - 3 u --- (1)

Total number of red buttons = 1 u + 1 p
1 u + 1 p = 216 + 400
1 u + 1 p = 616

1 p = 616 - 1 u --- (2)

(2) = (1)
616 - 1 u = 624 - 3 u
3 u - 1 u = 624 - 616
2 u = 8

1 u = 8 ÷ 2 = 4

Total number of red buttons and black buttons that must be moved from Packet L to Packet M
= 240 - 4 u
= 240 - 4 x 4
= 240 - 16
= 224

Answer(s): 224