Albert had three boxes, containing a total of 1050 plastic plates. The number of plastic plates in Box E to the total number of plastic plates was 3 : 7. He sold 410 plastic plates from Box F and sold ¹₄ of the plastic plates in Box G. The number of plastic plates left in Box F to the number of plastic plates left in Box G was 5 : 1. How many plastic plates were there in Box F at first?

	Box E	Box F	Box G	Total
Before	3 u	4 u (600)		7 u (1050)
		15 p + 410	4 p
Change		- 410	- 1 p
			3 p
After		5x3 = 15 p	1x3 = 3 p

7 u = 1050
1 u = 1050 ÷ 7 = 150

Number of plastic plates in Box F and Box G at first
= 4 u
= 4 x 150
= 600

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

15 p + 410 + 4 p = 600
15 p + 4 p = 600 - 410
19 p = 190

1 p = 190 ÷ 19 = 10

Number of plates in Box F at first
= 15 p + 410
= 15 x 10 + 410
= 150 + 410
= 560

Answer(s):560