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How many liters of pure water should be mixed with a 11-L solution of 60% acid to produce a mixture that is 90% water?

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Answer:

55 L of water

Explanation:

11-L of 60% acid contains 0.6 × 11 L = 6.6 L acid

Let x = amount of pure water.

Let y = total amount produced of 90% water solution.

A 90% water solution is a 10% acid solution.

Amounts of solutions:

x + 11 = y

Amounts of acid:

6.6 = 0.1y

y = 66

x + 11 = 66

x = 55

Answer: 55 L of water

55 L water has 0 acid.

11 L 60% acid solution has 6.6 L acid.

66 L of 10% acid solution has 6.6 L acid

User Nathan Roe
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