Daniel had three boxes, containing a total of 1120 paper bowls. The number of paper bowls in Box D to the total number of paper bowls was 3 : 8. He sold 420 paper bowls from Box E and sold ¹₃ of the paper bowls in Box F. The number of paper bowls left in Box E to the number of paper bowls left in Box F was 2 : 1. How many paper bowls were there in Box E at first?

	Box D	Box E	Box F	Total
Before	3 u	5 u (700)		8 u (1120)
		4 p + 420	3 p
Change		- 420	- 1 p
			2 p
After		2x2 = 4 p	1x2 = 2 p

8 u = 1120
1 u = 1120 ÷ 8 = 140

Number of paper bowls in Box E and Box F at first
= 5 u
= 5 x 140
= 700

The number of bowls left in F is the repeated identity.
LCM of 1 and 2 = 2

4 p + 420 + 3 p = 700
4 p + 3 p = 700 - 420
7 p = 280

1 p = 280 ÷ 7 = 40

Number of bowls in Box E at first
= 4 p + 420
= 4 x 40 + 420
= 160 + 420
= 580

Answer(s):580