There were some blue stickers and green stickers. The stickers were packed into 2 bags. At first, Bag P contained 130 stickers and 40% of them were green stickers. Bag Q contained 68 stickers and 75% of them were green stickers. How many blue stickers and green stickers in total must be moved from Bag P to Bag Q such that 30% of the stickers in Bag P are blue and 50% of the stickers in Bag Q are green?

	Bag P		Bag Q
Total	130		68
	Green stickers	Blue stickers	Green stickers	Blue stickers
Before	52	78	51	17
Change	- ?	- ?	+ ?	+ ?
After	7 u	3 u	1 p	1 p

Number of green stickers in Bag P at first
= 40% x 130
= ⁴⁰₁₀₀ x 130
= 52

Number of blue stickers in Bag P at first
= 130 - 52
= 78

Number of green stickers in Bag Q at first
= 75% x 68
= ⁷⁵₁₀₀ x 68
= 51

Number of blue stickers in Bag Q at first
= 68 - 51
= 17

Bag P in the end
30% = ³⁰₁₀₀ =³₁₀
Green stickers : Blue stickers = 7 : 3

Bag Q in the end
50% = ⁵⁰₁₀₀ =¹₂
Green stickers : Blue stickers = 1 : 1

Total number of green stickers = 7 u + 1 p

7 u + 1 p = 52 + 51
7 u + 1 p = 103 

1 p = 103 - 7 u --- (1)

Total number of blue stickers = 3 u + 1 p
3 u + 1 p = 78 + 17
3 u + 1 p = 95

1 p = 95 - 3 u --- (2)

(2) = (1)
95 - 3 u = 103 - 7 u
7 u - 3 u = 103 - 95
4 u = 8

1 u = 8 ÷ 4 = 2

Total number of blue stickers and green stickers that must be moved from Bag P to Bag Q
= 130 - 10 u
= 130 - 10 x 2
= 130 - 20
= 110

Answer(s): 110