Daniel had three boxes, containing a total of 990 plastic bowls. The number of plastic bowls in Box N to the total number of plastic bowls was 3 : 11. He sold 340 plastic bowls from Box P and sold ¹₄ of the plastic bowls in Box Q. The number of plastic bowls left in Box P to the number of plastic bowls left in Box Q was 2 : 1. How many plastic bowls were there in Box P at first?

	Box N	Box P	Box Q	Total
Before	3 u	8 u (720)		11 u (990)
		6 p + 340	4 p
Change		- 340	- 1 p
			3 p
After		2x3 = 6 p	1x3 = 3 p

11 u = 990
1 u = 990 ÷ 11 = 90

Number of plastic bowls in Box P and Box Q at first
= 8 u
= 8 x 90
= 720

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

6 p + 340 + 4 p = 720
6 p + 4 p = 720 - 340
10 p = 380

1 p = 380 ÷ 10 = 38

Number of bowls in Box P at first
= 6 p + 340
= 6 x 38 + 340
= 228 + 340
= 568

Answer(s):568