There were some brown pens and grey pens. The pens were packed into 2 bags. At first, Box J contained 220 pens and 30% of them were grey pens. Box K contained 530 pens and 60% of them were grey pens. How many brown pens and grey pens in total must be moved from Box J to Box K such that 20% of the pens in Box J are brown and 50% of the pens in Box K are grey?

	Box J		Box K
Total	220		530
	Grey pens	Brown pens	Grey pens	Brown pens
Before	66	154	318	212
Change	- ?	- ?	+ ?	+ ?
After	4 u	1 u	1 p	1 p

Number of grey pens in Box J at first
= 30% x 220
= ³⁰₁₀₀ x 220
= 66

Number of brown pens in Box J at first
= 220 - 66
= 154

Number of grey pens in Box K at first
= 60% x 530
= ⁶⁰₁₀₀ x 530
= 318

Number of brown pens in Box K at first
= 530 - 318
= 212

Box J in the end
20% = ²⁰₁₀₀ =¹₅
Grey pens : Brown pens = 4 : 1

Box K in the end
50% = ⁵⁰₁₀₀ =¹₂
Grey pens : Brown pens = 1 : 1

Total number of grey pens = 4 u + 1 p

4 u + 1 p = 66 + 318
4 u + 1 p = 384 

1 p = 384 - 4 u --- (1)

Total number of brown pens = 1 u + 1 p
1 u + 1 p = 154 + 212
1 u + 1 p = 366

1 p = 366 - 1 u --- (2)

(2) = (1)
366 - 1 u = 384 - 4 u
4 u - 1 u = 384 - 366
3 u = 18

1 u = 18 ÷ 3 = 6

Total number of brown pens and grey pens that must be moved from Box J to Box K
= 220 - 5 u
= 220 - 5 x 6
= 220 - 30
= 190

Answer(s): 190