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How many gallons of pure water must be added to 500 gallons of a 40% saline solution to reduce it to 25% saline solution

User Kiran P
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1 Answer

4 votes

Answer:

300 gallons

Explanation:

A saline solution is a solution of salt in water.

The percent is the percent of salt in the total amount of solution.

A 40% saline solution has 40% of salt out of the total volume.

Let the amount of pure water needed = x.

The amount of existing 40% solution is 500 gal.

Let the total amount of 25% saline solution made = y.

Equation of volumes of solutions:

500 + x = y

y = x + 500 Eq. 1

Equation of volume of salt:

500 gal of 40% saline solution has 0.4 × 500 gal = 200 gal of salt

Pure water has 0% salt.

The final product is a solution that is 25% saline. Its volume is y.

The volume of salt in the y gallons of 25% saline solution is 25% of y = 0.25y

The equation of salt content is:

200 + 0 = 0.25y

y = 800 Eq. 2

Eq. 1 and Eq. 2 form a system of equations.

y = x + 500

y = 800

Substitute 800 for y in Eq. 1.

y = x + 500

800 = x + 500

x = 300

Answer: 300 gallons of pure water

User Stephan Bijzitter
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