Bobby had three boxes, containing a total of 1100 paper cups. The number of paper cups in Box R to the total number of paper cups was 3 : 11. He sold 371 paper cups from Box S and sold ¹₃ of the paper cups in Box T. The number of paper cups left in Box S to the number of paper cups left in Box T was 4 : 1. How many paper cups were there in Box S at first?

	Box R	Box S	Box T	Total
Before	3 u	8 u (800)		11 u (1100)
		8 p + 371	3 p
Change		- 371	- 1 p
			2 p
After		4x2 = 8 p	1x2 = 2 p

11 u = 1100
1 u = 1100 ÷ 11 = 100

Number of paper cups in Box S and Box T at first
= 8 u
= 8 x 100
= 800

The number of cups left in T is the repeated identity.
LCM of 1 and 2 = 2

8 p + 371 + 3 p = 800
8 p + 3 p = 800 - 371
11 p = 429

1 p = 429 ÷ 11 = 39

Number of cups in Box S at first
= 8 p + 371
= 8 x 39 + 371
= 312 + 371
= 683

Answer(s):683