There were some gold pens and grey pens. The pens were packed into 2 bags. At first, Box E contained 110 pens and 40% of them were grey pens. Box F contained 72 pens and 75% of them were grey pens. How many gold pens and grey pens in total must be moved from Box E to Box F such that 25% of the pens in Box E are gold and 50% of the pens in Box F are grey?

	Box E		Box F
Total	110		72
	Grey pens	Gold pens	Grey pens	Gold pens
Before	44	66	54	18
Change	- ?	- ?	+ ?	+ ?
After	3 u	1 u	1 p	1 p

Number of grey pens in Box E at first
= 40% x 110
= ⁴⁰₁₀₀ x 110
= 44

Number of gold pens in Box E at first
= 110 - 44
= 66

Number of grey pens in Box F at first
= 75% x 72
= ⁷⁵₁₀₀ x 72
= 54

Number of gold pens in Box F at first
= 72 - 54
= 18

Box E in the end
25% = ²⁵₁₀₀ =¹₄
Grey pens : Gold pens = 3 : 1

Box F in the end
50% = ⁵⁰₁₀₀ =¹₂
Grey pens : Gold pens = 1 : 1

Total number of grey pens = 3 u + 1 p

3 u + 1 p = 44 + 54
3 u + 1 p = 98 

1 p = 98 - 3 u --- (1)

Total number of gold pens = 1 u + 1 p
1 u + 1 p = 66 + 18
1 u + 1 p = 84

1 p = 84 - 1 u --- (2)

(2) = (1)
84 - 1 u = 98 - 3 u
3 u - 1 u = 98 - 84
2 u = 14

1 u = 14 ÷ 2 = 7

Total number of gold pens and grey pens that must be moved from Box E to Box F
= 110 - 4 u
= 110 - 4 x 7
= 110 - 28
= 82

Answer(s): 82