Ken had three boxes, containing a total of 560 plastic bowls. The number of plastic bowls in Box C to the total number of plastic bowls was 3 : 8. He sold 255 plastic bowls from Box D and sold ¹₄ of the plastic bowls in Box E. The number of plastic bowls left in Box D to the number of plastic bowls left in Box E was 5 : 1. How many plastic bowls were there in Box D at first?

	Box C	Box D	Box E	Total
Before	3 u	5 u (350)		8 u (560)
		15 p + 255	4 p
Change		- 255	- 1 p
			3 p
After		5x3 = 15 p	1x3 = 3 p

8 u = 560
1 u = 560 ÷ 8 = 70

Number of plastic bowls in Box D and Box E at first
= 5 u
= 5 x 70
= 350

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

15 p + 255 + 4 p = 350
15 p + 4 p = 350 - 255
19 p = 95

1 p = 95 ÷ 19 = 5

Number of bowls in Box D at first
= 15 p + 255
= 15 x 5 + 255
= 75 + 255
= 330

Answer(s):330