Class 8 RD Sharma Solutions - Chapter 14 Compound Interest - Exercise 14.4 | Set 1

In Class 8 Mathematics, Chapter 14 of the RD Sharma textbook focuses on Compound Interest a crucial concept that students encounter in everyday financial transactions. Understanding compound interest helps in grasping how investments grow over time which is vital for financial literacy. Exercise 14.4 delves into the practical application of the compound interest formulas providing the students with the problems that enhance their problem-solving skills and comprehension of this essential mathematical topic.

Compound Interest

The Compound interest refers to the interest calculated on the initial principal which also includes the all the accumulated interest from the previous periods. Unlike simple interest where interest is only earned on the principal amount, compound interest allows money to grow at a faster rate as the interest is reinvested leading to exponential growth over time. This concept is widely used in finance, banking, and investment planning.

Question 1. The present population of the town is 28000. If it increases at the rate of 5% per annum, what will be its population after 2 years?

Solution:

We have,

Present population of town is = 28000

Rate of increase in population is = 5% per annum

Number of years = 2

By using the formula,

A = P (1 + \frac{R}{100}  )ⁿ

Substituting the values, we have 

Population of town after 2 years = 28000 (1 + \frac{5}{100}  )²

= 28000 (1.05)²

= 30870

Therefore,

Population of town after 2 years will be 30870.

Question 2. The population of a city is 125000. If the annual birth rate and death rate are 5.5% and 3.5% respectively, calculate the population of city after 3 years.

Solution:

We have,

Population of city (P) = 125000

Annual birth rate R1= 5.5 %

Annual death rate R2 = 3.5 %

Annual increasing rate R = (R1 – R2) = 5.5 – 3.5 = 2 %

Number of years = 3

By using the formula,

A = P (1 + \frac{R}{100}  )ⁿ

Substituting the values, we have 

So, population after two years is = 125000 (1 + \frac{2}{100}  )³

= 125000 (1.02)³

= 132651

Therefore,

Population after 3 years will be 132651.

Question 3. The present population of a town is 25000. It grows at 4%, 5% and 8% during first year, second year and third year respectively. Find its population after 3 years.

Solution:

We have,

Present population is = 25000

First year growth R1 = 4%

Second year growth R2 = 5%

Third year growth R3 = 8%

Number of years = 3

By using the formula,

A = P (1 + \frac{R}{100}  )ⁿ

Substituting the values, we have 

So, population after three years = P (1 + \frac{R_1}{100}  ) (1 + \frac{R_2}{100}  ) (1 + \frac{R_3}{100}  )

= 25000(1 + \frac{4}{100}  ) (1 + \frac{5}{100}  ) (1 + \frac{8}{100}  0)

= 25000 (1.04) (1.05) (1.08)

= 29484

Therefore,

Population after 3 years will be 29484.

Question 4. Three years ago, the population of a town was 50000. If the annual increase during three successive years be at the rate of 4%, 5% and 3% respectively, find the present population.

Solution:

We have,

Three years ago population of town was = 50000

Annual increasing in 3 years = 4%,5%, 3% respectively

So, let present population be = x

By using the formula,

A = P (1 + \frac{R}{100}  )ⁿ

Substituting the values, we have 

x = 50000 (1 + \frac{4}{100}  ) (1 + \frac{5}{100}  ) (1 + \frac{3}{100}  )

= 50000 (1.04) (1.05) (1.03)

= 56238

Therefore,

Present population of the town is 56238.

Question 5. There is a continuous growth in population of a village at the rate of 5% per annum. If its present population is 9261, what it was 3 years ago?

Solution:

We have,

Present population of town is = 9261

Continuous growth of population is = 5%

So, let population three years ago be = x

By using the formula,

A = P (1 + \frac{R}{100}  )ⁿ

Substituting the values, we have 

9261 = x (1 + \frac{5}{100}  ) (1 + \frac{5}{100}  ) (1 + \frac{5}{100}  )

9261 = x (1.05) (1.05) (1.05)

= 8000

Therefore,

Present population of the town is 8000.

Question 6. In a factory the production of scooters rose to 46305 from 40000 in 3 years. Find the annual rate of growth of the production of scooters.

Solution:

We have,

Initial production of scooters is = 40000

Final production of scooters is = 46305

Time = 3 years

Let annual growth rate be = R%

By using the formula,

A = P (1 + \frac{R}{100}  )ⁿ

Substituting the values, we have 

46305 = 40000 (1 + \frac{R}{100}  ) (1 + \frac{R}{100}  ) (1 + \frac{R}{100}  )

46305 = 40000 (1 + \frac{R}{100}  )³

(1 + \frac{R}{100}  )³ = 46305/40000

= 9261/8000

= (21/20)³

1 + \frac{R}{100}   = 21/20

\frac{R}{100}   = 21/20 – 1

\frac{R}{100}   = (21-20)/20

= 1/20

R = 100/20

= 5

Therefore,

Annual rate of growth of the production of scooters is 5%.

Question 7. The annual rate of growth in population of a certain city is 8%. If its present population is 196830, what it was 3 years ago?

Solution:

We have,

Annual growth rate of population of city is = 8%

Present population of city is = 196830

Let population of city 3 years ago be = x

By using the formula,

A = P (1 + \frac{R}{100}  )ⁿ

Substituting the values, we have 

196830 = x (1 + \frac{8}{100}  ) (1 + \frac{8}{100}  ) (1 + \frac{8}{100}  )

196830 = x (27/25) (27/25) (27/25)

196830 = x (1.08) (1.08) (1.08)

196830 = 1.259712x

x = 196830/1.259712

= 156250

Therefore,

Population 3 years ago was 156250.

Question 8. The population of a town increases at the rate of 50 per thousand. Its population after 2 years will be 22050. Find its present population.

Solution:

We have,

Growth rate of population of town is = 50/1000×100 = 5%

Population after 2 years is = 22050

So, let present population of town be = x

By using the formula,

A = P (1 + \frac{R}{100}  )ⁿ

Substituting the values, we have 

22050 = x (1 + \frac{5}{100}  ) (1 + \frac{5}{100}  )

22050 = x (\frac{105}{100}  ) (\frac{105}{100}  )

22050 = x (1.05) (1.05)

22050 = 1.1025x

x = 22050/1.1025

= 20000

Therefore,

Present population of the town is 20000.

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

Solution:

Given details are,

Count of bacteria in sample is = 13125000

The increase and decrease of growth rates are = 10%, -8%, 12%

So, let the count of bacteria after 3 hours be = x

By using the formula,

A = P (1 + \frac{R}{100}  )ⁿ

Substituting the values, we have 

x = 13125000 (1 + \frac{10}{100}  ) (1 – \frac{8}{100}  ) (1 + \frac{12}{100}  )

x = 13125000 (110/100) (92/100) (112/100)

x = 13125000 (1.1) (0.92) (1.12)

= 14876400

Therefore,

Count of bacteria after three hours will be 14876400.

Question 10. The population of a certain city was 72000 on the last day of the year 1998. During next year it increased by 7% but due to an epidemic it decreased by 10% in the following year. What was its population at the end of the year 2000?

Solution:

We have,

Population of city on last day of year 1998 = 72000

Increasing rate (R) in 1999 = 7%

Decreasing rate (R) in 2000 = 10 %

By using the formula,

A = P (1 + \frac{R}{100}  )ⁿ

Substituting the values, we have 

x = 72000 (1 + \frac{7}{100}  ) (1 – \frac{10}{100}  )

= 72000 (107/100) (90/100)

= 72000 (1.07) (0.9)

= 69336

Therefore,

Population at the end of the year 2000 will be 69336.

Read More:

• Compound Interest Formula
• Compound Interest – Aptitude Questions and Answers