There were some green cards and brown cards. The cards were packed into 2 bags. At first, Packet E contained 300 cards and 30% of them were brown cards. Packet F contained 795 cards and 60% of them were brown cards. How many green cards and brown cards in total must be moved from Packet E to Packet F such that 20% of the cards in Packet E are green and 50% of the cards in Packet F are brown?

	Packet E		Packet F
Total	300		795
	Brown cards	Green cards	Brown cards	Green cards
Before	90	210	477	318
Change	- ?	- ?	+ ?	+ ?
After	4 u	1 u	1 p	1 p

Number of brown cards in Packet E at first
= 30% x 300
= ³⁰₁₀₀ x 300
= 90

Number of green cards in Packet E at first
= 300 - 90
= 210

Number of brown cards in Packet F at first
= 60% x 795
= ⁶⁰₁₀₀ x 795
= 477

Number of green cards in Packet F at first
= 795 - 477
= 318

Packet E in the end
20% = ²⁰₁₀₀ =¹₅
Brown cards : Green cards = 4 : 1

Packet F in the end
50% = ⁵⁰₁₀₀ =¹₂
Brown cards : Green cards = 1 : 1

Total number of brown cards = 4 u + 1 p

4 u + 1 p = 90 + 477
4 u + 1 p = 567 

1 p = 567 - 4 u --- (1)

Total number of green cards = 1 u + 1 p
1 u + 1 p = 210 + 318
1 u + 1 p = 528

1 p = 528 - 1 u --- (2)

(2) = (1)
528 - 1 u = 567 - 4 u
4 u - 1 u = 567 - 528
3 u = 39

1 u = 39 ÷ 3 = 13

Total number of green cards and brown cards that must be moved from Packet E to Packet F
= 300 - 5 u
= 300 - 5 x 13
= 300 - 65
= 235

Answer(s): 235