70% of the caramel apples in Packet R were strawberry caramel apples and the rest were banana caramel apples. Packet S had 25% more strawberry caramel apples than Packet R and four times as many caramel apples than the total number of caramel apples in Packet R. Find the percentage of the banana caramel apples in Packet S that would need to be transferred into Packet R, so that there were an equal number of strawberry and banana caramel apples in Packet R.

	Packet R		Packet S
	Strawberry	Banana	Strawberry	Banana
	10 u		40 u
Before	7 u	3 u	8.75 u	31.25 u
Change		+ 4 u		- 4 u
After	7 u	7 u	8.75 u	27.25 u

70% = ⁷⁰₁₀₀ = ⁷₁₀
100 %+ 25% = 125%

 Total number of caramel apples in Packet R
= 7 u + 3 u
= 10 u

Total number of caramel apples in Packet S
= 4 x 10 u 
= 40 u

Number of strawberry caramel apples in Packet S
= 125% x 7 u
=  ¹²⁵₁₀₀ x 7 u
= 8.75 u

Number of strawberry caramel apples in Packet S
= 40 u - 8.75 u
= 31.25 u

Number of banana caramel apples to be transferred from Packet S to Packet R
= 7 u - 3 u
= 4 u

Percentage of caramel apples to be transferred from Packet S
= ⁴_31.25 x 100%
= 12.8%

Answer(s): 12.8%