60% of the candies in Packet U were banana candies and the rest were chocolate chip candies. Packet V had 25% more banana candies than Packet U and twice as many candies than the total number of candies in Packet U. Find the percentage of the chocolate chip candies in Packet V that would need to be transferred into Packet U, so that there were an equal number of banana and chocolate chip candies in Packet U.

	Packet U		Packet V
	Banana	Chocolate Chip	Banana	Chocolate Chip
	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 candies in Packet U
= 3 u + 2 u
= 5 u

Total number of candies in Packet V
= 2 x 5 u 
= 10 u

Number of banana candies in Packet V
= 125% x 3 u
=  ¹²⁵₁₀₀ x 3 u
= 3.75 u

Number of banana candies in Packet V
= 10 u - 3.75 u
= 6.25 u

Number of chocolate chip candies to be transferred from Packet V to Packet U
= 3 u - 2 u
= 1 u

Percentage of candies to be transferred from Packet V
= ¹_6.25 x 100%
= 16%

Answer(s): 16%