Container S and Container T were filled completely with sugar. The total mass of ¹₅ of the sugar in Container T and ¹₁₁ of the sugar in Container S was 460 g. If ⁴₁₁ of the sugar in Container S was poured out, the total mass of the sugar in both containers became 2.6 kg. How much sugar was in

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

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

¹₅ T + ¹₁₁ S = 460 --- (1)

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

1 kg = 1000 g
2.6 kg = 2600 g

1 T + ⁷₁₁ S = 2600
1 T = 2600 - ⁷₁₁ S --- (2)

Make T the same.

(1)_x5
⁵₅ T + ⁵₁₁ S = 2300
1 T + ⁵₁₁ S = 2300
1 T = 2300 - ⁵₁₁ S --- (3)

(3) = (2) 
2300 - ⁵₁₁ S = 2600 - ⁷₁₁ X
⁷₁₁ S - ⁵₁₁ S = 2600 - 2300
²₁₁ S = 300
¹₁₁ S = 300 ÷ 2 = 150

¹¹₁₁ S = 11 x 150 = 1650

1 S = 1650

Mass of Container S = 1650 g

(b)
From (1)
¹₅ T + 150 = 460
¹₅ T = 460 - 150 = 310

⁵₅ T = 5 x 310 = 1550

1 T = 1550

Mass of Container T = 1550 g

Answer(s): (a) 1650 g; (b) 1550 g