There were some red coins and brown coins. The coins were packed into 2 bags. At first, Box S contained 300 coins and 30% of them were brown coins. Box T contained 690 coins and 60% of them were brown coins. How many red coins and brown coins in total must be moved from Box S to Box T such that 20% of the coins in Box S are red and 50% of the coins in Box T are brown?

	Box S		Box T
Total	300		690
	Brown coins	Red coins	Brown coins	Red coins
Before	90	210	414	276
Change	- ?	- ?	+ ?	+ ?
After	4 u	1 u	1 p	1 p

Number of brown coins in Box S at first
= 30% x 300
= ³⁰₁₀₀ x 300
= 90

Number of red coins in Box S at first
= 300 - 90
= 210

Number of brown coins in Box T at first
= 60% x 690
= ⁶⁰₁₀₀ x 690
= 414

Number of red coins in Box T at first
= 690 - 414
= 276

Box S in the end
20% = ²⁰₁₀₀ =¹₅
Brown coins : Red coins = 4 : 1

Box T in the end
50% = ⁵⁰₁₀₀ =¹₂
Brown coins : Red coins = 1 : 1

Total number of brown coins = 4 u + 1 p

4 u + 1 p = 90 + 414
4 u + 1 p = 504 

1 p = 504 - 4 u --- (1)

Total number of red coins = 1 u + 1 p
1 u + 1 p = 210 + 276
1 u + 1 p = 486

1 p = 486 - 1 u --- (2)

(2) = (1)
486 - 1 u = 504 - 4 u
4 u - 1 u = 504 - 486
3 u = 18

1 u = 18 ÷ 3 = 6

Total number of red coins and brown coins that must be moved from Box S to Box T
= 300 - 5 u
= 300 - 5 x 6
= 300 - 30
= 270

Answer(s): 270