The mass of the mixture, consisting of two substances, was 800 g. After 5/8 of the first substance and 60%

The mass of the mixture, consisting of two substances, was 800 g. After 5/8 of the first substance and 60% of the second were isolated from it, 72 g less than the second remained in the mixture of the first substance. How many grams of each substance was in the mixture first?

1. The mass of the first substance in the mixture is: M1 kg;

2. The mass of the second substance in the mixture: M2 kg;

3. The total mass of the mixture is: M = 800 g;

M1 + M2 = 800;

M1 = 800 – M2;

4. We compose the equations according to the condition of the problem:

M1 – 5/8 * M1 = (M2 – 0.6 * M2) – 72;

0.375 * M1 = 0.4 * M2 – 72;

0.375 * (800 – M2) = 0.4 * M2 – 72;

0.775 * M2 = 300 + 72 = 372;

M2 = 372 / 0.775 = 485 g;

M1 = 800 – M2 = 800 – 485 = 325 g.

Answer: the mixture contained 325 g of the first substance and 485 g of the second substance.