Mixed 70% and 40% acid solution and added 15 kg of pure water received a 37% acid solution

Mixed 70% and 40% acid solution and added 15 kg of pure water received a 37% acid solution. if instead of 15 kg of water 15 kg of a 20% solution of the same acid were added, a 43% solution of acids would be obtained. how many kilograms of 70% solution were used to obtain the mixture?

1. The mass of the first solution of 70% concentration: M70 kg;

2. The mass of the second solution of 40% concentration: M40 kg;

3. The mass of pure water: Mb = 15 kg;

4. The mass of the third solution of 37% concentration: M37 = M70 + M40 + Mb kg;

5. The mass of the fourth solution of 20% concentration: M20 = 15 kg;

6. Weight of the fifth solution of 43% concentration: M43 kg;

7. Acid balance of the third solution:

0.7 * M70 + 0.4 * M40 + Mb = 0.37 * (M70 + M40 + Mb) kg;

0.7 * M70 + 0.4 * M40 = 0.37 * (M70 + M40 + 15) kg;

M70 * (0.7 – 0.37) + M40 * (0.4 – 0.37) = 0.37 * 15;

0.33 * M70 + 0.03 * M40 = 5.55;

8. Acid balance of the fifth solution:

0.7 * M70 + 0.4 * M40 + 0.2 * M20 = 0.43 * (M70 + M40 + M20) kg;

0.7 * M70 + 0.4 * M40 + 0.2 * 15 = 0.43 * (M70 + M40 + 15) kg;

M70 * (0.7 – 0.43) + M40 * (0.4 – 0.43) = 0.43 * 15 – 0.2 * 15;

0.27 * M70 – 0.03 * M40 = 3.45;

9. We solve the system of two equations:

0.33 * M70 + 0.03 * M40 = 5.55;

0.27 * M70 – 0.03 * M40 = 3.45;

10. Add up:

0.6 * M70 = 9;

M70 = 9 / 0.6 = 15 kg.

Answer: 15 kg of 70% solution was used.