Michael had three boxes, containing a total of 800 paper plates. The number of paper plates in Box E to the total number of paper plates was 3 : 10. He sold 384 paper plates from Box F and sold ¹₄ of the paper plates in Box G. The number of paper plates left in Box F to the number of paper plates left in Box G was 4 : 1. How many paper plates were there in Box F at first?

	Box E	Box F	Box G	Total
Before	3 u	7 u (560)		10 u (800)
		12 p + 384	4 p
Change		- 384	- 1 p
			3 p
After		4x3 = 12 p	1x3 = 3 p

10 u = 800
1 u = 800 ÷ 10 = 80

Number of paper plates in Box F and Box G at first
= 7 u
= 7 x 80
= 560

The number of plates left in G is the repeated identity.
LCM of 1 and 3 = 3

12 p + 384 + 4 p = 560
12 p + 4 p = 560 - 384
16 p = 176

1 p = 176 ÷ 16 = 11

Number of plates in Box F at first
= 12 p + 384
= 12 x 11 + 384
= 132 + 384
= 516

Answer(s):516