There were some gold pens and green pens. The pens were packed into 2 bags. At first, Box T contained 100 pens and 40% of them were green pens. Box U contained 88 pens and 75% of them were green pens. How many gold pens and green pens in total must be moved from Box T to Box U such that 25% of the pens in Box T are gold and 50% of the pens in Box U are green?

	Box T		Box U
Total	100		88
	Green pens	Gold pens	Green pens	Gold pens
Before	40	60	66	22
Change	- ?	- ?	+ ?	+ ?
After	3 u	1 u	1 p	1 p

Number of green pens in Box T at first
= 40% x 100
= ⁴⁰₁₀₀ x 100
= 40

Number of gold pens in Box T at first
= 100 - 40
= 60

Number of green pens in Box U at first
= 75% x 88
= ⁷⁵₁₀₀ x 88
= 66

Number of gold pens in Box U at first
= 88 - 66
= 22

Box T in the end
25% = ²⁵₁₀₀ =¹₄
Green pens : Gold pens = 3 : 1

Box U in the end
50% = ⁵⁰₁₀₀ =¹₂
Green pens : Gold pens = 1 : 1

Total number of green pens = 3 u + 1 p

3 u + 1 p = 40 + 66
3 u + 1 p = 106 

1 p = 106 - 3 u --- (1)

Total number of gold pens = 1 u + 1 p
1 u + 1 p = 60 + 22
1 u + 1 p = 82

1 p = 82 - 1 u --- (2)

(2) = (1)
82 - 1 u = 106 - 3 u
3 u - 1 u = 106 - 82
2 u = 24

1 u = 24 ÷ 2 = 12

Total number of gold pens and green pens that must be moved from Box T to Box U
= 100 - 4 u
= 100 - 4 x 12
= 100 - 48
= 52

Answer(s): 52