# Mixing 43% and 89% acid solution and adding 10 kg of pure water, we got 69%

**Mixing 43% and 89% acid solution and adding 10 kg of pure water, we got 69% acid solution. If instead of 10 kg of water 10 kg of 50% solution of the same acid were added, then we would get 73% solution of acid. How many kg of 43% solution were used to obtain the mixture?**

1. The mass of the first solution with a concentration of K1 = 0.43 is equal to: M1 kg;

2. The mass of the second solution with a concentration of K2 = 0.89 is equal to: M2 kg;

3. Water added: Mb = 10 kg;

4. Received the third solution with a concentration of K3 = 0.69;

5. In the fourth solution, instead of water, the same amount of acid solution with

concentration Ko = 0.5:

6. Its concentration is equal to: K4 = 0.73;

7. We make up the acid balance of the third solution:

K1 * M1 + K2 * M2 = K3 * (M1 + M2 + Mb);

0.43 * M1 + 0.89 * M2 = 0.69 * (M1 + M2 + 10);

(0.43 – 0.69) * M1 + (0.89 – 0.69) * M2 = 6.9;

0.2 * M2 – 0.26 * M1 = 6.9;

8. Acid balance of the fourth solution:

K1 * M1 + K2 * M2 + Ko * Mb = K4 * (M1 + M2 + Mb);

0.43 * M1 + 0.89 * M2 + 0.5 * 10 = 0.73 * (M1 + M2 + 10);

(0.43 – 0.73) * M1 + (0.89 – 0.73) * M2 = 7.3 – 5;

0.16 * M2 – 0.3 * M1 = 2.3;

9. Let’s solve the system of two equations:

0.2 * M2 – 0.26 * M1 = 6.9;

0.16 * M2 – 0.3 * M1 = 2.3;

M2 = (6.9 + 0.26 * M1) / 0.2;

0.16 * (6.9 + 0.26 * M1) / 0.2 – 0.3 * M1 = 2.3;

0.092 * M1 = 5.52 – 2.3 = 3.22;

M1 = 3 / 0.092 = 35 kg;

M2 = (6.9 + 0.26 * M1) / 0.2 = (6.9 + 0.26 * 35) / 0.2 = 80 kg

Answer: 35 kg of solution with a concentration of 43% was taken.