How many grams of a 20% solution and anhydrous sodium carbonate are needed to prepare 2 kg

How many grams of a 20% solution and anhydrous sodium carbonate are needed to prepare 2 kg of a 30% sodium carbonate solution?

Mass fraction of solute w = m (r.v.) / m (p-pa), where:
m (p-pa) is the mass of the solution, m (r.v.) is the mass of the solute.
The solution that we need to prepare must contain anhydrous sodium carbonate: m (r.v.) = W * m (p-pa) = 0.3 * 2 = 0.6 kg
We must obtain this amount from anhydrous salt and water, which is contained in a 20% solution.
The mass of water in the required solution m (H2O) = m (p-pa) – m (r.v) = 2 – 0.6 = 1.4 kg
Let us find the mass of a 20% solution containing so much water m (p-pa2) = m (H2O) / (1 – w2)), where w2 = 20% is the mass fraction of salt, m (p-pa2) = 1.4 / ( 1 – 0.2) = 1.75 kg
This solution contains m (r.v. 2) = w2 * m (p-pa2) = 0.2 * 1.75 = 0.35 kg
Find the missing amount of anhydrous sodium carbonate:
m (r.v. neobx) = m (r.v.) – m (r.v. 2) = 0.6 – 0.35 = 0.25 kg
That is, we need 0.25 kg of anhydrous salt and 1.75 kg of a 20% solution