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COMPOUND INTEREST Class 8 MATHS Ncert Solutions

Compound interest class 8 maths book solutions.

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  • Class 8 Maths Chapter 8

ML Aggarwal Solutions for Class 8 Maths Chapter 8 Simple and Compound Interest

ML Aggarwal Solutions for Class 8 Maths Chapter 8 Simple and Compound Interest has solutions prepared by experienced teachers at BYJU’S. Mathematics is one such subject where students score high marks with ease, if they practise on a regular basis. These solutions help them to improve conceptual knowledge, which  assists them in acing the exam. For a better understanding of the concepts, students can make use of ML Aggarwal Solutions for Class 8 Maths Chapter 8 Simple and Compound Interest PDF from the links given below.

Chapter 8 discusses problems on finding the rate of interest, principal amount and time period, as per the current ICSE syllabus. Once they are through with the ML Aggarwal Solutions, students can cross-check their answers while solving textbook questions and also clear their doubts instantly. It also enhances their speed of solving problems, which is essential for better performance in academics.

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Access ML Aggarwal Solutions for Class 8 Maths Chapter 8 Simple and Compound Interest

Exercise 8.1

1. Find the simple interest on Rs 4000 at 7.5% p.a. for 3 years 3 months. Also, find the amount.

Principal (P) = Rs 4000

Rate of interest (R) = 7.5% p.a.

= (15 / 2) % p.a.

Time (T) = 3 years 3 months

= 13 / 4 years

Simple Interest (I) = (P × R × T) / 100

= Rs {4000 × (15 / 2) × (13 / 4)} / 100

= Rs (4000 × 15 × 13) / (100 × 2 × 4)

On simplification, we get,

= Rs 5 × 15 × 13

Amount = P + I

= Rs 4000 + Rs 975

= RS 4, 975

2. What sum of money will yield Rs 170.10 as simple interest in 2 years 3 months at 6% per annum?

I = Rs 170.10

T = 2 years 3 months

= 9 / 4 years

P = (I × 100) / (R × T)

= Rs (170.10 × 100) / {6 × (9 / 4)}

On calculating further, we get,

= Rs (170.10 × 100 × 4) / (6 × 9)

= Rs (17010 × 4) / (6 × 9)

= Rs (17010 × 2) / (3 × 9)

= Rs 34020 / 27

3. Find the rate of interest when Rs 800 fetches Rs 130 as a simple interest in 2 years 6 months.

T = 2 years 6 months

= 5 / 2 years

R = (I × 100) / (P × T)

= (130 × 100) / {800 × (5 / 2)} % p.a.

= (130 × 100 × 2) / (800 × 5) % p.a.

= (130 × 2) / 40 %

= 130 / 20 % p.a.

= 6.5% p.a.

Therefore, the required rate of interest is 6.5% p.a.

4. Find the time when simple interest on Rs 3.3 lakhs at 6.5% per annum is Rs 75075.

P = 3.3 lakhs

= Rs 3.3 × 100000

= Rs 330000

R = 6.5% per annum

I = Rs 75075

T = (I × 100) / (P × R)

= (75075 × 100) / (330000 × 6.5) years

= (75075 × 100 × 10) / (330000 × 65) years

On further calculation, we get,

= (75075) / (330 × 65) years

= 1155 / 330 years

= 7 / 2 years

5. Find the sum of money when

(ii) the final amount is Rs 11300 at 4% p.a. for 3 years 3 months.

I = Rs 2356.25

= 29 / 4 % p.a.

= Rs (2356.25 × 100) / (29 / 4) × (5 / 2)

= Rs (2356.25 × 100 × 4 × 2) / (29 × 5)

= Rs (235625 × 8) / (29 × 5)

= Rs (47125 × 8) / 29

= Rs 1625 × 8

(ii) Amount (A) = Rs 11300

Rate (R) = 4% p.a.

Let the principal be Rs x

S.I. = (P × R × T) / 100

= Rs (x × 4 × 13) / (100 × 4)

= Rs 13x / 100

Amount = Principal + Simple Interest

= Rs x + Rs 13x / 100

= Rs (x + 13x) / 100

= Rs (100x + 13x) / 100

= Rs (113x / 100)

But, the amount given is Rs 11300

113x / 100 = 11300

x = 11300 × 100 / 113

x = 100 × 100

Therefore, principal (P) = Rs 10000

Let the sum of money be x

Amount = 3 × Rs x

Interest = Amount – Principal

= Rs 3x – Rs x

= 40 / 3 % p.a.

Time (T) = (I × 100) / (P × R)

= (2x × 100) / x × (40 / 3) years

= (2 × 100 × 3) / 40 years

= (100 × 3) / 20 years

= 5 × 3 years

7. At a certain rate of simple interest Rs 4050 amounts to Rs 4576.50 in 2 years. At the same rate of simple interest, how much would Rs 1 lakh amount to in 3 years?

P = Rs 40000

A = Rs 4576.50

T = 2 years

= Rs 4576.50 – Rs 4050

= Rs 526.50

Let the rate of simple interest = R% per annum

= (526.50 × 100) / (4050 × 2) % p.a.

= (526.50 × 10) / (405 × 2) % p.a.

= 5265 / 810 % p.a.

P = Rs 1 lakh

= Rs 100000

R = 6.5% p.a.

T = 3 years

I = (P × R × T) / 100

= Rs (100000 × 6.5 × 3) / 100

= RS 1000 × 6.5 × 3

Amount = Principal + Interest

= Rs 100000 + Rs 19500

= Rs 119500

8. What sum of money invested at 7.5% p.a. simple interest for 2 years produces twice as much interest as Rs 9600 in 3 years 6 months at 10% p.a. simple interest?

First Case:

Principal (P 1 ) = Rs 9600

Rate (R 1 ) = 10%

Period = (T) = 3 years 6 months

Simple interest = (P × R × T) / 100

= (9600 × 10 × 7) / (100 × 2)

Second case:

Simple interest = Rs 3360 × 2

Rate (R) = 7.5% p.a. and

Period (T) = 2 years

Principal = (S.I × 100) / (R × T)

= (6720 × 100) / (7.5 × 2)

= Rs (6720 × 100 × 10) / (75 × 2)

= 6720000 / 150

Exercise 8.2

1. Calculate the compound interest on Rs 6000 at 10% per annum for two years.

Rate of interest = 10% per annum

Principal for the first year = Rs 6000

Interest for the first year = Rs (6000 × 10 × 1) / 100

Amount at the end of first year = Rs 6000 + Rs 600

Principal for the second year = Rs 6600

Interest for the second year = Rs (6600 × 10 × 1) / 100

Amount for the second year = Rs 6600 + Rs 660

Therefore, compound interest for 2 years = final amount – (original) Principal

= Rs 7260 – Rs 6000

2. Salma borrowed from Mahila Samiti a sum of Rs 1875 to purchase a sewing machine. If the rate of interest is 4% per annum, what is the compound interest that she has to pay after 2 years?

Principal for the first year = Rs 1875

Rate of interest = 4% p.a.

Interest for the first year = Rs (1875 × 4 × 1) / 100

Amount at the end of first year = Rs 1875 + Rs 75

Principal for the second year = Rs 1950

Interest for the second year = Rs (1950 × 4 × 1) / 100

Amount at the end of second year = Rs 1950 + Rs 78

Compound interest paid by Salma = Final amount – (original) Principal

= Rs 2028 – Rs 1875

3. Jacob invests Rs 12000 for 3 years at 10% per annum. Calculate the amount and the compound interest that Jacob will get after 3 years.

Principal for the first year = Rs 12000

Rate of interest = 10% p.a.

Interest for the first year = Rs (12000 × 10 × 1) / 100

Amount at the end of first year = Rs 12000 + Rs 1200

Principal for the second year = Rs 13200

Interest for the second year = Rs (13200 × 10 × 1) / 100

Amount at the end of second year = Rs 13200 + Rs 1320

Principal for the third year = Rs 14520

Interest for the third year = Rs (14520 × 10 × 1) / 100

Amount at the end of third year = Rs 14520 + Rs 1452

Compound interest for 3 year = Final amount – (original) Principal

= Rs 15972 – Rs 12000

4. A man invests Rs 46875 at 4% per annum compound interest for 3 years.

(i) the interest for the first year

(ii) the amount standing to his credit at the end of second year

(iii) the interest for the third year

(i) Principal for the first year = Rs 46875

Rate of interest = 4% per annum

Interest for the first year = Rs (46875 × 4 × 1) / 100

= Rs 46875 / 25

Hence, interest for the first year is Rs 1875

(ii) Amount at the end of first year

= Rs 46875 + Rs 1875

Principal for the second year = Rs 48750

Interest for the second year = Rs (48750 × 4 × 1) / 100

= Rs 48750 / 25

Amount at the end of second year = Rs 48750 + Rs 1950

Hence, the amount at the end of second year is Rs 50700

(iii) Principal for the third year = Rs 50700

Interest for the third year = Rs (50700 × 4 × 1) / 100

= Rs 507 × 4

Hence, the interest for the third year is Rs 2028

5. Calculate the compound interest for the second year on Rs 6000 invested for 3 years at 10% p.a. Also find the sum due at the end of third year.

Amount at the end of second year = Rs 6600 + Rs 660

Compound interest for the second year = Final amount – (original) Principal

Principal for the third year = Rs 7260

Interest for the third year = Rs (7260 × 10 × 1) / 100

Amount at the end of third year = Rs 7260 + Rs 726

6. Calculate the amount and the compound interest on Rs 5000 in 2 years when the rate of interest for successive years is 6% and 8% respectively.

Principal for the first year = Rs 5000

Rate of interest = 6% p.a.

Interest for the first year = Rs (5000 × 6 × 1) / 100

= Rs 50 × 6

Amount at the end of first year = Rs 5000 + Rs 300

Principal for the second year = Rs 5300

Rate of interest = 8% p.a.

Interest for the second year = Rs (5300 × 8 × 1) / 100

= Rs 53 × 8

Amount for the second year = Rs 5300 + Rs 424

Compound interest for two years = Final amount – (original) Principal

= Rs 5724 – Rs 5000

7. Calculate the difference between the compound interest and the simple interest on Rs 20000 in 2 years at 8% per annum.

Principal (P) = Rs 20000

Rate (R) = 8% p.a.

Simple interest (S.I.) = PRT / 100

= Rs (20000 × 8 × 2) / 100

Amount on compound interest

A = P {1 + (R / 100)} n

= RS 20000 {1 + (8 / 100)} 2

On further calculation,

= Rs 20000 × (27 / 25) × (27 / 25)

= Rs 32 × 729

Compound interest = Final amount – (original) Principal

= Rs 23328 – Rs 20000

Difference in compound interest – simple interest

= Rs 3328 – Rs 3200

Exercise 8.3

1. Calculate the amount and compound interest on

(i) Rs 15000 for 2 years at 10% per annum compounded annually.

(iii) Rs 100000 for 9 months at 4% per annum compounded quarterly.

Principal (P) = Rs 15000

Rate (R) = 10% p.a.

Period (n) = 2 years

Amount (A) = P {1 + (R / 100)} n

= Rs 15000 {1 + (10 / 100)} 2

= Rs 15000 × (11 / 10) × (11 / 10)

Compound interest = Amount – Principal

= Rs 18150 – 15000

(ii) Principal (P) = Rs 156250

Rate (R) = 8% p.a. or 4% half-yearly

= 3 half-year

= Rs 156250 {1 + (4 / 100)} 3

= Rs 156250 × (26 / 25) 3

= Rs 156250 × (26 / 25) × (26 / 25) × (26 / 25)

= Rs 175760

= Rs 175760 – Rs 156250

2. Find the difference between the simple interest and compound interest on Rs 4800 for 2 years at 5% per annum, compound interest being reckoned annually.

Principal (P) = Rs 4800

Rate (R) = 5% p.a.

S.I. = PRT / 100

= (4800 × 5 × 2) / 100

And when interest is compounded annually

= Rs 4800 {1 + (5 / 100)} 2

= Rs 4800 × (21 / 20) × (21 / 20)

= Rs 5292 – Rs 4800

Difference in compound interest and simple interest = Rs 492 – Rs 480

3. Find the compound interest on Rs 3125 for 3 years if the rates of interest for the first, second and third year are respectively 4%, 5% and 6% per annum.

Principal (P) = Rs 3125

Rate of interest for continuous 3 years = 4%, 5%, 6%

Period (n) = 3 years

Amount = P {1 + (r / 100)} n

= 3125 {1 + (4 / 100)} {1 + (5 / 100)} {1 + (6 / 100)}

= 3125 × (26 / 25) × (21 / 20) × (53 / 50)

= Rs 14469 / 4

= Rs 3617.25

= Rs 3617. 25 – Rs 3125

= Rs 492. 25

4. Kamla borrowed Rs 26400 from a Bank to buy a scooter at a rate of 15% p.a. compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan?

Money borrowed (P) = Rs 26400

Rate (R) = 15% p.a.

Period (n) = 2 years 4 months

Amount = P {1 + (R / 100)} n

= Rs 26400 {1 + (5 / 100) 2 } × [1 + {15 / (3 × 100)}] 1

= Rs 26400 × (23 / 20) × (23 / 20) × (21 / 20)

= Rs 366597 / 10

= Rs 36659.70

5. Anil borrowed Rs 18000 from Rakesh at 8% per annum simple interest for 2 years. If Anil had borrowed this sum at 8% per annum compound interest, what extra amount would he has to pay?

Money borrowed (P) = Rs 18000

Time (n) = 2 years

Simple Interest = PRT / 100

= Rs (18000 × 8 × 2) / 100

In case of compound interest

= Rs 18000 {1 + (8 / 100)} 2

= Rs 18000 × (27 / 25) 2

= Rs 18000 × (27 / 25) × (27 / 25)

= Rs 104976 / 5

= Rs 20995.20

= Rs 20995.20 – Rs 18000

= Rs 2995.20

Difference between compound interest and simple interest

= Rs 2995.20 – Rs 2880

= Rs 115.20

(i) compounded annually

(ii) compounded half-yearly

Money borrowed (P) = Rs 75000

Rate (R) = 12% p.a. or 6% half- yearly

(i) When the interest compounded yearly

= Rs 75000 {1 + (12 / 100)} {1 + (6 / 100)}

= Rs 75000 × (28 / 25) × (53 / 50)

(ii) When the interest compounded half-yearly

Amount = Rs 75000 {1 + (6 / 100)} 3

= Rs 75000 × (53 / 50) 3

= Rs 75000 × (53 / 50) × (53 / 50) × (53 / 50)

= Rs 446631 / 5

= Rs 89326.20

7. Aryaman invested Rs 10000 in a company, he would be paid interest at 7% per annum compounded annually. Find

(i) the amount received by him at the end of 2 years

(ii) the interest for the 3 rd year

Investment to a company (P) = Rs 10000

Rate of interest (R) = 7% p.a.

= Rs 10000 {1 + (7 / 100)} 2

= Rs 10000 × (107 / 100) × (107 / 100)

(ii) Amount after 3 rd year = Rs 11449 × (107 / 100)

= Rs 12250.43

Interest on the 3 rd year = Rs 12250.43 – 11449

= Rs 801.43

8. What sum of money will amount to Rs 9261 in 3 years at 5% per annum compound interest?

Amount (A) = Rs 9261

Rate of interest = 5% p.a.

Time (T) = 3 years

Principal (P) =?

A = P {1 + (r / 100)} t

9261 = P {1 + (5 / 100)} 3

9261 = P (21 / 20) 3

P = (9261 × 20 × 20 × 20) / (21 × 21 × 21)

Therefore, the sum of money = Rs 8000

Amount (A) = Rs 140608

Rate (R) = 8% p.a. = 4% half-yearly

140608 = P {1 + (4 / 100)} 3

140608 = P (26 / 25) 3

P = 140608 × (25 / 26) × (25 / 26) × (25 / 26)

P = Rs 125000

Principal = Rs 125000

10. At what rate percent will Rs 2000 amount to Rs 2315.25 in 3 years at compound interest?

Principal (P) = Rs 2000

Amount (A) = Rs 2315.25

Let the rate of interest be r% p.a.

A / P = {1 + (r / 100)} n

2315.25 / 2000 = {1 + (r / 100)} 3

{1 + (r / 100)} 3 = (231525) / (100 × 2000)

On calculating, we get,

{1 + (r / 100)} 3 = 9261 / 8000

{1 + (r / 100)} 3 = (21 / 20) 3

1 + (r / 100) = 21 / 20

r / 100 = (21 / 20) – 1

r / 100 = 1 / 20

r = 100 / 20

Therefore, rate of interest = 5% p.a.

Principal (P) = Rs 40000

Amount (A) = Rs 46305

So half yearly, 2n = 2 × (3/2) = 3 years.

A / P = (1 + r / 100) n

46305 / 40000 = (1 + r / 100) 3

(1 + r / 100) 3 = 46305 / 40000

(1 + r / 100) 3 = 9261 / 8000

(1 + r / 100) 3 = (21 / 20) 3

(1 + r / 100) = (21 / 20)

Therefore, rate of interest = 5% for half year.

So, 2 × 5 = 10% per annum.

12. In what time will Rs 15625 amount to Rs 17576 at 4% per annum compound interest?

Amount (A) = Rs 17576

Principal (P) = Rs 15625

Let period be n years

17576 / 15625 = {1 + (4 / 100)} n

(26 / 25) 3 = (26 / 25) n

Therefore, time = 3 years

13. Rs 16000 invested at 10% p.a. compounded semi-annually, amounts to Rs 18522. Find the time period of investment.

Principal (P) = Rs 16000

Amount (A) = Rs 18522

Rate (R) = 10% p.a. or 5% semi-annually

Let period be n half-years

18522 / 16000 = {1 + (5 / 100)} n

9261 / 8000 = (21 / 20) n

(21 / 20) 3 = (21 / 20) n

n = 3 half years

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Class viii math, class 8 - compound interest worksheet - 1.

1. Find the simple interest and amount on Rs. 3000 at 7% per annum for 3 years.

2. What sum of money will yield Rs. 210 as simple interest in 2 years 4 months at 5% per annum?

3. Find the rate of interest when Rs. 750 fetches Rs. 150 as a simple interest in 2 years.

4. When simple interest on Rs 3 lakhs at 6% per annum is Rs. 6000, then find the time.

5. Find the sum of money when simple interest at 11% per annum for 2 years is Rs. 3300.

6. Find the sum of money when the final amount is Rs. 42000 at 5% per annum for 5years.

7. How long will it take a certain sum of money to 4 times of itself at 12 1 ⁄ 2 % per annum simple interest?

8. At a certain rate of simple interest Rs. 3050 amounts to Rs. 3576 in 2 years. At the same rate of simple interest, how much would Rs. 50,000 amount to in 3 years?

9. What sum of money invested at 5% per annum simple interest for 2 years produces twice as much interest as Rs. 8500 in 3 years at 7% per annum simple interest?

10. Calculate the compound interest on Rs. 7200 at 12% per annum for two years.

11. Sobha borrowed from bank a sum of Rs. 2000 to purchase a machine. If the rate of interest is 5% per annum, what is the compound interest that she has to pay after 2 years?

12. Nitin invests Rs. 14000 for 2years at 5% per annum. Calculate the amount and the compound interest that he will get after 2 years.

13. A investor invests Rs. 50,000 at 5% per annum compound interest for 3 years. Calculate (i) The interest for the first year (ii) The amount for his credit at the end of second year (iii) The interest for the third year

14. At what percent per annum compound interest for Rs. 15,625 amounts to Rs. 21,952 in 3 years?

15. Calculate the amount and the compound interest on Rs. 7000 in 2 years when the rate of interest for successive years is 5% and 4% respectively.

16. Calculate the difference between the compound interest and the simple interest on Rs. 25000 in 3 years at 5% per annum.

17. Calculate the amount and compound interest on Rs. 35000 for 3 years at 10% per annum compounded annually.

18. Sita borrowed Rs. 23,000 from a Bank to buy a scooter at a rate of 10% per annum compounded yearly. What amount will she pay at the end of 2 years and 3 months to clear the loan?

19. Shreyas borrowed Rs. 28000 from Raghav at 7% per annum simple interest for 2 years. If Shreyas had borrowed this sum at 7% per annum compound interest, what extra amount would he has to pay?

20. Ramesh borrowed Rs. 85000 from a bank. If the rate of interest is 10% per annum, find the amount he would be paying after 2 1 ⁄ 3 years if the interest is compounded annually.

If you want to download the above worksheet, please click below link.

Compound Interest Worksheet-1 Download the pdf

Class-8 Compound Interest Worksheets

Compound Interest Worksheet - 1

Compound Interest Worksheet - 2

Answer Sheet

Compound-Interest-Answer Download the pdf

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  • Introduction to Graphs | Class 8 Maths
  • What is Linear Graph? Definition, Equation, Examples
  • Cartesian Plane

Chapter 16: Playing with Numbers

  • Playing with Numbers
  • Letters for Digits
  • Divisibility Rules
  • CBSE Class 8 Maths Formulas
  • NCERT Solutions for Class 8 Maths - Chapter Wise
  • RD Sharma Class 8 Solutions for Maths

Compounding is a process of re-investing the earnings in your principal to get an exponential return as the coming next growth is on a bigger principal, following this process of adding earnings to the principal. In this passage of time, the principal will grow exponentially and produce unusual returns.

Sometimes we come across some statements like “one year interest for FD in the bank @ 11 % per annum.” or “Savings account with interest @ 8% per annum”.  When it comes to investment, there are usually two types of interests :

Compound Interest

We already know about Simple Interest(S.I), we will look at Compound Interest(C.I) in detail in this article. First, let’s understand what is compounding through a story. 

A Prisoner was once awaiting his death sentence when the king asked for his last wish. The Prisoner demanded grain of rice (foolish demand right?) but added that the number of grain should be doubled after moving to every square till the last square of the Chess Board ( that is 1 on first, 2 on second, 4 on third, 8 on fourth, 16 on fifth and so on, till the 64th square). The king thought that it is a very small demand and ordered his ministers to have that much amount of rice calculated and provided to the prisoner. The amount calculated was so big that the king lost his entire kingdom and was indebted to prisoner all of his life.

What the prisoner used was the idea of “ Compounding “. Now, let’s define Compound Interest. 

Compound interest (or compounding interest) is the interest on a loan or deposit calculated based on both the initial principal and the accumulated interest from previous periods. Thought to have originated in 17th-century Italy, compound interest can be thought of as “interest on interest,” and will make a sum grow at a faster rate than simple interest, which is calculated only on the principal amount. Let’s see an example before working out the formula, 

Question: Hema borrowed a sum of Rs. 2,00,000 for 2 years at an interest of 8% compounded annually from a bank. Find the Compound Interest and the amount she had to pay at the end of 2 years.

Answer : 

To do it, we need to find interest year by year. Step 1. First let’s find Simple Interest for the first year,           Here, principal P 1 = 2,00,000, R = 8% and T = 1.           SI 1 = SI at 8% on P 1 for one year =  Step 2. So, now the amount received at the end of the first year = SI 1 + P 1 = 16000 + 2,00,000 = 2,16,000. Now, this will become principal.             Thus, P 2 = 2,16,000, R = 8 and T = 1 Step 3. Now we will find simple interest for the second year by taking the total amount at the end of 1st year as principal P 2 .            SI 2 = SI at 8% on P 2 for one year =            This amount now at the end of 2nd year = SI 2 + P 2 = 17280 + 2,16,000 = 2,33,280            Total interest given = 17280 + 16000 = 33280., We need to notice that Principal remains the same in Simple Interest(SI), but in Compound Interest(CI) it recalculated and changes every year. 

Formula for Compound Interest

Let’s derive the formula for compound interest by taking the previous example only, but this time we will not use the values for the variables.     Now, the amount at the end of first year will the principal for the second year, i.e So, now SI for 2nd year Calculating the amount for the 2nd year, Now using the value of P2 in the above equation, Similar if we keep calculating for “n” years, We’ll end up with this formula of amount where P is the initial principal amount, R is the rate and n is the number of years after which the amount is calculated. 

Rate Compounded Annually or Half Yearly

You may notice that, in the beginning, we used “rate compounded yearly”. What does it mean?

It means that interest was compounded once a year. We can also have our interest compounded half-yearly or quarterly. What happens in such cases?

Let’s compare the two cases through an example to see the difference between rates compounded yearly and half-yearly. 

Suppose P = 1000, R = 5% and n = 2 years, Case 1: Interest compound annually. A = 50 + 1000 = 1050. Case 2: Interest Compounded Half-Yearly. P 2 = I + P = 1025 Final amount in this case A = P 2 + I 2 = 1025 + 25.625 = 1050.625  

We can that if interest is compounded half-yearly, compute the interest two times. So the time period becomes twice and the rate is taken half.

So the formula becomes, 

A  = P(1 + \frac{\frac{R}{2}}{100})^{2n}

Applications of Compound Interest Formula and Examples

Growth and decay.

The compound interest concept can be applied to any quantity which increases or decreases such that the amount at the end of each period bears a constant ratio to the amount at the beginning of that period.

(1+\frac{2}{100})

Question: The population of a town increases at the rate of 2.5% annually. If its present population is 3,26,40,000, find the population after 3 years.

We can apply compound interest formula here, Population “P” at the end of 3 years will be,

Appreciation and Depreciation

When the value of an article increases with the passage of time, the article is said to appreciate. When the value of an article decreases with the passage of time, the article is said to depreciate.

For example, if a man buys a car and uses it for two years, it is obvious that the car will not be worth it as a new one. The car will thus have depreciated in value. On the other hand, if a man buys a piece of land, he will probably find that in a few years he will be able to get a better price for it than the price he paid for it. The value of the land will thus have appreciated. When things are difficult to obtain, they have a rarity value and appreciation.

Question: The value of a residential flat constructed at a cost of Rs 10,00,000 is appreciating at the rate of 10% per year annum. What will be its value 3 years after construction?

Value of flat P = 1000000, rate of appreciation = 10, n = 3 After 3 years, let the value of flat be “A”. So, after 3 years value of flat will be 13,31,000. 

The growth of a bacteria if the rate of growth is known:

Question: In a certain experiment, the count of bacteria was increasing at the rate of 2.5% per hour. Initially, the count was 51,20,000. Find the bacteria at the end of 2 hours.

Answer:  

Initial count of bacteria = P = 51,20,000, Increase rate “r” = 2.5 per hour, We want to find the count after 2 hours, i.e; n = 2 Let the final count be “A” So, the final count of bacteria is 53,79,200. 

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  • RD Sharma Class 8 Solutions Chapter 14 - Compound Interest (Ex 14.3) Exercise 14.3
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RD Sharma Class 8 Solutions Chapter 14 - Compound Interest (Ex 14.3) Exercise 14.3 - Free PDF

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Free PDF Download of RD Sharma Class 8 Solutions Chapter 14 - Compound Interest Exercise 14.3 solved by Expert Mathematics Teachers on Vedantu.com. All Chapter 14 - Compound Interest Ex 14.3 Questions with Solutions for RD Sharma Class 8 Maths to help you to revise complete Syllabus and Score More marks. Register for online coaching for IIT JEE (Mains & Advanced) and other Engineering entrance exams. Register Online for Class 8 Science tuition on Vedantu.com to score more marks in CBSE board examination. Vedantu is a platform that provides free CBSE Solutions (NCERT) and other study materials for students.

Class 8 Chapter 14 RD Sharma Solutions - Compound Interest (Ex 14.3) Exercise 14.3

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Students of Class 8 who need access to the RD Sharma Class 8 Solutions Chapter 14 - Compound Interest (Ex 14.3) Exercise 14.3 - Free PDF can now get the same via Vedantu with just a click away. In this particular Exercise 14.3 of Chapter 14 of Class 8, you get to discuss how to compute each of the three quantities which are P, T, and R when there is enough data provided with respect to these. RD Sharma Class 8 Solutions Chapter 14 - Compound Interest (Ex 14.3) Exercise 14.3 - Free PDF is considered to be one of the best study materials when it comes to studying according to the CBSE pattern that is followed in the NCERT textbooks. Each solution will provide you with a detailed explanation of the numerical involved in Exercise 14.3 and hence you can understand the concept in a much easier way.

Features of the RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 Free PDF

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FAQs on RD Sharma Class 8 Solutions Chapter 14 - Compound Interest (Ex 14.3) Exercise 14.3

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RD Sharma Class 8 Solutions Chapter 14 - Compound Interest Ex 14.3 gives each and every question its importance that is needed. This allows the students to check each question in a different manner. For example, one question might have a particular way in which you can solve which cannot be applied to the other question that is mentioned, and hence a new method of solving the question is needed. This is taken care of by the Vedantu experts to cross-check whether the method and its explanation are correct or not in Chapter 14 of Class 8.

3. Is it better to access the RD Sharma Class 8 Solutions Chapter 14 - Compound Interest Exercise 14.3 or the RS Aggarwal version of the same?

RD Sharma Class 8 Solutions Chapter 14 - Compound Interest Exercise 14.3 has its own unique way of solving the questions than the RS Aggarwal solutions for the same Chapter. If you compare both of the solutions it is seen that RS Aggarwal is almost the same as the solutions that are needed for your basic studies in order to understand the questions. However when it comes to RD Sharma Class 8 Solutions Chapter 14 - Compound Interest Exercise 14.3 you see that it has a bit more complex way which is eventually beneficial to those students who have the interest to learn more regarding the topic.

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If you are referring to the RD Sharma Class 8 Solutions Chapter 14 Compound Interest Exercise 14.3 Free PDF then you need not worry as it will cover almost everything that is needed in the CBSE syllabus for you to learn at the level of grade 8. It might have a few complex things being explained but it will help you in the future to understand much more complex things with this base. Hence if you are worried about the coverage of topics you can also check the Vedantu NCERT Solutions for Class 8 Maths that will also have all the questions mentioned along with step-wise equations. 

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There are a number of reasons as to why you should be referring to the RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 Free PDF via Vedantu and can be provided as follows:

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You get a well-explained diagram that will provide a visual answer to the question and give a more detailed view.

The explanation of the chapters and formulas are included along with the questions and the solutions itself so it becomes easier to compare where to use what

You get the free RD Sharma Class 8 Chapter 14, 14.3 PDF to download. 

COMMENTS

  1. RD Sharma Solutions for Class 8 Maths Chapter 14 Compound Interest

    1. Find the compound interest when principal = Rs 3000, rate = 5% per annum and time = 2 years. Solution: Given details are, Principal (p) = Rs 3000 Rate (r) = 5% Time = 2years Interest for the first year = (3000×5×1)/100 = 150 Amount at the end of first year = Rs 3000 + 300 = Rs 3150 Principal interest for the second year = (3150×5×1)/100 = 157.5

  2. Compound Interest Questions with Solutions

    1. Find the compound interest (CI) on Rs. 12,600 for 2 years at 10% per annum compounded annually. Solution: Given, Principal (P) = Rs. 12,600 Rate (R) = 10 Number of years (n) = 2 A = P [1 + (R/100)] n = 12600 [1 + (10/100)] 2 = 12600 [1 + (1/10)] 2 = 12600 [ (10 + 1)/10] 2 = 12600 × (11/10) × (11/10) = 126 × 121 = 15246

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    1. Find the simple interest on Rs 4000 at 7.5% p.a. for 3 years 3 months. Also, find the amount. Solution: Here Principal (P) = Rs 4000 Rate of interest (R) = 7.5% p.a. = (15 / 2) % p.a. Time (T) = 3 years 3 months = years = years = 13 / 4 years Hence, Simple Interest (I) = (P × R × T) / 100 = Rs {4000 × (15 / 2) × (13 / 4)} / 100

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    This set of Class 8 Maths Chapter 8 Multiple Choice Questions & Answers (MCQs) focuses on "Compound Interest". 1. Calculate the simple interest if the principal amount is 50000 and the rate is 2% for 4 years. a) 4000 b) 400 c) 40000 d) 40 View Answer 2. Find the Compound Interest on Rs. 1000 for two years at 2% per annum. a) 20 b) 20.5 c) 20.4

  7. Class-8 Compound Interest and Worksheets

    Solution. Here, rate of interest = 4%, principal for the first year = Rs. 20,000. Interest for the first year = 20,000 × 4 × 1 ⁄ 100 = Rs. 800 Amount at the end of first year = 20,000 + 800 = Rs. 20,800. Principal for the second year = Rs. 20,800 Interest for the second = 20,800 × 4 × 1 ⁄ 100 = Rs. 832

  8. RD Sharma Class 8 Maths Solutions Chapter 14

    Vedantu has provided step-by-step solutions for all exercise questions given in the pdf of Class 8 RD Sharma Chapter 14 - Compound Interest. All the Exercise questions with solutions in Chapter 14 - Compound Interest are given below: Exercise 14.1 Exercise 14.2 Exercise 14.3 Exercise 14.4 Exercise 14.5

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    RS Aggarwal Class 8 Mathematics Solutions for Chapter-11 Compound Interest Last updated date: 26th Jan 2024 • Total views: 585.6k • Views today: 13.85k Download PDF NCERT Solutions CBSE CBSE Study Material Textbook Solutions CBSE Notes Class 8 RS Aggarwal Maths Compound Interest Solutions - Free PDF Download

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    RD Sharma solutions for Class 8 Maths chapter 14 - Compound Interest [Latest edition] Advertisement Solutions for Chapter 14: Compound Interest Below listed, you can find solutions for Chapter 14 of CBSE RD Sharma for Class 8 Maths. Exercise 14.1 Exercise 14.2 Exercise 14.3 Exercise 14.4 Exercise 14.5 Exercise 14.1 [Pages 4 - 5]

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    Key features of Aakash Institute RD Sharma solutions for class 8th Maths Chapter 14 - Compound Interest. Students will get a clear understanding of the concept of Class 8 Mathematics Chapter 14 Compound Interest and will cover all the topic areas with Aakash Institute's solutions. Using Aakash Institute's solutions, students can score a perfect ...

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    RS Aggarwal Solutions Class 8 Ex 11A. RS Aggarwal Class 8 Maths Exercise 11A is about compound interest. Compound interest is the method of addition of interest to the principal sum of money borrowed, i.e. loans or any credit deposit. It results from reinvestment of interests despite paying it out to the creditor or the lender.

  15. CBSE Class 8 Maths Compound Interest

    Compound Interest ,Comparing Quantities - Get topics notes, Online test, Video lectures, Doubts and Solutions for CBSE Class 8 on TopperLearning.

  16. Class-8 Compound Interest Worksheets

    12. Nitin invests Rs. 14000 for 2years at 5% per annum. Calculate the amount and the compound interest that he will get after 2 years. 13. A investor invests Rs. 50,000 at 5% per annum compound interest for 3 years. Calculate. (i) The interest for the first year. (ii) The amount for his credit at the end of second year.

  17. RD Sharma Class 8 Solutions Chapter 14

    Class 8 Chapter 14 RD Sharma Solutions - Compound Interest (Ex 14.2) Exercise 14.2. Class 7 had included simple interest and how to calculate its amount and formula. Even in Class 8 of Chapter 14, we will analyze concepts based on compound interest and the methods to calculate it. Additionally, we will examine the effects of population growth ...

  18. RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1

    Amount after one year = Rs 40,000 + Rs 2,800 = Rs 42,800. Or principal for the second year = Rs 42,800. Interest for the second year. Total interest paid after two years = Rs 2,800 + 2,996 = Rs 5,796. Hope given RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 are helpful to complete your math homework.

  19. NCERT Solutions for Class 8 Maths

    Maths Formulas for Class 8. NCERT Solutions for Class 8 Maths (Download PDF) With the help of NCERT solutions for class 8 maths, you get chapter wise solutions of each and every exercise. Thus, it helps you solve and practice the questions at your convenience and with ease.

  20. Compound Interest

    Answer : To do it, we need to find interest year by year. Step 1. First let's find Simple Interest for the first year, Here, principal P 1 = 2,00,000, R = 8% and T = 1. SI 1 = SI at 8% on P 1 for one year = Step 2. So, now the amount received at the end of the first year = SI 1 + P 1 = 16000 + 2,00,000 = 2,16,000. Now, this will become principal.

  21. RD Sharma Class 8 Solutions Chapter 14

    Exercise 14.5 consists of 8 questions, where the questions are based on the application of compound interest and depreciation-related problems. To have a step-by-step clear knowledge of the concepts, Vedantu provides RD Sharma Class 8 Solutions for Chapter 14, Exercise-14.1 PDF. It has a step-by-step solution for all the 22 questions on the ...

  22. Compound Interest Practice Questions

    Click here for Answers. Practice Questions. Previous: Increasing/Decreasing by a Percentage Practice Questions. Next: Percentages of an Amount (Calculator) Practice Questions. The Corbettmaths Practice Questions on Compound Interest.

  23. RD Sharma Class 8 Solutions Chapter 14

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