Chapter 1.7, Problem 31E

Solution

To calculate the quantity of solutions to make $100\text{ gal}$ of $25\%$ solution.

  $57.14\text{ gal}$ of $10\%$ solution and $42.86\text{ gal}$ of $45\%$ solution.

Given:

Total quantity of solution = $100\text{ gal}$ of $25\%$ solution

Calculation:

Let the $10\%$ solution be $x\text{ gal}\text{.}$ As the total quantity of the mixed solution is $\text{100 gal,}$ so the quantity of $45\%$ solution be $(100-x)\text{ gal}\text{.}$

Now, the total mixed solution should be $25\%$ , i.e.

  $\begin{array}{l}x\text{ of }10\%+(100-x)\text{ of }45\%=100\text{ of }25\%\\ \Rightarrow \frac{10x}{100}+\frac{(100-x)\times 45}{100}=\frac{100\times 25}{100}\\ \Rightarrow \frac{x}{10}+45-\frac{45x}{100}=25\\ \Rightarrow \frac{45x}{100}-\frac{x}{10}=45-25\\ \Rightarrow \frac{45x-10x}{100}=20\\ \Rightarrow \frac{35x}{100}=20\\ \Rightarrow x=\frac{400}{7}\\ \Rightarrow x=57.14\text{ gal}\end{array}$

So, the other part of the solution will be $(100-x)=100-57.14=42.86\text{ gal}\text{.}$

Hence, $57.14\text{ gal}$ of $10\%$ solution and $42.86\text{ gal}$ of $45\%$ solution should be mixed together to make $100\text{ gal}$ of $25\%$ solution.