Container R and Container S were filled completely with sugar. The total mass of ¹₄ of the sugar in Container S and ¹₁₁ of the sugar in Container R was 570 g. If ⁵₁₁ of the sugar in Container R was poured out, the total mass of the sugar in both containers became 2.58 kg. How much sugar was in

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

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

¹₄ S + ¹₁₁ R = 570 --- (1)

Fraction of the sugar left in Container R after ⁵₁₁ of it was poured out
= 1 - ⁵₁₁
= ⁶₁₁

1 kg = 1000 g
2.58 kg = 2580 g

1 S + ⁶₁₁ R = 2580
1 S = 2580 - ⁶₁₁ R --- (2)

Make S the same.

(1)_x4
⁴₄ S + ⁴₁₁ R = 2280
1 S + ⁴₁₁ R = 2280
1 S = 2280 - ⁴₁₁ R --- (3)

(3) = (2) 
2280 - ⁴₁₁ R = 2580 - ⁶₁₁ X
⁶₁₁ R - ⁴₁₁ R = 2580 - 2280
²₁₁ R = 300
¹₁₁ R = 300 ÷ 2 = 150

¹¹₁₁ R = 11 x 150 = 1650

1 R = 1650

Mass of Container R = 1650 g

(b)
From (1)
¹₄ S + 150 = 570
¹₄ S = 570 - 150 = 420

⁴₄ S = 4 x 420 = 1680

1 S = 1680

Mass of Container S = 1680 g

Answer(s): (a) 1650 g; (b) 1680 g