ML Aggarwal Compound Interest Exe-2.3 Class 9 ICSE Maths Solutions. We Provide Step by Step Answer of Exe-Exe-2.3 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-2.3 Class 9 ICSE Maths Solutions

Board | ICSE |

Subject | Maths |

Class | 9th |

Chapter-2 | Compound Interest |

Topics | Solution of Exe-2.3 Questions |

Edition | 2024-2025 |

**Compound Interest Exe-2.3**

**Question 1. The present population of a town is 200000. Its population increases by 10% in the first year and 15% in the second year. Find the population of the town at the end of two years.**

**Answer :**

We know that

Population after 2 years = Present population × (1 + r/100)^{n}

Here the present population = 200000

Population after first year = 200000 × (1 + 10/100)^{1}

By further calculation

= 200000 × 11/10

= 220000

Population after two years = 220000 × (1 + 15/100)^{1}

By further calculation

= 220000 × 23/20

= 253000

**Question 2. The present population of a town is 15625. If the population increases at the rate of 4% every year, what will be the increase in the population in next 3 years?**

**Answer :**

It is given that

Present population (P) = 15625

Rate of increase (r) = 4% p.a.

Period (n) = 3 years

We know that

Population after 3 years = P (1 + r/100)^{n}

Substituting the values

= 15625 (1 + 4/100)^{3}

By further calculation

= 15625 × 26/25 × 26/25 × 26/25

= 17576

So the increase = 17576 – 15625 = 1951

**Question 3. The population of a city increase each year by 4% of what it had been at the beginning of each year. If its present population is 6760000, find:**

**(i) its population 2 years hence**

**(ii) its population 2 years ago.**

**Answer :**

It is given that

Present population = 6760000

Increase percent = 4% p.a.

**(i)** We know that

Population 2 years hence = P (1 + r/100)^{2}

Substituting the values

= 6760000 (1 + 4/100)^{2}

By further calculation

= 6760000 × 26/25 × 26/25

= 7311616

**(ii)** We know that A = 6760000

Population 2 years ago P = A + (1 + r/100)^{2}

Substituting the values

= 6760000 + (1 + 4/100)^{2}

By further calculation

= 6760000 + (26/25)^{2}

= 6760000 × 25/26 × 25/26

= 6250000

**Question 4. The cost of a refrigerator is ₹ 9000. Its value depreciates at the rate of 5% ever year. Find the total depreciation in its value at the end of 2 years.**

**Answer :**

It is given that

Present value (P) = ₹ 9000

Rate of depreciation (r) = 5% p.a.

Period (n) = 2 years

We know that

Value after 2 years = P (1 – r/100)^{n}

Substituting the values

= 9000 (1 – 5/100)^{2}

By further calculation

= 9000 × 19/20 × 19/20

= ₹ 8122.50

So the total depreciation = 9000 – 8122.50 = ₹ 877.50

**Question 5. Dinesh purchased a scooter for ₹ 24000. The value of the scooter is depreciating at the rate of 5% per annum. Calculate its value after 3 years.**

**Answer :**

It is given that

Present value of scooter (P) = ₹ 24000

Rate of depreciation (r) = 5%

Period (n) = 3 years

We know that

Value after 3 years = P (1 – r/100)^{n}

Substituting the values

= 24000 (1 – 5/100)^{3}

By further calculation

= 24000 × 19/20 × 19/20 × 19/20

= ₹ 20577

**Question 6. A farmer increases his output of wheat in his farm every year by 8%. This year he produced 2187 quintals of wheat. What was the yearly produce of wheat two years ago?**

**Answer :**

It is given that

Present production of wheat = 2187 quintals

Increase in production = 8% p.a.

We know that

Production of wheat 2 years ago = A ÷ (1 + r/100)^{n}

Substituting the values

= 2187 ÷ (1 + 8/100)^{2}

By further calculation

= 2187 ÷ (27/25)^{2}

So we get

= 2187 × 25/27 × 25/27

= 1875 quintals

**Question 7. The value of a property decreases every year at the rate of 5%. If its present value is ₹ 411540, what was its value three years ago?**

**Answer :**

It is given that

Present value of property = ₹ 411540

Rate of decrease = 5% p.a.

We know that

Value of property 3 years ago = A ÷ (1 – r/100)^{n}

Substituting the values

= 411540 ÷ (1 – 5/100)^{3}

By further calculation

= 411540 ÷ (19/20)^{3}

So we get

= 411540 × 20/19 × 20/19 × 20/19

= ₹ 480000

**Question 8. Ahmed purchased an old scooter for ₹ 16000. If the cost of the scooter after 2 years depreciates to ₹ 14440, find the rate of depreciation.**

**Answer :**

It is given that

Present value = ₹ 16000

Value after 2 years = ₹ 14440

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

We know that

A/P = (1 – r/100)^{n}

Substituting the values

14440/16000 = (1 – r/100)^{2}

By further calculation

361/400 = (1 – r/100)^{2}

(19/20)^{2} = (1 – r/100)^{2}

We can write it as

1 – r/100 = 19/20

So we get

r/100 = 1 – 19/20 = 1/20

By cross multiplication

r = 1/20 × 100 = 5%

Hence, the rate of depreciation is 5%.

**Question 9. A factory increased its production of cars from 80000 in the year 2011-2012 to 92610 in 2014-15. Find the annual rate of growth of production of cars.**

**Answer :**

It is given that

Production of cars in 2011-2012 = 80000

Production of cars in 2014-2015 = 92610

Period (n) = 3 years

Consider r% as the rate of increase

We know that

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

Substituting the values

92610/80000 = (1 + r/100)^{3}

By further calculation

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

We can write it as

1 + r/100 = 21/20

r/100 = 21/20 – 1 = 1/20

By cross multiplication

r = 1/20 × 100 = 5

Hence, the annual rate of growth of production of cars is 5% p.a.

**Question 10. The value of a machine worth ₹ 500000 is depreciating at the rate of 10% every year. In how many years will its value be reduced to ₹ 364500?**

**Answer:**

It is given that

Present value = ₹ 500000

Reduced value = ₹ 364500

Rate of depreciation = 10% p.a.

Consider n years as the period

We know that

A/P = (1 – r/100)^{n}

Substituting the values

364500/500000 = (1 – 10/100)^{n}

By further calculation

(9/10)^{n} = 729/1000 = (9/10)^{3}

So we get

n = 3

Therefore, the period in which its value be reduced to ₹ 364500 is 3 years.

**Question 11. Afzal purchased an old motorbike for ₹ 16000. If the value of the motorbike after 2 years is ₹ 14440, find the rate of depreciation.**

**Answer :**

It is given that

CP of an old motorbike = ₹ 16000

Price after 2 years = ₹ 14440

Consider r% as the rate of depreciation

We know that

A/P = (1 – r/100)^{n}

Substituting the values

14440/16000 = (1 – r/100)^{2}

By further calculation

361/400 = (1 – r/100)^{2}

(19/20)^{2} = (1 – r/100)^{2}

So we get

19/20 = 1 – r/100

r/100 = 1 – 19/20 = (20 – 19)/ 20 = 1/20

By cross multiplication

r = 100/20 = 5

Hence, the rate of depreciation is 5%.

**Compound Interest Exe-2.3**

ML Aggarwal Class 9 ICSE Maths Solutions

Page 65

**Question 12. Mahindra set up a factory by investing ₹ 2500000. During the first two years, his profits were 5% and 10% respectively. If each year the profit was on previous year’s capital, calculate his total profit.**

**Answer :**

It is given that

Investment = ₹ 2500000

Rates of profit during first two years = 5% and 10%

We know that

Capital after two years (A) = P (1 + r/100)^{n}

Substituting the values

= 2500000 (1 + 5/100) (1 + 10/100)

By further calculation

= 2500000 × 21/20 × 11/10

= ₹ 2887500

So the net profit = A – P

Substituting the values

= 2887500 – 2500000

= ₹ 387500

**Question 13. The value of a property is increasing at the rate of 25% every year. By what percent will the value of the property increase after 3 years?**

**Answer :**

It is given that

Original price of the property (P) = ₹ 100

Rate of increase (r) = 25% p.a.

Period (n) = 3 years

We know that

Increased value after 3 years = P (1 + r/100)^{n}

Substituting the values

= 100 (1 + 25/100)^{3}

By further calculation

= 100 × 5/4 × 5/4 × 5/4

= ₹ 3125/16

Here

Increased value = 3125/16 – 100

Taking LCM

= (3125 – 1600)/ 16

= 1525/16

So the percent increase after 3 years = 1525/16 = 95 5/16%

**Question 14. Mr. Durani bought a plot of land for ₹ 180000 and a car for ₹ 320000 at the same time. The value of the plot of land grows uniformly at the rate of 30% p.a.., while the value of the car depreciates by 20% in the first year and by 15% p.a. thereafter. If he sells the plot of land as well as the car after 3 years, what will be his profit or loss?**

**Answer :**

It is given that

Price of plot of land = ₹ 180000

Growth rate = 30% p.a.

Period (n) = 3 years

We know that

Amount after 3 years = P (1 + R/100)^{n}

Substituting the values

= 180000 (1 + 30/100)^{3}

By further calculation

= 180000 × (13/10)^{3}

It can be written as

= 180000 × 13/10 × 13/10 × 13/10

= ₹ 395460

Here

Price of car = ₹ 320000

Rate of depreciation = 20% for the first year and 15% for next period

Period (n) = 3 years

We know that

Amount after 3 years = A (1 – R_{1}/100)^{n} × (1 – R_{2}/100)^{2}

Substituting the values

= 320000 (1 – 20/100) (1 – 15/100)^{2}

By further calculation

= 320000 × 4/5 × (17/20)^{2}

So we get

= 320000 × 4/5 × 17/20 × 17/20

= ₹ 184960

Here

Total cost of plot and car = 180000 + 320000 = ₹ 500000

Total sale price of plot and car = 395460 + 184960 = ₹ 580420

We know that

Profit = S.P. – C.P.

Substituting the values

= 580420 – 500000

= ₹ 80420

— : End of ML Aggarwal Compound Interest Exe-2.3 Class 9 ICSE Maths Solutions :–

Return to :- ** ML Aggarawal Maths Solutions for ICSE Class-9**

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