There were some white coins and brown coins. The coins were packed into 2 bags. At first, Packet Y contained 220 coins and 30% of them were brown coins. Packet Z contained 236 coins and 75% of them were brown coins. How many white coins and brown coins in total must be moved from Packet Y to Packet Z such that 20% of the coins in Packet Y are white and 50% of the coins in Packet Z are brown?

	Packet Y		Packet Z
Total	220		236
	Brown coins	White coins	Brown coins	White coins
Before	66	154	177	59
Change	- ?	- ?	+ ?	+ ?
After	4 u	1 u	1 p	1 p

Number of brown coins in Packet Y at first
= 30% x 220
= ³⁰₁₀₀ x 220
= 66

Number of white coins in Packet Y at first
= 220 - 66
= 154

Number of brown coins in Packet Z at first
= 75% x 236
= ⁷⁵₁₀₀ x 236
= 177

Number of white coins in Packet Z at first
= 236 - 177
= 59

Packet Y in the end
20% = ²⁰₁₀₀ =¹₅
Brown coins : White coins = 4 : 1

Packet Z in the end
50% = ⁵⁰₁₀₀ =¹₂
Brown coins : White coins = 1 : 1

Total number of brown coins = 4 u + 1 p

4 u + 1 p = 66 + 177
4 u + 1 p = 243 

1 p = 243 - 4 u --- (1)

Total number of white coins = 1 u + 1 p
1 u + 1 p = 154 + 59
1 u + 1 p = 213

1 p = 213 - 1 u --- (2)

(2) = (1)
213 - 1 u = 243 - 4 u
4 u - 1 u = 243 - 213
3 u = 30

1 u = 30 ÷ 3 = 10

Total number of white coins and brown coins that must be moved from Packet Y to Packet Z
= 220 - 5 u
= 220 - 5 x 10
= 220 - 50
= 170

Answer(s): 170