Pierre had three boxes, containing a total of 1200 plastic bowls. The number of plastic bowls in Box M to the total number of plastic bowls was 3 : 10. He sold 399 plastic bowls from Box N and sold ¹₃ of the plastic bowls in Box P. The number of plastic bowls left in Box N to the number of plastic bowls left in Box P was 3 : 1. How many plastic bowls were there in Box N at first?

	Box M	Box N	Box P	Total
Before	3 u	7 u (840)		10 u (1200)
		6 p + 399	3 p
Change		- 399	- 1 p
			2 p
After		3x2 = 6 p	1x2 = 2 p

10 u = 1200
1 u = 1200 ÷ 10 = 120

Number of plastic bowls in Box N and Box P at first
= 7 u
= 7 x 120
= 840

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

6 p + 399 + 3 p = 840
6 p + 3 p = 840 - 399
9 p = 441

1 p = 441 ÷ 9 = 49

Number of bowls in Box N at first
= 6 p + 399
= 6 x 49 + 399
= 294 + 399
= 693

Answer(s):693