Ivan had three boxes, containing a total of 1200 paper cups. The number of paper cups in Box X to the total number of paper cups was 3 : 8. He sold 291 paper cups from Box Y and sold ¹₅ of the paper cups in Box Z. The number of paper cups left in Box Y to the number of paper cups left in Box Z was 3 : 1. How many paper cups were there in Box Y at first?

	Box X	Box Y	Box Z	Total
Before	3 u	5 u (750)		8 u (1200)
		12 p + 291	5 p
Change		- 291	- 1 p
			4 p
After		3x4 = 12 p	1x4 = 4 p

8 u = 1200
1 u = 1200 ÷ 8 = 150

Number of paper cups in Box Y and Box Z at first
= 5 u
= 5 x 150
= 750

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

12 p + 291 + 5 p = 750
12 p + 5 p = 750 - 291
17 p = 459

1 p = 459 ÷ 17 = 27

Number of cups in Box Y at first
= 12 p + 291
= 12 x 27 + 291
= 324 + 291
= 615

Answer(s):615