There were some silver coins and brown coins. The coins were packed into 2 bags. At first, Box M contained 160 coins and 30% of them were brown coins. Box N contained 140 coins and 75% of them were brown coins. How many silver coins and brown coins in total must be moved from Box M to Box N such that 20% of the coins in Box M are silver and 50% of the coins in Box N are brown?

	Box M		Box N
Total	160		140
	Brown coins	Silver coins	Brown coins	Silver coins
Before	48	112	105	35
Change	- ?	- ?	+ ?	+ ?
After	4 u	1 u	1 p	1 p

Number of brown coins in Box M at first
= 30% x 160
= ³⁰₁₀₀ x 160
= 48

Number of silver coins in Box M at first
= 160 - 48
= 112

Number of brown coins in Box N at first
= 75% x 140
= ⁷⁵₁₀₀ x 140
= 105

Number of silver coins in Box N at first
= 140 - 105
= 35

Box M in the end
20% = ²⁰₁₀₀ =¹₅
Brown coins : Silver coins = 4 : 1

Box N in the end
50% = ⁵⁰₁₀₀ =¹₂
Brown coins : Silver coins = 1 : 1

Total number of brown coins = 4 u + 1 p

4 u + 1 p = 48 + 105
4 u + 1 p = 153 

1 p = 153 - 4 u --- (1)

Total number of silver coins = 1 u + 1 p
1 u + 1 p = 112 + 35
1 u + 1 p = 147

1 p = 147 - 1 u --- (2)

(2) = (1)
147 - 1 u = 153 - 4 u
4 u - 1 u = 153 - 147
3 u = 6

1 u = 6 ÷ 3 = 2

Total number of silver coins and brown coins that must be moved from Box M to Box N
= 160 - 5 u
= 160 - 5 x 2
= 160 - 10
= 150

Answer(s): 150