# RS Aggarwal Solutions for Class 8 Maths Chapter 11 - Compound Interest

RS Aggarwal Solutions For Class 8 Maths Chapter 11 Compound Interest is provided here. You can download the pdf of RS Aggarwal Solutions for Class 8 MathsÂ Chapter 11 Compound Interest from the given links. RS Aggarwal Solutions helps students to attain a good score in the examinations, while also providing extensive knowledge about the subject, as Class 8 is a critical stage in their academic career. Therefore, we at BYJUâ€™S provide answers to all questions uniquely and briefly.

RS Aggarwal Solutions For Class 8 Maths also provides the foundation to higher studies. RS Aggarwal Solutions For Class 8 Maths Chapter 11 Compound Interest provides the concepts in an effective way.Here we will learn about different methods of calculating the Compound Interest.

## Exercise 11A

1. Find the amount and the compound interest on â‚¹ 2500 for 2 years at 10% per annum, compounded annually.

Solution:

Given:

Present value= â‚¹ 2500

Interest rate= 10% per annum

Time=2 years

To find the amount we have the formula,

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

Where P is present value, r is rate of interest, n is time in years.

Now substituting the values in above formula we get,

âˆ´ A = 2500 (1 + 10/100)2

â‡’ A = 2500 (11/10)2

â‡’ A = 2500 (121/100)

â‡’ A = 25 (121)

â‡’ A = â‚¹ 3025

âˆ´ Compound interest = A â€“ P

= 3025 â€“ 2500= â‚¹ 525

2. Find the amount and the compound interest on â‚¹ 15625 for 3 years at 12% per annum, compounded annually.

Solution:

Given:

Present value= â‚¹ 15625

Interest rate= 12% per annum

Time=3 years

To find the amount we have the formula,

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

Where P is present value, r is rate of interest, n is time in years.

Now substituting the values in above formula we get,

âˆ´ A = 15625 (1 + 12/100)3

â‡’ A = 15625 (112/100)3

â‡’ A = 15625 (28/25)3

â‡’ A = 15625 (21952/15625)

â‡’ A = â‚¹ 21925

âˆ´ Compound interest = A â€“ P

= 21925 â€“ 15625 = â‚¹ 6327

3. Find the difference between the simple interest and the compound interest on â‚¹ 5000 for 2 years at 9% per annum.

Solution:

Given:

Present value= â‚¹ 5000

Interest rate= 9% per annum

Time=2 years

Now find compound interest,

To find the amount we have the formula,

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

Where P is present value, r is rate of interest, n is time in years.

Now substituting the values in above formula we get,

âˆ´ A = 5000 (1 + 9/100)2

â‡’ A = 5000 (109/100)2

â‡’ A = 5000 (1.09)2

â‡’ A = 5000 (1.1881)

â‡’ A = â‚¹ 5940.5

âˆ´ Compound interest = A â€“ P

= 5940.5 â€“ 5000 = â‚¹ 940.5

Now find the simple interest,

Simple interest (SI) = PTR/100

Where P is principle amount, T is time taken, R is rate per annum

SI = (5000 Ã— 2 Ã— 9) / 100

= 50 Ã— 2 Ã— 9

= â‚¹ 900

Difference between compound interest and simple interest is = (CI â€“ SI)

= 940.5 â€“ 900

= â‚¹ 40.5

4. Ratna obtained a loan of â‚¹ 25000 from the Syndicate bank to renovate her house. If the rate of interest is 8% per annum. What amount will she have to pay to the bank after 2 years to discharge her debt?

Solution:

Given:

Present value= â‚¹ 25000

Interest rate= 8% per annum

Time=2 years

To find the amount we have the formula,

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

Where P is present value, r is rate of interest, n is time in years.

Now substituting the values in above formula we get,

âˆ´ A = 25000 (1 + 8/100)2

â‡’ A = 25000 (108/100)2

â‡’ A = 25000 (1.08)2

â‡’ A = 25000(1.1664)

â‡’ A = â‚¹ 29160

Ratna has to pay â‚¹ 29160

## Exercise 11B

By using the formula, find the amount and compound interest on:

1. â‚¹ 6000 for 2 years at 9% per annum compounded annually.

Solution:

Given:

Present value= â‚¹ 6000

Interest rate= 9% per annum

Time=2 years

To find the amount we have the formula,

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

Where P is present value, r is rate of interest, n is time in years.

Now substituting the values in above formula we get,

âˆ´ A = 6000 (1 + 9/100)2

â‡’ A = 6000 (109/100)2

â‡’ A = 6000 (1.09)2

â‡’ A = 7128.6

â‡’ A = â‚¹ 7128.6

âˆ´ Compound interest = A â€“ P

= 7128.6 â€“ 6000= â‚¹ 1128.6

2. â‚¹ 10000 for 2 years at 11% per annum compounded annually.

Solution:

Given:

Present value= â‚¹ 10000

Interest rate= 11% per annum

Time=2 years

To find the amount we have the formula,

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

Where P is present value, r is rate of interest, n is time in years.

Now substituting the values in above formula we get,

âˆ´ A = 10000 (1 + 11/100)2

â‡’ A = 10000 (111/100)2

â‡’ A = 10000 (111/100) (111/100)

â‡’ A = 111 Ã— 111

â‡’ A = â‚¹ 12321

âˆ´ Compound interest = A â€“ P

= 12321 â€“ 10000= â‚¹ 2321

3. â‚¹ 31250 for 3 years at 8% per annum compounded annually.

Solution:

Given:

Present value= â‚¹ 31250

Interest rate= 8% per annum

Time=3 years

To find the amount we have the formula,

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

Where P is present value, r is rate of interest, n is time in years.

Now substituting the values in above formula we get,

âˆ´ A = 31250 (1 + 8/100)3

â‡’ A = 31250 (1+2/25)3

â‡’ A = 31250 (27/25)3

â‡’ A = 31250 Ã— 19683/15625 = 2 Ã— 19683

â‡’ A = â‚¹ 39366

âˆ´ Compound interest = A â€“ P

= 39366 â€“ 31250= â‚¹ 8116

4. â‚¹ 10240 for 3 years at 12 Â½ % per annum compounded annually.

Solution:

Given:

Present value= â‚¹ 10240

Interest rate= 12 Â½ % per annum = 25/2%

Time=3 years

To find the amount we have the formula,

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

Where P is present value, r is rate of interest, n is time in years.

Now substituting the values in above formula we get,

âˆ´ A = 10240 (1 + (25/2)/100)3

â‡’ A = 10240 (1+1/8)3

â‡’ A = 10240 (9/8)3

â‡’ A = 31250 Ã— 729/512 = 20 Ã— 729

â‡’ A = â‚¹ 14580

âˆ´ Compound interest = A â€“ P

= 14580 â€“ 10240= â‚¹ 4340

5. â‚¹ 62500 for 2 years 6 months at 12 % per annum compounded annually.

Solution:

Given:

Present value= â‚¹ 62500

Interest rate= 12 % per annum

Time=2 years 6 months = (2 + Â½) years= 5/2 years

To find the amount we have the formula,

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

Where P is present value, r is rate of interest, n is time in years.

Now substituting the values in above formula we get,

âˆ´ A = 62500 (1 + 12/100)2 [1 + (1/2 Ã— 12)/100]

â‡’ A = 62500 (1+3/25)2 (1+6/100)

â‡’ A = 62500 (28/25)2 (106/100)

â‡’ A = 625 Ã— 784/625 Ã— 106 = 784 Ã— 106

â‡’ A = â‚¹ 83104

âˆ´ Compound interest = A â€“ P

= 83104 â€“ 62500 = â‚¹ 20604

6. â‚¹ 9000 for 2 years 4 months at 10 % per annum compounded annually.

Solution:

Given:

Present value= â‚¹ 9000

Interest rate= 10 % per annum

Time=2 years 6 months = (2 + Â½) years= 5/2 years

To find the amount we have the formula,

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

Where P is present value, r is rate of interest, n is time in years.

Now substituting the values in above formula we get,

âˆ´ A = 9000 (1 + 10/100)2 [1 + (1/3 Ã— 10)/100]

â‡’ A = 9000 (1+1/10)2 (1+1/30)

â‡’ A = 9000 (11/10)2 (31/30)

â‡’ A = 9000 Ã— 121/100 Ã— 31/30 = 9 Ã— 121 Ã— 31/3

â‡’ A = â‚¹ 11253

âˆ´ Compound interest = A â€“ P

= 11253 â€“ 6000 = â‚¹ 2253

7. Find the amount of â‚¹ 8000 for 2 years compounded annually and the rates being 9% per annum during the first year and 10% per annum during the second year.

Solution:

Given:

Present value= â‚¹ 8000

Interest rate for first year, p= 9 % per annum

Interest rate for second year, q= 10 % per annum

Amount (A) = P Ã— (1+p/100) Ã— (1+q/100)

Now substituting the values in above formula we get,

âˆ´ A = 8000 Ã— (1+9/100) Ã— (1+10/100)

â‡’ A = 8000 Ã— (109/100) Ã— (11/10)

â‡’ A = 8 Ã— 109 Ã— 11

â‡’ A = â‚¹ 9592

8. Anand obtained a loan of â‚¹ 125000 from the Allahabad bank for buying computers. The bank charges compound interest at 8% per annum, compounded annually. What amount will have to pay after 3 years to clear the debt?

Solution:

Given:

Present value= â‚¹ 125000

Interest rate= 8 % per annum

Time= 3 years

To find the amount we have the formula,

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

Where P is present value, r is rate of interest, n is time in years.

Now substituting the values in above formula we get,

âˆ´ A = 125000 (1 + 8/100)3

â‡’ A = 125000 (1+108/100)3

â‡’ A = 125000 (108/100) (108/100) (108/100)

â‡’ A = 125000 Ã— 1259712/1000 = 1259712/8

â‡’ A = â‚¹ 157464

Anand has to pay â‚¹ 157464

9. Three years ago, Beeru purchased a buffalo from Surjeet for â‚¹ 11000. What payment

Will discharge his debt now, the rate of interest being 10% per annum, compounded annually?

Solution:

Given:

Present value= â‚¹ 11000

Interest rate= 10% per annum

Time=3 years

To find the amount we have the formula,

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

Where P is present value, r is rate of interest, n is time in years.

Now substituting the values in above formula we get,

âˆ´ A = 11000 (1 + 10/100)3

â‡’ A = 11000 (11/10)3

â‡’ A = 11000 (1331/1000)

â‡’ A = 11 (1331)

â‡’ A = â‚¹ 14614

âˆ´ Beeru has to pay â‚¹ 14614

10. Shubhalaxmi took a loan of â‚¹ 18000 from Surya Finance to purchase a TV set. If the company charges compound interest at 12% per annum during the first year and 12 Â½ % per annum during the second year, how much will she have to pay after 2 years?

Solution:

Given:

Present value= â‚¹ 18000

Interest rate for first year, p= 12 % per annum

Interest rate for second year, q= 12 Â½ % per annum

Amount (A) = P Ã— (1+p/100) Ã— (1+q/100)

Now substituting the values in above formula we get,

âˆ´ A = 18000 Ã— (1+12/100) Ã— (1 + (25/2)/100)

â‡’ A = 8000 Ã— (112/100) Ã— (1+25/200)

â‡’ A = 8000 Ã— (112/100) Ã— (1+1/8)

â‡’ A = 180 Ã— 14 Ã— 8

â‡’ A = â‚¹ 22680

Shubhalaxmi has to pay â‚¹ 22680

11. Neha borrowed â‚¹ 24000 from the state bank of India to buy a scooter. If the rate of interest be 10% per annum compounded annually, what payment will she have to make after 2 years 3 months?

Solution:

Given:

Present value= â‚¹ 24000

Interest rate= 10 % per annum

Time= 2 years 3 months = 2 Â¼ years

To find the amount we have the formula,

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

Where P is present value, r is rate of interest, n is time in years.

Now substituting the values in above formula we get,

âˆ´ A = 24000 (1 + 10/100)2 [1 + (1/4 Ã— 10)/100]

â‡’ A = 24000 (1+1/10)2 (1+1/40)

â‡’ A = 24000 (11/10)2 (41/40)

â‡’ A = 24000 Ã— 121/100 Ã— 41/40 = 24 Ã— 121 Ã— 41/4

â‡’ A = â‚¹ 29766

âˆ´ Neha has to pay â‚¹ 29766

12. Abhay borrowed â‚¹ 16000 at 7 Â½ % per annum simple interest. On the same day, he lent it to gurmeet at the same rate but compounded annually. What does he gain at the end of 2 years?

Solution:

Given:

Present value= â‚¹ 16000

Interest rate= 7 Â½ % per annum= 15/2 %

Time=2 years

Now find compound interest,

To find the amount we have the formula,

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

Where P is present value, r is rate of interest, n is time in years.

Now substituting the values in above formula we get,

âˆ´ A = 16000 (1 + (15/2)/100)2

â‡’ A = 16000 (1+3/40)2

â‡’ A =16000 (43/40)2

â‡’ A = 16000 (1894/1600)

â‡’ A = â‚¹ 18490

âˆ´ Compound interest = A â€“ P

= 18490 â€“ 16000 = â‚¹ 2490

Now find the simple interest,

Simple interest (SI) = PTR/100

Where P is principle amount, T is time taken, R is rate per annum

SI = (16000 Ã— (15/2) Ã— 2) / 100

= 160 Ã— 15

= â‚¹ 2400

Abhay gains at the end of 2 year= (CI â€“ SI)

= 2490 â€“ 2400

= â‚¹ 90

13. The simple interest on a sum of money for 2 years at 8% per annum is â‚¹ 2400. What will be the compound interest on that sum at the rate and for the same period?

Solution:

Given:

Simple interest= â‚¹ 2400

Interest rate= 8% per annum

Time=2 years

Simple interest (SI) = PTR/100

Where P is principle amount, T is time taken, R is rate per annum

2400 = (P Ã— 2Ã— 8)/ 100

On rearranging we get,

P = 2400 Ã— 25/4

= â‚¹ 15000

âˆ´ the sum is â‚¹ 15000.

To find the amount we have the formula,

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

Where P is present value, r is rate of interest, n is time in years.

Now substituting the values in above formula we get,

âˆ´ A = 15000 (1 + 8/100)2

â‡’ A = 15000 (108/100)2

â‡’ A = 600 (27/25)2

â‡’ A = 24 (27) (27)

â‡’ A = â‚¹ 17496

âˆ´ Compound interest = A â€“ P

= 17496 â€“ 15000 = â‚¹ 2496

14. The difference between the compound interest and simple interest on a certain sum for 2 years at 6% per annum is â‚¹ 90. Find the sum.

Solution:

Given:

Interest rate= 6% per annum

Time=2 years

Simple interest (SI) = PTR/100

Where P is principle amount, T is time taken, R is rate per annum

Let sum is P

SI = (P Ã— 2Ã— 6)/ 100

â‡’ SI = (12P)/ 100

â‡’ SI = (3P)/ 25 —————- equation 1

To find the amount we have the formula,

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

Where P is present value, r is rate of interest, n is time in years.

Now substituting the values in above formula we get,

âˆ´ CI= P (1 + 6/100)2-P

â‡’ CI = P (1+3/50)2-P

â‡’ CI = P (53/50)2-P

â‡’ CI = P (2809) / (2500) – P

â‡’ CI = 309P/2500——– equation 2

Now the difference is

(CI â€“ SI) = 309P/2500 – (3P)/ 25

â‡’ 90 = 309P/2500 – (3P)/ 25

â‡’ 90 = 309P – (300P)/2500

â‡’ 90 = 9P/2500

â‡’ P= 90 Ã— 2500/9

â‡’ P= 10 Ã— 2500

â‡’ P=25000

âˆ´ Sum = 25000

15. The difference between the compound interest and simple interest on a certain sum for 3 years at 10% per annum is â‚¹ 93. Find the sum.

Solution:

Given:

Interest rate= 10% per annum

Time=3 years

Simple interest (SI) = PTR/100

Where P is principle amount, T is time taken, R is rate per annum

Let sum is P

SI = (P Ã— 3Ã— 10)/ 100

â‡’ SI = (30P)/ 100

â‡’ SI = (3P)/ 10 —————- equation 1

To find the amount we have the formula,

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

Where P is present value, r is rate of interest, n is time in years.

Now substituting the values in above formula we get,

âˆ´ CI= P (1 + 10/100)3-P

â‡’ CI = P (1+1/10)3-P

â‡’ CI = P (11/10)3-P

â‡’ CI = P (1331) / (1000) – P

â‡’ CI = 331P/1000——– equation 2

Now the difference is

(CI â€“ SI) = 331P/1000 – (3P)/ 10

â‡’ 93 = 331P- 300P/1000

â‡’ 93 = 31P/1000

â‡’ P= 93 Ã— 1000/31

â‡’ P= 3 Ã— 1000

â‡’ P= 3000

âˆ´ Sum = â‚¹3000

## Exercise 11C

1. Find the amount and the compound interest on â‚¹ 8000 for 1 year at 10% per annum, compounded half-yearly.

Solution:

Given:

Present value= â‚¹ 8000

Interest rate= 10 % per annum

Time=1 year and compounded half yearly

To find the amount we have the formula,

Amount (A) = P (1+(R/100)2n

Where P is present value, r is rate of interest, n is time in years.

Now substituting the values in above formula we get,

âˆ´ A = 8000 (1 + (10/2)/100)2

â‡’ A = 8000 (1+5/100)2

â‡’ A = 8000 (1+1/20)2

â‡’ A = 8000 (21/20)2

â‡’ A = 20 Ã— 441

â‡’ A = â‚¹ 8820

âˆ´ Compound interest = A â€“ P

= 8820 â€“ 8000 = â‚¹ 820

2. Find the amount and the compound interest on â‚¹ 31250 for 1 Â½ year at 8% per annum, compounded half-yearly.

Solution:

Given:

Present value= â‚¹ 31250

Interest rate= 8 % per annum

Time=1 Â½ year= 3/2 year and compounded half yearly

To find the amount we have the formula,

Amount (A) = P (1+(R/100)2n

Where P is present value, r is rate of interest, n is time in years.

Now substituting the values in above formula we get,

âˆ´ A = 31250 (1 + (8/2) /100)3

â‡’ A = 31250 (1+4/100)3

â‡’ A = 31250 (1+1/25)3

â‡’ A = 31250 (26/25)3

â‡’ A = 31250 Ã— 17576/15625

â‡’ A = â‚¹ 35152

âˆ´ Compound interest = A â€“ P

= 35452 â€“ 31250 = â‚¹ 3902

3. Find the amount and the compound interest on â‚¹ 12800 for 1 year at 7 Â½ % per annum, compounded half-yearly.

Solution:

Given:

Present value= â‚¹ 12800

Interest rate= 7 Â½ % per annum

Time=1 year and compounded half yearly

To find the amount we have the formula,

Amount (A) = P (1+(R/100)2n

Where P is present value, r is rate of interest, n is time in years.

Now substituting the values in above formula we get,

âˆ´ A = 12800 (1 + (15/4)/100)2

â‡’ A = 12800 (1+3/80)2

â‡’ A = 12800 (83/80)2

â‡’ A = 128 Ã— 6889/64

â‡’ A = â‚¹ 13778

âˆ´ Compound interest = A â€“ P

= 13778 â€“ 12800 = â‚¹ 978

4. Find the amount and the compound interest on â‚¹ 160000 for 2 years at 10 % per annum, compounded half-yearly.

Solution:

Given:

Present value= â‚¹ 160000

Interest rate= 10 % per annum

Time=2 years and compounded half yearly

To find the amount we have the formula,

Amount (A) = P (1+(R/100)2n

Where P is present value, r is rate of interest, n is time in years.

Now substituting the values in above formula we get,

âˆ´ A = 160000 (1 + (10/2)/100)4

â‡’ A = 160000 (1+5/100)4

â‡’ A = 160000 (21/20)4

â‡’ A = 160000 Ã— 194481/160000

â‡’ A = â‚¹ 194481

âˆ´ Compound interest = A â€“ P

= 194481 â€“ 160000 = â‚¹ 34481

5. Swati borrowed â‚¹ 40960 from a bank to buy a piece of land. If the bank charges 12 Â½ % per annum, compounded half yearly. What amount will she have to pay after 1 Â½ years? Also, find the interest paid by her.

Solution:

Given: Present value= â‚¹ 40960

Interest rate= 12 Â½ % per annum

Time=1 Â½ = 3/2 years and compounded half yearly

To find the amount we have the formula,

Amount (A) = P (1+(R/100)2n

Where P is present value, r is rate of interest, n is time in years.

Now substituting the values in above formula we get,

âˆ´ A = 40960 (1 + (25/4)/100)3

â‡’ A = 40960 (1+1/16)3

â‡’ A = 40960 (17/16)3

â‡’ A = 40960 Ã— 4913/4096

â‡’ A = â‚¹ 49130

âˆ´ Compound interest = A â€“ P

= 49130 â€“ 40960 = â‚¹ 8170

## Exercise 11D Page No: 154

Select the correct answer in each of the following:

1. The compound interest on â‚¹ 5000 at 8% per annum for 2 years, compounded annually, is

(a) â‚¹ 800 (b) â‚¹ 825 (c) â‚¹ 832 (d) â‚¹ 850

Solution:

(c) â‚¹ 832

Explanation:

Present value= â‚¹ 5000

Interest rate= 8% per annum

Time=2 years

To find the amount we have the formula,

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

Where P is present value, r is rate of interest, n is time in years.

Now substituting the values in above formula we get,

âˆ´ A = 5000 (1 + 8/100)2

â‡’ A = 5000 (27/25)2

â‡’ A = 200 (27 Ã— 27/25)

â‡’ A = 27 (27) (8)

â‡’ A = â‚¹ 5832

âˆ´ Compound interest = A â€“ P

= 5832 â€“ 5000= â‚¹ 832

2. The compound interest on â‚¹ 10000 at 10% per annum for 3 years, compounded annually, is

(a) â‚¹ 1331 (b) â‚¹ 3310 (c) â‚¹ 3130 (d) â‚¹ 13310

Solution:

(b) â‚¹ 3310

Explanation:

Present value= â‚¹ 10000

Interest rate= 10% per annum

Time=3 years

To find the amount we have the formula,

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

Where P is present value, r is rate of interest, n is time in years.

Now substituting the values in above formula we get,

âˆ´ A = 10000 (1 + 10/100)3

â‡’ A = 10000 (11/10)3

â‡’ A = 121 (10) (11)

â‡’ A = 1331 (10)

â‡’ A = â‚¹ 13310

âˆ´ Compound interest = A â€“ P

= 13310 â€“ 10000= â‚¹ 3310

3. The compound interest on â‚¹ 10000 at 12% per annum for 1 Â½ years, compounded annually, is

(a) â‚¹ 1872 (b) â‚¹ 1720 (c) â‚¹ 1910.16 (d) â‚¹ 1782

Solution:

(a) â‚¹ 1872

Explanation:

Present value= â‚¹ 10000

Interest rate= 12% per annum

Time=1 Â½Â years

To find the amount we have the formula,

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

Where P is present value, r is rate of interest, n is time in years.

Now substituting the values in above formula we get,

âˆ´ A = 10000 (1 + 12/100)1 [1 + (12/2 )/100]

â‡’ A = 10000 (1+12/100)1(1+6/100)

â‡’ A = 10000 (112/100) (106/100)

â‡’ A = 1 Ã— 112 Ã— 106

â‡’ A = â‚¹ 11872

âˆ´ Compound interest = A â€“ P

= 11872 â€“ 10000= â‚¹ 1872

4. The compound interest on â‚¹ 4000 at 10% per annum for 2 years 3 months , compounded annually, is

(a) â‚¹ 916 (b) â‚¹ 900 (c) â‚¹ 961 (d) â‚¹ 896

Solution:

(c) â‚¹ 961

Explanation:

Present value= â‚¹ 4000

Interest rate= 10% per annum

Time=2 Â¼ years

To find the amount we have the formula,

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

Where P is present value, r is rate of interest, n is time in years.

Now substituting the values in above formula we get,

âˆ´ A = 4000 (1 + 10/100)2 [1 + (10/4)/100]

â‡’ A = 4000 (1+1/10)2(1+ (5/2)/100)

â‡’ A = 4000 (121/100) (41/40)

â‡’ A = 121 Ã— 41

â‡’ A = â‚¹ 4961

âˆ´ Compound interest = A â€“ P

= 4961 â€“ 4000= â‚¹ 961

5. A sum of â‚¹ 25000 was given as loan on compound interest for 3 years compounded annually at 5% per annum during the first year,6% per annum during the second year and 8% per annum during the third year. The compound interest is

(a) â‚¹ 5035 (b) â‚¹ 5051 (c) â‚¹ 5072 (d) â‚¹ 5150

Solution:

(b) â‚¹ 5051

Explanation:

Present value= â‚¹ 25000

Interest rate for first year, p= 5 % per annum

Interest rate for second year, q= 6 % per annum

Interest rate for second year, r= 8 % per annum

Amount (A) = P Ã— (1+p/100) Ã— (1+q/100) Ã— (1+r/100)

Now substituting the values in above formula we get,

âˆ´ A = 25000 Ã— (1+5/100) Ã— (1+6/100) Ã— (1+8/100)

â‡’ A = 25000 Ã— (105/100) Ã— (106/100) Ã— (108/100)

â‡’ A = 21 Ã— 53 Ã— 27

â‡’ A = â‚¹ 30051

âˆ´ Compound interest = A â€“ P

= 30051 â€“ 25000= â‚¹ 5051

6. The compound interest on â‚¹ 6250 at 8% per annum for 1 year, compounded half yearly, is

(a) â‚¹ 500 (b) â‚¹ 510 (c) â‚¹ 550 (d) â‚¹ 512.50

Solution:

(b) â‚¹ 510

Explanation:

Present value= â‚¹ 6250

Interest rate= 8% per annum

Time=1 year

To find the amount we have the formula,

Amount (A) = P (1+(R/100))2n

Where P is present value, r is rate of interest, n is time in years.

Now substituting the values in above formula we get,

âˆ´ A = 6250 (1 + (8/2)/100)2

â‡’ A = 6250 (1+4/100)2

â‡’ A = 6250 (1+1/25)2

â‡’ A = 6250 (26/25)2

â‡’ A = 10 Ã— 26 Ã— 26

â‡’ A = â‚¹ 6760

âˆ´ Compound interest = A â€“ P

= 6760 â€“ 6250 = â‚¹ 510

7. The compound interest on â‚¹ 40000 at 6% per annum for 6 months, compounded quarterly, is

(a) â‚¹ 1209 (b) â‚¹ 1902 (c) â‚¹ 1200 (d) â‚¹ 1306

Solution:

(a) â‚¹ 1209

Explanation:

Present value= â‚¹ 40000

Interest rate= 6% per annum

Time=6 months

To find the amount we have the formula,

Amount (A) = P (1+(R/100))4n

Where P is present value, r is rate of interest, n is time in years.

Now substituting the values in above formula we get,

âˆ´ A = 40000 (1 + (6/4)/100)2

â‡’ A = 40000 (1+3/200)2

â‡’ A = 40000 (203/200)2

â‡’ A = 40000 (203/200)2

â‡’ A = 203 Ã— 203

â‡’ A = â‚¹ 41209

âˆ´ Compound interest = A â€“ P

= 41209 â€“ 40000 = â‚¹ 1209

## RS Aggarwal Solutions for Class 8 Maths Chapter 11- Compound Interest

Chapter 11, Compound Interest, contains 4 Exercises. RS Aggarwal Solutions given here contains the answers to all the questions present in these exercises. Let us have a look at some of the concepts that are being discussed in this chapter.

• Introduction to compound interest
• Finding the Compound interest when interest is compounded annually
• Finding the compound interest when interest is compounded half-yearly
• Calculating compound interest by using formulae
• When the interest is compounded annually
• When the interest is compounded annually but rates are different for different years
• When interest is compounded annually but time is a fraction
• Applications of compound interest formula

Exercise 11A

Exercise 11B

Exercise 11C

Exercise 11D

### Chapter Brief of RS Aggarwal Solutions for Class 8 Maths Chapter 11 – Compound Interest

The RS Aggarwal Solutions for Class 8 Maths Chapter 11 â€“ Compound Interest deals with the definition of compound interest, finding the compound interest when interest is compounded annually and half-yearly using the formulae and without using the formulae. It also deals with applications of compound interest.