There were some brown beads and blue beads. The beads were packed into 2 bags. At first, Packet B contained 240 beads and 30% of them were blue beads. Packet C contained 236 beads and 75% of them were blue beads. How many brown beads and blue beads in total must be moved from Packet B to Packet C such that 25% of the beads in Packet B are brown and 50% of the beads in Packet C are blue?

	Packet B		Packet C
Total	240		236
	Blue beads	Brown beads	Blue beads	Brown beads
Before	72	168	177	59
Change	- ?	- ?	+ ?	+ ?
After	3 u	1 u	1 p	1 p

Number of blue beads in Packet B at first
= 30% x 240
= ³⁰₁₀₀ x 240
= 72

Number of brown beads in Packet B at first
= 240 - 72
= 168

Number of blue beads in Packet C at first
= 75% x 236
= ⁷⁵₁₀₀ x 236
= 177

Number of brown beads in Packet C at first
= 236 - 177
= 59

Packet B in the end
25% = ²⁵₁₀₀ =¹₄
Blue beads : Brown beads = 3 : 1

Packet C in the end
50% = ⁵⁰₁₀₀ =¹₂
Blue beads : Brown beads = 1 : 1

Total number of blue beads = 3 u + 1 p

3 u + 1 p = 72 + 177
3 u + 1 p = 249 

1 p = 249 - 3 u --- (1)

Total number of brown beads = 1 u + 1 p
1 u + 1 p = 168 + 59
1 u + 1 p = 227

1 p = 227 - 1 u --- (2)

(2) = (1)
227 - 1 u = 249 - 3 u
3 u - 1 u = 249 - 227
2 u = 22

1 u = 22 ÷ 2 = 11

Total number of brown beads and blue beads that must be moved from Packet B to Packet C
= 240 - 4 u
= 240 - 4 x 11
= 240 - 44
= 196

Answer(s): 196