There were some silver stickers and black stickers. The stickers were packed into 2 bags. At first, Packet R contained 180 stickers and 30% of them were black stickers. Packet S contained 192 stickers and 75% of them were black stickers. How many silver stickers and black stickers in total must be moved from Packet R to Packet S such that 20% of the stickers in Packet R are silver and 50% of the stickers in Packet S are black?

	Packet R		Packet S
Total	180		192
	Black stickers	Silver stickers	Black stickers	Silver stickers
Before	54	126	144	48
Change	- ?	- ?	+ ?	+ ?
After	4 u	1 u	1 p	1 p

Number of black stickers in Packet R at first
= 30% x 180
= ³⁰₁₀₀ x 180
= 54

Number of silver stickers in Packet R at first
= 180 - 54
= 126

Number of black stickers in Packet S at first
= 75% x 192
= ⁷⁵₁₀₀ x 192
= 144

Number of silver stickers in Packet S at first
= 192 - 144
= 48

Packet R in the end
20% = ²⁰₁₀₀ =¹₅
Black stickers : Silver stickers = 4 : 1

Packet S in the end
50% = ⁵⁰₁₀₀ =¹₂
Black stickers : Silver stickers = 1 : 1

Total number of black stickers = 4 u + 1 p

4 u + 1 p = 54 + 144
4 u + 1 p = 198 

1 p = 198 - 4 u --- (1)

Total number of silver stickers = 1 u + 1 p
1 u + 1 p = 126 + 48
1 u + 1 p = 174

1 p = 174 - 1 u --- (2)

(2) = (1)
174 - 1 u = 198 - 4 u
4 u - 1 u = 198 - 174
3 u = 24

1 u = 24 ÷ 3 = 8

Total number of silver stickers and black stickers that must be moved from Packet R to Packet S
= 180 - 5 u
= 180 - 5 x 8
= 180 - 40
= 140

Answer(s): 140