Container T and Container U were filled completely with sugar. The total mass of ¹₆ of the sugar in Container U and ¹₁₁ of the sugar in Container T was 590 g. If ²₁₁ of the sugar in Container T was poured out, the total mass of the sugar in both containers became 3.99 kg. How much sugar was in

1. Container T in grams?
2. Container U in grams?

(a)
Let the mass of the sugar in Container U be U.
Let the mass of the sugar in Container T be T.

¹₆ U + ¹₁₁ T = 590 --- (1)

Fraction of the sugar left in Container T after ²₁₁ of it was poured out
= 1 - ²₁₁
= ⁹₁₁

1 kg = 1000 g
3.99 kg = 3990 g

1 U + ⁹₁₁ T = 3990
1 U = 3990 - ⁹₁₁ T --- (2)

Make U the same.

(1)_x6
⁶₆ U + ⁶₁₁ T = 3540
1 U + ⁶₁₁ T = 3540
1 U = 3540 - ⁶₁₁ T --- (3)

(3) = (2) 
3540 - ⁶₁₁ T = 3990 - ⁹₁₁ X
⁹₁₁ T - ⁶₁₁ T = 3990 - 3540
³₁₁ T = 450
¹₁₁ T = 450 ÷ 3 = 150

¹¹₁₁ T = 11 x 150 = 1650

1 T = 1650

Mass of Container T = 1650 g

(b)
From (1)
¹₆ U + 150 = 590
¹₆ U = 590 - 150 = 440

⁶₆ U = 6 x 440 = 2640

1 U = 2640

Mass of Container U = 2640 g

Answer(s): (a) 1650 g; (b) 2640 g