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

	Packet S		Packet T
	Apple	Banana	Apple	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 sweets in Packet S
= 7 u + 3 u
= 10 u

Total number of sweets in Packet T
= 4 x 10 u 
= 40 u

Number of apple sweets in Packet T
= 125% x 7 u
=  ¹²⁵₁₀₀ x 7 u
= 8.75 u

Number of apple sweets in Packet T
= 40 u - 8.75 u
= 31.25 u

Number of banana sweets to be transferred from Packet T to Packet S
= 7 u - 3 u
= 4 u

Percentage of sweets to be transferred from Packet T
= ⁴_31.25 x 100%
= 12.8%

Answer(s): 12.8%