50 mg of salt was added to a brine solution weighing 250 mg, after which the salt content

50 mg of salt was added to a brine solution weighing 250 mg, after which the salt content in the solution increased by 10%. How much salt was in the solution initially?

1. According to the condition of the problem, a saline solution is given.

Its mass is 250 mg.

50 mg of salt was added to it.

Let’s calculate the weight of the new solution.

250 + 50 = 300 mg.

2. Let X mg be the initial weight of the salt in the solution.

Then the salt content was X / 250.

(X + 50) mg is the new weight of salt.

Its content became (X + 50) / 300.

3. According to the condition of the problem in the new solution, the content increased by 10% or 10/100 = 1/10 of the total.

(X + 50) / 300 – X / 250 = 1/10.

5 * (X + 50) – 6 * X = 1/10 * 1500.

250 – X = 150.

X = 100 mg.

Answer: The solution originally contained 100 mg of salt.