There were some red pens and blue pens. The pens were packed into 2 bags. At first, Box E contained 135 pens and 40% of them were blue pens. Box F contained 132 pens and 75% of them were blue pens. How many red pens and blue pens in total must be moved from Box E to Box F such that 20% of the pens in Box E are red and 50% of the pens in Box F are blue?

	Box E		Box F
Total	135		132
	Blue pens	Red pens	Blue pens	Red pens
Before	54	81	99	33
Change	- ?	- ?	+ ?	+ ?
After	4 u	1 u	1 p	1 p

Number of blue pens in Box E at first
= 40% x 135
= ⁴⁰₁₀₀ x 135
= 54

Number of red pens in Box E at first
= 135 - 54
= 81

Number of blue pens in Box F at first
= 75% x 132
= ⁷⁵₁₀₀ x 132
= 99

Number of red pens in Box F at first
= 132 - 99
= 33

Box E in the end
20% = ²⁰₁₀₀ =¹₅
Blue pens : Red pens = 4 : 1

Box F in the end
50% = ⁵⁰₁₀₀ =¹₂
Blue pens : Red pens = 1 : 1

Total number of blue pens = 4 u + 1 p

4 u + 1 p = 54 + 99
4 u + 1 p = 153 

1 p = 153 - 4 u --- (1)

Total number of red pens = 1 u + 1 p
1 u + 1 p = 81 + 33
1 u + 1 p = 114

1 p = 114 - 1 u --- (2)

(2) = (1)
114 - 1 u = 153 - 4 u
4 u - 1 u = 153 - 114
3 u = 39

1 u = 39 ÷ 3 = 13

Total number of red pens and blue pens that must be moved from Box E to Box F
= 135 - 5 u
= 135 - 5 x 13
= 135 - 65
= 70

Answer(s): 70