Bobby had three boxes, containing a total of 660 paper plates. The number of paper plates in Box C to the total number of paper plates was 3 : 11. He sold 291 paper plates from Box D and sold ¹₃ of the paper plates in Box E. The number of paper plates left in Box D to the number of paper plates left in Box E was 3 : 1. How many paper plates were there in Box D at first?

	Box C	Box D	Box E	Total
Before	3 u	8 u (480)		11 u (660)
		6 p + 291	3 p
Change		- 291	- 1 p
			2 p
After		3x2 = 6 p	1x2 = 2 p

11 u = 660
1 u = 660 ÷ 11 = 60

Number of paper plates in Box D and Box E at first
= 8 u
= 8 x 60
= 480

The number of plates left in E is the repeated identity.
LCM of 1 and 2 = 2

6 p + 291 + 3 p = 480
6 p + 3 p = 480 - 291
9 p = 189

1 p = 189 ÷ 9 = 21

Number of plates in Box D at first
= 6 p + 291
= 6 x 21 + 291
= 126 + 291
= 417

Answer(s):417