Solve Simple Interest Applications

Use the simple interest formula

Do you know that banks pay you to let them keep your money? The money you put in the bank is called the principal, $P$, and the bank pays you interest, $I$. The interest is computed as a certain percent of the principal, called the rate of interest, $r$. The rate of interest is usually expressed as a percent per year, and is calculated by using the decimal equivalent of the percent. The variable for time, $t$, represents the number of years the money is left in the account.

The formula we use to calculate simple interest is $I = Prt$. To use the simple interest formula, we substitute in the values for the variables that are given, and then solve for the unknown variable. It may be helpful to organize the information by listing all four variables and filling in the given information.

Example. Find the simple interest earned after $3$ years on $\text{\textdollar}500$ at an interest rate of $6\%$.

Organize the given information: $I = ?$, $P = \text{\textdollar}500$, $r = 6\%$, $t = 3$ years. Write the formula: $I = Prt$. Substitute the given information, remembering to write the percent in decimal form: $I = (500)(0.06)(3)$. Simplify: $I = 90$. Check: in $3$ years the money earned $18\%$; if we rounded to $20\%$, the interest would have been $500(0.20)$, or $\text{\textdollar}100$, so $\text{\textdollar}90$ is reasonable. The simple interest is $\text{\textdollar}90$.

In the next example, we will use the simple interest formula to find the principal.

Example. Find the principal invested if $\text{\textdollar}178$ interest was earned in $2$ years at an interest rate of $4\%$.

Organize the given information: $I = \text{\textdollar}178$, $P = ?$, $r = 4\%$, $t = 2$ years. Write the formula: $I = Prt$. Substitute the given information: $178 = P(0.04)(2)$. Divide: $\tfrac{178}{0.08} = \tfrac{0.08P}{0.08}$. Simplify: $2{,}225 = P$. Check: $178 \overset{?}{=} 2{,}225(0.04)(2)$; $178 = 178\ \checkmark$. The principal is $\text{\textdollar}2{,}225$.

Now we will solve for the rate of interest.

Example. Find the rate if a principal of $\text{\textdollar}8{,}200$ earned $\text{\textdollar}3{,}772$ interest in $4$ years.

Organize the given information: $I = \text{\textdollar}3{,}772$, $P = \text{\textdollar}8{,}200$, $r = ?$, $t = 4$ years. Write the formula: $I = Prt$. Substitute the given information: $3{,}772 = 8{,}200 r (4)$. Multiply: $3{,}772 = 32{,}800r$. Divide: $\tfrac{3{,}772}{32{,}800} = \tfrac{32{,}800r}{32{,}800}$. Simplify: $0.115 = r$. Write as a percent: $11.5\% = r$. Check: $3{,}772 \overset{?}{=} 8{,}200(0.115)(4)$; $3{,}772 = 3{,}772\ \checkmark$. The rate was $11.5\%$.

Solve simple interest applications

Applications with simple interest usually involve either investing money or borrowing money. To solve these applications, we continue to use the same strategy for applications that we have used earlier in this chapter. The only difference is that in place of translating to get an equation, we can use the simple interest formula.

We will start by solving a simple interest application to find the interest.

Example. Nathaly deposited $\text{\textdollar}12{,}500$ in her bank account where it will earn $4\%$ interest. How much interest will Nathaly earn in $5$ years?

We are asked to find the interest, $I$. Organize the given information: $I = ?$, $P = \text{\textdollar}12{,}500$, $r = 4\%$, $t = 5$ years. Write the formula: $I = Prt$. Substitute the given information: $I = (12{,}500)(0.04)(5)$. Simplify: $I = 2{,}500$. Check: at $4\%$ interest per year, in $5$ years the interest would be $20\%$ of the principal, and $20\%$ of $\text{\textdollar}12{,}500$ is $\text{\textdollar}2{,}500$, so this checks out. The interest is $\text{\textdollar}2{,}500$.

There may be times when you know the amount of interest earned on a given principal over a certain length of time, but you don’t know the rate. For instance, this might happen when family members lend or borrow money among themselves instead of dealing with a bank.

Example. Loren lent his brother $\text{\textdollar}3{,}000$ to help him buy a car. In $4$ years his brother paid him back the $\text{\textdollar}3{,}000$ plus $\text{\textdollar}660$ in interest. What was the rate of interest?

We are asked to find the rate of interest, $r$. Organize the given information: $I = 660$, $P = \text{\textdollar}3{,}000$, $r = ?$, $t = 4$ years. Write the formula: $I = Prt$. Substitute the given information: $660 = (3{,}000)r(4)$. Multiply: $660 = (12{,}000)r$. Divide: $\tfrac{660}{12{,}000} = \tfrac{(12{,}000)r}{12{,}000}$. Simplify: $0.055 = r$. Change to percent form: $5.5\% = r$. Check: $660 \overset{?}{=} (3{,}000)(0.055)(4)$; $660 = 660\ \checkmark$. The rate of interest was $5.5\%$.

There may be times when you take a loan for a large purchase and the amount of the principal is not clear — for instance, when a dealer adds the cost of a warranty to the price of a car.

Example. Eduardo noticed that his new car loan papers stated that with an interest rate of $7.5\%$, he would pay $\text{\textdollar}6{,}596.25$ in interest over $5$ years. How much did he borrow to pay for his car?

We are asked to find the principal, $P$. Organize the given information: $I = 6{,}596.25$, $P = ?$, $r = 7.5\%$, $t = 5$ years. Write the formula: $I = Prt$. Substitute the given information: $6{,}596.25 = P(0.075)(5)$. Multiply: $6{,}596.25 = 0.375P$. Divide: $\tfrac{6{,}596.25}{0.375} = \tfrac{0.375P}{0.375}$. Simplify: $17{,}590 = P$. Check: $6{,}596.25 \overset{?}{=} (17{,}590)(0.075)(5)$; $6{,}596.25 = 6{,}596.25\ \checkmark$. The amount borrowed was $\text{\textdollar}17{,}590$.

In the simple interest formula, the rate of interest is given as an annual rate, the rate for one year. So the units of time must be in years. If the time is given in months, we convert it to years.

Example. Caroline got $\text{\textdollar}900$ as graduation gifts and invested it in a $10$-month certificate of deposit that earned $2.1\%$ interest. How much interest did this investment earn?

We are asked to find the interest, $I$. Organize the given information: $I = ?$, $P = \text{\textdollar}900$, $r = 2.1\%$, $t = 10$ months. Write the formula: $I = Prt$. Substitute the given information, converting $10$ months to $\tfrac{10}{12}$ of a year: $I = 900(0.021)\left(\tfrac{10}{12}\right)$. Multiply: $I = 15.75$. Check: if Caroline had invested the $\text{\textdollar}900$ for a full year at $2\%$ interest, the amount of interest would have been $\text{\textdollar}18$, so $\text{\textdollar}15.75$ for $10$ months is reasonable. The interest earned was $\text{\textdollar}15.75$.

Key terms

principal — the amount of money invested or borrowed, $P$. rate of interest — the percent of the principal charged or earned as interest, usually expressed as an annual rate, $r$. simple interest — interest computed on the principal only, using $I = Prt$, where $I$ is the interest, $P$ is the principal, $r$ is the rate, and $t$ is the time in years.

Practice

Use the simple interest formula

Solve simple interest applications

This section is adapted from Prealgebra 2e, Section 6.4: Solve Simple Interest Applications by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: omitted the Be Prepared quiz, Self Check checklist, and media links; converted the practice problems (“Try Its”) into interactive exercises with instant feedback; and adapted selected end-of-section exercises into the interactive Practice block, restating the missing-value tables as prose prompts.