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Simple Interest Worksheets
The young bankers will immensely profit from our printable simple interest worksheets! Simple interest is the interest on a specific principal amount of money where some rate of interest is agreed upon. Our pdf mathematical and real-world problems on simple interest get the children in grade 6, grade 7, and grade 8 calculating the simple interest accrued over a period of time. Gather momentum finding the missing principal, the interest rate, or the term given the other values. Access our free simple interest worksheets!
Simple Interest | Basic Practice
Kick into gear computing the amount, principal, or interest when given any two values. All you need to know to complete this fundamental practice set is that the amount is the principal and interest put together.
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Finding Simple Interest and Total Amount | Level 1
Understand how the rate of interest (R) and time (T) affect the simple interest(I) and the amount (A) on a sum of money (P), and solve the problems using the formulas I = PRT/100 and A = P + I.
Finding Simple Interest and Total Amount | Level 2
Time to level up! Recall the formulas, and calculate the simple interest and amount to the nearest cent. The term of the deposit is given in months and half and quarter years to aid the 6th grade and 7th grade prepping.
Simple Interest | Missing Principal, Rate, or Time
Can you calculate the principal when you know the simple interest, time, and interest rate? Well, take up this task as a challenge, and solve for P, R, or T using the given values. Simple interest is the name of the game!
Simple Interest Word Problems | Level 1
Money is best learned using real-life scenarios, isn't it? This batch of printable simple interest worksheets for grade 6, grade 7, and grade 8 has exciting everyday simple-interest situations flowing through it.
Simple Interest Word Problems | Level 2
Bringing loads of decimal money and half and quarter years and months into play, these pdf resources with a variety of word problems help children speed and power through calculating simple interest!
Simple Interest - Missing Principal, Rate, or Time | Word Problems
Word problems that get the busy bees in the 7th grade and 8th grade enthusing! Find the sum of money deposited/borrowed, the interest rate charged by the bank or a scheme, and the period of deposit or loan.
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Simple Interest Worksheets With Answers
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Calculating simple interest is an essential skill for anyone who maintains a bank account, carries a credit card balance, or applies for a loan. The free printable worksheets in this lesson will improve your homeschool math lessons and help your students become better at calculations.
This collection of worksheets will also help students understand the process using word problems. Answers are provided for each of the five worksheets on the second page for ease of grading.
Lesson Introduction
Before having students start on the worksheets, explain that when you borrow money, you have to repay the amount you borrowed as well as any added interest charges, which represents the cost of borrowing. In the same way, explain to students that when you lend money or deposit funds in interest-bearing accounts, you typically earn interest income for making your money available to other people.
Simple Interest Worksheet 1
Print the PDF: Simple Interest Worksheet No. 1
In this exercise, students will answer 10 word problems about calculating interest. These exercises will help homeschoolers learn how to calculate the rate of return on investments and illustrate how interest can accrue over time.
Students will answer such questions as,
"How much interest does a $318 investment earn at 9 percent over one year?"
Explain to students that the answer would be $28.62 because $318 x 9 percent is the same as $318 x 0.09, which equals $28.62. Explain to students that they would have to pay this amount of interest in addition to repaying the principal, the amount of the original loan, $318.
Simple Interest Worksheet 2
Print the PDF: Simple Interest Worksheet No. 2
These 10 questions will reinforce the lessons from worksheet No. 1. Homeschoolers and other students will learn how to calculate rates and determine interest payments. For this PDF, students will answer word problem questions such as:
"If the balance at the end of eight years on an investment of $630 that has been invested at a rate of 9 percent is $1,083.60, how much was the interest?"
If students are struggling, explain that calculating this answer involves only simple subtraction, where you subtract the initial investment of $630 from the ending balance of $1,083.60. Students would set up the problem as follows:
$1,083.60 – $630 = $453.60
Explain that some of the information in the question was extraneous and not necessary to solve the problem. For this problem, you don't need to know the years of the loan (eight years) or even the interest rate; you only need to know the beginning and ending balance.
Simple Interest Worksheet 3
Print the PDF: Simple Interest Worksheet No. 3
Use these word questions to continue practicing how to calculate simple interest. Students can also use this exercise to learn about the principal, rate of return (the net gain or loss on an investment over a specified time), and other terms commonly used in finance.
Simple Interest Worksheet 4
Print the PDF: Simple Interest Worksheet No. 4
Teach your students the basics of investing and how to determine which investments will pay the most over time. This worksheet will help your homeschoolers polish their calculating skills.
Simple Interest Worksheet 5
Print the PDF: Simple Interest Worksheet No. 5
Use this final worksheet to review the steps for calculating simple interest. Take time to answer questions your homeschoolers may have about how banks and investors use interest calculations.
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Worksheets on Simple Interest | Word Problems on Simple Interest with Solutions
In this Worksheet on Simple Interest, we can see the questions on calculating simple interest, rate of interest, and the amount, which is useful to the students so that they can practice for their examinations or any competitive tests. Given below are different types of questions with solutions that help you to understand the chapter better. Assess your preparation level by solving the Simple Interest Worksheet Questions on your own.
Here to solve the questions on simple interest we will use the formula, Simple interest (S.I.) = (Principal × Rate× Time)/100 and to calculate the Amount Amount (A) = Principal +Interest
1. Find the simple interest and amount in each of the following:
(a) P = $30,000 R = 10% T = 5 years
Given: Principal= $30,000 Rate= 10% Time= 5 years. To find Simple interest we will use the formula, Simple interest (S.I.) = (Principal × Rate× Time)/100 = (30,000×10×5)/100 By solving we will get Simple interest = $15,000. (b) P= $17000 R= 20% T= 30 days Solution: Given: Principal= $17,000 Rate= 20% Time= 30 days Here we will convert days to years by dividing with 365 i.e 30/365, then substitute the given values in the formula, = (17000×20×30)/100×365 by solving we will get Simple interest = $279.45 (c) P= $6000 R= 15% T= 13 months Solution: Given: Principal= $6000 Rate= 15% Time= 13 months Here we have to convert months to years, so we will divide with 12 13÷12= 1.08 years Simple interest (S.I.) = (Principal × Rate× Time)/100 = $975. (d) P= $500 R= 3% T= 1 1/2 year Solution: Given: Principal= $500 Rate= 3% Time= 1 1/2 year= 1.5 years. Simple interest (S.I.) = (Principal × Rate× Time)/100 = $1,350.
2. What sum would yield an interest of $80 in 4 years at 4% p.a.?
Given: Interest= $80 Time= 4 years Rate= 4% Principal = Interest×100/ Rate×Time = 80×100/ 4×4 = $500.
3. At what rate percent per annum will $550 amount to $650 in 4 years?
Given: Amount= $650 Principal= $550 Time = 4 years Simple interest= Amount – Principal = $650 – $550 =$100 So Rate= Simple interest ×100 / Principal×Time = 100×100 / 550× 4 = 4.5%
4 . In what time will $800 amount to $1050 if the simple interest is calculated at 15% p.a.?
Given: Amount= $1050 Principal= $800 Rate = 15% Simple interest= Amount – Principal = $1050-$800 = $250 Time= Simple interest ×100 / Principal×Rate = 250×100 / 800×15 = 2 1/12 years.
5. A sum amount to $1500 at 5% simple interest per annum after 3 years. Find the sum.
Given: Amount payable = $1500 Rate= 5% Time= 3 years So amount payable= Principal(100+r×t)/100 1500= Principal(100+5×3)/100 By solving we will get Principal= $1304.34
6. Mr.Mike borrowed $5500 from Sam at 5% per annum. After 5 years he cleared the amount by giving $8,500 cash and a bag. Find the cost of the bag.
Given: Principal= $5,500 Rate= 5% Time= 5 years Simple interest (S.I.) = (Principal × Rate× Time)/100 = $1,375. The total amount to be cleared after 5 years= Principal+Interest = $1,375+$5,500 = $6,875 Mike cleared by paying $8,500 and a bag, So the cost of the bag is = $8,500-$6,875 = $1,625
8. In how many years will $7000 yield an interest of $4620 at 22% simple interest?
Given: Principal= $7,000 Simple interest= $4,620 Rate= 22% Time= $4,620= (7,000×22×Time)/100 By solving Time= 3.04 years
9. At what rate of simple interest will $2800 amount to $3500 in 2 years, 3 months?
Given: Principal= $2,800 Amount= $3,500 Time= 2 years 3 months = 9/4 years Simple interest= amount – principal = $3,500 – $2,800 = $700 Rate= (Simple interest × 100)/ Principal× time By substituting the values we will get rate= 11.1%
10. Find the simple interest at the rate of 8% p.a. for 4 years on that principal which in 6 years, 6 months at the rate of 4% p.a. gives $1600 as simple interest.
Given: Rate= 4% Time= 6 years 6 months which is 13/2 years Simple interest= $1600 As Principal = (Simple interest × 100)/ Rate× time = 1600×100×2 / 13×4 = $6153.84 Now Simple interest for 4 years is = (6153.84×4×8)/100 = $1969.22
11. Simple interest on a certain sum is 36/72 of the sum. Find the rate percent and time if both are numerically equal. [Hint: (T = R), P = x, S.I. = 36/72x]
Let Sum be X Time= Rate= R Simple interest= 36/72X Simple interest (S.I.) = (Principal × Rate× Time)/100 36/72X =(X × R × R)/100 on solving we will get 7% approx.
12. Simple interest on a sum of money at the end of 5 years is 2/5 of the sum itself. Find the rate percent p.a.
Let Principal be X Time= 5 years Simple interest= 2/5X Simple interest (S.I.) = (Principal × Rate× Time)/100 2/5X= ( X × R × 5)/100 on solving we will get Rate= 8%
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Simple Interest Word Problems
Give your seventh graders a leg up on managing their future finances with this math worksheet featuring simple interest word problems. Students will learn the simple interest formula and review an example of how to apply the formula to a real-world example. Then they will set out to find the simple interest earned given the principal, interest rate, and time elapsed across several word problems. The worksheet begins with problems that ask just for the simple interest, but progresses to problems that ask for students to find the balance of an account with the simple interest added.
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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.
Algebra: Interest Word ProblemsIn these lessons, we will how to solve • word problems that involve a single Simple Interest • word problems that involve more that one Simple Interest. • word problems that involve Simple Interest with discounted loan. Related Pages Simple Interest Formula Simple and Compound Interest Compound Interest Word Problems Interest Problems are word problems that use the formula for Simple Interest . There is also another type of interest word problems called Compound Interest Word Problems . The following tables give the formulas for Simple Interest, Compound Interest, and Continuously Compounded Interest. Scroll down the page for examples and solutions on how to use the Simple Interest Formula. Simple Interest Word ProblemsInterest represents a change of money. If you have a saving account, the interest will increase your balance based upon the interest rate paid by the bank. If you have a loan, the interest will increase the amount you owe based upon the interest rate charged by the bank. The formula for Simple Interest is: I = prt where I is the interest generated. p is the principal amount that is either invested or owed. r is the rate at which the interest is paid. t is the time that the principal amount is either invested or owed. This type of word problem is not difficult. Just remember the formula and make sure you plug in the right values. The rate is usually given in percent , which you will need to change to a decimal value. Word Problems With One Simple InterestExample 1: John wants to have an interest income of $3,000 a year. How much must he invest for one year at 8%? Solution: Step 1: Write down the formula I = prt Step 2: Plug in the values 3000 = p × 0.08 × 1 3000 = 0.08p p = 37,500 Answer: He must invest $37,500 Example 2: Jane owes the bank some money at 4% per year. After half a year, she paid $45 as interest. How much money does she owe the bank? Answer: She owes $2,250 How To Solve Simple Interest Word Problems (Investment Problems)?
How To Solve Interest Problems Using The Simple Interest Formula?
How to use the Simple Interest Formula to solve Word Problems? Example: Jenna invests $13,000 into separate bank accounts, one earning 6% simple interest and the other earning 3% simple interest. If at the end of one year she earns $682.50 in interest, how much did she invest in each account? Word Problems With More Than One Simple Interest RateHow To Solve Word Problems With More Than One Simple Interest? Example: Pam invested $5000. She earned 14% on part of her investment and 6% on the rest. If she earned a total of $396 in interest for the year, how much did she invest at each rate? Note that this problem requires a chart to organize the information. The chart is based on the interest formula, which states that the amount invested times the rate of interest = interest earned. The chart is then used to set up the equation. How To Solve Word Problems With Two Simple Interest Rates? Example: Johnny is a shrewd eight-year-old. For Christmas, his grandparents gave him ten thousand dollars. Johnny decides to invest some of the money in a savings account that pays two percent per annum and the rest in a stock fund that pays ten percent per annum. Johnny wants his investments to yield seven percent per annum. How much should he put in each account? How To Solve A Real Life Problem Involving Interest? 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)? Simple Interest Discounted LoanA discounted loan is a loan that collects interest from the amount of the loan or face value of the loan when the loan is made. The interest is deducted from the loan amount so you don’t receive the full loan amount or face value of the loan when you receive the loan. The deducted interest is the discount. Example: You borrow $2,000 on a 12% discount loan for 12 months. 1. What is the loan discount? 2. Determine the net amount of money that you will actually receive. 3. What is the loan’s actual annual simple interest rate? We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page. WORD PROBLEMS ON SIMPLE INTERESTProblem 1 : Find the simple interest for 2 years on $2000 at 6% per year. Formula for simple interest : Substitute P = 2000, t = 2 and r = 6% or 0.06. I = 2000 ⋅ 0.06 ⋅ 2 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. Let P be the sum of money invested. Given : Sum of money doubles itself in 10 years. Then, P will become 2P in 10 years. Now we can calculate interest for ten years as given below From the above calculation, P is the interest for the first 10 years. In simple interest, interest earned will be same for every year. So, interest earned in the next 10 years also will be P. It has been explained below. So, it will take 20 years for the principal to become triple itself. Problem 3 : In simple interest, a sum of money amounts to $ 6200 in 2 years and $ 7400 in 3 years. Find the principal. At the end of 2 years, we get $6200. At the end of 3 years, we get $7400. From these two information, we can get the interest earned in the 3rd year as given below. In simple interest, interest will be same for every year. Based on this, we can calculate the principal as given below So, the principal is $3800. Problem 4 : A sum of $46875 was lent out at simple interest and at the end of 1 year 8 months, the total amount was $50000.Find the rate of interest per year. From the given information, we have P = 46875 and A = 50000 Interest = Amount - Principal I = 50000 - 46875 The value of n must always be in years. But in the question, it is given in both years and months. To convert months to years, divide the given months by 12. 1 year 8 months = 1 ⁸⁄₁₂ years or 1 ⅔ years So, the value of t is 1 ⅔ or ⁵⁄₃ . Substitute P = 46875 and t = ⁵⁄₃ . 3125 = 46875 ⋅ r ⋅ ⁵⁄₃ 3125 = 78125 ⋅ r Divide both sides by 78125. 0.04 ⋅ 100% = r Problem 5 : Find the accumulated value of the deposit $2500 made in simple interest for 3.5 years at 5% rate of interest per year. First let us find the interest earned and then we can find the accumulated value. I = Pr t Substitute P = 2500, t = 3.5 and r = 0.05. I = 2500 ⋅ 0.05 ⋅ 3.5 Accumulated value : A = 2500 + 437.50 A = $2937.50 Problem 6 : How much interest will be earned on $3000 at 7% simple interest per year for 9 months ? I = Pr t ----(1) Substitute P = 3000 and r = 7%. The value of n must always be in years. But in the question, it is given in months. 9 months = ⁹⁄₁₂ years or ¾ years In (1), substitute P = 3000, r = 0.07 and t = ¾ . I = 3000 ⋅ 0.07 ⋅ ¾ I = $157.50 Problem 7 : What sum of money will produce $28600 as interest in 3 years and 3 months at 2.5% per year simple interest ? The value of n must always be in years. But in the question, it is given in both years and months. 3 years 3 months = 3 ³⁄₁₂ years or 3 ¼ years n = 3 ¼ or ¹³⁄₄ Substitute I = 28600, r = 0.025 and t = ¹³⁄₄ in (1) 28600 = P ⋅ 0.025 ⋅ ¹³⁄₄ 2860000 = P ⋅ 0.025 ⋅ 3.25 2860000 = P ⋅ 8.125 Divide both sides by 8.125. Required sum of money is $352,000. Problem 8 : Mr. Abraham invested an amount of $13900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be $ 3508, what was the amount invested in Scheme B ? Let m be the amount invested in scheme B. Then the amount invested in scheme A = 13900 - m. Interest in scheme (A) + Interest in scheme (B) = 3508 (13900 - m) ⋅ 0.14 ⋅ 2 + m ⋅ 0.11 ⋅ 2 = 3508 (13900 - m) ⋅ 0.28 + 0.22m = 3508 3892 - 0.28m + 0.22m = 3508 3892 - 0.06m = 3508 3892 - 3508 = 0.06m 384 = 0.06m m = 6400 The amount invested in scheme B is $6,400. Problem 9 : Lily took a loan of $1200 with simple interest for as many years as the rate of interest. If she paid $432 as interest at the end of the loan period, what was the rate of interest? Let m be the rate of interest. Given : The rate of interest and the number of years are same. Then, the number of years = m. Formula for simple interest : Substitute I = 432, P = 1200, r = 0.01m and t = m. 432 = 1200 ⋅ 0.01m ⋅ m 432 = 12m 2 Divide both sides by 12. The rate of interest is 6%. Problem 10 : A lent $5000 to B for 2 years and $3000 to C for 4 years on simple interest at the same rate of interest and received $2200 in all from both of them as interest. Find the rate of interest per year. Let m be the rate of interest. Interest from B + Interest from C = 2200 5000 ⋅ 0.01m ⋅ 2 + 3000 ⋅ 0.01m ⋅ 4 = 2200 100m + 120m = 2200 220m = 2200 Divide both sides by 220. The rate of interest is 10%. Kindly mail your feedback to [email protected] We always appreciate your feedback. © All rights reserved. onlinemath4all.com
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Simple Interest QuestionsSimple 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. In this article, you will get solved and practice questions on simple interest that cover all possible applications of simple interest. What is the simple interest formula? Simple Interest (SI) calculates the amount of interest for a certain principal amount of money at some interest rate and for a given period of time. The formula of simple interest is given by: SI = PTR/100 P = Principal amount T = Time (in years) R = Interest rate Click here to learn more about simple interest in mathematics. Simple Interest Questions and Answers1. Given that simple interest on a certain sum of money is Rs. 4016.25 at 9% per annum in 5 years. Find the sum of money. Let P be the principal amount or sum of money. Time (T) = 5 years Rate of interest (R) = 9% Simple interest earned (SI) = Rs. 4016.25 As we know, Rs. 4016.25 = (P × 5 × 9)/100 P = (Rs. 4016.25 × 100)/(5 × 9) Therefore, the sum of money is Rs. 8925. 2. Calculate the simple interest on Rs. 8000 for 15 months at 6 paise per rupee per month. Principal amount (P) = Rs. 8000 Time (T) = 15 months Rate (R) = 6 paise per rupee per month Here, the rate of interest and the time are given per month. So, we can use the simple interest formula as: = (Rs. 8000 × 15 × 6)/100 Therefore, the simple interest is Rs. 7200. 3. A sum of Rs. 25000 will become Rs. 31000 in 48 months at some rate of simple interest. Find the rate of interest per annum. Principal amount (P) = Rs. 25000 Time (T) = 48 months = (48/12) years = 4 years Total amount after 4 years (A) = Rs. 31000 Let R be the rate of interest. Rs. 31000 = Rs. 25000 + [(Rs. 25000 × 4 × R)/100] Rs. 31000 – Rs. 25000 = Rs. 1000 × R R = Rs. 6000/Rs. 1000 = 6 Therefore, the rate of interest per annum is 6%. 4. A sum of Rs. 12000 is lent out at 5% per annum simple interest for 5 years. What will be the amount after 5 years? Principal amount (P) = Rs. 12000 Rate of interest per annum (R) = 5% = (Rs. 12000 × 5 × 5)/100 Thus, the total amount after 5 years = P + SI = Rs. 12000 + Rs. 3000 = Rs. 15000 5. A sum of Rs 1750 is divided into two parts such that the interests on the first part at 8% simple interest per annum and that on the other part at 6% simple interest per annum are equal. What is the interest accumulated on each part (in Rs)? Principal (P) = Rs. 1750 Let x be the first part of the sum. So, the second part will be (Rs. 1750 – x). We know that SI = PTR/100 According to the given, ⇒ (x × 8 × 1)/100 = [(1750 – x) × 6 × 1]/100 ⇒ 8x = (1750 – x) × 6 ⇒ 4x = 5250 − 3x ⇒ 7x = 5250 Therefore, the first part of the sum = Rs. 750 Second part of the sum = Rs. (1750 – 750) = Rs. 1000 Hence, the interest on each part = Rs. 750 × (8/100) = Rs. 60 6. A person deposited some amount of money in the bank at simple interest. After 15 years, the amount became seven times. In how many years will the amount become ten times if the interest rate remains the same? Let P be the amount of money deposited in the bank. And let R be the rate of interest per annum. Time = 15 years 7P = P + (P × 15 × R)/100 7P – P = 15PR/100 6P = 15PR/100 ⇒ R = (6P × 100)/15P Thus, the rate of interest is 40%. Let T be the number of years in which the amount P becomes 10 times. So, 10P = P + (P × T × 40)/100 10P – P = 2PT/5 ⇒ T = (9P × 5)/2P Therefore, in 22.5 years, the amount will be 10 times the initial deposit. 7. A sum of Rs. 800 amounts to Rs. 920 in 3 years at simple interest. If the interest rate increases by 3%, what will be the amount? Principal (P) = Rs. 800 Time (T) = 3 years Amount after 3 years (A) = Rs. 920 SI = A – P = Rs. 920 – Rs. 800 Using the simple interest formula, Rs. 120 = (Rs. 800 × 3 × R)/100 Rs. 12000 = Rs. 2400 × R Thus, the rate of interest is 5%. New interest rate = (5 + 3)% = 8% New simple interest = (Rs. 800 × 3 × 8)/100 New amount = Principle + New simple interest = Rs. 800 + Rs. 192 8. A sum of money was lent at simple interest at 11% per annum for 7/2 years and 9/2 years, respectively. If the interest difference for two periods was Rs. 5500, find the sum. Let P be the sum of money lent. Rate of interest per annum (R) = 11% Interest earned in 9/2 years – Interest earned in 7/2 years = Rs. 5500 Using simple interest formula, we have; P × (9/2) × 11 × (1/100) – P × (7/2) × 11 × (1/100) = Rs. 5500 (P/200) [99 – 77] = Rs. 5500 22P/200 = Rs. 5500 P = (Rs. 5500 × 200)/22 = Rs. 50000 Therefore, the required sum is Rs. 50000. 9. Rs. 2379 is divided into 3 parts so that the amounts after 2, 3 and 4 years, respectively, are equal. What is the first part if the interest rate is 5% per annum on simple interest? Given that a sum of money Rs. 2379 is divided into three parts. Rate of interest (R) = 5% Let a, b, and c be the three parts. As we know, SI = PTR/100 a + (a × 2 × 5/100) = b + (b × 3 × 5/100) = c + (c × 4 × 5/100) a + (a/10) = b + (3b/20) = c + (c/5) 11a/10 = 23b/20 = 6c/5 Let 11a/10 = 23b/20 = 6c/5 = k Thus, a = 10k/11, b = 20k/23, c = 5k/6 a + b + c = Rs. 2379 (10k/11) + (20k/23) + (5k/6) = 2379 (10k × 23 × 6) + (20k × 11 × 6) + (5k × 11 × 23) = 2379 × 11 × 23 × 6 1380k + 1320k + 1265k = 2379 × 11 × 23 × 6 ⇒ 3965k = 2379 × 11 × 23 × 6 ⇒ k = (2379 × 11 × 23 × 6)/3965 Now, x = 10k/11 = (2379 × 11 × 23 × 6 × 10)/ (3965 × 11) = 828 Therefore, the first part of the sum is Rs. 828. 10. A simple interest of Rs. 2500 is earned by investing a sum of money for 13 years. The interest rate is charged at 4% for the first 3 years, 5% for the next 4 years and 8% beyond 7 years. Find the sum of money invested. Let P be the sum of money invested. Interest rate for the first 3 years = 4% Interest rate for the next 4 years = 5% Interest rate beyond 7 years, i.e. for (13 – 7) = 6 years = 8% Using the simple interest formula, we have; (P × 3 × 4/100) + (P × 4 × 5/100) + (P × 6 × 8/100) = Rs. 2500 12P + 20P + 48P = Rs. 2500 × 100 80P = Rs. 250000 P = Rs. 250000/80 = Rs. 3125 Therefore, the sum of money invested is Rs. 3125.
Practice Questions on Simple Interest
Leave a Comment Cancel replyYour Mobile number and Email id will not be published. Required fields are marked * Request OTP on Voice Call Post My Comment Register with BYJU'S & Download Free PDFsRegister with byju's & watch live videos. GCSE Tutoring Programme "Our chosen students improved 1.19 of a grade on average - 0.45 more than those who didn't have the tutoring." In order to access this I need to be confident with: This topic is relevant for: Simple InterestHere we will learn about simple interest including how to calculate simple interest for increasing and decreasing values, and how to set up, solve and interpret growth and decay problems. There are also simple interest worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck. What is simple interest?Simple interest is calculated by finding a percentage of the principal (original) amount and multiplying by the time period of the investment. The final value of the investment can then be found by adding the simple interest to the principal amount. Simple Interest Formula Simple interest can be calculated using the following formula: And we can calculate the value of the investment, A, after the time period with the formula:
Lets calculate the interest earned on £3000 with a simple interest rate of 5\% over 2 years. Using the formula I=Prt, remembering that 5\% = 0.05 To find the final value of the investment we can now add the interest to the principal amount. We could have calculated this directly using the formula A=P\left( 1+rt \right) How to calculate simple interestIn order to calculate simple interest:
Explain how to calculate simple interest in 3 steps.Simple interest worksheetGet your free simple interest worksheet of 20+ questions and answers. Includes reasoning and applied questions. Related lessons on simple and compound interestSimple interest is part of our series of lessons to support revision on simple interest and compound interest . You may find it helpful to start with the main simple interest and compound interest lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:
Simple interest examplesExample 1: finding the simple interest. £2100 is invested for 3 years at an annual interest rate of 2\% per year simple interest. Find the interest earned on the investment in that time? State the formula needed and the value of each variable We required the interest so we will use the formula I=Prt with: 2 Substitute the values into the formula . Substituting these values into the simple interest formula I=Prt, we get: 3 Solve the equation . £126 was earned on the investment. Example 2: Simple interest – finding the final amount after an increase£1500 is invested for 4 years at an annual interest rate of 5\% per year simple interest. What is the value of the investment after this time? We required final the value of the investment so we will use the formula A=P(1+rt) with: Substitute the values into the formula . Substituting these values into the simple interest formula A=P(1+rt), we get: A=1500(1+0.05\times{4}) Solve the equation . Example 3: Simple interest – finding the final amount after a decreaseA car is bought for £10,000 and loses 9\% of its value per annum, simple interest. What is the value of the car after 8 years? Here we use the formula A=P(1+rt) with: A=10000(1-0.09\times{8}) Example 4: Simple interest – different time scale£7600 is borrowed for 2 years on a credit card. The cost of borrowing is a 1\% interest payment per month simple interest for the life of the loan. What is the total cost to pay off after this time?
A=7600(1+0.01\times{24}) Example 5: Simple interest – different percentagesA house is currently valued at £175,000. For the first 3 years, the value of the house increases by the rate of simple interest of 0.2\% per annum. For the following 4 years, the value of the house decreases in value by a simple interest rate of 0.18\% per annum. Calculate the value of the house after these 7 years. Here we use the formula A=P\left(1+r_{1} t_{1}+r_{2} t_{2}\right)
Substituting these values into the simple interest formula A=P\left(1+r_{1} t_{1}+r_{2} t_{2}\right), A=175,000(1+0.002\times{3}-0.0018\times{4}) Common misconceptions
This is a very common mistake where the simple interest on an amount is calculated instead of using the compound interest formula.
In example 4 “ £7600 is invested for 2 years at 1\% per month simple interest. What is the value of the investment after this time?”, the value of t is written as 2 and not 24. A=7600(1+0.01\times{2})
Using the percentage as the value for r (therefore not dividing the percentage by 100 ). For example when using simple interest to increase £100 by 2\% for 5 years, this calculation is made:
If a value is depreciating (going down), the value of r is negative whereas it is incorrectly used as a positive and so the answer will be larger than the original amount. Simple Interest practice questions1. A technology store has a back to school offer: save 20\% on all full price laptops. Paula buys a laptop that was £689 full price. After 3 years, the value of the purchased laptop has decreased by 4\% per year, simple interest. What is the value of the laptop after these 3 years? The original price was £689 . The sale price was 80 % of this. 80 % of £689 is £551.20 . The value of the laptop has decreased by 4 % of £551.20 per year for three years. We can work out 4 % of 551.20 and multiply by 3 ; this is £66.14 . Subtracting this from the purchase price gives the new value. 2. A saxophone that costs £999 is bought using a finance deal. The total price is divided by the number of months with 7 % of the original amount added as an extra charge. The finance deal is spread over 2 years. How much money has been spent on the saxophone after 6 months? The amount to be paid back is given by 1.07\times999=1068.93 Six months is a quarter of the finance period, so a quarter of the full amount will have been spent on the saxophone, therefore \frac{1}{4}\times1068.93=267.23 3. Two shops have a sale. Shop A advertises a 10 % reduction on all items. Shop B has reduced all items by 8 % with a further 2 % reduction on the sale price of t-shirts. Josie can buy the same t-shirt from each shop that originally costs £16 . Which shop is selling the item cheapest? Shop A: After a 10 % reduction, the t-shirt will cost £14.40 Shop B: The 8 % reduction means the t-shirt will be £14.72 , with the additional 2 % reduction making the price £14.43 4. £7342 is invested in a savings account with a 0.4 % simple interest rate per month. What is the value of the investment after 4 years? The initial amount is 7342 , the simple interest rate is 0.4 % and the number of months is 48 , which results in 7342(1+0.004\times48) 5. A house is valued at £365,500 . The value of the house increases by an average of 0.25 % per year, simple interest. How much is the house worth after 12 years? The initial amount is 365500 , the simple interest rate is 0.25 % and the number of years is 12 , which results in 365500(1+0.0025\times12) Simple Interest GCSE Exam Questions:1.(a) £850 is invested for 6 years at 2\% simple interest per year. Work out the total interest earned? (b) The interest rate changes to 0.15\% simple interest per month. How does this affect the final value of the investment over the same time period of 6 years? Decreased by £10.20 over 6 years 2. (a) Freya invests \$6700 for 2 years. The simple interest rate is 1.2\% per year. Which calculation below works out the total value after 2 years? Circle your answer. (b) Euan invests \$6400 for 2 years. The simple interest rate is 1.4\% for the first year 1.1\% for the second year. Whose investment is worth more after 2 years and by how much? You must show your working. (a) \$6700 \times 1.024 3. (a) Lauren places £300 into a new bank account with a simple interest rate of 3\%. After the second year, she withdraws £30. How much money would she have after 5 years? (b) The interest rate after 5 years halves. How many years will it now take Lauren to save £400? Learning checklistYou have now learned how to:
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Seventh grade math - Simple interestSimple interest – seventh grade math. Simple interest is the interest paid only on the original amount of money(Principal). The amount of interest is given by I = Prt, where P is the principal, r is the annual interest rate in decimal form, and t is the loan period expressed in years. The videos, worksheets, and apps on this page will help you learn simple interest formula, calculate simple interest and solve simple interest problems. The apps, sample questions, videos, and worksheets listed below will help you learn Simple Interest.Sample questions on simple interest, simple interest worksheets, educational videos related to simple interest. Simple Interest What is Simple Interest Educational Apps related to Simple InterestMath ELA Grade 7 – Common Core Simple Interest Calculator Find More… Related TopicsHow do you calculate simple interest. Amount after adding Simple Interest can be calculated using the formula, A = P(1 + rt) where P is the Principal amount of money to be invested at an Interest Rate R% per period for t Number of Time Periods. How do you find the monthly interest rate?You can calculate monthly interest rate by dividing the annual interest rate by 12 What is the definition of compound interest?Compound interest is the addition of interest to the principal sum of a loan or deposit, or in other words, interest on interest. It is the result of reinvesting interest, rather than paying it out, so that interest in the next period is then earned on the principal sum plus previously-accumulated interest.
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Read More Louisiana Department of Education is not affiliated to Lumos Learning. Louisiana department of education, was not involved in the production of, and does not endorse these products or this site. Report an Error Word Problems on Simple InterestWord 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 he gets after 1 year, 2 years and 3 years? Solution: In every $ 100, Robert gets $ 8. (Since rate is 8% → 8 for every 100) Therefore, for $ 1 he gets = $ 8/100 And for $ 2000 he gets = 3000 x 8/100 = $ 240 Simple Interest for 1 year = $ 240. Simple Interest for 2 year = $ 240 x 2 = $ 480 Simple Interest for 3 year = $ 240 x 3 = $ 720 Therefore, Amount after 1 year = Principal (P) + Simple Interest (SI) = 3000 + 240 = $ 3240 Amount after 2 years = Principal (P) + Simple Interest (SI) = 3000 + 480 = $ 3480 Amount after 3 years = Principal (P) + Simple Interest (SI) = 3000 + 720 = $ 3720 We observe from the above example that, the Interest cannot be calculated without Principal, Rate and Time. Therefore, we can conclude that Simple Interest (S.I.) depends upon: (i) Principal (P) (ii) Rate (R) (iii) Time (T) And therefore, the formula for calculating the simple interest is Simple Interest (SI) = {Principal (P) × Rate (R) × Time (T)}/100 Amount (A) = Principal (P) + Interest (I) Principal (P) = Amount (A) – Interest (I) Interest (I) = Amount (A) – Principal (P) 2. Richard deposits $ 5400 and got back an amount of $ 6000 after a year. Find the simple interest he got. Solution: Principal (P) = $ 5400, Amount (A) = $ 6000 Simple Interest (SI) = Amount (A) – Principal (P) = 6000 - 5400 = 600 Therefore, Richard got an interest of $ 600. 3. Seth invested a certain amount of money and got back an amount of $ 8400. If the bank paid an interest of $ 700, find the amount Sam invested. Solution: Amount (A) = $ 8400, Simple Interest (SI) = $ 700 Principal (P) = Amount (A) – Interest (I) = 8400 - 700 = 7700 Therefore, Seth invested $ 7700. 4. Diego deposited $ 10000 for 4 year at a rate of 6% p.a. Find the interest and amount Diego got. Solution: Principal (P) = $ 10000, Time (T) = 4 years, Rate (R) = 6% p.a. Simple Interest (SI) = {Principal (P) × Rate (R) × Time (T)}/100 = (10000 x 6 x 4)/100 = $ 2400 Amount (A) = Principal (P) + Interest (I) = 10000 + 2400 = $ 12400 The interest Diego got = $ 2400. Therefore, the amount Diego got $ 12400. ● Simple Interest. Word Problems on Simple Interest. Factors Affecting Interest In Simple Interest when the Time is given in Months and Days. To find Principal when Time Interest and Rate are given. To find Rate when Principal Interest and Time are given . To find Time when Principal Interest and Rate are given. Worksheet on Simple Interest. Worksheet on Factors affecting Interest 5th Grade Numbers Page 5th Grade Math Problems From Word Problems on Simple Interest to HOME PAGE Didn't find what you were looking for? Or want to know more information about Math Only Math . Use this Google Search to find what you need. New! CommentsShare this page: What’s this?
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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:
Simple Interest Determine the simple interest for these loans. ... Solve each simple interest word problem. 11) A new car, valued at $28,000, depreciates at 9% per year. What is the value of the ... Math Worksheets Name: _____ Date: _____ … So Much More Online! Please visit: EffortlessMath.com Answers Simple Interest 1) $63.00 2) $624.00 ...
Practicing questions on simple interest worksheet help the children to calculate simple interest (S.I.) and amount (A). We will recapitulate the same and know more about it and practice more questions given in the worksheet on simple interest.. While solving the questions on simple interest worksheet we need to remember; the formula for calculating simple interest (S.I.) = (P × R × T)/100 ...
Simple Interest Worksheets. The young bankers will immensely profit from our printable simple interest worksheets! Simple interest is the interest on a specific principal amount of money where some rate of interest is agreed upon. Our pdf mathematical and real-world problems on simple interest get the children in grade 6, grade 7, and grade 8 ...
Simple Interest To remember the calculations for Simple Interest, remember I = Prt I = Interest rate, P = Principal amount, r = rate in percentage, t = time in years. Solve the Simple Interest Problems: 1. $400 If you put money into a savings account that earns $84.00 over seven years at a rate of 3%, how much money did you put into the account? 2.
Simple Interest Practice Questions - Corbettmaths. Welcome. Videos and Worksheets. Primary. 5-a-day.
Explain that some of the information in the question was extraneous and not necessary to solve the problem. For this problem, you don't need to know the years of the loan (eight years) or even the interest rate; you only need to know the beginning and ending balance. ... Deb. "Simple Interest Worksheets With Answers." ThoughtCo. https://www ...
Solution: 9. At what rate of simple interest will $2800 amount to $3500 in 2 years, 3 months? Solution: 10. Find the simple interest at the rate of 8% p.a. for 4 years on that principal which in 6 years, 6 months at the rate of 4% p.a. gives $1600 as simple interest. Solution: 11. Simple interest on a certain sum is 36/72 of the sum.
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?
Use simple interest to find the ending balance. 1) $34,100 at 4% for 3 years $38,192.00 2) $210 at 8% for 7 years $327.60 3) $4,000 at 3% for 4 years ... Create your own worksheets like this one with Infinite Pre-Algebra. Free trial available at KutaSoftware.com. Title: Simple and Compound Interest
Simple Interest Word Problems. Give your seventh graders a leg up on managing their future finances with this math worksheet featuring simple interest word problems. Students will learn the simple interest formula and review an example of how to apply the formula to a real-world example. Then they will set out to find the simple interest earned ...
Solving simple interest word problems involves using the simple interest formula and understanding the given information. The formula for calculating simple interest is: 𝐼 = 𝑃 × 𝑟 × 𝑡. Where: 𝐼 is the interest earned or paid. 𝑃 is the principal amount (the initial amount of money). 𝑟 is the annual interest rate (expressed ...
SIMPLE INTEREST PROBLEMS WITH SOLUTIONS. 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.
In worksheet on simple interest we will solve 10 different types of question. 1. Fill in the Blank: 2. Fill in the Blank: 3. Maria borrowed $ 1400 at the rate of 6% (p.a.) for 8 years and 5 months. Find the amount she paid back. 4.
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%.
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. Problem: To buy a computer, Raquel borrowed $3,000 at 9% ...
The formula for Simple Interest is: I = prt where. I is the interest generated. p is the principal amount that is either invested or owed. r is the rate at which the interest is paid. t is the time that the principal amount is either invested or owed. This type of word problem is not difficult. Just remember the formula and make sure you plug ...
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 :
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.
Here we will learn about simple interest including how to calculate simple interest for increasing and decreasing values, and how to set up, solve and interpret growth and decay problems. There are also simple interest worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you're still stuck.
The amount of interest is given by I = Prt, where P is the principal, r is the annual interest rate in decimal form, and t is the loan period expressed in years. The videos, worksheets, and apps on this page will help you learn simple interest formula, calculate simple interest and solve simple interest problems.
176. $1.50. PDF. Google Apps™. Activity. Printable PDF, Google Slides & Easel by TPT Versions are included in this distance learning ready activity which consists of 11 simple & compound interest problems. It is a self-checking worksheet that allows students to strengthen their skills at calculating both simple and compound interest.
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 he gets after 1 year, 2 years and 3 years? ... Worksheet on Simple Interest. Worksheet on Factors affecting Interest.