Ivan had three boxes, containing a total of 770 paper bowls. The number of paper bowls in Box C to the total number of paper bowls was 3 : 7. He sold 389 paper bowls from Box D and sold ¹₅ of the paper bowls in Box E. The number of paper bowls left in Box D to the number of paper bowls left in Box E was 3 : 1. How many paper bowls were there in Box D at first?

	Box C	Box D	Box E	Total
Before	3 u	4 u (440)		7 u (770)
		12 p + 389	5 p
Change		- 389	- 1 p
			4 p
After		3x4 = 12 p	1x4 = 4 p

7 u = 770
1 u = 770 ÷ 7 = 110

Number of paper bowls in Box D and Box E at first
= 4 u
= 4 x 110
= 440

The number of bowls left in E is the repeated identity.
LCM of 1 and 4 = 4

12 p + 389 + 5 p = 440
12 p + 5 p = 440 - 389
17 p = 51

1 p = 51 ÷ 17 = 3

Number of bowls in Box D at first
= 12 p + 389
= 12 x 3 + 389
= 36 + 389
= 425

Answer(s):425