60% of the caramel apples in Packet Y were cranberry caramel apples and the rest were apricot caramel apples. Packet Z had 25% more cranberry caramel apples than Packet Y and twice as many caramel apples than the total number of caramel apples in Packet Y. Find the percentage of the apricot caramel apples in Packet Z that would need to be transferred into Packet Y, so that there were an equal number of cranberry and apricot caramel apples in Packet Y.

	Packet Y		Packet Z
	Cranberry	Apricot	Cranberry	Apricot
	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 caramel apples in Packet Y
= 3 u + 2 u
= 5 u

Total number of caramel apples in Packet Z
= 2 x 5 u 
= 10 u

Number of cranberry caramel apples in Packet Z
= 125% x 3 u
=  ¹²⁵₁₀₀ x 3 u
= 3.75 u

Number of cranberry caramel apples in Packet Z
= 10 u - 3.75 u
= 6.25 u

Number of apricot caramel apples to be transferred from Packet Z to Packet Y
= 3 u - 2 u
= 1 u

Percentage of caramel apples to be transferred from Packet Z
= ¹_6.25 x 100%
= 16%

Answer(s): 16%