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Joy is preparing 11 Liters of a 30% saline solution she only has 50% and a 10% solution in her lab how many liters of the 50% and how many liters of a 10% should she mix to make the 30% solution 

User Courtland
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1 Answer

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Answer:

Explanation:

Let x be the amount of 50% saline solution needed, and y be the amount of 10% saline solution needed.

To make 11 liters of a 30% saline solution, we know that:

The total amount of solution is 11 liters, so x + y = 11.

The total amount of salt in the solution is 30% of 11 liters, or 0.3 x 11 = 3.3 liters of salt.

The amount of salt contributed by the 50% saline solution is 0.5x liters of salt.

The amount of salt contributed by the 10% saline solution is 0.1y liters of salt.

Therefore, we can set up a system of two equations:

x + y = 11 (equation 1)

0.5x + 0.1y = 3.3 (equation 2)

To solve for x and y, we can use any method of solving systems of equations. Here, we'll use substitution:

From equation 1, we know that y = 11 - x. Substituting this into equation 2, we get:

0.5x + 0.1(11 - x) = 3.3

0.5x + 1.1 - 0.1x = 3.3

0.4x = 2.2

x = 5.5

So Joy needs 5.5 liters of the 50% saline solution. Substituting this into equation 1, we get:

5.5 + y = 11

y = 5.5

So Joy needs 5.5 liters of the 10% saline solution.

Therefore, to make 11 liters of a 30% saline solution, Joy needs to mix 5.5 liters of 50% saline solution and 5.5 liters of 10% saline solution.

User Odgiiv
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