Question

Завтра сдавать! Прошу, помогите.Решите задачу с помощью уравнения.Смешали два раствора соли 5%-ной и 9%-ной концентраций и получили 12л. раствора 8%-ной концентрации. Сколько литров было в каждом из первоначальных растворов?

Answer 1

Пусть х л  - 1-й раствор, тогда (12-х) л - 2-й раствор, по условию
$0,05x+0,09(12-x)=0,08*12;0,04x=0,12;x=3$
 тогда 3 л  - 1-й раствор, тогда 9 л - 2-й раствор,

Answer 2

Solving the Salt Solution Problem

To solve this problem, we can use the method of mixtures, which involves setting up a system of equations based on the amount and concentration of the solutions being mixed.

Let's denote: - \( x \) as the amount of the 5% solution - \( y \) as the amount of the 9% solution

We know that the total amount of the resulting 8% solution is 12 liters. Therefore, we can set up the following equation based on the amount of salt in the resulting solution:

Equation 1: \( x + y = 12 \)

We also know that the resulting solution is 8% salt. This allows us to set up the following equation based on the concentration of salt in the resulting solution:

Equation 2: \( 0.05x + 0.09y = 0.08 \times 12 \)

Now, we can solve this system of equations to find the values of \( x \) and \( y \).

Solution

Using the system of equations, we can solve for the values of \( x \) and \( y \).

Equation 1: \( x + y = 12 \) Equation 2: \( 0.05x + 0.09y = 0.96 \)

Solving for \( x \) and \( y \) gives: - \( x = 6 \) liters - \( y = 6 \) liters

Therefore, there were 6 liters of the 5% solution and 6 liters of the 9% solution in the initial mix.

I hope this helps! If you have any further questions or need additional assistance, feel free to ask.