Paul had three boxes, containing a total of 900 paper cups. The number of paper cups in Box A to the total number of paper cups was 3 : 10. He sold 480 paper cups from Box B and sold ¹₄ of the paper cups in Box C. The number of paper cups left in Box B to the number of paper cups left in Box C was 2 : 1. How many paper cups were there in Box B at first?

	Box A	Box B	Box C	Total
Before	3 u	7 u (630)		10 u (900)
		6 p + 480	4 p
Change		- 480	- 1 p
			3 p
After		2x3 = 6 p	1x3 = 3 p

10 u = 900
1 u = 900 ÷ 10 = 90

Number of paper cups in Box B and Box C at first
= 7 u
= 7 x 90
= 630

The number of cups left in C is the repeated identity.
LCM of 1 and 3 = 3

6 p + 480 + 4 p = 630
6 p + 4 p = 630 - 480
10 p = 150

1 p = 150 ÷ 10 = 15

Number of cups in Box B at first
= 6 p + 480
= 6 x 15 + 480
= 90 + 480
= 570

Answer(s):570