How many grams of a 10% sodium hydroxide solution should be added to 300 g

How many grams of a 10% sodium hydroxide solution should be added to 300 g of a 40% solution of the same substance to obtain a 20% solution?

Given:
ω1 (NaOH) = 10%
m2 solution (NaOH) = 300 g
ω2 (NaOH) = 40%
ω3 (NaOH) = 20%

To find:
m1 solution (NaOH) -?

Decision:
1) Let m1 solution (NaOH) = (x) g;
2) m1 (NaOH) = ω1 (NaOH) * m1 solution (NaOH) / 100% = 10% * x / 100% = (0.1x) g;
3) m2 (NaOH) = ω2 (NaOH) * m2 solution (NaOH) / 100% = 40% * 300/100% = 120 g;
4) m3 (NaOH) = m1 (NaOH) + m2 (NaOH) = (0.1x + 120) g;
5) m3 solution (NaOH) = m1 solution (NaOH) + m2 solution (NaOH) = (x + 300) g;
6) ω3 (NaOH) = m3 (NaOH) * 100% / m3 solution (NaOH);
20% = (0.1x + 120) * 100% / (x + 300);
x = 600;
7) m1 solution (NaOH) = x = 600 g.

Answer: The mass of the NaOH solution is 600 g.