Cody had three boxes, containing a total of 1320 paper bowls. The number of paper bowls in Box N to the total number of paper bowls was 3 : 11. He sold 330 paper bowls from Box P and sold ¹₄ of the paper bowls in Box Q. The number of paper bowls left in Box P to the number of paper bowls left in Box Q was 2 : 1. How many paper bowls were there in Box P at first?

	Box N	Box P	Box Q	Total
Before	3 u	8 u (960)		11 u (1320)
		6 p + 330	4 p
Change		- 330	- 1 p
			3 p
After		2x3 = 6 p	1x3 = 3 p

11 u = 1320
1 u = 1320 ÷ 11 = 120

Number of paper bowls in Box P and Box Q at first
= 8 u
= 8 x 120
= 960

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

6 p + 330 + 4 p = 960
6 p + 4 p = 960 - 330
10 p = 630

1 p = 630 ÷ 10 = 63

Number of bowls in Box P at first
= 6 p + 330
= 6 x 63 + 330
= 378 + 330
= 708

Answer(s):708