Cole had three boxes, containing a total of 840 paper cups. The number of paper cups in Box C to the total number of paper cups was 3 : 7. He sold 420 paper cups from Box D and sold ¹₄ of the paper cups in Box E. The number of paper cups left in Box D to the number of paper cups left in Box E was 2 : 1. How many paper cups were there in Box D at first?

	Box C	Box D	Box E	Total
Before	3 u	4 u (480)		7 u (840)
		6 p + 420	4 p
Change		- 420	- 1 p
			3 p
After		2x3 = 6 p	1x3 = 3 p

7 u = 840
1 u = 840 ÷ 7 = 120

Number of paper cups in Box D and Box E at first
= 4 u
= 4 x 120
= 480

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

6 p + 420 + 4 p = 480
6 p + 4 p = 480 - 420
10 p = 60

1 p = 60 ÷ 10 = 6

Number of cups in Box D at first
= 6 p + 420
= 6 x 6 + 420
= 36 + 420
= 456

Answer(s):456