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Questions: Algebra BusinessCalculus
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$10100 is invested in two different accounts. One account pays 13% simple interest per year and the other account pays 11% simple interest per year. If the total interest earn in one year is $1259.00, how much is invested at each rate?
Let \(\displaystyle x\) be the amount invested at 13%. Then \(\displaystyle 10100-x\) is amount invested at 11%. This gives the equation \(\displaystyle 0.13 x+0.11(10100 - x)=1259\). Solving for \(\displaystyle x\) gives \(\displaystyle x = 7400\). So the amount in the other account is $2700.
\begin{question}\$10100 is invested in two different accounts. One account pays 13\% simple interest per year and the other account pays 11\% simple interest per year. If the total interest earn in one year is \$1259.00, how much is invested at each rate?
\soln{10cm}{Let $x$ be the amount invested at 13\%. Then $10100-x$ is amount invested at 11\%. This gives the equation $0.13 x+0.11(10100 - x)=1259$. Solving for $x$ gives $x = 7400$. So the amount in the other account is \$2700.}
\end{question}
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\end{question}\end{document}<p> <p>$10100 is invested in two different accounts. One account pays 13% simple interest per year and the other account pays 11% simple interest per year. If the total interest earn in one year is $1259.00, how much is invested at each rate? </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 invested at 13%. Then <img class="equation_image" title=" \displaystyle 10100-x " src="/equation_images/%20%5Cdisplaystyle%2010100-x%20" alt="LaTeX: \displaystyle 10100-x " data-equation-content=" \displaystyle 10100-x " /> is amount invested at 11%. This gives the equation <img class="equation_image" title=" \displaystyle 0.13 x+0.11(10100 - x)=1259 " src="/equation_images/%20%5Cdisplaystyle%200.13%20x%2B0.11%2810100%20-%20x%29%3D1259%20" alt="LaTeX: \displaystyle 0.13 x+0.11(10100 - x)=1259 " data-equation-content=" \displaystyle 0.13 x+0.11(10100 - x)=1259 " /> . 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 <img class="equation_image" title=" \displaystyle x = 7400 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%207400%20" alt="LaTeX: \displaystyle x = 7400 " data-equation-content=" \displaystyle x = 7400 " /> . So the amount in the other account is $2700.</p> </p>