6.4 Solve Simple Interest Applications

Learning objectives

Learning objectives

• Use the simple interest formula

• 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 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 6.33

Solution

1. Identify the simple-interest variables. The principal is $500$, the annual rate is $6\%=0.06$, and the time is $3$ years.
2. Write the simple-interest formula. The formula is $I=Prt$, where $I$ is the interest earned over the given time.
3. Substitute the known values. Replacing the symbols gives $I=(500)(0.06)(3)$, so multiplication will give the interest earned.
4. Evaluate the product. Multiplying $500(0.06)$ gives $30$, and $30(3)$ gives $I=90$.
5. Check the reasonableness. Over three years the account earns about $18\%$ of the principal, and $18\%$ of $500$ is $90$, so the result is reasonable.

Takeaway: Using $I=Prt$ in Example 6.33 gives $500(0.06)(3)=90$ dollars of simple interest over three years.

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

Example 6.34

Solution

1. List the known quantities. The interest is $178$, the annual rate is $4\%=0.04$, and the time is $2$ years; the principal is unknown.
2. Write the simple-interest formula. The relationship $I=Prt$ connects the interest, principal, decimal rate, and time.
3. Substitute and isolate the principal. The equation is $178=P(0.04)(2)=0.08P$.
4. Divide by the combined factor. Dividing both sides by $0.08$ gives $P=\frac{178}{0.08}=2,225$.
5. Check by substitution. The amount $2,225(0.04)(2)$ equals $178$, confirming that the principal is $2,225$ dollars.

Takeaway: Solving $178=P(0.04)(2)$ in Example 6.34 gives a principal of $2,225$ dollars, verified by recomputing the interest.

Now we will solve for the rate of interest.

Example 6.35

Solution

1. Identify the known values. The interest is $3,772$, the principal is $8,200$, and the time is $4$ years; the rate is unknown.
2. Substitute into the formula. Using $I=Prt$ gives $3,772=8,200r(4)$.
3. Combine the factors. Multiplying $8,200$ by $4$ gives $32,800$, so $3,772=32,800r$.
4. Solve for the decimal rate. Dividing by $32,800$ gives $r=0.115$, and converting to a percent gives $11.5\%$.
5. Check the result. Substituting gives $8,200(0.115)(4)=3,772$, so the computed rate produces the stated interest.

Takeaway: Dividing the $3,772$ interest by $8,200(4)$ in Example 6.35 yields the decimal rate $0.115$, or $11.5\%$.

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 6.36

Solution

1. Convert the rate to a decimal. The annual rate $4\%$ becomes $0.04$, while the principal is $12,500$ dollars and the time is $5$ years.
2. Write the simple-interest formula. The amount earned is $I=Prt$.
3. Substitute and evaluate. The equation $I=(12,500)(0.04)(5)$ gives $30,000(0.04)=1,200$, and then $1,200(5)=2,500$.
4. Check the result. Five years at $4\%$ per year earns $20\%$ of the principal, and $20\%$ of $12,500$ is $2,500$ dollars.

Takeaway: The $12,500$ deposit at $4\%$ for five years earns $12,500(0.04)(5)=2,500$ dollars in Example 6.36.

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. In the next example, we'll show how to solve for the rate.

Example 6.37

Solution

1. Separate principal from interest. Loren lent $3,000$, and the separate interest payment was $660$ after $4$ years.
2. Substitute into the formula. The equation $I=Prt$ becomes $660=(3,000)r(4)$.
3. Combine and solve. Multiplying gives $660=12,000r$, so dividing by $12,000$ gives $r=0.055$.
4. Convert the rate to a percent. Moving the decimal two places right gives $5.5\%$.
5. Check the interest. Substituting gives $(3,000)(0.055)(4)=660$, confirming the rate that was calculated.

Takeaway: The $660$ interest on the $3,000$ loan for four years in Example 6.37 corresponds to the annual rate $5.5\%$.

There may be times when you take a loan for a large purchase and the amount of the principal is not clear. This might happen, for instance, in making a car purchase when the dealer adds the cost of a warranty to the price of the car. In the next example, we will solve a simple interest application for the principal.

Example 6.38

Solution

1. Identify the known quantities. The total interest is $6,596.25$, the annual rate is $7.5\%=0.075$, and the time is $5$ years; the principal is unknown.
2. Substitute into the formula. The simple-interest equation is $6,596.25=P(0.075)(5)$.
3. Combine the rate and time factors. Multiplying $0.075$ by $5$ gives $0.375$, so $6,596.25=0.375P$.
4. Solve for the principal. Dividing both sides by $0.375$ gives $P=17,590$.
5. Check by substitution. The amount $17,590(0.075)(5)$ equals $6,596.25$, so the borrowed principal is $17,590$ dollars.

Takeaway: Dividing $6,596.25$ by $0.075(5)$ in Example 6.38 gives the car-loan principal of $17,590$ dollars.

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 6.39

Solution

1. Convert the time to years. Since simple interest uses an annual rate, $10$ months equals $\frac{10}{12}$ year.
2. Convert the rate to a decimal. The $2.1\%$ annual rate becomes $0.021$, and the principal is $900$ dollars.
3. Substitute into the formula. Using $I=Prt$ gives $I=900(0.021)\left(\frac{10}{12}\right)$.
4. Evaluate the product. The calculation is $900(0.021)=18.9$, and $18.9\left(\frac{10}{12}\right)=15.75$.
5. Check the reasonableness. The interest is less than the $900$ principal and is close to one year's interest, as expected for ten months.

Takeaway: Converting ten months to $\frac{10}{12}$ year gives Example 6.39 an interest amount of $900(0.021)(10/12)=15.75$ dollars.

Section 6.4 Exercises

Section 6.4 Exercises

Use the Simple Interest Formula

In the following exercises, use the simple interest formula to fill in the missing information.

In the following exercises, solve the problem using the simple interest formula.

Solve Simple Interest Applications

In the following exercises, solve the problem using the simple interest formula.

Everyday Math

Writing Exercises

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ On a scale of 1–10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

1. Find the simple interest earned after $4$ years on $\text{\textdollar 800}$ at an interest rate of $\mathrm{5\%.}$

2. Find the simple interest earned after $2$ years on $\text{\textdollar 700}$ at an interest rate of $\mathrm{4\%.}$

3. Find the principal invested if $\text{\textdollar 495}$ interest was earned in $3$ years at an interest rate of $\mathrm{6\%.}$

1. Find the principal invested if $\text{\textdollar 1,246}$ interest was earned in $5$ years at an interest rate of $\mathrm{7\%}.$

2. Find the rate if a principal of $\text{\textdollar 5,000}$ earned $\text{\textdollar 1,350}$ interest in $6$ years.

3. Find the rate if a principal of $\text{\textdollar 9,000}$ earned $\text{\textdollar 1,755}$ interest in $3$ years.

1. Areli invested a principal of $\text{\textdollar 950}$ in her bank account with interest rate $\mathrm{3\%.}$ How much interest did she earn in $5$ years?

2. Susana invested a principal of $\text{\textdollar 36,000}$ in her bank account with interest rate $\mathrm{6.5\%}.$ How much interest did she earn in $3$ years?

3. Jim lent his sister $\text{\textdollar 5,000}$ to help her buy a house. In $3$ years, she paid him the $\text{\textdollar 5,000},$ plus $\text{\textdollar 900}$ interest. What was the rate of interest?

1. Hang borrowed $\text{\textdollar 7,500}$ from her parents to pay her tuition. In $5$ years, she paid them $\text{\textdollar 1,500}$ interest in addition to the $\text{\textdollar 7,500}$ she borrowed. What was the rate of interest?

2. Sean's new car loan statement said he would pay $\text{\textdollar 4,866.25}$ in interest from an interest rate of $\mathrm{8.5\%}$ over $5$ years. How much did he borrow to buy his new car?

3. In $5$ years, Gloria's bank account earned $\text{\textdollar 2,400}$ interest at $\mathrm{5\%.}$ How much had she deposited in the account?

1. Adriana invested $\text{\textdollar 4,500}$ for $8$ months in an account that paid $\mathrm{1.9\%}$ interest. How much interest did she earn?

2. Milton invested $\text{\textdollar 2,460}$ for $20$ months in an account that paid $\mathrm{3.5\%}$ interest How much interest did he earn?

3. | Interest | Principal | Rate | Time (years) |

| --- | --- | --- | --- |

| | $\text{\textdollar 1200}$ | $\mathrm{3\%}$ | $5$ |

Table

6.5

1. | Interest | Principal | Rate | Time (years) |

| --- | --- | --- | --- |

| $\$4410$ | | $\mathrm{4.5\%}$ | $7$ |

Table

6.7

2. | Interest | Principal | Rate | Time (years) |

| --- | --- | --- | --- |

| $\$577.08$ | $\text{\textdollar 4580}$ | | $2$ |

Table

6.9

3. Find the simple interest earned after $5$ years on $\text{\textdollar 600}$ at an interest rate of $\mathrm{3\%.}$

1. Find the simple interest earned after $2$ years on $\text{\textdollar 8,950}$ at an interest rate of $\mathrm{3.24\%}.$

2. Find the simple interest earned after $8$ years on $\text{\textdollar 15,500}$ at an interest rate of $\mathrm{11.425\%}.$

3. Find the principal invested if $\text{\textdollar 656}$ interest was earned in $5$ years at an interest rate of $\mathrm{4\%}.$

1. Find the principal invested if $\text{\textdollar 70.95}$ interest was earned in $3$ years at an interest rate of $\mathrm{2.75\%.}$

2. Find the principal invested if $\text{\textdollar 15,222.57}$ interest was earned in $6$ years at an interest rate of $\mathrm{10.28\%}.$

3. Find the rate if a principal of $\text{\textdollar 5,400}$ earned $\text{\textdollar 432}$ interest in $2$ years.

1. Find the rate if a principal of $\text{\textdollar 11,000}$ earned $\text{\textdollar 1,815}$ interest in $3$ years.

2. Casey deposited $\text{\textdollar 1,450}$ in a bank account with interest rate $\mathrm{4\%.}$ How much interest was earned in $2$ years?

3. Robin deposited $\text{\textdollar 31,000}$ in a bank account with interest rate $\mathrm{5.2\%}.$ How much interest was earned in $3$ years?

1. Hilaria borrowed $\text{\textdollar 8,000}$ from her grandfather to pay for college. Five years later, she paid him back the $\text{\textdollar 8,000},$ plus $\text{\textdollar 1,200}$ interest. What was the rate of interest?

2. Lebron lent his daughter $\text{\textdollar 20,000}$ to help her buy a condominium. When she sold the condominium four years later, she paid him the $\text{\textdollar 20,000},$ plus $\text{\textdollar 3,000}$ interest. What was the rate of interest?

3. In $10$ years, a bank account that paid $\mathrm{5.25\%}$ earned $\text{\textdollar 18,375}$ interest. What was the principal of the account?

1. Joshua's computer loan statement said he would pay $\text{\textdollar 1,244.34}$ in interest for a $3$ year loan at $\mathrm{12.4\%}.$ How much did Joshua borrow to buy the computer?

2. Caitlin invested $\text{\textdollar 8,200}$ in an $\mathrm{18-month}$ certificate of deposit paying $\mathrm{2.7\%}$ interest. How much interest did she earn form this investment?

3. Airin borrowed $\text{\textdollar 3,900}$ from her parents for the down payment on a car and promised to pay them back in $15$ months at a $\mathrm{4\%}$ rate of interest. How much interest did she owe her parents?

1. Interest on savings Find the interest rate your local bank pays on savings accounts.

1. ⓐ What is the interest rate?

2. ⓑ Calculate the amount of interest you would earn on a principal of $\text{\textdollar 8,000}$ for $5$ years.

2. Why do banks pay interest on money deposited in savings accounts?