Chapter 1.7, Problem 32E

Solution

To calculate the quantity of solutions to make $\text{25 L}$ of an $26\%$ acid solution.

  $\text{15 L}$ of $20\%$ solution and $\text{10 L}$ of $35\%$ solution

Given:

Total quantity of acid solution = $\text{25 L}$ of $26\%$ solution

Calculation:

Let the $20\%$ solution be $x\text{ L}\text{.}$ As the total quantity of the mixed solution is $\text{25 L,}$ so the quantity of $35\%$ solution be $(25-x)\text{ L}\text{.}$

Now, the total mixed solution should be $26\%$ , i.e.

  $\begin{array}{l}x\text{ of 2}0\%+(25-x)\text{ of 3}5\%=\text{ 25 of }26\%\\ \Rightarrow \frac{20x}{100}+\frac{(25-x)\times 35}{100}=\frac{25\times 26}{100}\\ \Rightarrow \frac{x}{5}+\frac{35}{4}-\frac{7x}{20}=\frac{13}{2}\\ \Rightarrow \frac{7x}{20}-\frac{x}{5}=\frac{35}{4}-\frac{13}{2}\\ \Rightarrow \frac{7x-4x}{20}=\frac{35-26}{4}\\ \Rightarrow \frac{3x}{20}=\frac{9}{4}\\ \Rightarrow x=\frac{180}{12}\\ \Rightarrow x=15\text{ L}\end{array}$

So, the other part of the solution will be $(25-x)=25-15=10\text{ L}\text{.}$

Hence, $\text{15 L}$ of $20\%$ acid solution and $\text{10 L}$ of $35\%$ acid solution should be mixed together to make $\text{25 L}$ of $26\%$ acid solution.