Question

a chemist has three different acid solutions. the first acid solution contains 20% acid, the second contains 40% and the third contains 60%. they want to use all three solutions to obtain a mixture of 280 liters containing 30% acid, using 3 times as much of the 60% solution as the 40% solution. how many liters of each solution should be used? the chemist should use liters of 20% solution, liters of 40% solution, and liters of 60% solution.

Explanation:

Step1: Set up equations

Let \(x\) be the liters of \(20\%\) solution, \(y\) be the liters of \(40\%\) solution, and \(z\) be the liters of \(60\%\) solution.
We have the following equations:

• Total volume: \(x + y+z = 280\)
• Acid - content: \(0.2x+0.4y + 0.6z=0.3\times280\) (since the final mixture is \(280\) liters of \(30\%\) acid)
• Relationship between \(z\) and \(y\): \(z = 3y\)

Step2: Substitute \(z = 3y\) into the other equations

Substitute \(z = 3y\) into \(x + y+z = 280\), we get \(x+y + 3y=280\), so \(x=280 - 4y\)
Substitute \(z = 3y\) and \(x = 280 - 4y\) into \(0.2x+0.4y + 0.6z=84\)

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Step3: Find \(x\) and \(z\)

Since \(y = 20\), then \(z=3y = 60\)
Since \(x=280 - 4y\), substitute \(y = 20\) into it, \(x=280-4\times20=200\)

Answer:

The chemist should use \(200\) liters of \(20\%\) solution, \(20\) liters of \(40\%\) solution, and \(60\) liters of \(60\%\) solution.