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

	Packet V		Packet W
	Banana	Cherry	Banana	Cherry
	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 V
= 3 u + 2 u
= 5 u

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

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

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

Number of cherry candies to be transferred from Packet W to Packet V
= 3 u - 2 u
= 1 u

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

Answer(s): 16%