Intermediate Algebra, Chapter 2, 2.3, Section 2.3, Problem 68

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.