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.

**Answer :**

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**.**

**Answer :**

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

**(iii)** Interest for the third year = (50700 × 4 × 1)/ 100 = ₹ 2028

**Question 3. Calculate the compound interest for the second year on ₹ 8000 for three years at 10% p.a.**

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

**Answer :**

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

**(ii)** Amount due at the end of the third year = 9680 + 968

= ₹ 10648

**Question 4. Ramesh invests ₹ 12800 for three years at the rate of 10% per annum compound interest.**

**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.

**Answer :**

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.

**Answer :**

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.

**Answer :**

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.

**Answer :**

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.

**Answer :**

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

In, Ex 2.1 here it is given Me Lalit invested ₹75000 but in the solutions there is written ₹5000……… I think the question is wrong as I have the same ML Aggarwal book of same edition

is this book applicable for 2023-24 session if yes send screen shot on 8948221203 whatsapp number