80% of the candies in Packet B were vanilla candies and the rest were apricot candies. Packet C had 30% more vanilla candies than Packet B and twice as many candies than the total number of candies in Packet B. Find the percentage of the apricot candies in Packet C that would need to be transferred into Packet B, so that there were an equal number of vanilla and apricot candies in Packet B.

	Packet B		Packet C
	Vanilla	Apricot	Vanilla	Apricot
	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 candies in Packet B
= 4 u + 1 u
= 5 u

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

Number of vanilla candies in Packet C
= 130% x 4 u
=  ¹³⁰₁₀₀ x 4 u
= 5.2 u

Number of vanilla candies in Packet C
= 10 u - 5.2 u
= 4.8 u

Number of apricot candies to be transferred from Packet C to Packet B
= 4 u - 1 u
= 3 u

Percentage of candies to be transferred from Packet C
= ³_4.8 x 100%
= 62.5%

Answer(s): 62.5%