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

Answer: 300 gallons Step-by-step 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 Reply

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