There were some black pens and green pens. The pens were packed into 2 bags. At first, Box A contained 250 pens and 10% of them were green pens. Box B contained 420 pens and 75% of them were green pens. How many black pens and green pens in total must be moved from Box A to Box B such that 25% of the pens in Box A are black and 50% of the pens in Box B are green?

	Box A		Box B
Total	250		420
	Green pens	Black pens	Green pens	Black pens
Before	25	225	315	105
Change	- ?	- ?	+ ?	+ ?
After	3 u	1 u	1 p	1 p

Number of green pens in Box A at first
= 10% x 250
= ¹⁰₁₀₀ x 250
= 25

Number of black pens in Box A at first
= 250 - 25
= 225

Number of green pens in Box B at first
= 75% x 420
= ⁷⁵₁₀₀ x 420
= 315

Number of black pens in Box B at first
= 420 - 315
= 105

Box A in the end
25% = ²⁵₁₀₀ =¹₄
Green pens : Black pens = 3 : 1

Box B in the end
50% = ⁵⁰₁₀₀ =¹₂
Green pens : Black pens = 1 : 1

Total number of green pens = 3 u + 1 p

3 u + 1 p = 25 + 315
3 u + 1 p = 340 

1 p = 340 - 3 u --- (1)

Total number of black pens = 1 u + 1 p
1 u + 1 p = 225 + 105
1 u + 1 p = 330

1 p = 330 - 1 u --- (2)

(2) = (1)
330 - 1 u = 340 - 3 u
3 u - 1 u = 340 - 330
2 u = 10

1 u = 10 ÷ 2 = 5

Total number of black pens and green pens that must be moved from Box A to Box B
= 250 - 4 u
= 250 - 4 x 5
= 250 - 20
= 230

Answer(s): 230