Frank Solutions for Class 9 Maths Chapter 3 Compound Interest

Frank Solutions for Class 9 Maths Chapter 3 Compound Interest has problems prepared by experienced teachers at BYJU’S. In a subject like Mathematics, it’s important for students to understand the concepts in depth. Practising the problems on a regular basis is the main key to secure more marks in the examination. For more conceptual knowledge, students can make use of Frank Solutions for Class 9 Maths Chapter 3 Compound Interest PDF, from the links available below.

Chapter 3 consists of problems in finding the amount and compound interest, according to the latest ICSE board guidelines. The solutions are available in PDF format to help students in solving the Frank textbook with ease. This also helps them to improve their problem solving and time management skills, which are important from the exam perspective.

Frank Solutions for Class 9 Maths Chapter 3 Compound Interest Download PDF

 

frank solutions class 9 maths chapter 3 01
frank solutions class 9 maths chapter 3 02
frank solutions class 9 maths chapter 3 03
frank solutions class 9 maths chapter 3 04
frank solutions class 9 maths chapter 3 05
frank solutions class 9 maths chapter 3 06
frank solutions class 9 maths chapter 3 07
frank solutions class 9 maths chapter 3 08
frank solutions class 9 maths chapter 3 09
frank solutions class 9 maths chapter 3 10
frank solutions class 9 maths chapter 3 11
frank solutions class 9 maths chapter 3 12

 

Access Frank Solutions for Class 9 Maths Chapter 3 Compound Interest

1. Find the amount and the compound interest payable annually on the following:

(i) Rs 25000 for FRANK Solutions Class 9 Maths Chapter 3 - 1 years at 10% per annum.

(ii) Rs 32000 for 2 years at FRANK Solutions Class 9 Maths Chapter 3 - 2% per annum.

(iii) Rs 10000 for FRANK Solutions Class 9 Maths Chapter 3 - 3 years at 6% per annum.

(iv) Rs 24000 for FRANK Solutions Class 9 Maths Chapter 3 - 4 years at FRANK Solutions Class 9 Maths Chapter 3 - 5% per annum.

Solution:

(i) Rs 25000 for
FRANK Solutions Class 9 Maths Chapter 3 - 6years at 10% per annum.

Here,

P = Rs 25000, t =
FRANK Solutions Class 9 Maths Chapter 3 - 7years, r = 10%

Now,

Amount after 1 year = P (1+ r / 100)

= 25000 (1 + 10 / 100)

= 25000 (1 + 1 / 10)

On further calculation, we get,

= 25000 (11 / 10)

= 27500

Hence, principle for the next 6 months = Rs 27500

Interest for the next 6 months = (27500 × 6 × 10) / (100 × 12)

= 1375

Hence, amount after
FRANK Solutions Class 9 Maths Chapter 3 - 8years = Rs 27500 + Rs 1375

= Rs 28875

And CI = A – P

= Rs 28875 – Rs 25000

= Rs 3875

(ii) Rs 32000 for 2 years at
FRANK Solutions Class 9 Maths Chapter 3 - 9% per annum.

Here,

P1 = Rs 32000 and r =
FRANK Solutions Class 9 Maths Chapter 3 - 10% = (15 / 2)%

So, Amount after 1 year = P (1 + r / 100)

= 32000 {1 + 15 / (2 × 100)}

= 32000 (1 + 3 / 40)

= 32000 (43 / 40)

We get,

= 34400

Therefore,

P2 = Rs 34400 and r = (15 / 2)%

So, Amount after 2 year = P (1 + r / 100)

= 34400 {1 + 15 / (2 × 100)}

On further calculation, we get,

= 34400 (1 + 3 / 40)

= 34400 (43 / 40)

We get,

= 36980

Hence, Amount = Rs 36980

And CI = A – P

= Rs 36980 – Rs 32000

We get,

= Rs 4980

(iii) Rs 10000 for
FRANK Solutions Class 9 Maths Chapter 3 - 11years at 6% per annum.

Here,

P1 = Rs 10000 and r = 6%

So, Amount after 1 year = P (1 + r / 100)

= 10000 (1 + 6 / 100)

= 10000 (106 / 100)

On simplification, we get,

= 10600

Hence, P2 = Rs 10600 and r = 6%

Amount after 2 year = P (1 + r / 100)

= 10600 (1 + 6 / 100)

= 10600 (106 / 100)

On simplification, we get,

= Rs 11236

Hence, principle for the next 6 months = Rs 11236

Interest for the next 6 months = (11236 × 6 × 6) / (100 × 12)

= 337.08

Hence, amount after
FRANK Solutions Class 9 Maths Chapter 3 - 12years = Rs 11236 + Rs 337.08

= Rs 11573.08

And CI = A – P

= Rs 11573.08 – Rs 10000

= Rs 1573.08

(iv) Rs 24000 for
FRANK Solutions Class 9 Maths Chapter 3 - 13years at
FRANK Solutions Class 9 Maths Chapter 3 - 14% per annum.

Here,

P = Rs 24000, t =
FRANK Solutions Class 9 Maths Chapter 3 - 15years, r =
FRANK Solutions Class 9 Maths Chapter 3 - 16= (15 / 2)%

Now,

Amount after 1 year = P (1 + r / 100)

= 24000 {1 + 15 / (2 × 100)}

= 24000 (1 + 3 / 40)

= 24000 (43 / 40)

We get,

= 25800

Hence, principle for the next 6 months = Rs 25800

Interest for the next 6 months = (25800 × 15 × 6) / (200 × 12)

= 967.50

Hence, amount after
FRANK Solutions Class 9 Maths Chapter 3 - 17years = Rs 25800 + Rs 967.50

= 26767.50

And CI = A – P

= Rs 26767.50 – Rs 24000

= Rs 2767.50

2. Find the amount and the compound interest payable annually on:

(a) Rs 16000 for 2 years at 15% and 12% for the successive years.

(b) Rs 17500 for 3 years at 8%, 10% and 12% for the successive years

Solution:

(a) For first year: P = Rs 16000, R = 15% and T = 1 year

Therefore, interest = Rs (16000 × 15 × 1) / 100

= Rs 2400

And, amount = Rs 16000 + Rs 2400

= Rs 18400

For second year: P = Rs 18400, R = 12% and T = 1 year

Therefore, interest = Rs (18400 × 12 × 1) / 100

= Rs 2208

And, amount = Rs 18400 + Rs 2208

= Rs 20608

Hence, required amount = Rs 20608

And, Compound Interest = A – P

= Rs 20608 – Rs 16000

We get,

= Rs 4608

(b) For first year: P = Rs 17500, R = 8% and T = 1 year

Therefore, interest = Rs (17500 × 8 × 1) / 100

= Rs 1400

And, amount = Rs 17500 + Rs 1400

= Rs 18900

For second year: P = Rs 18900, R = 10% and T = 1 year

Therefore, interest = Rs (18900 × 10 × 1) / 100

= Rs 1890

And, amount = Rs 18900 + Rs 1890

= Rs 20790

For third year: P = Rs 20790, R = 12% and T = 1 year

Therefore, interest = Rs (20790 × 12 × 1) / 100

= Rs 2494.80

And, amount = Rs 20790 + Rs 2494.80

= Rs 23284.80

Hence, required amount = Rs 23284.80

And, Compound Interest = A – P

= Rs 23284.80 – Rs 17500

= Rs 5784.80

3. Calculate the amount and compound interest on Rs 20000 for 3 years at 10% per annum, interest being payable annually.

Solution:

Here,

P1 = Rs 20000 and r = 10%

So, amount after 1 year = P (1 + r / 100)

= 20000 (1 + 10 / 100)

= 20000 (110 / 100)

We get,

= 22000

Hence, P2 = Rs 22000 and r = 10%

Amount after 2 year = P (1 + r / 100)

= 22000 (1 + 10 / 100)

= 22000 (110 / 100)

We get,

= 24200

Hence, P3 = Rs 24200 and r = 10%

Amount after 3 year = P (1 + r / 100)

= 24200 (1 + 10 / 100)

= 24200 (110 / 100)

We get,

= 26620

Therefore, amount = Rs 26620

Also, CI = A – P

= Rs 26620 – Rs 20000

= Rs 6620

4. Compute the compound interest for the third year on Rs 5000 invested for 5 years at 10% per annum, the interest being payable annually.

Solution:

For first year: P = Rs 5000, R = 10% and T = 1 year

Therefore, interest = Rs (5000 × 10 × 1) / 100

= Rs 500

And, amount = Rs 5000 + Rs 500

= Rs 5500

For second year: P = Rs 5500, R = 10% and T = 1 year

Therefore, interest = Rs (5500 × 10 × 1) / 100

= Rs 550

And, amount = Rs 5500 + Rs 550

= Rs 6050

For third year: P = Rs 6050, R = 10% and T = 1 year

Therefore, interest = Rs (6050 × 10 × 1) / 100

= Rs 605

Hence, Compound Interest for third year is Rs 605

5. Rakesh invests Rs 25600 at 5% per annum compound interest payable annually for 3 years. Find the amount standing to his credit at the end of the second year.

Solution:

For first year: P = Rs 25600, R = 5% and T = 1 year

Therefore, interest = Rs (25600 × 5 × 1) / 100

= Rs 1280

And, amount = Rs 25600 + Rs 1280

= Rs 26880

For second year: P = Rs 26880, R = 5% and T = 1 year

Therefore, interest = Rs (26880 × 5 × 1) / 100

= Rs 1344

And, amount = Rs 26880 + Rs 1344

= Rs 28224

Therefore, amount at the end of second year is Rs 28224

6. Find the amount and compound interest on Rs 7500 for FRANK Solutions Class 9 Maths Chapter 3 - 18 years at 8%, payable semi-annually.

Solution:

Here,

P1 = Rs 7500 and rate of interest for half year (r) = 4%

So, amount after half year = P (1 + r / 100)

= 7500 (1 + 4 / 100)

= 7500 (104 / 100)

We get,

= 7800

Hence, P2 = Rs 7800 and r = 4%

Amount after 1 year = P (1 + r / 100)

= 7800 (1 + 4 / 100)

= 7800 (104 / 100)

We get,

= 8112

Hence, P3 = Rs 8112 and r = 4%

Amount after
FRANK Solutions Class 9 Maths Chapter 3 - 19year = P (1 + r / 100)

= 8112 (1 + 4 / 100)

= 8112 (104 / 100)

We get,

= 8436.48

Therefore, amount = Rs 8436.48

Also, CI = A – P

= Rs 8436.48 – Rs 7500

= Rs 936.48

7. A man invests Rs 24000 for two years at compound interest, if his money amounts to Rs 27600 after one year, find the amount at the end of second year.

Solution:

Amount after 1 year = P (1 + r / 100)

27600 = 24000 (1 + r / 100)

(1 + r / 100) = 27600 / 24000

We get,

(1 + r / 100) = 23 / 20

On further calculation, we get,

r / 100 = (23 / 20) – 1

We get,

r / 100 = 3 / 20

r = (100 × 3) / 20

r = 15

Amount after 2 year = P (1 + r / 100)

= 27600 {1 + (15 / 100)}

= 27600 (115 / 100)

We get,

= 31740

Therefore, the amount at the end of second year is Rs 31740

8. How much will Rs 14000 amounts to 2 years at compound interest, if the rates for the successive years be 5% and 8% respectively?

Solution:

Here,

P1 = Rs 14000 and r = 5%

So, Amount after 1 year = P (1 + r / 100)

= 14000 (1 + 5 / 100)

= 14000 (105 / 100)

We get,

= 14700

Hence, P2 = Rs 14700 and r = 8%

Amount after 2 year = P (1 + r / 100)

= 14700 (1 + 8 / 100)

= 14700 (108 / 100)

We get,

= 15876

Therefore, amount = Rs 15876

9. Find the amount and the compound interest on the following:

(i) Rs 8000 for 3 years at 10% per annum compounded annually

(ii) Rs 15000 for 2 years at 8% per annum compounded semi-annually

(iii) Rs 12000 for FRANK Solutions Class 9 Maths Chapter 3 - 20 years at 5% per annum compounded annually

(iv) Rs 25000 for 2 years at 6% per annum compounded semi-annually

(v) Rs 16000 for 3 years at 10%, 8% and 6% for successive years

Solution:

(i) Rs 8000 for 3 years at 10% per annum compounded annually

Here,

P = Rs 8000, t = 3 years, r = 10%

Now,

Amount = P (1 + r / 100)t

= 8000 (1 + 10 / 100)3

= 8000 (11 / 10)3

= 8000 × (1331 / 1000)

We get,

= 10648

Therefore, amount = Rs 10648

Also, CI = A – P

= Rs 10648 – Rs 8000

= Rs 2648

(ii) Rs 15000 for 2 years at 8% per annum compounded semi-annually

Here,

P = Rs 15000, t = 2 years, r = 8%

Since interest is compounded semi-annually, so

Amount = P (1 + r / 200)2t

= 15000 (1 + 8 / 200)4

= 15000 (26 / 25)4

= 15000 × (26 / 25) × (26 / 25) × (26 / 25) × (26 / 25)

On simplification, we get,

= 17547.88

Therefore, amount = Rs 17547.88

Also, CI = A – P

= Rs 17547.88 – Rs 15000

= Rs 2547.88

(iii) Rs 12000 for
FRANK Solutions Class 9 Maths Chapter 3 - 21years at 5% per annum compounded annually

Here,

P = Rs 12000, t =
FRANK Solutions Class 9 Maths Chapter 3 - 22years, r = 5%

Now,

Amount after 1 year = P (1 + r / 100)t

= 12000 (1 + 5 / 100)

On simplification, we get,

= 12000 (105 / 100)

= 12600

Now, interest for the next half year = (12600 × 5) / (100 × 2)

= 315

Therefore, amount = Rs 12600 + Rs 315

= Rs 12915

Also, CI = A – P

= Rs 12915 – Rs 12000

= Rs 915

(iv) Rs 25000 for 2 years at 6% per annum compounded semi-annually

Here,

P = Rs 25000, t = 2 years, r = 6%

Since interest is compounded semi-annually,

Amount = P (1 + r / 200)2t

= 25000 (1 + 6 / 200)4

= 25000 (103 / 100)4

On simplification, we get,

= 28137.72

Hence, amount = Rs 28137.72

Also, CI = A – P

= Rs 28137.72 – Rs 25000

= Rs 3137.72

(v) Rs 16000 for 3 years at 10%, 8% and 6% for successive years

Here,

P = Rs 16000, t = 3 years, r = 10%, 8%, 6% successively

Now,

Amount = P (1 + r1 / 100) (1 + r2 / 100) (1 + r3 / 100)

= 16000 (1 + 10 / 100) (1 + 8 / 100) (1 + 6 / 100)

On simplification, we get,

= 16000 (11 / 10) (108 / 100) (106 / 100)

= 20148.48

Therefore, Amount = Rs 20148.48

Also, CI = A – P

= Rs 20148.48 – Rs 16000

= Rs 4148.48

10. Find the amount and compound interest on Rs 15000 in FRANK Solutions Class 9 Maths Chapter 3 - 23 years at 10% p.a. compounded annually

Solution:

Here,

P = Rs 15000, t =
FRANK Solutions Class 9 Maths Chapter 3 - 24years, r = 10%

Now, Amount after 2 year = P (1 + r / 100)t

= 15000 (1 + 10 / 100)2

= 15000 (11 / 10)2

We get,

= 18150

Now, interest for the next half year = (18150 × 10) / (100 × 2)

= 907.5

Therefore, Amount = Rs 18150 + Rs 907.50

= Rs 19057.50

Also, CI = A – P

= Rs 19057.50 – Rs 15000

= Rs 4057.50

11. Find the amount on Rs 36000 in 2 years 15% p.a. compounded annually.

Solution:

Here,

P = Rs 36000, t = 2 years, r = 15%

Now,

Amount = P (1 + r / 100)t

= 36000 (1 + 15 / 100)2

= 36000 (115 / 100)2

We get,

= 47610

Therefore, amount = Rs 47610

12. Find the amount and compound interest on Rs 50000 in FRANK Solutions Class 9 Maths Chapter 3 - 25 years at 8% p.a. compounded half-yearly.

Solution:

Here,

P = Rs 50000, t =
FRANK Solutions Class 9 Maths Chapter 3 - 26years, r = 8%

Since interest is compounded half-yearly,

So, amount = P (1 + r / 200)2t

= 50000 (1 + 8 / 200)3

= 50000 (104 / 100)3

We get,

= 56243.20

Therefore, Amount = Rs 56243.20

Also, CI = A – P

= Rs 56243.20 – Rs 50000

= Rs 6243.20

13. How much will Rs 25000 amount to in 2 years at compound interest, if the rates for 1st and 2nd years be 4% and 5% p.a. respectively?

Solution:

Here,

P = Rs 25000, t = 2 years, r = 4%, 5% successively

Now,

Amount = P (1 + r1 / 100) (1 + r2 / 100)

= 25000 (1 + 4 / 100) (1 + 5 / 100)

On further calculation, we get,

= 25000 (104 / 100) (105 / 100)

= 27300

Therefore, Amount = Rs 27300

14. Find compound interest on Rs 31250 for 3 years, if the rates of interest for 1st, 2nd and 3rd years be 8%, 10% and 12% respectively.

Solution:

Here,

P = Rs 31250, t = 3 years, r = 8%, 10%, 12% successively

Now,

Amount = P (1 + r1 / 100) (1 + r2 / 100) (1 + r3 / 100)

= 31250 (1 + 8 / 100) (1 + 10 / 100) (1 + 12 / 100)

On further calculation, we get,

= 31250 (108 / 100) (110 / 100) (112 / 100)

= 41580

Therefore, Amount = Rs 41580

15. Calculate the rate percent when Rs 28000 amount to Rs 30870 in 2 years at compounded annually.

Solution:

Here,

P = Rs 28000, A = 30870, t = 2 years

Now,

Amount = P (1 + r / 100)t

30870 = 28000 (1 + r / 100)2

(1 + r / 100)2 = (30870 / 28000)

We get,

(1 + r / 100)2 = (441 / 400)

(1 + r / 100)2 = (21 / 20)2

Hence,

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

r / 100 = (21 / 20) – 1

r / 100 = 1 / 20

r = 100 / 20

r = 5

Therefore, rate of interest is 5%

Leave a Comment

Your Mobile number and Email id will not be published.

*

*