An ingot of a silver-zinc alloy weighing 7 kg contained 76% silver. It was alloyed with another ingot and got

An ingot of a silver-zinc alloy weighing 7 kg contained 76% silver. It was alloyed with another ingot and got a 21 kg ingot containing 80% silver. How much silver was contained in the second ingot?

1. Determine how many kilograms of silver were contained in the first ingot. We take this weight as a and make up the proportion:

7 kg – 100%, and kg – 76%;

7: a = 100: 76;

a = (7 x 76): 100;

a = 5.32 kg;

2. Determine how many kilograms of silver are contained in the new ingot. We take this weight as a and make up the proportion:

21 kg – 100%, and kg – 80%;

21: a = 100: 80;

a = (21 x 80): 100;

a = 16.8 kg;

3. Determine the mass of the second ingot:

21 – 7 = 14 kg;

4. Determine the mass of silver in the second ingot:

16.8 – 5.32 = 11.48 kg;

4. Determine the percentage of silver in the second ingot. We take it for a and make up the proportion:

14 kg – 100%, 11.48 kg – a%;

14: 11.48 = 100: a;

a = (11.48 x 100): 14;

a = 82%.

Answer: 82 percent of the silver was contained in the second bar.