Ken had three boxes, containing a total of 1040 paper bowls. The number of paper bowls in Box P to the total number of paper bowls was 3 : 8. He sold 468 paper bowls from Box Q and sold ¹₅ of the paper bowls in Box R. The number of paper bowls left in Box Q to the number of paper bowls left in Box R was 2 : 1. How many paper bowls were there in Box Q at first?

	Box P	Box Q	Box R	Total
Before	3 u	5 u (650)		8 u (1040)
		8 p + 468	5 p
Change		- 468	- 1 p
			4 p
After		2x4 = 8 p	1x4 = 4 p

8 u = 1040
1 u = 1040 ÷ 8 = 130

Number of paper bowls in Box Q and Box R at first
= 5 u
= 5 x 130
= 650

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

8 p + 468 + 5 p = 650
8 p + 5 p = 650 - 468
13 p = 182

1 p = 182 ÷ 13 = 14

Number of bowls in Box Q at first
= 8 p + 468
= 8 x 14 + 468
= 112 + 468
= 580

Answer(s):580