Saturday, May 11, 2013

Intermediate Algebra, Chapter 2, 2.3, Section 2.3, Problem 50

Sheryl Zavertnik won $\$60,000$ on a slot machine in Las Vegas. She invested part of the money at $2\%$ simple
interest and the rest at $3\%$. In one year, she earned a total of $\$1,600$ in interest. How much was invested at each rate?

Step 1: Read the problem, we are asked to find the amount invested on each rate.
Step 2 : Assign the variable. Then organize the information in the table.
Let $x = $ amount invested in $2\%$ interest rate
Then, $60,000 - x= $ amount invested in $3\%$ interest rate


$
\begin{array}{|c|c|c|c|c|c|}
\hline
& \rm{Principal} & \cdot & \text{Interest Rate} & = & \rm{Interest} \\
\hline
2\% & x & \cdot & 0.02 & = & 0.02x \\
\hline
3\% & 60,000 - x & \cdot & 0.03 & = & 0.03(60,000 - x) \\
\hline
\end{array}
$


The total interest earned is equal to the sum of the interests at each rate.

Step 3: Write an equation from the last column of the table
$0.02x + 0.03(60,000 - x) = 1,600$

Step 4: Solve

$
\begin{equation}
\begin{aligned}
0.02x + 1,800 - 0.03x &= 1,600\\
\\
-0.01x + 1,800 &= 1,600 \\
\\
-0.01x &= -200\\
\\
x &= 20,000
\end{aligned}
\end{equation}
$


Then by substitution,
$60,000 - x = 60,000 - 20,000 = 40,000$

Step 5: State the answer
In other words, the amount invested in $2\%$ and $3\%$ interest rate is $\$20,000$ and $\$40,000$ respectively.

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