Question

к 40% раствору соляной кислоты добавили 50 г чистой кислоты,после чего стала равной 60%.Найдите первоначальный вес раствора.Алгебра

Answer 1

Х- было р-ра
0,4х было в нём кислоты
х+50 стало р-ра
0,6(х+50) стало в нём кислоты
0,4х+50=0,6(х+50)
0,4х+50+0,6х+30
50-30=0,6х-0,4х
20=0,2х
х=100г было раствора

Answer 2

Problem Analysis

We are given that a 40% solution of hydrochloric acid was mixed with 50 grams of pure acid, resulting in a solution that is now 60% acid. We need to find the initial weight of the solution.

Solution

Let's assume the initial weight of the solution is x grams.

The amount of acid in the 40% solution is 40% of x grams, which is 0.4x grams.

The amount of acid in the 50 grams of pure acid is 50 grams.

After mixing, the total amount of acid in the solution is 60% of the new total weight, which is 0.6(x + 50) grams.

According to the problem, the amount of acid in the solution before and after mixing is the same. Therefore, we can set up the following equation:

0.4x + 50 = 0.6(x + 50)

Now, let's solve this equation to find the value of x.

0.4x + 50 = 0.6x + 30

0.4x - 0.6x = 30 - 50

-0.2x = -20

x = -20 / -0.2

x = 100

Therefore, the initial weight of the solution is 100 grams.

Answer

The initial weight of the solution is 100 grams.