There were some green coins and red coins. The coins were packed into 2 bags. At first, Box A contained 200 coins and 30% of them were red coins. Box B contained 535 coins and 60% of them were red coins. How many green coins and red coins in total must be moved from Box A to Box B such that 20% of the coins in Box A are green and 50% of the coins in Box B are red?

	Box A		Box B
Total	200		535
	Red coins	Green coins	Red coins	Green coins
Before	60	140	321	214
Change	- ?	- ?	+ ?	+ ?
After	4 u	1 u	1 p	1 p

Number of red coins in Box A at first
= 30% x 200
= ³⁰₁₀₀ x 200
= 60

Number of green coins in Box A at first
= 200 - 60
= 140

Number of red coins in Box B at first
= 60% x 535
= ⁶⁰₁₀₀ x 535
= 321

Number of green coins in Box B at first
= 535 - 321
= 214

Box A in the end
20% = ²⁰₁₀₀ =¹₅
Red coins : Green coins = 4 : 1

Box B in the end
50% = ⁵⁰₁₀₀ =¹₂
Red coins : Green coins = 1 : 1

Total number of red coins = 4 u + 1 p

4 u + 1 p = 60 + 321
4 u + 1 p = 381 

1 p = 381 - 4 u --- (1)

Total number of green coins = 1 u + 1 p
1 u + 1 p = 140 + 214
1 u + 1 p = 354

1 p = 354 - 1 u --- (2)

(2) = (1)
354 - 1 u = 381 - 4 u
4 u - 1 u = 381 - 354
3 u = 27

1 u = 27 ÷ 3 = 9

Total number of green coins and red coins that must be moved from Box A to Box B
= 200 - 5 u
= 200 - 5 x 9
= 200 - 45
= 155

Answer(s): 155