80% of the mochi balls in Packet T were apple mochi balls and the rest were vanilla mochi balls. Packet U had 30% more apple mochi balls than Packet T and twice as many mochi balls than the total number of mochi balls in Packet T. Find the percentage of the vanilla mochi balls in Packet U that would need to be transferred into Packet T, so that there were an equal number of apple and vanilla mochi balls in Packet T.

	Packet T		Packet U
	Apple	Vanilla	Apple	Vanilla
	5 u		10 u
Before	4 u	1 u	5.2 u	4.8 u
Change		+ 3 u		- 3 u
After	4 u	4 u	5.2 u	1.8 u

80% = ⁸⁰₁₀₀ = ⁴₅
100 %+ 30% = 130%

 Total number of mochi balls in Packet T
= 4 u + 1 u
= 5 u

Total number of mochi balls in Packet U
= 2 x 5 u 
= 10 u

Number of apple mochi balls in Packet U
= 130% x 4 u
=  ¹³⁰₁₀₀ x 4 u
= 5.2 u

Number of apple mochi balls in Packet U
= 10 u - 5.2 u
= 4.8 u

Number of vanilla mochi balls to be transferred from Packet U to Packet T
= 4 u - 1 u
= 3 u

Percentage of mochi balls to be transferred from Packet U
= ³_4.8 x 100%
= 62.5%

Answer(s): 62.5%