There were some brown coins and red coins. The coins were packed into 2 bags. At first, Bag Q contained 220 coins and 30% of them were red coins. Bag R contained 204 coins and 75% of them were red coins. How many brown coins and red coins in total must be moved from Bag Q to Bag R such that 25% of the coins in Bag Q are brown and 50% of the coins in Bag R are red?

	Bag Q		Bag R
Total	220		204
	Red coins	Brown coins	Red coins	Brown coins
Before	66	154	153	51
Change	- ?	- ?	+ ?	+ ?
After	3 u	1 u	1 p	1 p

Number of red coins in Bag Q at first
= 30% x 220
= ³⁰₁₀₀ x 220
= 66

Number of brown coins in Bag Q at first
= 220 - 66
= 154

Number of red coins in Bag R at first
= 75% x 204
= ⁷⁵₁₀₀ x 204
= 153

Number of brown coins in Bag R at first
= 204 - 153
= 51

Bag Q in the end
25% = ²⁵₁₀₀ =¹₄
Red coins : Brown coins = 3 : 1

Bag R in the end
50% = ⁵⁰₁₀₀ =¹₂
Red coins : Brown coins = 1 : 1

Total number of red coins = 3 u + 1 p

3 u + 1 p = 66 + 153
3 u + 1 p = 219 

1 p = 219 - 3 u --- (1)

Total number of brown coins = 1 u + 1 p
1 u + 1 p = 154 + 51
1 u + 1 p = 205

1 p = 205 - 1 u --- (2)

(2) = (1)
205 - 1 u = 219 - 3 u
3 u - 1 u = 219 - 205
2 u = 14

1 u = 14 ÷ 2 = 7

Total number of brown coins and red coins that must be moved from Bag Q to Bag R
= 220 - 4 u
= 220 - 4 x 7
= 220 - 28
= 192

Answer(s): 192