There were some green erasers and red erasers. The erasers were packed into 2 bags. At first, Packet U contained 300 erasers and 10% of them were red erasers. Packet V contained 1260 erasers and 60% of them were red erasers. How many green erasers and red erasers in total must be moved from Packet U to Packet V such that 30% of the erasers in Packet U are green and 50% of the erasers in Packet V are red?

	Packet U		Packet V
Total	300		1260
	Red erasers	Green erasers	Red erasers	Green erasers
Before	30	270	756	504
Change	- ?	- ?	+ ?	+ ?
After	7 u	3 u	1 p	1 p

Number of red erasers in Packet U at first
= 10% x 300
= ¹⁰₁₀₀ x 300
= 30

Number of green erasers in Packet U at first
= 300 - 30
= 270

Number of red erasers in Packet V at first
= 60% x 1260
= ⁶⁰₁₀₀ x 1260
= 756

Number of green erasers in Packet V at first
= 1260 - 756
= 504

Packet U in the end
30% = ³⁰₁₀₀ =³₁₀
Red erasers : Green erasers = 7 : 3

Packet V in the end
50% = ⁵⁰₁₀₀ =¹₂
Red erasers : Green erasers = 1 : 1

Total number of red erasers = 7 u + 1 p

7 u + 1 p = 30 + 756
7 u + 1 p = 786 

1 p = 786 - 7 u --- (1)

Total number of green erasers = 3 u + 1 p
3 u + 1 p = 270 + 504
3 u + 1 p = 774

1 p = 774 - 3 u --- (2)

(2) = (1)
774 - 3 u = 786 - 7 u
7 u - 3 u = 786 - 774
4 u = 12

1 u = 12 ÷ 4 = 3

Total number of green erasers and red erasers that must be moved from Packet U to Packet V
= 300 - 10 u
= 300 - 10 x 3
= 300 - 30
= 270

Answer(s): 270