# ML Aggarwal Compound Interest Exe-2.1 Class 9 ICSE Maths Solutions

ML Aggarwal Compound Interest Exe-2.1 Class 9 ICSE Maths Solutions. We Provide Step by Step Answer of Exe-2.1 Questions for Compound Interest council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-9.

## ML Aggarwal Compound Interest  Exe-2.1 Class 9 ICSE Maths Solutions

 Board ICSE Subject Maths Class 9th Chapter-2 Compound Interest Topics Solution of Exe-2.1 Questions Edition 2024-2025

### Compound Interest Exe-2.1

#### Question 1. Find the amount and the compound interest on ₹ 8000 at 5% per annum for 2 years.

It is given that

Principal = ₹ 8000

Rate of interest = 5% p.a.

We know that

Interest for the first year = Prt/100

Substituting the values

= (8000 × 5 × 1)/ 100

= ₹ 400

So the amount for the first year or principal for the second year = 8000 + 400 = ₹ 8400

Here

Interest for the second year = (8400 × 5 × 1)/ 100

So we get

= ₹ 420

We know that

Amount after the second year = 8400 + 420

= ₹ 8820

Total compound interest = 8820 + 8000

= ₹ 820

#### Question 2. A man invests ₹ 46875 at 4% per annum compound interest for 3 years. Calculate:

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

(ii) the interest for the third year.

(iii) the interest for the first year.

It is given that

Principal = ₹ 46875

Rate of interest = 4% p.a.

##### (i) Interest for the first year = Prt/100

Substituting the values

= (46875 × 4 × 1)/ 100

= ₹ 1875

So the amount after first year or principal for the second year = 46875 + 1875 = ₹ 48750

Here

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

So we get

= ₹ 1950

##### (ii) We know that

Amount at the end of second year = 48750 + 1950

= ₹ 50700

#### Also find the sum due at the end of third year.

It is given that

Principal = ₹ 8000

Rate of interest = 10% p.a.

We know that

Interest for the first year = Prt/100

Substituting the values

= (8000 × 10 × 1)/ 100

= ₹ 800

So the amount after the first year or principal for the second year = 8000 + 800 = ₹ 8800

##### (i) Interest for the second year = (8800 × 10 × 1)/ 100

= ₹ 880

So the amount after second year or principal for the third year = 8800 + 880 = ₹ 9680

Interest for the third year = (9680 × 10 × 1)/ 100

= ₹ 968

= ₹ 10648

#### Find:

(i) the sum due to Ramesh at the end of the first year.

(ii) the interest he earns for the second year.

(iii) the total amount due to him at the end of three years.

It is given that

Principal = ₹ 12800

Rate of interest = 10% p.a.

##### (i) We know that

Interest for the first year = (12800 × 10 × 1)/ 100

= ₹ 1280

So the sum due at the end of first year = 12800 + 1280

= ₹ 14080

##### (ii) Principal for second year = ₹ 14080

So the interest for the second year = (14080 × 10 × 1)/ 100

= ₹ 1408

##### (iii) We know that

Sum due at the end of second year = 14080 + 1408

= ₹ 15488

Here

Principal for third year = ₹ 15488

Interest for the third year = (15488 × 10 × 1)/ 100

= ₹ 1548.80

So the total amount due to him at the end of third year = 15488 + 1548.80

= ₹ 17036.80

#### Question 5. The simple interest on a sum of money for 2 years at 12% per annum is ₹ 1380. Find:

(i) the sum of money.

(ii) the compound interest on this sum for one year payable half-yearly at the same rate.

It is given that

Simple Interest (SI) = ₹ 1380

Rate of interest (R) = 12% p.a.

Period (T) = 2 years

##### (i) We know that

Sum (P) = (SI × 100)/ (R × T)

Substituting the values

= (1380 × 100)/ (12 × 2)

= ₹ 5750

##### (ii) Here

Principal (P) = ₹ 5750

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

Period (n) = 1 year – 2 half years

So we get

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

Substituting the values

= 5750 (1 + 6/100)2

By further calculation

= 5750 × (53/50)2

So we get

= 5750 × 53/50 × 53/50

= ₹ 6460.70

Here

Compound Interest = A – P

Substituting the values

= 6460.70 – 5750

= ₹ 710.70

#### Question 6. A person invests ₹ 10000 for two years at a certain rate of interest, compounded annually. At the end of one year this sum amounts to ₹ 11200. Calculate:

(i) the rate of interest per annum.

(ii) the amount at the end of second year.

It is given that

Principal (P) = ₹ 10,000

Period (T) = 1 year

Sum amount (A) = ₹ 11,200

Rate of interest = ?

##### (i) We know that

Interest (I) = 11200 – 10000 = ₹ 1200

So the rate of interest

R = (I × 100)/ (P × T)

Substituting the values

R = (1200 × 100)/ (10000 × 1)

So we get

R = 12% p.a.

Therefore, the rate of interest per annum is 12% p.a.

##### (ii) We know that

Period (T) = 2 years

Rate of interest (R) = 12% p.a.

Here

A = P (1 + R/100)t

Substituting the values

A = 10000 (1 + 12/100)2

By further calculation

A = 10000 (28/25)2

We can write it as

A = 10000 × 28/25 × 28/25

So we get

A = 16 × 28 × 28

A = ₹ 12544

Therefore, the amount at the end of second year is ₹ 12544.

#### Question 7. Mr. Lalit invested ₹ 75000 at a certain rate of interest, compounded annually for two years. At the end of first year it amounts to ₹ 5325. Calculate

(i) the rate of interest.

(ii) the amount at the end of second year, to the nearest rupee.

It is given that

Investment of Mr. Lalit = ₹ 5000

Period (n) = 2 years

##### (i) We know that

Amount after one year = ₹ 5325

So the interest for the first year = A – P

Substituting the values

= 5325 – 5000

= ₹ 325

Here

Rate = (SI × 100)/ (P × T)

Substituting the values

= (325 × 100)/ (5000 × 1)

So we get

= 13/2

= 6.5 % p.a.

##### (ii) We know that

Interest for the second year = (5325 × 13 × 1)/ (100 × 2)

By further calculation

= (213 × 13)/ (4 × 2)

So we get

= 2769/8

= ₹ 346.12

So the amount after second year = 5325 + 346.12

We get

= ₹ 5671.12

= ₹ 5671 (to the nearest rupee)

#### Question 8. A man invests ₹ 5000 for three years at a certain rate of interest, compounded annually. At the end of one year it amounts to ₹ 5600. Calculate:

(i) the rate of interest per annum

(ii) the interest accrued in the second year.

(iii) the amount at the end of the third year.

It is given that

Principal = ₹ 5000

Consider r% p.a. as the rate of interest

##### (i) We know that

At the end of one year

Interest = Prt/100

Substituting the values

= (5000 × r × 1)/ 100

= 50r

Here

Amount = 5000 + 50r

We can write it as

5000 + 50r = 5600

By further calculation

50r = 5600 – 5000 = 600

So we get

r = 600/50 = 12

Hence, the rate of interest is 12% p.a.

##### (ii) We know that

Interest for the second year = (5600 × 12 × 1)/ 100

= ₹ 672

So the amount at the end of second year = 5600 + 672

= ₹ 6272

##### (iii) We know that

Interest for the third year = (6272 × 12 × 1)/ 100

= ₹ 752.64

So the amount after third year = 6272 + 752.64

= ₹ 7024.64

#### Question 9. Find the amount and the compound interest on ₹ 2000 at 10% p.a. for 2 years, compounded annually.

It is given that

Principal (P) = ₹ 2000

Rate of interest (r) = 10% p.a.

Period (n) = 2 ½ years

We know that

Amount = P (1 + r/100)n

Substituting the values

= 2000 (1 + 10/100)2 (1 + 10/(2 × 100))

By further calculation

= 2000 × 11/10 × 11/10 × 21/20

So we get

= ₹ 2541

Here

Interest = A – P

Substituting the values

= 2541 – 2000

= ₹ 541

#### Question 10. Find the amount and the compound interest on ₹ 50000 for 1 ½ years at 8% per annum, the interest being compounded semi-annually.

It is given that

Principal (P) = ₹ 50000

Rate of interest (r) = 8% p.a. = 4% semi-annually

Period (n) = 1 ½ years = 3 semi-annually

We know that

Amount = P (1 + r/100)n

Substituting the values

= 50000 (1 + 4/100)3

By further calculation

= 50000 (26/25)3

= 50000 × 26/25 × 26/25 × 26/25

= ₹ 56243.20

Here

Compound Interest = A – P

Substituting the values

= 56243.20 – 50000

= ₹ 6243.20

#### Question 11. Calculate the amount and the compound interest on ₹ 5000 in 2 years when the rate of interest for successive years is 6% and 8%, respectively.

It is given that

Principal = ₹ 5000

Period = 2 years

Rate of interest for the first year = 6%

Rate of interest for the second year = 8%

We know that

Amount for two years = P (1 + r/100)n

Substituting the values

= 5000 (1 + 6/100) (1 + 8/100)

By further calculation

= 5000 × 53/50 × 27/25

= ₹ 5724

Here

Interest = A – P

Substituting the values

= 5724 – 5000

= ₹ 724

#### Question 12. Calculate the amount and the compound interest on ₹ 17000 in 3 years when the rate of interest for successive years is 10%, 10% and 14%, respectively.

It is given that

Principal = ₹ 17000

Period = 3 years

Rate of interest for 3 successive years = 10%, 10% and 14%

We know that

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

Substituting the values

= 17000 (1 + 10/100) (1 + 10/100) (1 + 14/100)

By further calculation

= 17000 × 11/10 × 11/10 × 57/50

= ₹ 23449.80

Here

Amount of compound interest = A – P

Substituting the values

= 23449.80 – 17000

= ₹ 6449.80

#### Question 13. A sum of ₹ 9600 is invested for 3 years at 10% per annum at compound interest.

(i) What is the sum due at the end of the first year?

(ii) What is the sum due at the end of the second year?

(iii) Find the compound interest earned in 2 years.

(iv) Find the difference between the answers in (ii) and (i) and find the interest on this sum for one year.

(v) Hence, write down the compound interest for the third year.

It is given that

Principal = ₹ 9600

Rate of interest = 10% p.a.

Period = 3 years

We know that

Interest for the first year = Prt/100

Substituting the values

= (9600 × 10 × 1)/ 100

= ₹ 960

(i) Amount after one year = 9600 – 960 = ₹ 10560So the principal for the second year = ₹ 10560

Here the interest for the second year = (10560 × 10 × 1)/ 100

= ₹ 1056

##### (iv) Difference between the answers in (ii) and (i) = 11616 – 10560 = ₹ 1056

We know that

Interest on ₹ 1056 for 1 year at the rate of 10% p.a. = (1056 × 10 × 1)/ 100

= ₹ 105.60

##### (v) Here

Principal for the third year = ₹ 11616

So the interest for the third year = (11616 × 10 × 1)/ 100

= ₹ 1161.60

#### Question 14. The simple interest on a certain sum of money for 2 years at 10% p.a. is ₹ 1600. Find the amount due and the compound interest on this sum of money at the same rate after 3 years, interest being reckoned annually.

It is given that

Period = 2 years

Rate = 10% p.a.

We know that

Sum = (SI × 100)/ (r × n)

Substituting the values

= (1600 × 100)/ (10 × 2)

= ₹ 8000

Here

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

Substituting the values

= 8000 (1 + 10/100)3

By further calculation

= 8000 × 11/10 × 11/10 × 11/10

= ₹ 10648

So the compound interest = A – P

Substituting the values

= 10648 – 8000

= ₹ 2648

#### Question 15. Vikram borrowed ₹ 20000 from a bank at 10% per annum simple interest. He lent it to his friend Venkat at the same rate but compounded annually. Find his gain after 2 ½ years.

First case-

Principal = ₹ 20000

Rate = 10% p.a.

Period = 2 ½ = 5/2 years

We know that

Simple interest = Prt/100

Substituting the values

= (20000 × 10 × 5)/ (100 × 2)

= ₹ 5000

Second case-

Principal = ₹ 20000

Rate = 10% p.a.

Period = 2 ½ years at compound interest

We know that

Amount = P (1 + r/100)n

Substituting the values

= 20000 (1 + 10/100)2 (1 + 10/ (2 × 100))2

By further calculation

= 20000 × 11/10 × 11/10 × 21/20

= ₹ 25410

Here

Compound Interest = A – P

Substituting the values

= 25410 – 20000

= ₹ 5410

So his gain after 2 years = CI – SI

We get

= 5410 – 5000

= ₹ 410

#### Question 16. A man borrows ₹ 6000 at 5% compound interest. If he repays ₹ 1200 at the end of each year, find the amount outstanding at the beginning of the third year.

It is given that

Principal = ₹ 6000

Rate of interest = 5% p.a.

We know that

Interest for the first year = Prt/100

Substituting the values

= (6000 × 5 × 1)/ 100

= ₹ 300

So the amount after one year = 6000 + 300 = ₹ 6300

Principal for the second year = ₹ 6300

Amount paid = ₹ 1200

So the balance = 6300 – 1200 = ₹ 5100

Here

Interest for the second year = (5100 × 5 × 1)/ 100 = ₹ 255

Amount for the second year = 5100 + 255 = ₹ 5355

Amount paid = ₹ 1200

So the balance = 5355 – 1200 = ₹ 4155

#### Question 17. Mr. Dubey borrows ₹ 100000 from State Bank of India at 11% per annum compound interest. He repays ₹ 41000 at the end of first year and ₹ 47700 at the end of second year. Find the amount outstanding at the beginning of the third year.

It is given that

Borrowed money (P) = ₹ 100000

Rate = 11% p.a.

Time = 1 year

We know that

Amount after first year = Prt/100

Substituting the values

= (100000 × 11 × 1)/ 100

By further calculation

= 100000 + 11000

= ₹ 111000

Amount paid at the end of first year = ₹ 41000

So the principal for second year = 111000 – 41000

= ₹ 70000

We know that

Amount after second year = P + (70000 × 11)/ 100

By further calculation

= 70000 + 700

= ₹ 77700

So the amount paid at the end of second year = ₹ 47700

Here the amount outstanding at the beginning year = 77700 – 47700

= ₹ 30000

#### Question 18. Jaya borrowed ₹ 50000 for 2 years. The rates of interest for two successive years are 12% and 15% respectively. She repays ₹ 33000 at the end of first year. Find the amount she must pay at the end of second year to clear her debt.

It is given that

Amount borrowed by Jaya = ₹ 50000

Period (n) = 2 years

Rate of interest for two successive years are 12% and 15% respectively

We know that

Interest for the first year = Prt/100

Substituting the values

= (50000 × 12 × 1)/ 100

= ₹ 6000

So the amount after first year = 50000 + 6000 = ₹ 56000

Amount repaid = ₹ 33000

Here

Balance amount for the second year = 56000 – 33000 = ₹ 23000

Rate = 15%

So the interest for the second year = (230000 × 15 × 1)/ 100

= ₹ 3450

Amount paid after second year = 23000 + 3450 = ₹ 26450

—  : End of ML Aggarwal Compound Interest Exe-2.1 Class 9 ICSE Maths Solutions :–

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