Container V and Container W were filled completely with sugar. The total mass of ¹₃ of the sugar in Container W and ¹₁₁ of the sugar in Container V was 530 g. If ⁴₁₁ of the sugar in Container V was poured out, the total mass of the sugar in both containers became 2.35 kg. How much sugar was in

1. Container V in grams?
2. Container W in grams?

(a)
Let the mass of the sugar in Container W be W.
Let the mass of the sugar in Container V be V.

¹₃ W + ¹₁₁ V = 530 --- (1)

Fraction of the sugar left in Container V after ⁴₁₁ of it was poured out
= 1 - ⁴₁₁
= ⁷₁₁

1 kg = 1000 g
2.35 kg = 2350 g

1 W + ⁷₁₁ V = 2350
1 W = 2350 - ⁷₁₁ V --- (2)

Make W the same.

(1)_x3
³₃ W + ³₁₁ V = 1590
1 W + ³₁₁ V = 1590
1 W = 1590 - ³₁₁ V --- (3)

(3) = (2) 
1590 - ³₁₁ V = 2350 - ⁷₁₁ X
⁷₁₁ V - ³₁₁ V = 2350 - 1590
⁴₁₁ V = 760
¹₁₁ V = 760 ÷ 4 = 190

¹¹₁₁ V = 11 x 190 = 2090

1 V = 2090

Mass of Container V = 2090 g

(b)
From (1)
¹₃ W + 190 = 530
¹₃ W = 530 - 190 = 340

³₃ W = 3 x 340 = 1020

1 W = 1020

Mass of Container W = 1020 g

Answer(s): (a) 2090 g; (b) 1020 g