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

	Packet V		Packet W
	Apple	Cheese	Apple	Cheese
	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 candies in Packet V
= 7 u + 3 u
= 10 u

Total number of candies in Packet W
= 4 x 10 u 
= 40 u

Number of apple candies in Packet W
= 125% x 7 u
=  ¹²⁵₁₀₀ x 7 u
= 8.75 u

Number of apple candies in Packet W
= 40 u - 8.75 u
= 31.25 u

Number of cheese candies to be transferred from Packet W to Packet V
= 7 u - 3 u
= 4 u

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

Answer(s): 12.8%