There were some blue stickers and black stickers. The stickers were packed into 2 bags. At first, Packet J contained 230 stickers and 10% of them were black stickers. Packet K contained 940 stickers and 60% of them were black stickers. How many blue stickers and black stickers in total must be moved from Packet J to Packet K such that 25% of the stickers in Packet J are blue and 50% of the stickers in Packet K are black?

	Packet J		Packet K
Total	230		940
	Black stickers	Blue stickers	Black stickers	Blue stickers
Before	23	207	564	376
Change	- ?	- ?	+ ?	+ ?
After	3 u	1 u	1 p	1 p

Number of black stickers in Packet J at first
= 10% x 230
= ¹⁰₁₀₀ x 230
= 23

Number of blue stickers in Packet J at first
= 230 - 23
= 207

Number of black stickers in Packet K at first
= 60% x 940
= ⁶⁰₁₀₀ x 940
= 564

Number of blue stickers in Packet K at first
= 940 - 564
= 376

Packet J in the end
25% = ²⁵₁₀₀ =¹₄
Black stickers : Blue stickers = 3 : 1

Packet K in the end
50% = ⁵⁰₁₀₀ =¹₂
Black stickers : Blue stickers = 1 : 1

Total number of black stickers = 3 u + 1 p

3 u + 1 p = 23 + 564
3 u + 1 p = 587 

1 p = 587 - 3 u --- (1)

Total number of blue stickers = 1 u + 1 p
1 u + 1 p = 207 + 376
1 u + 1 p = 583

1 p = 583 - 1 u --- (2)

(2) = (1)
583 - 1 u = 587 - 3 u
3 u - 1 u = 587 - 583
2 u = 4

1 u = 4 ÷ 2 = 2

Total number of blue stickers and black stickers that must be moved from Packet J to Packet K
= 230 - 4 u
= 230 - 4 x 2
= 230 - 8
= 222

Answer(s): 222