Solving Simple Interest Applications
Solving Simple Interest Applications
Do you know that banks pay you to keep your money? The money a customer puts in the bank is called the principal, P, and the money the bank pays the customer is called the interest. The interest is computed as a certain percent of the principal; called the rate of interest, r. We usually express rate of interest as a percent per year, and we calculate it by using the decimal equivalent of the percent. The variable t, (for time) represents the number of years the money is in the account.
To find the interest we use the simple interest formula, \(I=Prt.\)
Simple Interest
If an amount of money, P, called the principal, is invested for a period of t years at an annual interest rate r, the amount of interest, I, earned is
Interest earned according to this formula is called simple interest.
Interest may also be calculated another way, called compound interest. This type of interest will be covered in later math classes.
The formula we use to calculate simple interest is \(I=Prt.\) To use the formula, we substitute in the values the problem gives us for the variables, and then solve for the unknown variable. It may be helpful to organize the information in a chart.
Example
Nathaly deposited $12,500 in her bank account where it will earn 4% interest. How much interest will Nathaly earn in 5 years?
\(\begin{array}{ccc}\hfill I& =\hfill & ?\hfill \\ \hfill P& =\hfill & $12,500\hfill \\ \hfill r& =\hfill & 4\text{%}\hfill \\ \hfill t& =\hfill & 5\phantom{\rule{0.2em}{0ex}}\text{years}\hfill \end{array}\)
Solution
\(\begin{array}{cccccc}\mathbf{\text{Step 1. Read}}\phantom{\rule{0.2em}{0ex}}\text{the problem.}\hfill & & & & & \\ \mathbf{\text{Step 2. Identify}}\phantom{\rule{0.2em}{0ex}}\text{what we are looking for.}\hfill & & & & & \text{the amount of interest earned}\hfill \\ \begin{array}{c}\mathbf{\text{Step 3. Name}}\phantom{\rule{0.2em}{0ex}}\text{what we are looking for.}\hfill \\ \text{Choose a variable to represent that quantity}\hfill \end{array}\hfill & & & & & \text{Let}\phantom{\rule{0.2em}{0ex}}I=\text{the amount of interest.}\hfill \\ \mathbf{\text{Step 4. Translate}}\phantom{\rule{0.2em}{0ex}}\text{into an equation.}\hfill & & & & & \\ \phantom{\rule{2.5em}{0ex}}\text{Write the formula.}\hfill & & & & & I=Prt\hfill \\ \phantom{\rule{2.5em}{0ex}}\text{Substitute in the given information.}\hfill & & & & & I=\left(12,500\right)\left(.04\right)\left(5\right)\hfill \\ \mathbf{\text{Step 5. Solve}}\phantom{\rule{0.2em}{0ex}}\text{the equation.}\hfill & & & & & I=2,500\hfill \\ \mathbf{\text{Step 6. Check}}\text{:}\phantom{\rule{0.2em}{0ex}}\text{Does this make sense?}\hfill & & & & & \\ \begin{array}{}\\ \phantom{\rule{2.5em}{0ex}}\text{Is \$2,500 is a reasonable interest on}\hfill \\ \phantom{\rule{2.5em}{0ex}}\text{\$12,500? Yes.}\hfill \end{array}\hfill & & & & & \\ \begin{array}{c}\mathbf{\text{Step 7. Answer}}\phantom{\rule{0.2em}{0ex}}\text{the question with a}\hfill \\ \text{complete sentence.}\hfill \end{array}\hfill & & & & & \text{The interest is \$2,500.}\hfill \end{array}\)
There may be times when we know the amount of interest earned on a given principal over a certain length of time, but we don’t know the rate. To find the rate, we use the simple interest formula, substitute in the given values for the principal and time, and then solve for the rate.
Example
Loren loaned his brother $3,000 to help him buy a car. In 4 years his brother paid him back the $3,000 plus $660 in interest. What was the rate of interest?
Solution
\(\begin{array}{cccc}\mathbf{\text{Step 1. Read}}\phantom{\rule{0.2em}{0ex}}\text{the problem.}\hfill & & & \\ \mathbf{\text{Step 2. Identify}}\phantom{\rule{0.2em}{0ex}}\text{what we are looking for.}\hfill & & & \text{the rate of interest}\hfill \\ \begin{array}{c}\mathbf{\text{Step 3. Name}}\phantom{\rule{0.2em}{0ex}}\text{what we are looking for. Choose}\hfill \\ \text{a variable to represent that quantity}.\hfill \end{array}\hfill & & & \text{Let}\phantom{\rule{0.2em}{0ex}}r=\text{rate of interest.}\hfill \\ \begin{array}{c}\mathbf{\text{Step 4. Translate}}\phantom{\rule{0.2em}{0ex}}\text{into an equation.}\hfill \\ \\ \phantom{\rule{2.5em}{0ex}}\text{Write the formula.}\hfill \\ \phantom{\rule{2.5em}{0ex}}\text{Substitute in the given information.}\hfill \\ \mathbf{\text{Step 5. Solve}}\phantom{\rule{0.2em}{0ex}}\text{the equation.}\hfill \\ \\ \phantom{\rule{2.5em}{0ex}}\text{Divide.}\hfill \\ \phantom{\rule{2.5em}{0ex}}\text{Change to percent form.}\hfill \end{array}\hfill & & & \hfill \begin{array}{}\\ \hfill I& =\hfill & Prt\hfill \\ \hfill 660& =\hfill & \left(3,000\right)r\left(4\right)\hfill \\ \hfill 660& =\hfill & \left(12,000\right)r\hfill \\ \hfill 0.055& =\hfill & r\hfill \\ \hfill 5.5\text{%}& =\hfill & r\hfill \end{array}\hfill \\ \mathbf{\text{Step 6. Check}}\text{:}\phantom{\rule{0.2em}{0ex}}\text{Does this make sense?}\hfill & & & \\ \begin{array}{ccc}\hfill \phantom{\rule{2.5em}{0ex}}I& =\hfill & Prt\hfill \\ \hfill \phantom{\rule{2.5em}{0ex}}660& \stackrel{?}{=}\hfill & \left(3,000\right)\left(0.055\right)\left(4\right)\hfill \\ \hfill \phantom{\rule{2.5em}{0ex}}660& =\hfill & 660✓\hfill \end{array}\hfill & & & \\ \mathbf{\text{Step 7. Answer}}\phantom{\rule{0.2em}{0ex}}\text{the question with a complete sentence.}\hfill & & & \text{The rate of interest was 5.5%.}\hfill \end{array}\)
Notice that in this example, Loren’s brother paid Loren interest, just like a bank would have paid interest if Loren invested his money there.
Example
Eduardo noticed that his new car loan papers stated that with a 7.5% interest rate, he would pay $6,596.25 in interest over 5 years. How much did he borrow to pay for his car?
Solution
\(\begin{array}{cccc}\mathbf{\text{Step 1. Read}}\phantom{\rule{0.2em}{0ex}}\text{the problem.}\hfill & & & \\ \mathbf{\text{Step 2. Identify}}\phantom{\rule{0.2em}{0ex}}\text{what we are looking for.}\hfill & & & \text{the amount borrowed (the principal)}\hfill \\ \begin{array}{c}\mathbf{\text{Step 3. Name}}\phantom{\rule{0.2em}{0ex}}\text{what we are looking for.}\hfill \\ \text{Choose a variable to represent that quantity}.\hfill \end{array}\hfill & & & \text{Let}\phantom{\rule{0.2em}{0ex}}P=\text{principal borrowed.}\hfill \\ \begin{array}{c}\mathbf{\text{Step 4. Translate}}\phantom{\rule{0.2em}{0ex}}\text{into an equation.}\hfill \\ \\ \phantom{\rule{2.5em}{0ex}}\text{Write the formula.}\hfill \\ \phantom{\rule{2.5em}{0ex}}\text{Substitute in the given information.}\hfill \\ \mathbf{\text{Step 5. Solve}}\phantom{\rule{0.2em}{0ex}}\text{the equation.}\hfill \\ \\ \text{Divide.}\hfill \end{array}\hfill & & & \begin{array}{}\\ \hfill I& =\hfill & Prt\hfill \\ \hfill 6,596.25& =\hfill & P\left(0.075\right)\left(5\right)\hfill \\ \hfill 6,596.25& =\hfill & 0.375P\hfill \\ \hfill 17,590& =\hfill & P\hfill \end{array}\hfill \\ \mathbf{\text{Step 6. Check}}\text{:}\phantom{\rule{0.2em}{0ex}}\text{Does this make sense?}\hfill & & & \\ \begin{array}{ccc}\hfill \phantom{\rule{2.5em}{0ex}}I& =\hfill & Prt\hfill \\ \hfill \phantom{\rule{2.5em}{0ex}}6,596.25& \stackrel{?}{=}\hfill & \left(17,590\right)\left(0.075\right)\left(5\right)\hfill \\ \hfill \phantom{\rule{2.5em}{0ex}}6,596.25& =\hfill & 6,596.25✓\hfill \end{array}\hfill & & & \\ \begin{array}{c}\mathbf{\text{Step 7. Answer}}\phantom{\rule{0.2em}{0ex}}\text{the question with a}\hfill \\ \text{complete sentence.}\hfill \end{array}\hfill & & & \text{The principal was \$17,590.}\hfill \end{array}\)
This lesson is part of:
Math Models and Geometry II