Problem A:
Jack has $\$800$ invested in two accounts. One pays $5\%$ interest per year and other pays $10\%$
interest per year. The amount of yearly interest is the same as he would get if the entire $\$800$ was
invested at $8.75\%$. How much does he have invested at each rate?
Problem B:
Jill has $800$ L of acid solution. She obtained it by mixing some $5\%$ acid with some $10\%$ acid. Her final
mixture of $800$ L is $8.75\%$ acid. How much of each of the $5\%$ and $10\%$ solutions did she use to get her final
mixture?
a.) Solve Problem A
b.) Solve Problem B
c.) Explain the similarities between the processes used in solving Problems A and B
In problem (a), let $x$ represent the amount invested at $5\%$ interest, and in problem (b), let $y$ represent
the amount of $5\%$ acid used.
a.) Step 1: Read the problem, we are required to determine the amount invested at each account.
Step 2 : Assign the variable. Then organize the information in the table.
Let $x = $ amount invested at $5\%$ interest.
Then, $800 -x =$ amount invested at $10\%$ interest.
$
\begin{array}{|c|c|c|c|c|c|}
\hline
& \rm{Principal} & \cdot & \text{Interest rate} & = & \rm{Interest} \\
\hline
5 \% & x & \cdot & 0.05 & = & 0.05x \\
\hline
10\% & 800 - x & \cdot & 0.10 & = & 0.10(800 - x) \\
\hline
\end{array}
$
Step 3: Write an equation from the last column of the table
$0.05x + 0.10 (800 -x) = 0.0875(800)$
Step 4: Solve
$
\begin{equation}
\begin{aligned}
0.05x + 80 - 0.10x &= 70\\
\\
-0.05x &= 70 - 80\\
\\
-0.05x &= -10\\
\\
x &= 200
\end{aligned}
\end{equation}
$
Then, by substitution
$800 - x = 800 - 200 = 600$
Step 5: State the answer
In other words, the amount invested at $5\%$ and $10\%$ interest rates is $\$250$ and $\$600$ respectively.
b.) Step 1: Read the problem, we are required to determine quantity of each solutions used.
Step 2 : Assign the variable. Then organize the information in the table.
Let $x = $ amount of $5\%$ acid used.
Then, $800 - y =$ amount of $10\%$ acid used.
$
\begin{array}{|c|c|c|c|c|c|}
\hline
& \text{Liters of solution} & \cdot & \text{Percent Concentration} & = & \rm{Quantity} \\
\hline
5\% & y & \cdot & 0.05 & = & 0.05y \\
\hline
10\% & 800 - y & \cdot & 0.10 & = & 0.10(800 - y) \\
\hline
\end{array}
$
Step 3: Write an equation from the last column of the table
$0.05y + 0.10 (800 - y) = 0.0875(800)$
Step 4: Solve
$
\begin{equation}
\begin{aligned}
0.05y+ 80 - 0.10y &= 70\\
\\
-0.05y &= 70 - 80\\
\\
-0.05y &= -10\\
\\
y &= 200
\end{aligned}
\end{equation}
$
Then, by substitution
$800 - y = 800 - 200 = 600$
Step 5: State the answer
In other words, the final mixture she must use is $200$ L of $5\%$ acid solution and $600$ L
of $10\%$ acid solution.
c.) In general, solving part A and B is similar in a way that the total amount or quantity
is equated with the sum of the individual amounts or quantities of each condition.
Friday, October 11, 2013
Intermediate Algebra, Chapter 2, 2.3, Section 2.3, Problem 68
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