Lesson 4: Solve Simple Interest Applications

New Concept

Master the simple interest formula, $I=Prt$, to calculate money earned on investments or paid on loans. You'll learn how to solve for any missing variable—interest, principal, rate, or time—to handle various financial situations.

What’s next

Next, you'll work through interactive examples and a series of practice cards, applying the formula to calculate interest, principal, and rates in different scenarios.

Property

If an amount of money, $P$, the principal, is invested for a period of $t$ years at an annual interest rate $r$, the amount of interest, $I$, earned is

where

• $I$ = interest
• $P$ = principal
• $r$ = rate
• $t$ = time

Interest earned according to this formula is called simple interest.

Examples

• To find the interest earned on 1,200 dollars at a 3% rate for 5 years, calculate $I = (1200)(0.03)(5)$, which equals 180 dollars.

Property

The simple interest formula, $I = Prt$, can be rearranged to solve for the principal, $P$. Substitute the given values for interest ($I$), rate ($r$), and time ($t$) to find the unknown principal.

Examples

• If 240 dollars in interest was earned in 4 years at a 3% rate, the principal was $P = \frac{240}{(0.03)(4)} = 2000$ dollars.

• Jessica paid 4,500 dollars in interest over 5 years on a student loan with a 6% rate. She borrowed $P = \frac{4500}{(0.06)(5)} = 15000$ dollars.

Property

The simple interest formula, $I = Prt$, can be rearranged to solve for the rate, $r$. Substitute the given values for interest ($I$), principal ($P$), and time ($t$) to find the unknown rate.

Examples

• An investment of 10,000 dollars earned 1,500 dollars in interest in 3 years. The rate was $r = \frac{1500}{(10000)(3)} = 0.05$, or 5%.

• A loan of 4,000 dollars accrued 800 dollars in interest over 5 years. The interest rate was $r = \frac{800}{(4000)(5)} = 0.04$, or 4%.

Property

In the simple interest formula, the rate of interest $r$ is an annual rate, so the time $t$ must be in years. If the time is given in months, convert it to years by dividing the number of months by 12.

Examples

• An investment of 1,800 dollars for 6 months at 3.5% interest earns $I = 1800(0.035)(\frac{6}{12}) = 31.5$ dollars. Notice we used $t = \frac{6}{12}$ years.

• To find the interest on a 4,000 dollar loan for 20 months at 5.4%, use $t = \frac{20}{12}$ years. The interest is $I = 4000(0.054)(\frac{20}{12}) = 360$ dollars.