Pierre had three boxes, containing a total of 840 plastic bowls. The number of plastic bowls in Box A to the total number of plastic bowls was 3 : 7. He sold 285 plastic bowls from Box B and sold ¹₄ of the plastic bowls in Box C. The number of plastic bowls left in Box B to the number of plastic bowls left in Box C was 3 : 1. How many plastic bowls were there in Box B at first?

	Box A	Box B	Box C	Total
Before	3 u	4 u (480)		7 u (840)
		9 p + 285	4 p
Change		- 285	- 1 p
			3 p
After		3x3 = 9 p	1x3 = 3 p

7 u = 840
1 u = 840 ÷ 7 = 120

Number of plastic bowls in Box B and Box C at first
= 4 u
= 4 x 120
= 480

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

9 p + 285 + 4 p = 480
9 p + 4 p = 480 - 285
13 p = 195

1 p = 195 ÷ 13 = 15

Number of bowls in Box B at first
= 9 p + 285
= 9 x 15 + 285
= 135 + 285
= 420

Answer(s):420