Compound Interest Class 9 RS Aggarwal Exe-2C Goyal Brothers ICSE Foundation Maths Solutions. In this article you will learn how to solve growth and depreciation word problems easily**.** Visit official Website **CISCE** for detail information about ICSE Board Class-9 Mathematics.

## Compound Interest Class 9 RS Aggarwal Exe-2C Goyal Brothers ICSE Maths Solutions

Board | ICSE |

Publications | Goyal brothers Prakashan |

Subject | Maths |

Class | 9th |

Chapter-2 | Compound Interest |

Writer | RS Aggrawal |

Book Name | Foundation |

Topics | Growth and Depreciation |

Academic Session | 2024-2025 |

### How to Solve Growth and Depreciation Word Problems Easily

**Growth:** If the numeric value increase of a original value P at rate r then after n years the final value A is as below

Value after n year = P ((1 + r/100) nth year

Growth only = Value after n year – original value

**Depreciation** : If the numeric value decrease of a original value P at rate r then after n years the final value A is as below

Value after n year = P ((1 – r/100) nth year

Depreciation only = original value – Value after n year

**Exercise- 2C**

Compound Interest Class 9 RS Aggarwal Exe-2C Goyal Brothers ICSE Foundation Maths Solutions

Page- 40,41

**Que-1: A town has 15625 inhabitants. If the population of this increases at the rate of 4% per annum, find the number of inhabitants of the town at the end of 3 years.**

**Solution- **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 Ans.**

**Que-2: The population of a town is increasing at the rate of 10% per annum. If its present population is 36300, find : (i) its population after 2 years, (ii) its population 2 years ago.**

**Solution- **Present population = 36300

r = 10%

(i) n = after 2 years

= 36300×(1+10/100)^2

= 36300 × 11/10 × 11/10

= 36300 × 121/100

**= 43923 Ans.**

(ii) n = 2 year ago

= 36300/(1-10/100)^2

= 36300 × 10/11 × 10/11

= 36300 × 100/121

**= 30000 Ans.**

**Que-3: The present population of a town is 176400. If the rate of growth in its population is 5% per annum, find : (i) its population 2 years hence, (ii) its population one year ago.**

**Solution-**(i) Present population (P0) = 176,400

Annual growth rate (r) = 5%

Number of years (t) = 2

Using the formula:

𝑃(2) = 176,400(1+5/100)²

𝑃(2) = 176,400(1+0.05)²

𝑃(2) = 176,400(1.05)²

𝑃(2) = 176,400 × 1.1025

P(2) = Rs194481 Ans.

(ii) Present population (P0) = 176,400

Annual growth rate (r) = 5%

Number of years (t) = -1 (since we are going back one year)

Using the formula:

𝑃(−1) = 176,400(1+5/100)‾¹

𝑃(−1) = 176,400(1+0.05)‾¹

𝑃(−1) = 176,400(1.05)‾¹

𝑃(−1) = 176,400 × (1/1.05)

𝑃(−1) = 176,400 × 0.95238

**P(-1) = 168000 Ans.**

**Que-4: Three years ago, the population of a city was 50000. If the annual increase during three successive years be 5%, 8% and 10% respectively, find the present population of the city.**

**Solution-**Initial Population: 50,000

The population increases by 5% during the first year.

Population after 1 year = 50,000×(1+5/100)

= 50,000 × 1.05

= 52,500

The population increases by 8% during the second year.

Population after 2 years = 52,500×(1+8/100)

= 52,500 × 1.08

= 56,700

The population increases by 10% during the third year.

Population after 3 years = 56,700×(1+10/100)

= 56,700 × 1.10

= 62,370 Ans.

**Que-5: A farmer has an increase of 12.5% in the output of wheat in his farm every year. This year, he produced 2916 quintals of wheat. What was his annual production of wheat 2 years ago ?**

**Solution- **Let us Assume 2yrs ago as X, then next years as Y

And the current year as Z, which outcome is = 2916 Quintals

Y’th year outcome will be increased by 12.5% from X’th year

Then, Y = (12.5*X/100) + X

= 0.125*X + X

= 1.125 X ——> 1

Z’th Year outcome will be increase by 12.5% from Y’th year

Then, Z = (12.5*Y/100) + Y —–> 2

Now Substitute 1 in 2 we get

Z = ((12.5 * 1.125 * X) / 100) + 1.125 * X

2916 = (0.140625 * X) + 1.125 * X

2916 = 1.265625 X

**X = 2304 Quintals Ans.**

**Que-6: The population of a town is 64000. If the annual birth rate is 11.7% and the annual death rate is 4.2%, calculate the population of the town after 3 years.**

**Solution- **Present population = 64,000

Birth rate = 10.7% (increase), death rate = 3.2% (decrease)

R = total increase in population (%) = 10.7−3.2 = 7.5

Population after n years = p(1+R/100)^n

After 3 years = 64,0000(1+7.5/100)^3

∴ Population = 79507 Ans.

**Que-7: A mango tree was planted 2 years ago. The rate of its growth is 20% per annum. If at present, the height of the tree is 162cm, what it was when the tree was planted ?**

**Solution-**Let the height of the tree when it was planted be H cm.

The tree grows by 20%, so after one year the height becomes:

= H×1.20

The tree grows again by 20%, so after the second year the height becomes:

= H×1.20×1.20 = H × (1.20)²

Given that the current height is 162 cm, we can set up the equation:

= H × (1.20)² = 162

Solving for H:

= H×1.44 = 162

= H = 162/1.44

H = 112.5 Ans.

**Que-8: Two years ago, the population of a village was 4000. During next year it increased by 6% but due to an epidemic, it decreased by 5% in the following year. What is its population now ?**

**Solution- **Initial Population (2 years ago): 4000

The population increased by 6% during the first year.

Population after 1 year = 4000×(1+6/100)

= 4000×1.06

= 4240

The population decreased by 5% during the second year.

Population after 2 years = 4240×(1−5/100)

= 4240×0.95

= 4028

Thus, the current population of the village is: 4028 Ans.

**Que-9: The count of bacteria in a culture grows by 10% during first hour, decreases by 8% during second hour and again increases by 12% during third hour. If the count of bacteria in the sample is 13125000, what will be the count of bacteria after 3 hours ?**

**Solution- **𝑅1 = 10

𝑅2 = −8

𝑅3 = 12

P = Original count of bacteria = 13, 125, 000

We know that:

𝑃(1+𝑅1/100)(1−𝑅21/00)(1+𝑅3/100)

∴ Bacteria count after three hours = 13,125,000(1+10/100)(1−8/100)(1+12/100)

= 13,125,000(1.10)(0.92)(1.12)

= 14,876,400

Thus, the bacteria count after three hours will be 14, 876, 400.

**Que-10: In a factory, the production of scooters was 40000 per year, which rose to 57600 in 2 years. Find the rate of growth per annum.**

**Solution- **We will use the compound growth formula:

P = P0(1+r)^n

P is the final production amount (57,600),

P0 is the initial production amount (40,000),

r is the annual growth rate,

n is the number of years (2)

57600 = 40000(1+r)²

57600/40000 = (1+r)²

1.44 = (1+r)²

√1.44 = 1+r

1.2 = 1+r

r = 1.2-1

r = 0.2

Convert the decimal to a percentage:

r = 0.2×100% **= 20% Ans.**

**Que-11: Amit started a shop by investing Rs500000. In the first year, he incurred a loss of 5%. However, during the second the year, he earned a profit of 10% which in the third year rose to 12%. Calculate his net profit for the entire period of three years.**

**Solution- **Initial Investment: Rs500,000

Value after 1 year = 500,000×(1−5/100)

= 500,000 × 0.95 = 4,75,000

Value after 2 years = 475,000×(1+10/100)

= 475,000 × 1.10 = 5,22,500

Value after 3 years = 522,500×(1+12/100)

= 522,500 × 1.12

= **5,85,200** **Ans.**

**Que-12: The value of a machine depreciates 10% annually. Its present value is Rs64800. Find : (i) its value after 2 years (ii) its value 2 year ago.**

**Solution- **value of machine (V) = 64800

r = 10%, n = 2yrs

(i) value of machine after n yrs = V(1-R/100)^n

value of machines after 2 yrs = 64800(1-10/100)²

= 64800 X 9/10 X 9/10

**= 52,488 Ans.**

(ii) value of machine n yrs ago = V(1-r /100)^-n

= 64800(1-10/100)¯²

= 64800 x 10/9 x 10 /9

**= 80000 Ans.**

**Que-13: A refrigerator was purchased one year ago for Rs20000. Its value depreciates at the rate of 15% per annum. Find : (i) its present value (ii) its value after 1 year.**

**Solution- **Cost price of Refrigerator 20,000

Rate of decrease is 15%

(i) Decreased amount is (15/100) × 20,000 = 3,000

So the present price is 20,000 − 3,000 **= 17,000 Ans.**

(ii) Decrease amount a year after is (15/100) × 17,000 = 2550

So the price is 17,000 − 2,550** = 14,450Rs Ans.**

**Que-14: A machine depreciates each year at 8% of its value in the beginning of the year. If its value be Rs57500 at the end of the year 2015, find : (i) its value at the end of the year 2014 (ii) ****its value at the end of the year 2016**

**Solution- **(i) Let the value of machine at the end of 2014 be x

We are given that A machine depreciates each year at 8% of its value in the beginning of the year.

So, Value of machine in 2015 = x – 8%x = x – 0.08x = 0.92x

We are given that its value be 57500 at the end of the year 2015.

So, 0.92x = 57500

x = 57500/0.92

**= Rs62500 Ans.**

(ii) So, its value at the end of the year 2016.

= 57500 – (8%)(57500)

=57500 – (0.08)(57500 )

**= Rs52900 Ans.**

**Que-15: The value of a machine depreciates at the rate of 16*(2/3)% per annum. It was purchased 3 years ago. If its present value is Rs62500, find its purchased value.**

**Solution- **r = 16*(2/3)% = 50/3%

Next, convert this percentage into a decimal for ease of calculation.

50/3% = (50/3) x (1/100) = 1/6

This means that each year the machine retains 1−1/6 = 5/6 of its value from the previous year.

Let’s denote the purchase value of the machine as P. The value after 3 years, denoted as V, can be given by:

𝑉 = 𝑃(5/6)³

We know that:

V = 62500

Therefore: 62500 = P(5/6)³

Calculate (5/6)³:

(5/6)³ = 5³/6³ = 125/216

Now, substituting back:

62500 = P × 125/216

Solving for P:

𝑃 = 62500 × 216/125

Simplify the multiplication:

P = 62500 × 1.728

P = 108000 Ans.

– : End of Compound Interest Class 9 RS Aggarwal Exe-2C Goyal Brothers ICSE Foundation Maths Solutions : –

Return to :- **RS Aggarwal Solutions for ICSE Class-9 Mathematics (Goyal Brother Prakashan)**

Thanks

Please Share with your friends if helpful