Question

Determine the number of liters of a 30% saline solution and the number of liters of a 60% saline solution that are required to make 10 liters of a 50% saline solution.

Answer 1

Answer:

Number of liters of 30% solution = x = 3.3 liters

Number of liters of 60% solution = y

= 6.67 liters

Step-by-step explanation:

Let us represent:

Number of liters of 30% solution = x

Number of liters of 60% solution = y

Determine the number of liters of a 30% saline solution and the number of liters of a 60% saline solution that are required to make 10 liters of a 50% saline solution.

30% × x + 60% × y = 10 × 50%

0.3x + 0.6y = 5...... Equation 1

x + y = 10...... Equation 2

x = 10 - y

We substitute 10 - y for x in Equation 1

Hence

0.3(10 - y) + 0.6y = 5

3 - 0.3y + 0.6y = 5

- 0.3y + 0.6y = 5 - 3

0.3y = 2

y = 2/0.3

y = 6.67 liters

Solving for x

x = 10 - y

x = 10 - 6.67

x = 3.33 liters

Hence:

Number of liters of 30% solution = x = 3.3 liters

Number of liters of 60% solution = y

= 6.67 liters