Question

A manufacturer has $600$ liters of an $12\%$ acid solution. The number of liters of a $30\%$ acid solution to be added to it so that acid content in the resulting mixture will be more than $15\%$ but less $18\%$ is in the interval

• A. $(100,150)$
• B. $(120,180)$
• C. $(120,300)$
• D. $(100,180)$

Answer 1

The correct option is C $(120,300)$

Let $x$ liters of $30\%$ acid solution is added to the $600$ liters of $12\%$ acid solution.
Total acid content in the mixture is $\frac{30}{100}\cdot x+\frac{12}{100}\cdot \left(600\right)$
Now acid content of the mixture lies in the between $15\%$ to $18\%$ and has volume of $(x+600)$ liters
$\therefore \frac{15}{100}(x+600)<\frac{30x}{100}+\frac{7200}{100}<\frac{18}{100}(x+600)$
$\Rightarrow 30x+7200>15(x+600)$ and $30x+7200<18(x+600)$
$\Rightarrow 2x+480>x+600$ and $\frac{5}{3}x+400<x+600$
$\Rightarrow 120<x<300$