simple interest problem solving worksheet

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 Problems

In 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 Problems

Interest 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 Interest

Example 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)?

• Find the amount of interest earned by $8000 invested at 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 two 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. He paid his father $360 in interest after 3 years, how much did he borrow?
• A couple invested $6,000 of his $20,000 lottery earning in bonds. How much do they have left to invest in stocks?
• A college student wants to invest the $12,000 inheritance he received and use the annual interest earned to pay for his tuition cost of $945. The highest interest offered by a bank is 6% annual simple interest. At this rate, he cannot earn the needed $945, so he decides to invest some of the money in a riskier, but more profitable, investment offering a 9% return. How much should he invest in 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 interest was $3,600 that year. How much did the credit union loan out at each rate?

How To Solve Interest Problems Using The Simple Interest Formula?

• If you invest $3,500 in a savings account that pays 4% in simple interest, how much interest will you earn after 3 years? What will the new balance be?
• You borrow $6,000 from a loan shark. If you owe $7,200 in 18 months, what would be the simple interest rate?

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 Rate

How 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 Loan

A 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 INTEREST

Problem 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

• Sat Math Practice
• SAT Math Worksheets
• PEMDAS Rule
• BODMAS rule
• GEMDAS Order of Operations
• Math Calculators
• Transformations of Functions
• Order of rotational symmetry
• Lines of symmetry
• Compound Angles
• Quantitative Aptitude Tricks
• Trigonometric ratio table
• Word Problems
• Times Table Shortcuts
• 10th CBSE solution
• PSAT Math Preparation
• Privacy Policy
• Laws of Exponents

Recent Articles

Sat math resources (videos, concepts, worksheets and more).

Sep 16, 24 11:07 AM

Digital SAT Math Problems and Solutions (Part - 42)

Sep 16, 24 09:42 AM

Digital SAT Math Problems and Solutions (Part - 41)

Sep 14, 24 05:55 AM

• Math Article

Simple Interest Questions

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. 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 Answers

1. 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

• What will be the ratio of simple interest earned by a certain amount at the same interest rate for 4 years and that for 12 years?
• A sum of money amounts to Rs. 1008 in two years and to Rs. 1164 in three and half years. Estimate the sum and the rate of interest.
• Suppose you borrowed Rs. 10,000 from the bank at a 13% rate for 4 years. Find the interest you will pay on this loan.
• What will be simple interest for 1 yr and 8 months on a sum of Rs. 28500 at the rate of 12% per annum?
• The simple interest on a sum of money will be Rs.6000 after ten years. If the principal is triple after five years, what will be the total interest at the end of the tenth year?

Leave a Comment Cancel reply

Your 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 PDFs

Register 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 Interest

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.

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:

• I represents the simple interest
• A represents the final amount.
• P represents the original amount. 
• r is the percentage rate (written as a decimal). 
• t represents the time period (number of intervals).

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 interest

In order to calculate simple interest:

• State the formula needed and the value of each variable.
• Substitute the values into the formula.
• Solve the equation.

Explain how to calculate simple interest in 3 steps.

Simple interest worksheet

Get your free simple interest worksheet of 20+ questions and answers. Includes reasoning and applied questions.

Related lessons on simple and compound interest

Simple 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 and compound interest
• Compound interest
• Depreciation

Simple interest examples

Example 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 decrease

A 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?

• t= 2 \times 12 = 24 (remember there are 12 months in 1 year)

A=7600(1+0.01\times{24})

Example 5: Simple interest – different percentages

A 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)

• r_{1}=0.002
• r_{2}=-0.0018

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

• Applying the incorrect formula to the question

This is a very common mistake where the simple interest on an amount is calculated instead of using the compound interest formula.

• Incorrect percentage change due to different time scales

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 incorrect value for the percentage change

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:

• Is the value increasing or decreasing?

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 questions

1. 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 checklist

You have now learned how to:

• Solve problems involving percentage change, including: percentage increase, decrease and original value problems and simple interest in financial mathematics
• Set up, solve and interpret the answers in growth and decay problems, including compound interest {and work with general iterative processes}

The next lessons are

• Types of numbers
• Standard form

Still stuck?

Prepare your KS4 students for maths GCSEs success with Third Space Learning. Weekly online one to one GCSE maths revision lessons delivered by expert maths tutors.

Find out more about our GCSE maths tuition programme.

Privacy Overview

#1 State Test Prep Blended & Online Programs

• 888-309-8227
• 732-384-0146

Seventh grade math - Simple interest

Simple 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 Interest

Math ELA Grade 7 – Common Core

Simple Interest Calculator

Find More…

Related Topics

How 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.

• 2024 © Lumos Learning
• Privacy Policy - Terms of Service - Disclaimers

PARCC® is a registered trademark of PARCC, Inc. Lumos Learning, is not owned by or affiliated in any fashion with PARCC, Inc... Read More

PARCC® is a registered trademark of PARCC, Inc. Lumos Learning, is not owned by or affiliated in any fashion with PARCC, Inc., the Partnership for the Assessment of Readiness for College and Careers, nor any state of the Union. Neither PARCC, Inc., nor The Partnership for the Assessment of Readiness for College and Careers, nor any member state has endorsed this product. No portion of any fees or charges paid for any products or services Lumos Learning offers will be paid or inure to the benefit of PARCC, Inc., or any state of the Union

SBAC is a copyright of The Regents of the University of California – Smarter Balanced Assessment Consortium, which is not aff... Read More

SBAC is a copyright of The Regents of the University of California – Smarter Balanced Assessment Consortium, which is not affiliated to Lumos Learning. The Regents of the University of California – Smarter Balanced Assessment Consortium, was not involved in the production of, and does not endorse these products or this site.

ACT® Aspire™ is a registered trademark of ACT Aspire LLC., which is not affiliated to Lumos Learning. ACT Aspire LLC, was not... Read More

ACT® Aspire™ is a registered trademark of ACT Aspire LLC., which is not affiliated to Lumos Learning. ACT Aspire LLC,was not involved in the production of, and does not endorse these products or this site.

Florida Department of Education is not affiliated to Lumos Learning. Florida department of education, was not involved in the... Read More

Florida Department of Education is not affiliated to Lumos Learning. Florida department of education, was not involved in the production of, and does not endorse these products or this site.

Indiana Department of Education is not affiliated to Lumos Learning. Indiana department of education, was not involved in the... Read More

Indiana Department of Education is not affiliated to Lumos Learning. Indiana department of education, was not involved in the production of, and does not endorse these products or this site.

Mississippi Department of Education is not affiliated to Lumos Learning. Mississippi department of education, was not involved... Read More

Mississippi Department of Education is not affiliated to Lumos Learning. Mississippi department of education, was not involved in the production of, and does not endorse these products or this site.

Ohio Department of Education is not affiliated to Lumos Learning. Ohio department of education, was not involved in the prod... Read More

Ohio Department of Education is not affiliated to Lumos Learning. Ohio department of education, was not involved in the production of, and does not endorse these products or this site.

Tennessee Department of Education is not affiliated to Lumos Learning. Tennessee department of education, was not involved... Read More

Tennessee Department of Education is not affiliated to Lumos Learning. Tennessee department of education, was not involved in the production of, and does not endorse these products or this site.

Georgia Department of Education is not affiliated to Lumos Learning. Georgia department of education, was not involved... Read More

Georgia Department of Education is not affiliated to Lumos Learning. Georgia department of education, was not involved in the production of, and does not endorse these products or this site.

Missouri Department of Education is not affiliated to Lumos Learning. Missouri department of education, was not involved... Read More

Missouri Department of Education is not affiliated to Lumos Learning. Missouri department of education, was not involved in the production of, and does not endorse these products or this site.

Louisiana Department of Education is not affiliated to Lumos Learning. Louisiana department of education, was not involved... 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 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? 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! Comments

Share this page: What’s this?

• Preschool Activities
• Kindergarten Math
• 1st Grade Math
• 2nd Grade Math
• 3rd Grade Math
• 4th Grade Math
• 5th Grade Math
• 6th Grade Math
• 7th Grade Math
• 8th Grade Math
• 9th Grade Math
• 10th Grade Math
• 11 & 12 Grade Math
• Concepts of Sets
• Probability
• Boolean Algebra
• Math Coloring Pages
• Multiplication Table
• Cool Maths Games
• Math Flash Cards
• Online Math Quiz
• Math Puzzles
• Binary System
• Math Dictionary
• Conversion Chart
• Homework Sheets
• Math Problem Ans
• Free Math Answers
• Printable Math Sheet
• Funny Math Answers
• Employment Test
• Math Patterns
• Link Partners
• Privacy Policy

Recent Articles

Adding 1-digit number | understand the concept one digit number.

Sep 17, 24 02:25 AM

Counting Before, After and Between Numbers up to 10 | Number Counting

Sep 17, 24 01:47 AM

Worksheet on Three-digit Numbers | Write the Missing Numbers | Pattern

Sep 17, 24 12:10 AM

Arranging Numbers | Ascending Order | Descending Order |Compare Digits

Sep 16, 24 11:24 PM

Worksheet on Tens and Ones | Math Place Value |Tens and Ones Questions

Sep 16, 24 02:40 PM

© and ™ math-only-math.com. All Rights Reserved. 2010 - 2024.