Adam has some blue stickers and green stickers.
If 52 blue stickers are removed, 75% of the stickers will be green stickers.
If 592 green stickers are added, 30% of the stickers will be blue stickers.

1. How many blue stickers are there?
2. How many green stickers are there?

	Scenario 1		Scenario 2
	Blue	Green	Blue	Green
Before	1 u + 52	3 u	3 p	7 p - 592
Change	- 52	No change	No change	+ 592
After	1 u	3 u	3 p	7 p

(a)
75% = ⁷⁵₁₀₀
= ³₄
30% = ³⁰₁₀₀ = ³₁₀

Scenario 1 
Fraction of the stickers that are blue in the end
= 1 - ³₄
= ¹₄ 

Scenario 2
Fraction of the stickers that are blue in the end
= 1 - ³₁₀
= ⁷₁₀ 

1 u + 52 = 3 p --- (1)
3 u = 7 p - 592 --- (2)

(1) _{x 3}
3 u + 156 = 3 p
3 u = 3 p - 156 --- (3)

(2) = (3)
7 p - 592 = 3 p - 156
7 p - 3 p = 592 - 156
4 p = 436

1 p = 436 ÷ 4 = 109

Number of blue stickers
= 3 p
= 3 x 109
= 327

(b)
Number of green stickers
= 7 p - 592
= 7 x 109 - 592
= 763 - 592
= 171

Answer(s): (a) 327; (b) 171