Xavier had three boxes, containing a total of 660 paper cups. The number of paper cups in Box E to the total number of paper cups was 3 : 11. He sold 375 paper cups from Box F and sold ¹₅ of the paper cups in Box G. The number of paper cups left in Box F to the number of paper cups left in Box G was 4 : 1. How many paper cups were there in Box F at first?

	Box E	Box F	Box G	Total
Before	3 u	8 u (480)		11 u (660)
		16 p + 375	5 p
Change		- 375	- 1 p
			4 p
After		4x4 = 16 p	1x4 = 4 p

11 u = 660
1 u = 660 ÷ 11 = 60

Number of paper cups in Box F and Box G at first
= 8 u
= 8 x 60
= 480

The number of cups left in G is the repeated identity.
LCM of 1 and 4 = 4

16 p + 375 + 5 p = 480
16 p + 5 p = 480 - 375
21 p = 105

1 p = 105 ÷ 21 = 5

Number of cups in Box F at first
= 16 p + 375
= 16 x 5 + 375
= 80 + 375
= 455

Answer(s):455