How to calculate simple interest

How to Calculate Simple Interest

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Simple interest is interest that is calculated only on the original amount of money, called the principal. It does not add previous interest back into the principal. On the ACT, simple interest questions usually ask you to find the interest, the total amount, the rate, the time, or the starting principal.

The Simple Interest Formula

The formula is $I=Prt$, where $I$ is the interest, $P$ is the principal, $r$ is the annual interest rate written as a decimal, and $t$ is the time in years.

Change a percent to a decimal before substituting: $6\%=0.06$.
Change months to years when the rate is annual: $9$ months is $\frac{9}{12}=0.75$ years.
The total amount after interest is $A=P+I$, or $A=P(1+rt)$.

Example 1: Find the Interest

Find the simple interest on $\$800$ at $5\%$ per year for $3$ years.

Substitute into $I=Prt$: $I=800(0.05)(3)=120$. The interest is $\$120$.

Example 2: Find the Total Amount

If $\$2{,}000$ is borrowed at $4\%$ simple interest for $18$ months, first convert $18$ months to $1.5$ years. Then $I=2000(0.04)(1.5)=120$. The total amount is $2000+120=\$2{,}120$.

Simple Interest

Think of this lesson as more than a rule to memorize. Simple Interest is about comparisons, scaling, and equal ratios. A strong student does not rush to the first formula on the page; they pause, identify the structure of the problem, and then choose the tool that matches that structure. That pause is what prevents most mistakes.

A ratio compares quantities, and a proportion says two ratios are equal. Cross products help because $\frac{a}{b}=\frac{c}{d}$ implies $ad=bc$.

Here is the teacher way to approach the topic. First, read the problem slowly and underline the information that is actually given. Next, name the target: are you finding a value, simplifying an expression, comparing two quantities, solving for a variable, or interpreting a graph? Once the target is clear, the calculation becomes much less mysterious because every step has a job.

Read what is given and what is being asked.
Choose the rule that connects them.
Substitute carefully and simplify in small steps.
Check the final answer against the original question.

A helpful habit is to say what each number represents before using it. For example, if a number is a denominator, a radius, a slope, a common difference, or a coefficient, it should not be treated like an ordinary loose number. Its role tells you where it belongs in the formula. This is especially important on ACT-style questions because many wrong answer choices come from using the right numbers in the wrong places.

Another good habit is to keep the work organized vertically. Write one clean line for substitution, one line for simplifying, and one line for the final answer. If the problem has units, keep the units attached. If the problem has signs, exponents, or parentheses, copy them carefully from one line to the next. Most errors in this topic are not caused by a hard idea; they are caused by dropping a negative sign, combining unlike terms, using the wrong denominator, or skipping a check.

When you finish, ask a quick reasonableness question. Should the answer be positive or negative? Should it be larger or smaller than the original number? Does it fit the graph, table, shape, or equation? Can you plug it back into the original problem? This final check turns the lesson from a procedure into understanding.

On a test, the goal is not to write the longest solution. The goal is to write enough clear work that you can see the structure, avoid traps, and recover quickly if one line goes wrong. Practice the examples below with that mindset: identify the type of problem, choose the matching rule, show the substitution, simplify carefully, and check the answer before moving on.

Free printable Worksheets

Exercises for Simple Interest

1) Find the simple interest on $\$500$ at $6\%$ per year for $2$ years.

2) Find the simple interest on $\$1{,}200$ at $5\%$ per year for $3$ years.

3) Find the simple interest on $\$750$ at $8\%$ per year for $18$ months.

4) Find the simple interest on $\$2{,}400$ at $4.5\%$ per year for $5$ years.

5) Find the total amount owed after $4$ years if $\$3{,}000$ is borrowed at $7\%$ simple interest.

6) A savings account earns $3.2\%$ simple interest. How much interest does $\$1{,}500$ earn in $30$ months?

7) How much principal is needed to earn $\$180$ in simple interest at $6\%$ per year for $2$ years?

8) At what annual simple interest rate will $\$2{,}000$ earn $\$300$ in $5$ years?

9) How long will it take $\$900$ to earn $\$216$ at $8\%$ simple interest?

10) Find the simple interest on $\$4{,}800$ at $5.5\%$ per year for $9$ months.

11) A loan of $\$6{,}250$ earns $\$937.50$ simple interest in $3$ years. Find the annual rate.

12) Find the maturity value of $\$8{,}000$ invested at $4.25\%$ simple interest for $6$ years.

13) Which earns more interest: $\$2{,}500$ at $6\%$ for $4$ years, or $\$3{,}000$ at $5\%$ for $3$ years?

14) A $10$-month loan of $\$1{,}680$ has simple interest of $\$98$. Find the annual interest rate.

15) A student pays $\$1{,}092$ total after borrowing $\$1{,}000$ for $18$ months. Find the simple annual interest rate.

16) How much should be invested at $3.75\%$ simple interest for $4$ years to have $\$5{,}750$ at the end?

17) A bank charges $9\%$ annual simple interest. If the interest on a loan is $\$405$ for $15$ months, find the principal.

18) An investment grows from $\$7{,}200$ to $\$8{,}172$ in $3$ years using simple interest. Find the annual rate.

19) A $\$12{,}500$ loan at $6.8\%$ simple interest is repaid after $2$ years and $4$ months. Find the total amount repaid.

20) Two investments total $\$10{,}000$. One earns $4\%$ simple interest and the other earns $7\%$. After one year, the total interest is $\$580$. How much was invested at $7\%$?

Answers

1) Use $I=Prt$. Here $P=500$, $r=0.06$, and $t=2$. So $I=500(0.06)(2)=60$. The simple interest is $\$60$.

2) Use $I=Prt$: $I=1200(0.05)(3)=180$. The simple interest is $\$180$.

3) Convert $18$ months to $1.5$ years. Then $I=750(0.08)(1.5)=90$. The interest is $\$90$.

4) Write $4.5\%=0.045$. Then $I=2400(0.045)(5)=540$. The simple interest is $\$540$.

5) First find interest: $I=3000(0.07)(4)=840$. Add principal: $A=P+I=3000+840=3840$. The amount owed is $\$3{,}840$.

6) Convert $30$ months to $2.5$ years. Then $I=1500(0.032)(2.5)=120$. The account earns $\$120$.

7) Solve $I=Prt$ for $P$: $P=\frac{I}{rt}=\frac{180}{0.06\cdot2}=1500$. The principal is $\$1{,}500$.

8) Solve for $r$: $r=\frac{I}{Pt}=\frac{300}{2000\cdot5}=0.03$. The rate is $3\%$ per year.

9) Solve for $t$: $t=\frac{I}{Pr}=\frac{216}{900\cdot0.08}=3$. It takes $3$ years.

10) Convert $9$ months to $\frac{9}{12}=0.75$ years. Then $I=4800(0.055)(0.75)=198$. The interest is $\$198$.

11) Use $r=\frac{I}{Pt}$. Then $r=\frac{937.50}{6250\cdot3}=0.05$. The annual rate is $5\%$.

12) Interest is $I=8000(0.0425)(6)=2040$. Maturity value is $A=8000+2040=10040$. The value is $\$10{,}040$.

13) First: $2500(0.06)(4)=600$. Second: $3000(0.05)(3)=450$. The first investment earns more by $\$150$.

14) Convert $10$ months to $\frac{5}{6}$ year. Then $r=\frac{98}{1680\cdot(5/6)}=0.07$. The annual rate is $7\%$.

15) Interest is $1092-1000=92$. Time is $18$ months, or $1.5$ years. Then $r=\frac{92}{1000\cdot1.5}=0.061333\ldots$, so the rate is about $6.13\%$.

16) The amount formula is $A=P(1+rt)$. Here $5750=P(1+0.0375\cdot4)=1.15P$. Thus $P=\frac{5750}{1.15}=5000$. Invest $\$5{,}000$.

17) Time is $15$ months, or $1.25$ years. Solve $P=\frac{I}{rt}=\frac{405}{0.09\cdot1.25}=3600$. The principal is $\$3{,}600$.

18) Interest is $8172-7200=972$. Then $r=\frac{972}{7200\cdot3}=0.045$. The annual rate is $4.5\%$.

19) Time is $2+\frac{4}{12}=\frac{7}{3}$ years. Interest is $12500(0.068)(\frac{7}{3})=1983.33\ldots$. Total repaid is $12500+1983.33\ldots=\$14{,}483.33$ to the nearest cent.

20) Let $x$ be the amount at $7\%$. Then $10000-x$ is at $4\%$. Equation: $0.07x+0.04(10000-x)=580$. So $0.03x+400=580$, $0.03x=180$, and $x=6000$. The amount at $7\%$ is $\$6{,}000$.

How to calculate simple interest