60% of the mochi balls in Packet T were banana mochi balls and the rest were cranberry mochi balls. Packet U had 25% more banana 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 cranberry mochi balls in Packet U that would need to be transferred into Packet T, so that there were an equal number of banana and cranberry mochi balls in Packet T.

	Packet T		Packet U
	Banana	Cranberry	Banana	Cranberry
	5 u		10 u
Before	3 u	2 u	3.75 u	6.25 u
Change		+ 1 u		- 1 u
After	3 u	3 u	3.75 u	5.25 u

60% = ⁶⁰₁₀₀ = ³₅
100 %+ 25% = 125%

 Total number of mochi balls in Packet T
= 3 u + 2 u
= 5 u

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

Number of banana mochi balls in Packet U
= 125% x 3 u
=  ¹²⁵₁₀₀ x 3 u
= 3.75 u

Number of banana mochi balls in Packet U
= 10 u - 3.75 u
= 6.25 u

Number of cranberry mochi balls to be transferred from Packet U to Packet T
= 3 u - 2 u
= 1 u

Percentage of mochi balls to be transferred from Packet U
= ¹_6.25 x 100%
= 16%

Answer(s): 16%