Lee had three boxes, containing a total of 560 paper plates. The number of paper plates in Box D to the total number of paper plates was 3 : 8. He sold 312 paper plates from Box E and sold ¹₄ of the paper plates in Box F. The number of paper plates left in Box E to the number of paper plates left in Box F was 5 : 1. How many paper plates were there in Box E at first?

	Box D	Box E	Box F	Total
Before	3 u	5 u (350)		8 u (560)
		15 p + 312	4 p
Change		- 312	- 1 p
			3 p
After		5x3 = 15 p	1x3 = 3 p

8 u = 560
1 u = 560 ÷ 8 = 70

Number of paper plates in Box E and Box F at first
= 5 u
= 5 x 70
= 350

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

15 p + 312 + 4 p = 350
15 p + 4 p = 350 - 312
19 p = 38

1 p = 38 ÷ 19 = 2

Number of plates in Box E at first
= 15 p + 312
= 15 x 2 + 312
= 30 + 312
= 342

Answer(s):342