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5. Exponential Functions
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United States of AmericaFL
Grade 912

5.07 Simple interest

Lesson
Practice
Lesson

Concept summary

Simple interest is a method for computing interest where the interest is computed from the original principal alone, no matter how much money has accrued so far.

\displaystyle A=P\left(1+rt\right)
\bm{A}
future value, or final amount
\bm{P}
principal, or present value
\bm{t}
number of years
\bm{r}
rate of interest per year

If we want to determine the total amount of interest, we can use the formula:

\displaystyle I=Prt
\bm{I}
Interest
\bm{P}
principal, or present value
\bm{r}
rate of interest per year
\bm{t}
number of years

Because P and r are constant and t, the time, is variable, this is a linear function. Therefore, the growth of a sum of money by simple interest is linear growth. We can see this in the fact that simple interest causes an account balance to grow by a constant amount per interval of time.

Worked examples

Example 1

Taylor's investment of \$ 5000 earns interest at a simple interest rate of 2.5\% per year for 5 years.

a

Use the simple interest formula to find the final balance of their investment.

Approach

To apply the simple interest formula we need to identify which variables we have given information for. In this example we know that the principal is P=5000, the rate is r=0.025 and the time is t=5 years.

Solution

Substituting all given values into the formula gives us:

\displaystyle A\displaystyle =\displaystyle P\left(1+rt\right)Simple interest formula
\displaystyle A\displaystyle =\displaystyle 5000\left(1+0.025\left(5\right)\right)Substitute in values for the variables
\displaystyle A\displaystyle =\displaystyle 5000\left(1.125\right)Simplify
\displaystyle A\displaystyle =\displaystyle 5625Evaluate the product

Taylor will have \$5625 after 5 years.

Reflection

When solving using the simple interest formulas, the rate must be in decimal form. To find the decimal form of a percentage rate, divide the percentage value by 100.

In this example, we found the rate was 0.025 because \dfrac{2.5}{100}=0.025.

b

Taylor wanted to have \$6000 at the end of their investment term. Find the initial amount they would need to invest at the same rate for the same length of time to have an ending value of \$6000.

Approach

In this example we know the ending balance is A=6000, the rate is still r=0.025, the time is still t=5, and the unknown value is the principal P.

Solution

Substituting all given values into the formula gives us:

\displaystyle A\displaystyle =\displaystyle P\left(1+rt\right)Simple interest formula
\displaystyle 6000\displaystyle =\displaystyle P\left(1+0.025\left(5\right)\right)Substitute in values for the variables
\displaystyle 6000\displaystyle =\displaystyle 1.125PSimplify
\displaystyle 5333.33\displaystyle =\displaystyle PDivide

Taylor would need to invest \$533.33 to have an ending balance of \$6000 after investing their money for 5 years at a simple interest rate of 2.5\%.

Example 2

Rei borrows \$1200 at a simple interest rate of 8.2\%, to help buy a new car. She doesn't want to pay more than \$300 in interest over the duration of the loan. Calculate the maximum number of years Rei can take to repay the loan while keeping her total interest at \$300 or less.

Approach

We can use either formula to help solve this, but because we know the total maximum interest Rei wants to pay, we will use the formula I=Prt. In this example we know the maximum total interest is I=300, the rate is 0.082 and the unknown value is the number of years t.

Since we don't want the interest to exceed \$300, we will setup an inequality.

Solution

Substituting all given values into the formula gives us:

\displaystyle I\displaystyle \geq\displaystyle PrtSimple interest formula
\displaystyle 300\displaystyle \geq\displaystyle 1200\left(0.082\right)tSubstitute in values for the variables
\displaystyle 300\displaystyle \geq\displaystyle 98.4tSimplify
\displaystyle 3.048\displaystyle \geq\displaystyle tDivide

Rei would need to repay the loan within approximately 3 years in order to pay no more than \$300 in interest.

Outcomes

MA.912.AR.1.1

Identify and interpret parts of an equation or expression that represent a quantity in terms of a mathematical or real-world context, including viewing one or more of its parts as a single entity.

MA.912.AR.1.2

Rearrange equations or formulas to isolate a quantity of interest.

MA.912.AR.2.5

Solve and graph mathematical and real-world problems that are modeled with linear functions. Interpret key features and determine constraints in terms of the context.

MA.912.FL.3.2

Solve real-world problems involving simple, compound and continuously compounded interest.

MA.912.FL.3.4

Explain the relationship between simple interest and linear growth. Explain the relationship between compound interest and exponential growth and the relationship between continuously compounded interest and exponential growth.

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