There were some yellow buttons and black buttons. The buttons were packed into 2 bags. At first, Packet R contained 300 buttons and 30% of them were black buttons. Packet S contained 270 buttons and 80% of them were black buttons. How many yellow buttons and black buttons in total must be moved from Packet R to Packet S such that 20% of the buttons in Packet R are yellow and 50% of the buttons in Packet S are black?

	Packet R		Packet S
Total	300		270
	Black buttons	Yellow buttons	Black buttons	Yellow buttons
Before	90	210	216	54
Change	- ?	- ?	+ ?	+ ?
After	4 u	1 u	1 p	1 p

Number of black buttons in Packet R at first
= 30% x 300
= ³⁰₁₀₀ x 300
= 90

Number of yellow buttons in Packet R at first
= 300 - 90
= 210

Number of black buttons in Packet S at first
= 80% x 270
= ⁸⁰₁₀₀ x 270
= 216

Number of yellow buttons in Packet S at first
= 270 - 216
= 54

Packet R in the end
20% = ²⁰₁₀₀ =¹₅
Black buttons : Yellow buttons = 4 : 1

Packet S in the end
50% = ⁵⁰₁₀₀ =¹₂
Black buttons : Yellow buttons = 1 : 1

Total number of black buttons = 4 u + 1 p

4 u + 1 p = 90 + 216
4 u + 1 p = 306 

1 p = 306 - 4 u --- (1)

Total number of yellow buttons = 1 u + 1 p
1 u + 1 p = 210 + 54
1 u + 1 p = 264

1 p = 264 - 1 u --- (2)

(2) = (1)
264 - 1 u = 306 - 4 u
4 u - 1 u = 306 - 264
3 u = 42

1 u = 42 ÷ 3 = 14

Total number of yellow buttons and black buttons that must be moved from Packet R to Packet S
= 300 - 5 u
= 300 - 5 x 14
= 300 - 70
= 230

Answer(s): 230