Intermediate Algebra, Chapter 2, 2.3, Section 2.3, Problem 64

Write, solve and explain why the given problem below has no solution:
How much $30\%$ acid should be mixed with $15$ L of $50\%$ acid to obtain a mixture that is $50\%$ acid?

Step 1: Read the problem, we are asked to find the quantity of $30\%$ acid solutions.
Step 2 : Assign the variable. Then organize the information in the table.
Let $x = $ quantity of $30\%$ acid solution.


$
\begin{array}{|c|c|c|c|c|c|}
\hline
& \text{Liters of solution} & \cdot & \text{Percent concentration} & = & \text{Quantity} \\
\hline
\text{$30\%$ acid} & x & \cdot & 0.30 & = & 0.30x \\
\hline
\text{$50\%$ acid} & 15 & \cdot & 0.50 & = & 0.50(15) \\
\hline
\text{$60\%$ acid} & x + 15 & \cdot & 0.60 & = & 0.60(x + 15) \\
\hline
\end{array}
$

The sum of the quantities of each solution is equal to the quantity of the resulting solution

Step 3: Write an equation from the last column of the table
$0.30x + 0.50(15) = 0.60(x + 15)$

Step 4: Solve

$
\begin{equation}
\begin{aligned}
0.30x + 7.5 &= 0.60x + 9 \\
\\
0.30x - 0.60x &= 9 - 7.5\\
\\
-0.30x &= 1.5\\
\\
x &= -5
\end{aligned}
\end{equation}
$


Step 5: State the answer
We can say that the problem has no solution because the concentration of the resulting mixture
must be less than the highest concentration of the added solution in the mixture.