- + ACCUPLACER Mathematics
- + ACT Mathematics
- + AFOQT Mathematics
- + ALEKS Tests
- + ASVAB Mathematics
- + ATI TEAS Math Tests
- + Common Core Math
- + DAT Math Tests
- + FSA Tests
- + FTCE Math
- + GED Mathematics
- + Georgia Milestones Assessment
- + GRE Quantitative Reasoning
- + HiSET Math Exam
- + HSPT Math
- + ISEE Mathematics
- + PARCC Tests
- + Praxis Math
- + PSAT Math Tests
- + PSSA Tests
- + SAT Math Tests
- + SBAC Tests
- + SIFT Math
- + SSAT Math Tests
- + STAAR Tests
- + TABE Tests
- + TASC Math
- + TSI Mathematics
- + ACT Math Worksheets
- + Accuplacer Math Worksheets
- + AFOQT Math Worksheets
- + ALEKS Math Worksheets
- + ASVAB Math Worksheets
- + ATI TEAS 6 Math Worksheets
- + FTCE General Math Worksheets
- + GED Math Worksheets
- + 3rd Grade Mathematics Worksheets
- + 4th Grade Mathematics Worksheets
- + 5th Grade Mathematics Worksheets
- + 6th Grade Math Worksheets
- + 7th Grade Mathematics Worksheets
- + 8th Grade Mathematics Worksheets
- + 9th Grade Math Worksheets
- + HiSET Math Worksheets
- + HSPT Math Worksheets
- + ISEE Middle-Level Math Worksheets
- + PERT Math Worksheets
- + Praxis Math Worksheets
- + PSAT Math Worksheets
- + SAT Math Worksheets
- + SIFT Math Worksheets
- + SSAT Middle Level Math Worksheets
- + 7th Grade STAAR Math Worksheets
- + 8th Grade STAAR Math Worksheets
- + THEA Math Worksheets
- + TABE Math Worksheets
- + TASC Math Worksheets
- + TSI Math Worksheets
- + AFOQT Math Course
- + ALEKS Math Course
- + ASVAB Math Course
- + ATI TEAS 6 Math Course
- + CHSPE Math Course
- + FTCE General Knowledge Course
- + GED Math Course
- + HiSET Math Course
- + HSPT Math Course
- + ISEE Upper Level Math Course
- + SHSAT Math Course
- + SSAT Upper-Level Math Course
- + PERT Math Course
- + Praxis Core Math Course
- + SIFT Math Course
- + 8th Grade STAAR Math Course
- + TABE Math Course
- + TASC Math Course
- + TSI Math Course
- + Number Properties Puzzles
- + Algebra Puzzles
- + Geometry Puzzles
- + Intelligent Math Puzzles
- + Ratio, Proportion & Percentages Puzzles
- + Other Math Puzzles
![](http://farmaciacoslada.online/777/templates/cheerup1/res/banner1.gif)
How to Solve Simple Interest Problems? (+FREE Worksheet!)
Learn how to solve simple interest problems using simple interest formulas.
![problem solving about simple interest How to Solve Simple Interest Problems? (+FREE Worksheet!)](https://www.effortlessmath.com/wp-content/uploads/2019/11/Simple-Interest-Cover-512x240.jpg)
Related Topics
- How to Find Percent of Increase and Decrease
- How to Solve Percent Problems
- How to Do Percentage Calculations
- How to Find Discount, Tax, and Tip
Step-by-step guide to solve simple interest
- Simple Interest: The charge for borrowing money or the return for lending it.
- To solve a simple interest problem, use this formula: Interest \(=\) principal \(\times\) rate \(\times\) time \( ⇒ \color{blue}{I=p \ × \ r \ × \ t}\)
Simple Interest – Example 1:
Find simple interest for \( $450\) investment at \(7\%\) for \(8\) years.
Use Interest formula: \(\color{ blue }{I=prt }\) \(P=$450\), \(r=7\%=\frac{7}{100}=0.07\) and \(t=8\) Then: \(I=450 \ × \ 0.07 \ × \ 8=$252\)
The Absolute Best Books to Ace Pre-Algebra to Algebra II
The Ultimate Algebra Bundle From Pre-Algebra to Algebra II
Simple interest – example 2:.
Find simple interest for \( $5,200\) investment at \(4\%\) for \(3\) years.
Use simple interest formula: \(\color{ blue }{I=prt }\) \(P=$5,200\), \(r=4\%=\frac{4}{100}=0.04\) and \(t=3\) Then: \(I=5,200 \ × \ 0.04 \ × \ 3=$624\)
Simple Interest – Example 3:
Find simple interest for \($5,000\) investment at \(3\%\) for \(4\) years.
Use simple interest formula: \(\color{ blue }{I=prt }\) \( P=$5,000, r=3\%=\frac{3}{100}=0.03\) and \(t=4\) Then: \(I= 5,000 ×0.03×4=$600\)
The Absolute Best Book to Ace the Algebra Test
Algebra I for Beginners The Ultimate Step by Step Guide to Acing Algebra I
Simple interest – example 4:.
Find simple interest for \($900\) at \(3.5\%\) for \(5\) years.
Use simple interest formula: \(\color{ blue }{I=prt }\) \( P=$900, r=3.5\%=\frac{3.5}{100}=0.035\) and \(t=5\) Then: \(I=900×0.035×5=$157.50\)
Exercises for Solving Simple Interest
Use simple interest to find the ending balance.
- \($1,300\) at \(5\%\) for \(6\) years.
- \($5,400\) at \(7.5\%\) for \(6\) months.
- \($25,600\) at \(9.2\%\) for \(5\) years.
- \($240\) interest is earned on a principal of \($1500\) at a simple interest rate of \(4\%\) interest per year. For how many years was the principal invested?
- A new car, valued at \($28,000\), depreciates at \(9\%\) per year from original price. Find the value of the car \(3\) years after purchase.
- Sara puts \($2,000\) into an investment yielding \(5\%\) annual simple interest; she left the money in for five years. How much interest does Sara get at the end of those five years?
Download Simple Interest Worksheet
- \(\color{blue}{$1,690.00}\)
- \(\color{blue}{$5,602.50}\)
- \(\color{blue}{$37,376.00}\)
- \(\color{blue}{4 \ years}\)
- \(\color{blue}{$20,440.00}\)
- \(\color{blue}{$500.00}\)
The Greatest Books for Students to Ace the Algebra
Pre-Algebra Exercise Book A Comprehensive Workbook + PreAlgebra Practice Tests
Pre-algebra in 10 days the most effective pre-algebra crash course, college algebra practice workbook the most comprehensive review of college algebra, high school algebra i a comprehensive review and step-by-step guide to mastering high school algebra 1, 10 full length clep college algebra practice tests the practice you need to ace the clep college algebra test.
by: Effortless Math Team about 5 years ago (category: Articles , Free Math Worksheets )
Effortless Math Team
Related to this article, more math articles.
- Full-Length ATI TEAS 7 Math Practice Test-Answers and Explanations
- What does TSI Test Stand for?
- The Ultimate SHSAT Math Course (+FREE Worksheets & Tests)
- A Comprehensive Collection of Free CHSPE Math Practice Tests
- Estimating and Rounding
- Top Proven Strategies To Increase Your SAT Math Score
- How to Determine Limits Involving Floor and Absolute Value Functions
- 10 Most Common 3rd Grade SBAC Math Questions
- Sum and Difference of Trigonometric Functions Formulas
- How Is the HiSET Test Scored?
What people say about "How to Solve Simple Interest Problems? (+FREE Worksheet!) - Effortless Math: We Help Students Learn to LOVE Mathematics"?
No one replied yet.
Leave a Reply Cancel reply
You must be logged in to post a comment.
Pre-Algebra Practice Workbook The Most Comprehensive Review of Pre-Algebra
Algebra i practice workbook the most comprehensive review of algebra 1, algebra ii practice workbook the most comprehensive review of algebra 2, algebra ii for beginners the ultimate step by step guide to acing algebra ii, pre-algebra for beginners the ultimate step by step guide to preparing for the pre-algebra test, pre-algebra tutor everything you need to help achieve an excellent score.
- ATI TEAS 6 Math
- ISEE Upper Level Math
- SSAT Upper-Level Math
- Praxis Core Math
- 8th Grade STAAR Math
Limited time only!
Save Over 45 %
It was $89.99 now it is $49.99
Login and use all of our services.
Effortless Math services are waiting for you. login faster!
Register Fast!
Password will be generated automatically and sent to your email.
After registration you can change your password if you want.
- Math Worksheets
- Math Courses
- Math Topics
- Math Puzzles
- Math eBooks
- GED Math Books
- HiSET Math Books
- ACT Math Books
- ISEE Math Books
- ACCUPLACER Books
- Premium Membership
- Youtube Videos
Effortless Math provides unofficial test prep products for a variety of tests and exams. All trademarks are property of their respective trademark owners.
- Bulk Orders
- Refund Policy
MathBootCamps
Simple interest formula and examples.
Simple interest is when the interest on a loan or investment is calculated only on the amount initially invested or loaned. This is different from compound interest, where interest is calculated on on the initial amount and on any interest earned. As you will see in the examples below, the simple interest formula can be used to calculate the interest earned, the total amount, and other values depending on the problem. [adsenseWide]
Examples of finding the interest earned with the simple interest formula
In many simple interest problems, you will be finding the total interest earned over a set period, which is represented as \(I\). The formula for this is:
![problem solving about simple interest graphic showing the simple interest formula for interest earned](https://i0.wp.com/www.mathbootcamps.com/wp-content/uploads/Simple-Interest-formula-interest-earned-2.png?resize=560%2C315)
Let’s use an example to see how this formula works. Remember that in the formula, the principal \(P\) is the initial amount invested.
A 2-year loan of $500 is made with 4% simple interest. Find the interest earned.
Always take a moment to identify the values given in the problem. Here we are given:
- Time is 2 years: \(t = 2\)
- Initial amount is $500: \(P = 500\)
- The rate is 4%. Write this as a decimal: \(r = 0.04\)
Now apply the formula:
\(\begin{align}I &= Prt \\ &= 500(0.04)(2) \\ &= \bbox[border: 1px solid black; padding: 2px]{40}\end{align}\)
Answer : The interest earned is $40.
In this example, the time given was in years, just as in the formula. But what if you are only given a number of months? Let’s use another example to see how this might be different.
A total of $1,200 is invested at a simple interest rate of 6% for 4 months. How much interest is earned on this investment?
Before we can apply the formula, we will need to write the time of 4 months in terms of years. Since there are 12 months in a year:
\(\begin{align}t &= \dfrac{4}{12} \\ &= \dfrac{1}{3}\end{align}\)
With this adjusted to years, we can now apply the formula with \(P = 1200\) and \(r = 0.06\).
\(\begin{align}I &= Prt \\ &= 1200(0.06)\left(\dfrac{1}{3}\right) \\ &= \bbox[border: 1px solid black; padding: 2px]{24}\end{align}\)
Answer : The interest earned is $24.
If you hadn’t converted here, you would have found the interest for 4 years, which would be much higher. So, always make sure to check that the time is in years before applying the formula.
Important! The time must be in years to apply the simple interest formula. If you are given months, use a fraction to represent it as years.
Another type of problem you might run into when working with simple interest is finding the total amount owed or the total value of an investment after a given amount of time. This is known as the future value, and can be calculated in a couple of different ways.
Finding the future value for simple interest
One way to calculate the future value would be to just find the interest and then add it to the principal. The quicker method however, is to use the following formula.
![problem solving about simple interest future value of simple interest formula](https://i0.wp.com/www.mathbootcamps.com/wp-content/uploads/future-value-simple-interest-formula.png?resize=560%2C315)
You know to use this formula when you are asked questions like “what is the total amount to be repaid” or “what is the value of the investment” -anything that seems to refer to the overall total after interest is considered.
A business takes out a simple interest loan of $10,000 at a rate of 7.5%. What is the total amount the business will repay if the loan is for 8 years?
The total amount they will repay is the future value, \(A\). We are also given that:
- \(r = 0.075\)
- \(P = 10\,000\)
Using the simple interest formula for future value:
\(\begin{align}A &= P(1 + rt)\\ &= 10\,000(1 + 0.075(8)) \\ &= \bbox[border: 1px solid black; padding: 2px]{16\,000}\end{align}\)
Answer : The business will pay back a total of $16,000.
This may seem high, but remember that in the context of a loan, interest is really just a fee for borrowing the money. The larger the interest rate and the longer the time period, the more expensive the loan.
Also note that you could calculate this by first finding the interest, I = Prt = 10000(0.075(8)) = $6000, and adding it to the principal of $10000. The final answer is the same using either method.
[adsenseLargeRectangle]
Continue your study of interest
Now that you have studied the simple interest formula, you can learn the more advanced idea of compound interest. Most savings accounts, credit cards, and loans are based on compound instead of simple interest. You can review this idea here:
- Compound interest
Share this:
- Click to share on Twitter (Opens in new window)
- Click to share on Facebook (Opens in new window)
6.4 Solve Simple Interest Applications
Learning objectives.
- Use the simple interest formula
- Solve simple interest applications
Be Prepared 6.4
Before you get started, take this readiness quiz.
- Solve 0.6 y = 45 . 0.6 y = 45 . If you missed this problem, review Example 5.43 .
- Solve n 1.45 = 4.6 . n 1.45 = 4.6 . If you missed this problem, review Example 5.44 .
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 , P , and the bank pays you interest , I . I . The interest is computed as a certain percent of the principal; called the rate of interest , r . 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 , t , represents the number of years the money is left in the account.
Simple Interest
If an amount of money, P , P , the principal, is invested for a period of t t years at an annual interest rate r , r , the amount of interest, I , I , earned is
Interest earned according to this formula is called simple interest .
The formula we use to calculate simple interest is I = P r t . I = P r t . 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
Find the simple interest earned after 3 3 years on $500 $500 at an interest rate of 6%. 6%.
Organize the given information in a list.
I = ? P = $500 r = 6% t = 3 years I = ? P = $500 r = 6% t = 3 years
We will use the simple interest formula to find the interest.
Write the formula. | |
Substitute the given information. Remember to write the percent in decimal form. | |
Simplify. | |
Check your answer. Is $90 a reasonable interest earned on $500 in 3 years? | |
In 3 years the money earned 18%. If we rounded to 20%, the interest would have been 500(0.20) or $100. Yes, $90 is reasonable. | |
Write a complete sentence that answers the question. | The simple interest is $90. |
Try It 6.65
Find the simple interest earned after 4 4 years on $800 $800 at an interest rate of 5%. 5%.
Try It 6.66
Find the simple interest earned after 2 2 years on $700 $700 at an interest rate of 4%. 4%.
In the next example, we will use the simple interest formula to find the principal.
Example 6.34
Find the principal invested if $178 $178 interest was earned in 2 2 years at an interest rate of 4%. 4%.
I = $178 P = ? r = 4% t = 2 years I = $178 P = ? r = 4% t = 2 years
We will use the simple interest formula to find the principal.
Write the formula. | |
Substitute the given information. | |
Divide. | |
Simplify. | |
Check your answer. Is it reasonable that $2,225 would earn $178 in 2 years? | |
Write a complete sentence that answers the question. | The principal is $2,225. |
Try It 6.67
Find the principal invested if $495 $495 interest was earned in 3 3 years at an interest rate of 6%. 6%.
Try It 6.68
Find the principal invested if $1,246 $1,246 interest was earned in 5 5 years at an interest rate of 7% . 7% .
Now we will solve for the rate of interest.
Example 6.35
Find the rate if a principal of $8,200 $8,200 earned $3,772 $3,772 interest in 4 4 years.
Organize the given information.
I = $3,772 P = $8,200 r = ? t = 4 years I = $3,772 P = $8,200 r = ? t = 4 years
We will use the simple interest formula to find the rate.
Write the formula. | |
Substitute the given information. | |
Multiply. | |
Divide. | |
Simplify. | |
Write as a percent. | |
Check your answer. Is 11.5% a reasonable rate if $3,772 was earned in 4 years? | |
Write a complete sentence that answers the question. | The rate was 11.5%. |
Try It 6.69
Find the rate if a principal of $5,000 $5,000 earned $1,350 $1,350 interest in 6 6 years.
Try It 6.70
Find the rate if a principal of $9,000 $9,000 earned $1,755 $1,755 interest in 3 3 years.
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
Nathaly deposited $12,500 $12,500 in her bank account where it will earn 4% 4% interest. How much interest will Nathaly earn in 5 5 years?
We are asked to find the Interest, I . I .
I = ? P = $12,500 r = 4% t = 5 years I = ? P = $12,500 r = 4% t = 5 years
Write the formula. | |
Substitute the given information. | |
Simplify. | |
Check your answer. Is $2,500 a reasonable interest on $12,500 over 5 years? | |
At 4% interest per year, in 5 years the interest would be 20% of the principal. Is 20% of $12,500 equal to $2,500? Yes. | |
Write a complete sentence that answers the question. | The interest is $2,500. |
Try It 6.71
Areli invested a principal of $950 $950 in her bank account with interest rate 3%. 3%. How much interest did she earn in 5 5 years?
Try It 6.72
Susana invested a principal of $36,000 $36,000 in her bank account with interest rate 6.5% . 6.5% . How much interest did she earn in 3 3 years?
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
Loren lent his brother $3,000 $3,000 to help him buy a car. In 4 years 4 years his brother paid him back the $3,000 $3,000 plus $660 $660 in interest. What was the rate of interest?
We are asked to find the rate of interest, r . r .
I = 660 P = $3,000 r = ? t = 4 years I = 660 P = $3,000 r = ? t = 4 years
Write the formula. | |
Substitute the given information. | |
Multiply. | |
Divide. | |
Simplify. | |
Change to percent form. | |
Check your answer. Is 5.5% a reasonable interest rate to pay your brother? | |
Write a complete sentence that answers the question. | The rate of interest was 5.5%. |
Try It 6.73
Jim lent his sister $5,000 $5,000 to help her buy a house. In 3 3 years, she paid him the $5,000 , $5,000 , plus $900 $900 interest. What was the rate of interest?
Try It 6.74
Hang borrowed $7,500 $7,500 from her parents to pay her tuition. In 5 5 years, she paid them $1,500 $1,500 interest in addition to the $7,500 $7,500 she borrowed. What was the rate of interest?
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
Eduardo noticed that his new car loan papers stated that with an interest rate of 7.5% , 7.5% , he would pay $6,596.25 $6,596.25 in interest over 5 5 years. How much did he borrow to pay for his car?
We are asked to find the principal, P . P .
I = 6,596.25 P = ? r = 7.5% t = 5 years I = 6,596.25 P = ? r = 7.5% t = 5 years
Write the formula. | |
Substitute the given information. | |
Multiply. | |
Divide. | |
Simplify. | |
Check your answer. Is $17,590 a reasonable amount to borrow to buy a car? | |
Write a complete sentence that answers the question. | The amount borrowed was $17,590. |
Try It 6.75
Sean's new car loan statement said he would pay $4,866.25 $4,866.25 in interest from an interest rate of 8.5% 8.5% over 5 5 years. How much did he borrow to buy his new car?
Try It 6.76
In 5 5 years, Gloria's bank account earned $2,400 $2,400 interest at 5%. 5%. How much had she deposited in the account?
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
Caroline got $900 $900 as graduation gifts and invested it in a 10-month 10-month certificate of deposit that earned 2.1% 2.1% interest. How much interest did this investment earn?
We are asked to find the interest, I . I .
I = ? P = $900 r = 2.1% t = 10 months I = ? P = $900 r = 2.1% t = 10 months
Write the formula. | |
Substitute the given information, converting 10 months to of a year. | |
Multiply. | |
Check your answer. Is $15.75 a reasonable amount of interest? | |
If Caroline had invested the $900 for a full year at 2% interest, the amount of interest would have been $18. Yes, $15.75 is reasonable. | |
Write a complete sentence that answers the question. | The interest earned was $15.75. |
Try It 6.77
Adriana invested $4,500 $4,500 for 8 8 months in an account that paid 1.9% 1.9% interest. How much interest did she earn?
Try It 6.78
Milton invested $2,460 $2,460 for 20 20 months in an account that paid 3.5% 3.5% interest How much interest did he earn?
Section 6.4 Exercises
Practice makes perfect.
In the following exercises, use the simple interest formula to fill in the missing information.
Interest | Principal | Rate | Time (years) |
---|---|---|---|
Interest | Principal | Rate | Time (years) |
---|---|---|---|
Interest | Principal | Rate | Time (years) |
---|---|---|---|
Interest | Principal | Rate | Time (years) |
---|---|---|---|
Interest | Principal | Rate | Time (years) |
---|---|---|---|
Interest | Principal | Rate | Time (years) |
---|---|---|---|
In the following exercises, solve the problem using the simple interest formula.
Find the simple interest earned after 5 5 years on $600 $600 at an interest rate of 3%. 3%.
Find the simple interest earned after 4 4 years on $900 $900 at an interest rate of 6%. 6%.
Find the simple interest earned after 2 2 years on $8,950 $8,950 at an interest rate of 3.24% . 3.24% .
Find the simple interest earned after 3 3 years on $6,510 $6,510 at an interest rate of 2.85% . 2.85% .
Find the simple interest earned after 8 8 years on $15,500 $15,500 at an interest rate of 11.425% . 11.425% .
Find the simple interest earned after 6 6 years on $23,900 $23,900 at an interest rate of 12.175% . 12.175% .
Find the principal invested if $656 $656 interest was earned in 5 5 years at an interest rate of 4% . 4% .
Find the principal invested if $177 $177 interest was earned in 2 2 years at an interest rate of 3% . 3% .
Find the principal invested if $70.95 $70.95 interest was earned in 3 3 years at an interest rate of 2.75%. 2.75%.
Find the principal invested if $636.84 $636.84 interest was earned in 6 6 years at an interest rate of 4.35%. 4.35%.
Find the principal invested if $15,222.57 $15,222.57 interest was earned in 6 6 years at an interest rate of 10.28% . 10.28% .
Find the principal invested if $10,953.70 $10,953.70 interest was earned in 5 5 years at an interest rate of 11.04%. 11.04%.
Find the rate if a principal of $5,400 $5,400 earned $432 $432 interest in 2 2 years.
Find the rate if a principal of $2,600 $2,600 earned $468 $468 interest in 6 6 years.
Find the rate if a principal of $11,000 $11,000 earned $1,815 $1,815 interest in 3 3 years.
Find the rate if a principal of $8,500 $8,500 earned $3,230 $3,230 interest in 4 4 years.
Casey deposited $1,450 $1,450 in a bank account with interest rate 4%. 4%. How much interest was earned in 2 2 years?
Terrence deposited $5,720 $5,720 in a bank account with interest rate 6%. 6%. How much interest was earned in 4 4 years?
Robin deposited $31,000 $31,000 in a bank account with interest rate 5.2% . 5.2% . How much interest was earned in 3 3 years?
Carleen deposited $16,400 $16,400 in a bank account with interest rate 3.9% . 3.9% . How much interest was earned in 8 8 years?
Hilaria borrowed $8,000 $8,000 from her grandfather to pay for college. Five years later, she paid him back the $8,000 , $8,000 , plus $1,200 $1,200 interest. What was the rate of interest?
Kenneth lent his niece $1,200 $1,200 to buy a computer. Two years later, she paid him back the $1,200 , $1,200 , plus $96 $96 interest. What was the rate of interest?
Lebron lent his daughter $20,000 $20,000 to help her buy a condominium. When she sold the condominium four years later, she paid him the $20,000 , $20,000 , plus $3,000 $3,000 interest. What was the rate of interest?
Pablo borrowed $50,000 $50,000 to start a business. Three years later, he repaid the $50,000 , $50,000 , plus $9,375 $9,375 interest. What was the rate of interest?
In 10 10 years, a bank account that paid 5.25% 5.25% earned $18,375 $18,375 interest. What was the principal of the account?
In 25 25 years, a bond that paid 4.75% 4.75% earned $2,375 $2,375 interest. What was the principal of the bond?
Joshua's computer loan statement said he would pay $1,244.34 $1,244.34 in interest for a 3 3 year loan at 12.4% . 12.4% . How much did Joshua borrow to buy the computer?
Margaret's car loan statement said she would pay $7,683.20 $7,683.20 in interest for a 5 5 year loan at 9.8%. 9.8%. How much did Margaret borrow to buy the car?
Caitlin invested $8,200 $8,200 in an 18-month 18-month certificate of deposit paying 2.7% 2.7% interest. How much interest did she earn form this investment?
Diego invested $6,100 $6,100 in a 9-month 9-month certificate of deposit paying 1.8% 1.8% interest. How much interest did he earn form this investment?
Airin borrowed $3,900 $3,900 from her parents for the down payment on a car and promised to pay them back in 15 15 months at a 4% 4% rate of interest. How much interest did she owe her parents?
Yuta borrowed $840 $840 from his brother to pay for his textbooks and promised to pay him back in 5 5 months at a 6% 6% rate of interest. How much interest did Yuta owe his brother?
Everyday Math
Interest on savings Find the interest rate your local bank pays on savings accounts.
- ⓐ What is the interest rate?
- ⓑ Calculate the amount of interest you would earn on a principal of $8,000 $8,000 for 5 5 years.
Interest on a loan Find the interest rate your local bank charges for a car loan.
- ⓑ Calculate the amount of interest you would pay on a loan of $8,000 $8,000 for 5 5 years.
Writing Exercises
Why do banks pay interest on money deposited in savings accounts?
Why do banks charge interest for lending money?
ⓐ 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?
As an Amazon Associate we earn from qualifying purchases.
This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.
Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.
Access for free at https://openstax.org/books/prealgebra/pages/1-introduction
- Authors: Lynn Marecek, MaryAnne Anthony-Smith
- Publisher/website: OpenStax
- Book title: Prealgebra
- Publication date: Sep 25, 2015
- Location: Houston, Texas
- Book URL: https://openstax.org/books/prealgebra/pages/1-introduction
- Section URL: https://openstax.org/books/prealgebra/pages/6-4-solve-simple-interest-applications
© Feb 9, 2022 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.
Simple Interest Lesson
Analysis: When money is borrowed, interest is charged for the use of that money over a certain period of time. The amount of interest charged depends on the amount of money borrowed, the interest rate and the length of time for which the money is borrowed.
Definitions: Principal is the amount of money borrowed. The interest rate is given as a percent . Time is the length of time in years for which the money was borrowed.
Procedure: To find interest, take the product of the principal, the interest rate and the time. Thus, the formula for finding interest is:
Interest = Principal * Rate * Time which is also written as I = P*R*T
Now that we have a procedure and a formula, we can solve the problem above.
Solution: Principal = $3,000, Interest rate = 0.09 and Time = 4
I = (3000)*(0.09)*(4) = $1,080.00
Answer: Raquel had to pay back $3,000 in principal plus $1,080 in interest for a total of $4,080.00.
Remember that interest is the charge for borrowing the money. So Raquel had to pay back the original amount borrowed (principal) AND the interest. Let’s look at some more examples of interest.
Solution: P = $1,200, R = 0.18 and T = 0.75
Remember that the interest formula asks for the time in years. However, the time was given in months. So to get the time in years we represent 9 months as 9/12 of a year, or 0.75.
I = (1200)*(0.18)*(0.75) = 162.00
Answer: Kevin paid $162.00 in interest.
In the problem and example above, money was borrowed and interest was paid for borrowing that money. A person can also earn interest on money invested. Let’s look at an example of this.
Solution: P = $500, R = 0.055 and T = 1
I = (500)*(0.055)*(1) = $27.50
Answer: Isabella earns $27.50 per year in interest from her local bank.
In Example 2, the bank was the borrower and Isabella was the lender. Let’s revise our definition of interest so that it applies to all of these problems.
Interest is the amount of money the lender is paid for the use of his/her money. Interest is the money you pay to use someone else’s money. In either case, the more money being used and the longer it is used for, the more interest must be paid. Let’s look at some more examples of interest.
P = $38,000, R = 7.25% and T = 10
I = (38000)*(.0725)*(10) = $27,550.00
Answer: Jodi will have to pay $38,000 in principal plus $27,550 in interest for a total of $65,550.00.
Solution: P = $1000, I = 0.04 and T = 0.25
Remember that the interest formula asks for the time in years. However, the time was given in months. So to get the time in years we represent 3 months as 3/12 of a year, or 0.25.
I = (1000)*(0.04)*(0.25) = $10.00
Answer: Julia will have $1,000 in principal plus $10 of interest earned for a total of $1,010.00.
In each of the examples above, the interest rate was applied only to the original principal amount in computing the amount of interest. This is known as simple interest . When the interest rate is applied to the original principal and any accumulated interest, this is called compound interest . Simple and compound interest are compared in the tables below. In both cases, the principal is $100.00 is and the interest rate is 7%.
As you can see, compound interest can end up being higher than simple interest for the same principal and the same rate . If you were borrowing money, would you want to pay simple interest or compound interest? If you were lending or investing money, would you want to earn simple interest or compound interest? Summary: Interest is the amount of money the lender is paid for the use of his/her money. Interest is the money you pay to use someone else’s money. In either case, the more money being used and the longer it is used for, the more interest must be paid. So whether you are borrowing or lending (investing) money, interest is found by taking the product of the principal, the interest rate and the time in years. The formula for finding simple interest is: Directions: Each problem below involves simple interest. Solve each problem below by entering a dollar amount with cents. For each exercise below, click once in the ANSWER BOX, type in your answer and then click ENTER. After you click ENTER, a message will appear in the RESULTS BOX to indicate whether your answer is correct or incorrect. To start over, click CLEAR.
![]() Simple Interest Practice QuestionsClick here for Questions Click here for Answers GCSE Revision Cards![]() 5-a-day Workbooks![]() Primary Study Cards![]() Privacy Policy Terms and Conditions Corbettmaths © 2012 – 2024 ![]()
Select a Course Below
"Investment" Word ProblemsExplanation Examples What is "interest"?Interest is the price of money. If I borrow money from you, with a promise to pay you back later, that money is no longer available for you to use as you please. In return for the loss of your use of that money, I would pay you a certain amount (usually being a percentage of the amount that I borrowed), over and above the amount what I'd borrowed, in order to pay you for my use of your money. Content Continues Below MathHelp.com![]() Investment Word Problems What is an "interest rate"?An interest rate is a rate (usually expressed as a percentage) of the money borrowed, which is to be paid back over and above the borrowed amount. For instance, if I borrow $100 from you and pay you back $108 at the end of a year's time, then the interest rate "per annum" (that is, per year) is 8%. Interest rates are almost always expressed in terms of years. If I borrowed the $100 from you and paid you back in six months, then I would owe you $104, because six months is half of a year. What is "simple" interest?Advertisement Simple interest is an amount of money that is a fixed percentage of the amount borrowed, and which is added to the borrowed amount once each interest-rate period. So if I'd borrowed $100 from you and didn't pay it back until two years later, I would owe you an extra $8 for the first year and another extra $8 for the second year, for a total of $116 that I need to pay you. In pre-algebra or beginning algebra, "investment" word problems usually involve simple annual interest — as opposed to compounded interest . Simple interest is earned on the entire investment amount for a given period of time. This differs from compounded interest, where simple interest is earned for a smaller amount of time (for instance, for one month). Then, over the next period of time, the interest is earned on the original investment *plus* on the interest that was earned on that first time period. Then, during the third time period, interest is earned on the initial investment amount plus the interest earned during the first two periods. And so forth. What is an example of a simple-interest situation?One example of a simple-interest situation is getting a loan for the car you're buying. Another example would be investing in a certificate of deposit (that is, a CD ) at your bank. Mortgages and student loans can also be issued in terms of simple interest. What is the formula for simple interest?The formula for the simple interest I earned on the amount of an investment (that is, the "principle") P with an interest rate of r over a time period t is I = Prt . In the simple-interest formula I = Prt , the variable I stands for the interest earned on the original investment, P stands for the amount of the original investment, r is the interest rate (expressed in decimal form), and t is the time (usually in terms of years). ![]() The time units, in all cases, must match the interest-rate units. For instance, if you got a loan from your friendly neighborhood loan shark, where the interest rate is monthly, rather than yearly, then your time must be measured in terms of months. When working on investment word problems, you will want to substitute all given information into the I = Prt equation, and then solve for whatever is left.
The invested amount (that is, the principal) of my investment is P = $1000 , the interest rate (expressed in decimal form) is r = 0.06 per year, and the amount of time is t = 2 . Substituting these values into the simple-interest formula, I get: I = (1000)(0.06)(2) = 120 I will get $120 in interest.
For this exercise, I first need to find the amount of the interest. Since simple interest is added to the principal, and since the principal was P = $500, then the interest is I = $650 − 500 = $150 . The time is t = 3 . Substituting all of these values into the simple-interest formula, I get: 150 = (500)( r )(3) 150 = 1500 r 150 / 1500 = r = 0.10 Of course, I need to remember to convert this decimal to a percentage. I was getting 10% interest. When you have just one account, one simple situation, requiring one use of the simple-interest formula, it's pretty easy to set up and solve the exercises. The hard part comes when the exercises involve multiple investments or some other complication. But there is a trick to these that makes them fairly easy to handle. You use a simple table or grid.
The problem here comes from the fact that I'm splitting the $50,000 in principal into two smaller amounts. Here's how to handle this: I will make a table. The top row has, as its entries, the variables in the simple-interest formula. The left-hand column labels the two funds — and, in case I need it, a "totals" row. (Not all exercises will need the "totals" row.) I can't add rates — that's just a thing; rates can't be added — and each of the investments is for one year, so there are no "total" values for the interest rates or the years. Hence, the dashes, to remind me not to try to put anything in there.
I know the interest total that I'm aiming for, and I know the total amount that I'm investing, so I can enter "total" values for the "interest" and "investment" columns. I know the interest rates and the time (namely, one year) for the two investments, so I can enter these values in the "rate" and "time" columns, in each fund's row. Putting it all together, I get the following start to my set-up:
How do I fill in for those question marks? I'll start with the principal P . Let's say that I put x dollars into Fund X, and y dollars into Fund Y. Then x + y = 50,000 . But this doesn't help much, since I only know how to solve equations in one variable. However, I then notice that I can solve x + y = 50,000 to get y = 50,000 − x . THIS TECHNIQUE IS IMPORTANT! The amount left for me to put into Fund Y is (the total that I invested) less (what I've already put into Fund X), or 50,000 − x . Any time you have a total that is divided into two parts, you can designate one part as (one part), and the rest will be (the total) minus the (one part), because the second part is whatever is left, from the total, after (one part) is accounted for. You will need this technique — this "how much is left" construction — in the future, so make sure you understand it now. So now I have a variable for the Fund X part, and an expression for however much was left to go into the Fund Y part. I can add these to my table:
Now I will show you why I set up the table like this. By organizing the columns according to the interest formula, I can now multiply across (in this case, I will multiply the three right-hand columns to get expressions for I in the second column) to fill in the "interest" column.
The interest from Fund X and the interest from Fund Y will add up to $4,500 . As a result, I can add down the "interest" column, set the sum of the two interest expressions equal to the total interest, and solve the resulting equation for the value of the variable: 0.06 x + 0.14(50,000 − x ) = 4,500 0.06 x + 7,000 − 0.14 x = 4,500 7,000 − 0.08 x = 4,500 −0.08 x = −2,500 The value of x stands for the amount invested in Fund X. So the amount that is left, from the total invested, is given by 50,000 − 31,250 = 18,750 . And this (being the amount that is left after I'd put money into Fund X) is the amount that is invested in Fund Y. I should put $31,250 into Fund X, and $18,750 into Fund Y. Note that the answer did not involve nice, neat values like $10,000 or $35,000 . You should understand that this means that you cannot always expect to be able to use guess-n-check to find your answers. You really do need to know how to do these exercises. URL: https://www.purplemath.com/modules/investmt.htm Page 1 Page 2 Standardized Test PrepCollege math, homeschool math, share this page.
Visit Our Profiles![]() Simple Interest Word Problems These lessons, with videos, examples and step-by-step solutions, help Algebra students learn how to solve word problems that involve interest. Related Pages Simple Interest Compound Interest More Lessons for Grade 9 Math Math Worksheets How to solve word problems using the simple interest formula? The simple interest formula is interest = principal × rate × time or I = Prt How to use the Simple Interest Formula? Example: How much interest is earned in five years on $3,000 invested at an interest rate of 9% per year? Interest Word Problems - Algebra Help How to solve simple interest word problems? Example: Pam invested $5000. She earned 14% on part of her investment and 6% on the rest. If she earned a total of $396 for the year, how much did she invest at each rate? Find the amount of interest earned by $8,000 invested at a 5% annual simple interest rate for 1 year. To start a mobile dog-grooming service, a woman borrowed $2,500. If the loan was for 2 years and the amount of interest was $175, what simple interest rate was she charged? A student borrowed some money from his father at 2% simple interest to buy a car. If he paid his father $360 in interest after 3 years, how much did he borrow? A couple invested $6,000 of their $20,000 lottery winnings in bonds. How much do they have left to in stocks? A college student wants to invest the $12,000 inheritance he received and use the annual interest earned to pay his tuition cost of $945. The highest rate offered by a bank is 6% annual simple interest. At this rate, he cannot earn the needed $945, so he decided to invest some of the money in a riskier, but more profitable, investment offering a 9% return. How much should he invest at each rate? A credit union loaned out $50,000, part at an annual rate of 6% and the rest at an annual rate of 12%. The collected combined simple interest was $3600 that year. How much did the credit union loan out at each rate? Example: Suppose $7,000 is divided into two bank accounts. One account pays 10% simple interest per year and the other pays 5%. After three years there is a total of $1451.25 in interest between the two accounts. How much was invested into each account (rounded to the nearest cent)? ![]() We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.
Simple Interest ProblemsSimple interest is the interest calculated on the principal amount which is borrowed. While learning how simple interest is calculated, the main terms are – principal, denoted by P; rate of interest, denoted by R and time in years, denoted by T. The branch of commercial mathematics has one of the most important concepts, which is interest. The two types of interest are simple interest and compound interest. The idea of simple interest is based on the time value of money, which has a current value, present value and future value. If invested in a deposit, it earns an amount called interest. In this article, we will learn about simple interest, compound interest and how to solve simple interest problems. The following example will make these terms more clear: Malini said that she is going to buy a new refrigerator. Her father asked whether she had the money to buy it. She said to her father that she is planning to take a loan from the bank. The money she borrows is the sum borrowed or the principal. To keep this money for some time, she needs to pay some extra money to the bank. That amount is the interest. So, at the end of the year, she has to pay back the money borrowed and the interest. This is the amount denoted by A. Therefore, Amount = Principal + Interest. Interest is given in percentage for a time of one year. For example, 12% per annum or 12% p.a. This means that for every Rs. 100 you borrow, you have to pay Rs. 12 as interest for one year. Example: Rajiv took a loan of Rs. 7000 from a bank at 10% as a rate of interest. Find the interest he has to pay at the end of one year. Solution: Here, sum borrowed, P = 7000 Rate of interest, R = 10% This means if he borrowed Rs. 100, he had to pay Rs. 10 as interest. So, for Rs. 7000, the interest he has to pay for one year is 7000×10/100 = Rs. 700. So, at the end of the year, the amount he has to pay back = 7000 + 700 = Rs. 7700 Interest for multiple years: If money is borrowed for multiple years, the interest is calculated for the period of time it is kept. For example, if Rajiv returns money at the end of 2 years, he has to pay twice the interest. Rs. 700 for the first year and Rs. 700 for the second year. As the number of years increases, the interest to be paid also increases. If Rs. 100 is borrowed for 4 years at 5%, then the interest to be paid at the end of 4 years is 5 + 5 + 5 + 5 = 4 × 5 = 20. So, the interest paid for T years for a principal P at the rate R% is SI = PTR/100 How Simple Interest is Different from Compound InterestThe interest is calculated on the principal amount for a fixed period of time, and the rate of interest is called simple interest. It is used for a single period. The interest calculated at the end of a certain fixed period and which adds to the principal so that interest can be earned in the next compounding period is called compound interest. SI and CI Formulas
CI: Compound Interest P: Principal R: Rate of interest per annum T: Time in years SI: Simple Interest Solved ExamplesExample 1: An amount of Rs. 12800 was invested by Mr Rohan, dividing it into two different investment schemes, A and B, at a simple interest rate of 11% and 14%. What was the amount in plan B if the amount of interest earned in two years was Rs. 3508? Let the sum invested in Scheme A be Rs. x and that in Scheme B be Rs.(12800 – x). Then, [x . 14 . 2]/100 + [(12800 – x) . 11 . 2]/100 = 3508 28x – 22x = 350800 – (12800 × 22) x = 11533.33 So, the sum invested in Scheme B = Rs. (12800 – 11533.33) = Rs. 1266.67. Example 2: A lender claims to be lending at simple interest, but he adds the interest every 6 months in the calculation of the principal. The rate of interest charged by him is 8%. What will be the effective rate of interest? Let the sum be Rs. 100. Simple interest for 1 st 6 months = Rs. [100 × 8 × 1]/[100 × 2] = Rs. 4 Simple interest for last 6 months = Rs. [104 × 8 × 1]/[100 × 2] = Rs.4.16 So, amount at the end of 1 year = Rs. (100 + 4 + 4.16) = Rs. 108.16 Effective rate = (108.16 – 100) = 8.16% Example 3: A town has a population of 20,000. The population increases by 10% per year. What will be the population after 2 years? Here, R = 10/100 Population after 2 years will be = P[1 + (R/100)] T = 20000[1 + (10/100)] 2 = 20000(1.1) 2 Example 4: The time required for a sum of money to amount to five times itself at 16% simple interest p.a. will be: Let the sum of money be Rs. x and the time required to amount to five times itself be T years. Principal amount = Rs. x Amount after T years = Rs. 5x So, the interest in ‘T’ year should be Rs.(5x – x) = Rs. 4x. Using simple interest formula, (P × T × R)/100 = SI Where, P = Principal amount, T = Duration in years, R = Interest rate per year, SI = Simple interest (x × T × 16)/100 = 4x ⇒ T × (16/100) = 4 ⇒ T = 400/16 = 25 ∴ The required time = 25 years. Example 5: The rate of simple interest per annum at which a sum of money doubles itself in 16⅔ years is: Let the principal amount be P. Now, the amount A after 16⅔ years is doubled. Hence, amount is 2P. I = P × R × T/100 P = principal amount R = rate of interest T = time in years = 16 2 / 3 = 50/3 I = simple interest Amount A = I + P According to question, A = P + (P × R × T/100) 2P = P + (P × R × T/100) P = P × R × T/100 R = 100 × 3/50 Example 6: In which year will the amount on a sum of Rs. 800 at 20% compounded half-yearly exceed Rs.1000? Let the time taken for this amount to reach Rs. 1000 be X. The important thing to note is that this sum is compounded half-yearly. Hence, we use the formula: Where, A = Amount P = Principal r = Interest rate m = No. of periods within a year T = No. of years We need to obtain T such that the RHS should be greater than the LHS: \(\begin{array}{l}\therefore A < P\left(1 + [\frac{r}{m*100}]^{mT} \right)\end{array} \) In this case, m = 2 (since it is half-yearly) Substituting these values, we have ⇒ 1.25 < (1.21) T Now, we need to use trial and error to check for the values of T. For T = 1, 1.25 > 1.21; hence, the condition is not satisfied. For T = 2, 1.25 < 1.4641, hence the condition is met. Therefore, it is the second year in which the amount would be greater than Rs. 1000. Example 7: In how many years will a sum of Rs. 4,000 yield a simple interest of Rs. 1,440 at 12% per annum? We know that the formula for simple interest: SI = [ P × R × T] / 100 SI = Simple Interest = 1440 P = Principal = 4000 T = Time = ? R = Rate of Interest = 12% Substituting the values in the formula ⇒ 1440 = [ 4000 × 12 × T] / 100 ⇒ 1440 = 480T ⇒ T = 1440/480 = 3 yrs Quadratic equations Linear equations Frequently Asked QuestionsWhat is the difference between simple interest and compound interest. Simple interest is based on the principal amount of a loan. Compound interest is based on the principal amount and the interest which adds on it in every period. Give the formula for simple interest.Simple interest, I = PRT, where P is the principal amount, R is the rate of interest, and T is the time in years. How to find the amount if the principal and simple interest are given?We use the formula, amount = principal + simple interest, to find the amount. ![]() Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin! Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz Visit BYJU’S for all Maths related queries and study materials Your result is as below Request OTP on Voice Call Leave a Comment Cancel replyYour Mobile number and Email id will not be published. Required fields are marked * Post My Comment ![]() Register with Aakash BYJU'S & Download Free PDFsRegister with byju's & watch live videos. ![]() In order to continue enjoying our site, we ask that you confirm your identity as a human. Thank you very much for your cooperation. Simple Interest CalculatorThe Simple Interest Calculator calculates the interest and end balance based on the simple interest formula. Click the tabs to calculate the different parameters of the simple interest formula. In real life, most interest calculations involve compound Interest. To calculate compound interest, use the Interest Calculator .
Related Interest Calculator | Compound Interest Calculator What is Simple Interest?Interest is the cost you pay to borrow money or the compensation you receive for lending money. You might pay interest on an auto loan or credit card, or receive interest on cash deposits in interest-bearing accounts, like savings accounts or certificates of deposit (CDs). Simple interest is interest that is only calculated on the initial sum (the "principal") borrowed or deposited. Generally, simple interest is set as a fixed percentage for the duration of a loan. No matter how often simple interest is calculated, it only applies to this original principal amount. In other words, future interest payments won't be affected by previously accrued interest. Simple Interest FormulaThe basic simple interest formula looks like this: Simple Interest = Principal Amount × Interest Rate × Time Our calculator will compute any of these variables given the other inputs. Simple Interest Calculated Using YearsYou may also see the simple interest formula written as: In this formula:
Under this formula, you can manipulate "t" to calculate interest according to the actual period. For instance, if you wanted to calculate interest over six months, your "t" value would equal 0.5. Simple Interest for Different Frequencies
Under this formula, you can calculate simple interest taken over different frequencies, like daily or monthly. For instance, if you wanted to calculate monthly interest taken on a monthly basis, then you would input the monthly interest rate as "r" and multiply by the "n" number of periods. ![]() Simple Interest ExamplesLet's review a quick example of both I=Prt and I=Prn. For example, let's say you take out a $10,000 loan at 5% annual simple interest to repay over five years. You want to know your total interest payment for the entire loan. To start, you'd multiply your principal by your annual interest rate, or $10,000 × 0.05 = $500. Then, you'd multiply this value by the number of years on the loan, or $500 × 5 = $2,500. Now that you know your total interest, you can use this value to determine your total loan repayment required. ($10,000 + $2,500 = $12,500.) You can also divide the value to determine how much interest you'd pay daily or monthly. Alternatively, you can use the simple interest formula I=Prn if you have the interest rate per month. If you had a monthly rate of 5% and you'd like to calculate the interest for one year, your total interest would be $10,000 × 0.05 × 12 = $6,000. The total loan repayment required would be $10,000 + $6,000 = $16,000. What Financial Instruments Use Simple Interest?Simple interest works in your favor as a borrower, since you're only paying interest on the original balance. That contrasts with compound interest, where you also pay interest on any accumulated interest. You may see simple interest on short-term loans. For this same reason, simple interest does not work in your favor as a lender or investor. Investing in assets that don't offer compound growth means you may miss out on potential growth. However, some assets use simple interest for simplicity — for example bonds that pay an interest coupon. Investments may also offer a simple interest return as a dividend. To take advantage of compounding you would need to reinvest the dividends as added principal. By contrast, most checking and savings accounts, as well as credit cards, operate using compound interest. Simple Interest Versus Compound InterestCompound interest is another method of assessing interest. Unlike simple interest, compound interest accrues interest on both an initial sum as well as any interest that accumulates and adds onto the loan. (In other words, on a compounding schedule, you pay interest not just on the original balance, but on interest, too.) Over the long run, compound interest can cost you more as a borrower (or earn you more as an investor). Most credit cards and loans use compound interest. Savings accounts also offer compounding interest schedules. You can check with your bank on the compounding frequency of your accounts. Compound Interest FormulaThe basic formula for compound interest is:
|
COMMENTS
Problem 1 : A person deposits $5,000 in a bank account which pays 6% simple interest per year. Find the value of his deposit after 4 years. Solution : Formula for simple interest is. I = Prt. Substitute P = 5000, t = 4, r = 6%. I = 5000 ⋅ 6/100 ⋅ 4. I = 1200.
Step-by-step guide to solve simple interest Simple Interest: The charge for borrowing money or the return for lending it. To solve a simple interest problem, use this formula:
Examples of finding the interest earned with the simple interest formula. In many simple interest problems, you will be finding the total interest earned over a set period, which is represented as \(I\). The formula for this is: Let's use an example to see how this formula works. Remember that in the formula, the principal \(P\) is the ...
We will start by solving a simple interest application to find the interest. Example 6.36. ... In the following exercises, solve the problem using the simple interest formula. 208. Find the simple interest earned after 5 5 years on $600 $600 at an interest rate of 3%. 3%. 209.
In the following exercises, solve the problem using the simple interest formula. Find the simple interest earned after 5 years on $600 at an interest rate of 3%. Find the simple interest earned after 4 years on $900 at an interest rate of 6%. Find the simple interest earned after 2 years on $8,950 at an interest rate of 3.24%.
Simple Interest Problems. Let us see some simple interest examples using the simple interest formula in maths. Example 1: Rishav takes a loan of Rs 10000 from a bank for a period of 1 year. The rate of interest is 10% per annum. Find the interest and the amount he has to pay at the end of a year.
Simple Interest Lesson. Problem: To buy a computer, Raquel borrowed $3,000 at 9% interest for 4 years. How much money did she have to pay back? Analysis: When money is borrowed, interest is charged for the use of that money over a certain period of time. The amount of interest charged depends on the amount of money borrowed, the interest rate and the length of time for which the money is borrowed.
Simple interest questions are available here to help students learn the formula and how to apply the simple interest formula in various problems, including real-life scenarios. We know that "interest" is the most commonly used word when dealing with financial matters. Also, different types of interests exist, such as simple interest, compound interest, etc.
The Corbettmaths Practice Questions on Simple Interest. Previous: Similar Shapes Area Volume Textbook Exercise
How To Solve Simple Interest Problems, Compound Interest Problems, Continuously Compounded Interest Problems, And Determining The Effective Rate Of Return? Examples Of Simple Interest Problems: Joseph buys a new home using an interest only loan where he pays only the interest on the value of the home each month. The home is valued at $200,000 ...
To solve a simple-interest word problem, use the simple-interest formula I = Prt. Using a table can help you easily set up and solve these exercises. ... set the sum of the two interest expressions equal to the total interest, and solve the resulting equation for the value of the variable: 0.06x + 0.14(50,000 − x) = 4,500. 0.06x + 7,000 − 0 ...
Simple Interest Word Problems. Examples: Find the amount of interest earned by $8,000 invested at a 5% annual simple interest rate for 1 year. To start a mobile dog-grooming service, a woman borrowed $2,500. If the loan was for 2 years and the amount of interest was $175, what simple interest rate was she charged?
The two types of interest are simple interest and compound interest. The idea of simple interest is based on the time value of money, which has a current value, present value and future value. If invested in a deposit, it earns an amount called interest. In this article, we will learn about simple interest, compound interest and how to solve ...
The calculation for finding the amount of interest owed with simple interest is: {eq}I = P \times R \times T {/eq} Where: I is the interest owed. P is the principal (the initial amount of money ...
Hint: You must do the following. a) Find the present value of $1000. b) Find the present value of the $30 payments. c) The fair market value of the bond = a + b. 16) Find the fair market value of the twenty-year $1,000 bond that pays $30 every six months, if the current interest rate has increased to 7.5%.
In this video, I teach you how to solve simple interest problems by using the formula I = PRT. I go over a few different word problem examples.I = InterestP ...
Alternatively, you can use the simple interest formula I=Prn if you have the interest rate per month. If you had a monthly rate of 5% and you'd like to calculate the interest for one year, your total interest would be $10,000 × 0.05 × 12 = $6,000. The total loan repayment required would be $10,000 + $6,000 = $16,000.
Simple interest. Simple interest is based only on the principal amount, unlike compound interest, where we use previously earned interest to calculate the next interest. For example if we invest \$100 for 3 years at an interest rate of 10% than we earn $10 each year. This calculator uses the following formula to solve simple interest problems:
Word Problems on Simple Interest are solved here: 1. Robert deposits $ 3000 in State Bank of India for 3 year which earn him an interest of 8%.What is the amount. Word Problems on Simple Interest. ... In the worksheet on triangle we will solve 12 different types of questions. 1. Take three non - collinear points L, M, N. Join LM, MN and NL.
This calculator for simple interest-only finds I, the simple interest where P is the Principal amount of money to be invested at an Interest Rate R% per period for t Number of Time Periods. Where r is in decimal form; r=R/100. r and t are in the same units of time. Calculate Interest, solve for I. I = Prt. Calculate Principal Amount, solve for P.
The simple interest earned from the two investments is 2610. If we solve the simple interest formula for r (rate): I = prt r = I pt Since time is 1, we can rewrite our equation as: r = I p Since his goal is 3% interest, let's plug everything in and check: .03 = 2610 29, 000 + 58, 000 .03 = 2610 87, 000 .03 = .03 .
Step 1: Multiply the interest by 12 to get the interest for 1 year. 20 × 12 = $240. Interest to be paid in two years = 240 × 2. = $480. Step 2: Use the percent equation to find the interest rate. We know that, part = percent × whole. Here we understand that, part = interest, whole = principal and percent rate = p.
Problem 1 : Find the simple interest for 2 years on $2000 at 6% per year. Solution : Formula for simple interest : I = Pr t. Substitute P = 2000, t = 2 and r = 6% or 0.06. I = 2000 ⋅ 0.06 ⋅ 2. I = $240. Problem 2 : In simple interest, a sum of money doubles itself in 10 years. Find the number of years it will take to triple itself. Solution :
Good self-care is key to an improved mood and reduced anxiety. A self-care activity can be as simple as taking the time to enjoy a cup of tea, listen to your favourite music or go outside for a walk. ... Positivity, problem-solving and play . When you are facing challenging times, it can be difficult to feel hopeful that things can improve. But ...
Generative AI and rising GPU shipments is pushing data centers to scale to 100,000-plus accelerators, putting emphasis on power as a mission-critical problem to solve.
As a generalization of the popular NP-hard maximum clique problem, the maximum IUC problem is of great practical importance for social network analysis and network-based data mining. In this work, we present the first learning driven three-phase search algorithm for this relevant problem.