# 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.

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}

## 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:

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

## 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? 1. 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?

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

• P=1,500
• r=0.05
• t=4

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

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

\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?

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

• P=10,000
• r=0.09
• t=8

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

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

\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?

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)

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

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

\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.

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)

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})

\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

1. Freya invests \$6,700 for 2 years. The simple interest rate is 1.2\% per year. Which calculation below works out the total value after 2 years? \$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} 2. A technology store has a back to school offer: save 20\% on all full price laptops. Paula buys a laptop that was \689 full price. After 3 years, the value of the purchased laptop has decreased by 4\% per year, simple interest. What is the value of the laptop after these 3 years?

\$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

3. \$7,342 is invested in a savings account with a 0.4\% simple interest rate per month. What is the total interest earned after 4 years? \$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

4. A boat is valued at \$365,500. The value of the boat decreases by an average of 0.25\% per year, simple interest. How much is the boat worth after 12 years? \$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

5. To buy a new car, Jeff gets an auto loan of \$22,000 that he will pay off over five years. On the loan, the lender charges a 0.5\% simple interest rate per month. How much will Jeff pay for the loan in total? \$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

• Percent increase and decrease
• Percent error

## Still stuck?

At Third Space Learning, we specialize in helping teachers and school leaders to provide personalized math support for more of their students through high-quality, online one-on-one math tutoring delivered by subject experts.

Each week, our tutors support thousands of students who are at risk of not meeting their grade-level expectations, and help accelerate their progress and boost their confidence.