Simple Interest - Math Steps, Examples & Questions

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Simple interest

Here you will learn about simple interest, including how to calculate simple interest for increasing and decreasing values, and set-up, solve and interpret growth and decay problems.

Students will first learn about simple interest as part of Ratios and Proportional Thinking in 7th grade.

What is simple interest?

Simple interest is calculated by finding a percent of the principal (original) amount and multiplying by the time period of the investment. The final value of an investment can then be found by adding/subtracting the simple interest to the principal amount.

\text { Simple interest }=\text { Principal amount } \times \text { rate of interest } \times \text { time period }

You can simply this by using the simple interest formulas.


To calculate the interest $(\, \!I)$	$I=Prt$
To calculate the total $(\, \!A)$ after an increase	$\begin{aligned} A & =P+P r t \\ & =P(1+r t) \end{aligned}$
To calculate the total $(\, \!A)$ after a decrease	$\begin{aligned} A & =P+P r t \\ & =P(1-r t) \end{aligned}$

Where:

$I =$ simple interest
$A =$ total amount,
$P =$ principal (original amount),
$r =$ rate of interest (written as a decimal),
t = time period (number of intervals)
- ex. Annual percentage rate (once per year): $t=1$
- ex. Monthly interest rate (once per month): $t=12$

For example, calculate the interest earned on $\$3,000$ with a simple interest rate of $5\%$ over $2$ years.

Using the formula $I=Prt:$

$P=3,000$
$r=0.05$ (remember to write $5\%$ as a decimal)
$t=2$

\begin{aligned} I & =3,000 \times 0.05 \times 2 \\\\ & =\$ 300 \end{aligned}

To find the final value of the investment you can now add the interest to the principal amount.

\begin{aligned} A & =3,000+300 \\\\ & =\$ 3,300 \end{aligned}

You could have calculated this directly using the formula $A=P\left( 1+rt \right)$

\begin{aligned} A & =3,000(1+0.05 \times 2) \\\\ & =\$ 3,300 \end{aligned}

What is simple interest?

Common Core Standards

How does this relate to 7th grade math?

7th Grade: Ratios and Proportional Relationships (7.RP.A.3)
Use proportional relationships to solve multistep ratio and percent problems.
Examples: simple interest, tax, markups and markdowns, gratuities and commissions, fees, percent increase and decrease, percent error.

How to calculate simple interest

In order to calculate simple interest:

Identify the value of each known variable in $\textbf{I = Prt, A = P(1 + rt)}$ or $\textbf{A = P(1 − rt)}.$
Substitute the values into the formula.
Solve the equation.

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Simple interest examples

Example 1: finding the simple interest

$\$2,100$ is invested for $3$ years at an annual percentage rate of $2\%$ per year simple interest. Find the interest earned on the investment in that time?

Identify the value of each known variable in $\textbf{I = Prt, A = P(1 + rt)}$ or $\textbf{A = P(1 − rt)}.$

To find the interest, you use $I=Prt.$

$P=2,100$
$r=0.02$
$t=3$

2Substitute the values into the formula.

Substituting these values into the simple interest formula $I=Prt,$ you get:

I=2,100\times 0.02\times 3

3Solve the equation.

I=\$ 126

$\$126$ was earned on the investment.

Example 2: finding the total amount after an increase

An investment of $\$1,500$ is made at a simple interest rate of $5\%$ per year for $4$ years. What is the value of the investment after this time?

Identify the value of each known variable in $\textbf{I = Prt, A = P(1 + rt)}$ or $\textbf{A = P(1 − rt)}.$

To find the total after the increase, you use $A=P(1+rt).$

$P=1,500$
$r=0.05$
$t=4$

Substitute the values into the formula.

Substituting these values into the simple interest formula $A=P(1+rt),$ you get:

$A=1,500(1+0.05\times{4})$

Solve the equation.

\begin{aligned} A&=1,500(1+0.20)\\\\ A&=1,500(1.20)\\\\ A&=1,500\times{1.2}\\\\ A&=\$1,800 \end{aligned}

Example 3: finding the total amount after a decrease

A car is bought for $\$10,000$ and loses $9\%$ of its value per year, simple interest. What is the value of the car after $8$ years?

Identify the value of each known variable in $\textbf{I = Prt, A = P(1 + rt)}$ or $\textbf{A = P(1 − rt)}.$

To find the total after the decrease, you use $A=P(1-rt).$

$P=10,000$
$r=0.09$
$t=8$

Substitute the values into the formula.

Substituting these values into the simple interest formula $A=P(1-rt),$ you get:

$A=10,000(1-0.09\times{8})$

Solve the equation.

\begin{aligned} A&=10,000(1-0.72)\\\\ A&=10,000(0.28)\\\\ A&=10,000\times{0.28}\\\\ A&=\$2,800 \end{aligned}

Example 4: borrowing money on different time scales

$\$7,600$ is borrowed for $2$ years on a credit card. The cost of borrowing is a $1\%$ interest payment per month simple interest for the life of the loan. What is the total cost to pay off after this time?

Identify the value of each known variable in $\textbf{I = Prt, A = P(1 + rt)}$ or $\textbf{A = P(1 − rt)}.$

To find the total after the increase, you use $A=P(1+rt).$

$P=7,600$
$r=0.01$
$t=2\times{12}=24$ (remember there are $12$ months in $1$ year)

Substitute the values into the formula.

Substituting these values into the simple interest formula $A=P(1+rt),$ you get:

$A=7,600(1+0.01\times{24})$

Solve the equation.

\begin{aligned} A&=7,600(1+0.24)\\\\ A&=7,600(1.24)\\\\ A&=7,600\times{1.24}\\\\ A&=\$9,424 \end{aligned}

Example 5: different percents

A house is currently valued at $\$175,000.$ For the first $3$ years, the value of the house increases by the rate of simple interest of $0.2\%$ per year.

For the following $4$ years, the value of the house decreases in value by a simple interest rate of $0.18\%$ per annum. Calculate the value of the house after these $7$ years.

Identify the value of each known variable in $\textbf{I = Prt, A = P(1 + rt)}$ or $\textbf{A = P(1 − rt)}.$

To find the total after the increase, you use $A=P\left(1+r_{1} t_{1}-r_{2} t_{2}\right).$

$P=175,000$
$r_1=0.002$ (the first interest rate)
$t_1=3$ (the first time period)
$r_2=0.0018$ (the second interest rate)
$t_2=4$ (the second time period)

Substitute the values into the formula.

Substituting these values into the simple interest formula $A=P\left(1+r_{1} t_{1}-r_{2} t_{2}\right),$ you get:

$A=175,000(1+0.002\times{3}-0.0018\times{4})$

Solve the equation.

\begin{aligned} A&=175,000(1+0.006-0.0072)\\\\ A&=175,000(0.9988)\\\\ A&=175,000\times{0.9988}\\\\ A&=\$174,790 \end{aligned}

Teaching tips for simple interest

Assign students a personal finance project that requires them to use what they are learning about simple interest in the real world.

Let students spend time using a simple interest calculator, trying different values for each part of the equation and writing or talking about what patterns they notice.

Although students will not learn the compound interest formula until algebra, spend time explaining the differences so that students understand what simple interest does.

Easy mistakes to make

Applying the incorrect formula to the question
Pay attention to whether the question is asking for the interest or the total after an increase or a loss. This will determine which of the following formulas you need to use: $I=Prt, \; A=P(1+rt)$ or $A=P(1-rt).$
Additionally, these formulas cannot be used to calculate compound interest or other problems that do not involve simple interest.

Inputting the incorrect time scale
Example 3: $\$7,600$ is invested for $2$ years at $1\%$ per month simple interest.
What is the value of the investment after this time?”
The interest rate is monthly, but the period of time on the investment is years.
It is important to convert the $2$ years into months ( $2 \times 12=24$ months) before multiplying by the principal balance. This will show you the correct amount for monthly payments.

Not converting the percent to a decimal
Using the whole number from the percent as the value for $r$ (therefore not converting the percent to a decimal). For example, when using simple interest to increase $\$100$ by $2\%$ for $5$ years, this calculation is incorrectly made:
$A=100(1+2\times{5})$
$A=100\times{11}$
$A=\$1,100$
Notice that the value increased from $\$100$ to $\$1,100.$ This is an extremely large increase, and clearly does not show $2\%.$

Simple interest practice questions

\$6,700\times1.012\times2

\$6,700\times1.2\times2

\$6,700\times1.012^2

\$6,700 \times 1.024

Since the investment is an increase, use the equation $I=P(1 +rt):$

\begin{aligned} P&=\$6,700 \\ r&=0.012 \\ t&=2 \end{aligned}

\begin{aligned} & \$6,700\times(1+0.012\times2) \\\\ & =\$6,700\times(1+0.024) \\\\ & =\$6,700\times1.024 \end{aligned}

\$485.06

\$551.20

\$485.06

\$529.15

The original price was $\$689.$ The sale price was $80\%$ of this:
$\$689\times0.80=\$551.20$

Since the value decreased, use the equation $I=P(1-rt):$

\begin{aligned} P&=\$551.20 \\ r&=0.04 \\ t&=3 \end{aligned}

\$551.20 \; (1-0.04\times3)=\$485.056

$\$485.056 \,$ rounds to $\, \$485.06$

\$7,459.47

\$1,694.42

\$1,409.66

\$8,751.66

Use the equation $I=Prt:$

\begin{aligned} P&=\$7,342 \\ r&=0.004 \\ t&=4 \text{ years} \times12=48 \text{ months} \end{aligned}

\$7,342 \; (0.004\times48)=\$1,409.664

$\$1,409.664 \,$ rounds to $\, \$1,409.66$

\$255,850

\$376,465

\$354,535

\$233,920

Since the value is decreasing, use the equation $I=P(1-rt):$

\begin{aligned} P&=\$365,500 \\ r&=0.0025 \\ t&=12 \end{aligned}

\$365,500 \; (1-0.0025\times12)=\$354,535

\$28,600

\$15,400

\$37,400

\$22,550

Since the interest is being added to the original loan amount, use the equation $I=P(1+rt):$

\begin{aligned} P&=\$22,000 \\ r&=0.005 \\ t&=5 \text{ years} \times12=60 \text{ months} \end{aligned}

\$22,000 \; (1+0.005\times60)=\$28,600

Simple interest FAQs

What is simple interest used for in the real world?

Simple interest is used to calculate growth or decay, in terms of money. For example, you can use it to calculate the interest charges based on a loan amount or it can be used to calculate the amount of interest you can earn if you invest your money.

The next lessons are

Compound measures
Converting fractions, decimals, and percentages
Exponents
Algebraic expressions
Ratio

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