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Questions: Algebra BusinessCalculus
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A 20% acid solution is mixed with 290 ounces of a 43% acid solution to make a 25% acid mixture. How many ounces of the 20% acid solution were used?
Let \(\displaystyle x\) be the amount of the 20% acid solution. This gives the equation \(\displaystyle 0.2(x)+0.43(290)=0.25(x + 290)\). Solving for \(\displaystyle x\) gives 1044 ounces.
\begin{question}A 20\% acid solution is mixed with 290 ounces of a 43\% acid solution to make a 25\% acid mixture. How many ounces of the 20\% acid solution were used?
\soln{10cm}{Let $x$ be the amount of the 20\% acid solution. This gives the equation $0.2(x)+0.43(290)=0.25(x + 290)$. Solving for $x$ gives 1044 ounces.}
\end{question}
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\begin{document}\begin{question}(10pts) The question goes here!
\soln{9cm}{The solution goes here.}
\end{question}\end{document}<p> <p>A 20% acid solution is mixed with 290 ounces of a 43% acid solution to make a 25% acid mixture. How many ounces of the 20% acid solution were used?</p> </p>
<p> <p>Let <img class="equation_image" title=" \displaystyle x " src="/equation_images/%20%5Cdisplaystyle%20x%20" alt="LaTeX: \displaystyle x " data-equation-content=" \displaystyle x " /> be the amount of the 20% acid solution. This gives the equation <img class="equation_image" title=" \displaystyle 0.2(x)+0.43(290)=0.25(x + 290) " src="/equation_images/%20%5Cdisplaystyle%200.2%28x%29%2B0.43%28290%29%3D0.25%28x%20%2B%20290%29%20" alt="LaTeX: \displaystyle 0.2(x)+0.43(290)=0.25(x + 290) " data-equation-content=" \displaystyle 0.2(x)+0.43(290)=0.25(x + 290) " /> . Solving for <img class="equation_image" title=" \displaystyle x " src="/equation_images/%20%5Cdisplaystyle%20x%20" alt="LaTeX: \displaystyle x " data-equation-content=" \displaystyle x " /> gives 1044 ounces.</p> </p>