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RD Sharma Solutions for Class 8 Maths Chapter 14 – 
Compound Interest 
EXERCISE 14.4                                                  PAGE NO: 14.27 
 
1. The present population of a town is 28000. If it increases at the rate of 5% per 
annum, what will be its population after 2 years? 
Solution: 
Given details are, 
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 + R/100)
n
 
Population of town after 2 years = 28000 (1 + 5/100)
2
 
                                                    = 28000 (1.05)
2
 
                                                    = 30870 
 ? Population of town after 2 years will be 30870 
 
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: 
Given details are, 
Population of city (P) = 125000 
Annual birth rate R1= 5.5 % 
Annual death rate R
2
 = 3.5 % 
Annual increasing rate R = (R
1
 – R
2
) = 5.5 – 3.5 = 2 % 
Number of years = 3 
By using the formula, 
A = P (1 + R/100)
n
 
So, population after two years is = 125000 (1 + 2/100)
3
 
                                                     = 125000 (1.02)
3
 
                                                     = 132651 
? Population after 3 years will be 132651 
 
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: 
Given details are, 
Present population is = 25000 
First year growth R
1
 = 4% 
Second year growth R
2
 = 5% 
Page 2


 
 
 
 
RD Sharma Solutions for Class 8 Maths Chapter 14 – 
Compound Interest 
EXERCISE 14.4                                                  PAGE NO: 14.27 
 
1. The present population of a town is 28000. If it increases at the rate of 5% per 
annum, what will be its population after 2 years? 
Solution: 
Given details are, 
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 + R/100)
n
 
Population of town after 2 years = 28000 (1 + 5/100)
2
 
                                                    = 28000 (1.05)
2
 
                                                    = 30870 
 ? Population of town after 2 years will be 30870 
 
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: 
Given details are, 
Population of city (P) = 125000 
Annual birth rate R1= 5.5 % 
Annual death rate R
2
 = 3.5 % 
Annual increasing rate R = (R
1
 – R
2
) = 5.5 – 3.5 = 2 % 
Number of years = 3 
By using the formula, 
A = P (1 + R/100)
n
 
So, population after two years is = 125000 (1 + 2/100)
3
 
                                                     = 125000 (1.02)
3
 
                                                     = 132651 
? Population after 3 years will be 132651 
 
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: 
Given details are, 
Present population is = 25000 
First year growth R
1
 = 4% 
Second year growth R
2
 = 5% 
 
 
 
 
RD Sharma Solutions for Class 8 Maths Chapter 14 – 
Compound Interest 
Third year growth R
3
 = 8% 
Number of years = 3 
By using the formula, 
A = P (1 + R/100)
n
 
So, population after three years = P (1 + R
1
/100) (1 + R
2
/100) (1 + R
3
/100) 
                                                   = 25000(1 + 4/100) (1 + 5/100) (1 + 8/100) 
                                                   = 25000 (1.04) (1.05) (1.08)  
                                                   = 29484 
? Population after 3 years will be 29484 
 
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: 
Given details are, 
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 + R/100)
n
 
x = 50000 (1 + 4/100) (1 + 5/100) (1 + 3/100) 
  = 50000 (1.04) (1.05) (1.03) 
  = 56238 
? Present population of the town is 56238 
 
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: 
Given details are, 
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 + R/100)
n
 
9261 = x (1 + 5/100) (1 + 5/100) (1 + 5/100) 
9261 = x (1.05) (1.05) (1.05) 
  = 8000 
? Present population of the town is 8000 
 
Page 3


 
 
 
 
RD Sharma Solutions for Class 8 Maths Chapter 14 – 
Compound Interest 
EXERCISE 14.4                                                  PAGE NO: 14.27 
 
1. The present population of a town is 28000. If it increases at the rate of 5% per 
annum, what will be its population after 2 years? 
Solution: 
Given details are, 
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 + R/100)
n
 
Population of town after 2 years = 28000 (1 + 5/100)
2
 
                                                    = 28000 (1.05)
2
 
                                                    = 30870 
 ? Population of town after 2 years will be 30870 
 
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: 
Given details are, 
Population of city (P) = 125000 
Annual birth rate R1= 5.5 % 
Annual death rate R
2
 = 3.5 % 
Annual increasing rate R = (R
1
 – R
2
) = 5.5 – 3.5 = 2 % 
Number of years = 3 
By using the formula, 
A = P (1 + R/100)
n
 
So, population after two years is = 125000 (1 + 2/100)
3
 
                                                     = 125000 (1.02)
3
 
                                                     = 132651 
? Population after 3 years will be 132651 
 
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: 
Given details are, 
Present population is = 25000 
First year growth R
1
 = 4% 
Second year growth R
2
 = 5% 
 
 
 
 
RD Sharma Solutions for Class 8 Maths Chapter 14 – 
Compound Interest 
Third year growth R
3
 = 8% 
Number of years = 3 
By using the formula, 
A = P (1 + R/100)
n
 
So, population after three years = P (1 + R
1
/100) (1 + R
2
/100) (1 + R
3
/100) 
                                                   = 25000(1 + 4/100) (1 + 5/100) (1 + 8/100) 
                                                   = 25000 (1.04) (1.05) (1.08)  
                                                   = 29484 
? Population after 3 years will be 29484 
 
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: 
Given details are, 
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 + R/100)
n
 
x = 50000 (1 + 4/100) (1 + 5/100) (1 + 3/100) 
  = 50000 (1.04) (1.05) (1.03) 
  = 56238 
? Present population of the town is 56238 
 
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: 
Given details are, 
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 + R/100)
n
 
9261 = x (1 + 5/100) (1 + 5/100) (1 + 5/100) 
9261 = x (1.05) (1.05) (1.05) 
  = 8000 
? Present population of the town is 8000 
 
 
 
 
 
RD Sharma Solutions for Class 8 Maths Chapter 14 – 
Compound Interest 
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: 
Given details are, 
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 + R/100)
n
 
46305 = 40000 (1 + R/100) (1 + R/100) (1 + R/100) 
46305 = 40000 (1 + R/100)
3
 
(1 + R/100)
3
 = 46305/40000 
                     = 9261/8000 
                     = (21/20)
3
 
1 + R/100 = 21/20 
R/100 = 21/20 – 1 
R/100 = (21-20)/20 
          = 1/20 
R = 100/20 
    = 5 
? Annual rate of growth of the production of scooters is 5% 
 
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: 
Given details are, 
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 + R/100) 
196830 = x (1 + 8/100) (1 + 8/100) (1 + 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 
? Population 3 years ago was 156250 
Page 4


 
 
 
 
RD Sharma Solutions for Class 8 Maths Chapter 14 – 
Compound Interest 
EXERCISE 14.4                                                  PAGE NO: 14.27 
 
1. The present population of a town is 28000. If it increases at the rate of 5% per 
annum, what will be its population after 2 years? 
Solution: 
Given details are, 
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 + R/100)
n
 
Population of town after 2 years = 28000 (1 + 5/100)
2
 
                                                    = 28000 (1.05)
2
 
                                                    = 30870 
 ? Population of town after 2 years will be 30870 
 
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: 
Given details are, 
Population of city (P) = 125000 
Annual birth rate R1= 5.5 % 
Annual death rate R
2
 = 3.5 % 
Annual increasing rate R = (R
1
 – R
2
) = 5.5 – 3.5 = 2 % 
Number of years = 3 
By using the formula, 
A = P (1 + R/100)
n
 
So, population after two years is = 125000 (1 + 2/100)
3
 
                                                     = 125000 (1.02)
3
 
                                                     = 132651 
? Population after 3 years will be 132651 
 
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: 
Given details are, 
Present population is = 25000 
First year growth R
1
 = 4% 
Second year growth R
2
 = 5% 
 
 
 
 
RD Sharma Solutions for Class 8 Maths Chapter 14 – 
Compound Interest 
Third year growth R
3
 = 8% 
Number of years = 3 
By using the formula, 
A = P (1 + R/100)
n
 
So, population after three years = P (1 + R
1
/100) (1 + R
2
/100) (1 + R
3
/100) 
                                                   = 25000(1 + 4/100) (1 + 5/100) (1 + 8/100) 
                                                   = 25000 (1.04) (1.05) (1.08)  
                                                   = 29484 
? Population after 3 years will be 29484 
 
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: 
Given details are, 
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 + R/100)
n
 
x = 50000 (1 + 4/100) (1 + 5/100) (1 + 3/100) 
  = 50000 (1.04) (1.05) (1.03) 
  = 56238 
? Present population of the town is 56238 
 
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: 
Given details are, 
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 + R/100)
n
 
9261 = x (1 + 5/100) (1 + 5/100) (1 + 5/100) 
9261 = x (1.05) (1.05) (1.05) 
  = 8000 
? Present population of the town is 8000 
 
 
 
 
 
RD Sharma Solutions for Class 8 Maths Chapter 14 – 
Compound Interest 
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: 
Given details are, 
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 + R/100)
n
 
46305 = 40000 (1 + R/100) (1 + R/100) (1 + R/100) 
46305 = 40000 (1 + R/100)
3
 
(1 + R/100)
3
 = 46305/40000 
                     = 9261/8000 
                     = (21/20)
3
 
1 + R/100 = 21/20 
R/100 = 21/20 – 1 
R/100 = (21-20)/20 
          = 1/20 
R = 100/20 
    = 5 
? Annual rate of growth of the production of scooters is 5% 
 
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: 
Given details are, 
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 + R/100) 
196830 = x (1 + 8/100) (1 + 8/100) (1 + 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 
? Population 3 years ago was 156250 
 
 
 
 
RD Sharma Solutions for Class 8 Maths Chapter 14 – 
Compound Interest 
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: 
Given details are, 
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 + R/100) 
22050 = x (1 + 5/100) (1 + 5/100) 
22050 = x (105/100) (105/100) 
22050 = x (1.05) (1.05) 
22050 = 1.1025x 
x = 22050/1.1025 
   = 20000 
? Present population of the town is 20000 
 
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 + R/100) 
x = 13125000 (1 + 10/100) (1 – 8/100) (1 + 12/100) 
x = 13125000 (110/100) (92/100) (112/100) 
x = 13125000 (1.1) (0.92) (1.12)  
   = 14876400 
? Count of bacteria after three hours will be 14876400 
 
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: 
Given details are, 
Population of city on last day of year 1998 = 72000 
Page 5


 
 
 
 
RD Sharma Solutions for Class 8 Maths Chapter 14 – 
Compound Interest 
EXERCISE 14.4                                                  PAGE NO: 14.27 
 
1. The present population of a town is 28000. If it increases at the rate of 5% per 
annum, what will be its population after 2 years? 
Solution: 
Given details are, 
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 + R/100)
n
 
Population of town after 2 years = 28000 (1 + 5/100)
2
 
                                                    = 28000 (1.05)
2
 
                                                    = 30870 
 ? Population of town after 2 years will be 30870 
 
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: 
Given details are, 
Population of city (P) = 125000 
Annual birth rate R1= 5.5 % 
Annual death rate R
2
 = 3.5 % 
Annual increasing rate R = (R
1
 – R
2
) = 5.5 – 3.5 = 2 % 
Number of years = 3 
By using the formula, 
A = P (1 + R/100)
n
 
So, population after two years is = 125000 (1 + 2/100)
3
 
                                                     = 125000 (1.02)
3
 
                                                     = 132651 
? Population after 3 years will be 132651 
 
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: 
Given details are, 
Present population is = 25000 
First year growth R
1
 = 4% 
Second year growth R
2
 = 5% 
 
 
 
 
RD Sharma Solutions for Class 8 Maths Chapter 14 – 
Compound Interest 
Third year growth R
3
 = 8% 
Number of years = 3 
By using the formula, 
A = P (1 + R/100)
n
 
So, population after three years = P (1 + R
1
/100) (1 + R
2
/100) (1 + R
3
/100) 
                                                   = 25000(1 + 4/100) (1 + 5/100) (1 + 8/100) 
                                                   = 25000 (1.04) (1.05) (1.08)  
                                                   = 29484 
? Population after 3 years will be 29484 
 
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: 
Given details are, 
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 + R/100)
n
 
x = 50000 (1 + 4/100) (1 + 5/100) (1 + 3/100) 
  = 50000 (1.04) (1.05) (1.03) 
  = 56238 
? Present population of the town is 56238 
 
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: 
Given details are, 
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 + R/100)
n
 
9261 = x (1 + 5/100) (1 + 5/100) (1 + 5/100) 
9261 = x (1.05) (1.05) (1.05) 
  = 8000 
? Present population of the town is 8000 
 
 
 
 
 
RD Sharma Solutions for Class 8 Maths Chapter 14 – 
Compound Interest 
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: 
Given details are, 
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 + R/100)
n
 
46305 = 40000 (1 + R/100) (1 + R/100) (1 + R/100) 
46305 = 40000 (1 + R/100)
3
 
(1 + R/100)
3
 = 46305/40000 
                     = 9261/8000 
                     = (21/20)
3
 
1 + R/100 = 21/20 
R/100 = 21/20 – 1 
R/100 = (21-20)/20 
          = 1/20 
R = 100/20 
    = 5 
? Annual rate of growth of the production of scooters is 5% 
 
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: 
Given details are, 
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 + R/100) 
196830 = x (1 + 8/100) (1 + 8/100) (1 + 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 
? Population 3 years ago was 156250 
 
 
 
 
RD Sharma Solutions for Class 8 Maths Chapter 14 – 
Compound Interest 
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: 
Given details are, 
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 + R/100) 
22050 = x (1 + 5/100) (1 + 5/100) 
22050 = x (105/100) (105/100) 
22050 = x (1.05) (1.05) 
22050 = 1.1025x 
x = 22050/1.1025 
   = 20000 
? Present population of the town is 20000 
 
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 + R/100) 
x = 13125000 (1 + 10/100) (1 – 8/100) (1 + 12/100) 
x = 13125000 (110/100) (92/100) (112/100) 
x = 13125000 (1.1) (0.92) (1.12)  
   = 14876400 
? Count of bacteria after three hours will be 14876400 
 
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: 
Given details are, 
Population of city on last day of year 1998 = 72000 
 
 
 
 
RD Sharma Solutions for Class 8 Maths Chapter 14 – 
Compound Interest 
Increasing rate (R) in 1999 = 7% 
Decreasing rate (R) in 2000 = 10 % 
By using the formula, 
A = P (1 + R/100) 
x = 72000 (1 + 7/100) (1 – 10/100) 
   = 72000 (107/100) (90/100) 
   = 72000 (1.07) (0.9) 
   = 69336 
? Population at the end of the year 2000 will be 69336 
 
11. 6400 workers were employed to construct a river bridge in four years. At the 
end of the first year, 25% workers were retrenched. At the end of the second year, 
25% of those working at that time were retrenched. However, to complete the 
project in time, the number of workers was increased by 25% at the end of the third 
year. How many workers were working during the fourth year? 
Solution: 
Given details are, 
Initial number of workers are = 6400 
At the end of first year = 25% retrenched 
At the end of second year = 25% retrenched 
At the end of third year = 25% increased 
By using the formula, 
A = P (1 + R/100) 
x = 6400 (1 – 25/100) (1 – 25/100) (1 + 25/100) 
   = 6400 (75/100) (75/100) (125/100) 
   = 6400 (0.75) (0.75) (1.25) 
   = 4500 
? Number of workers working during the fourth year is 4500 
 
12. Aman started a factory with an initial investment of Rs 100000. In the first year, 
he incurred a loss of 5%. However, during the second 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: 
Given details are, 
Initial investment by Aman = Rs.100000 
In first year = incurred a loss of 5% 
In second year = earned a profit of 10% 
In third year = earned a profit of 12 %

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Document Description: Compound Interest - EXERCISE 14.4 for Class 8 2024 is part of Mathematics (Maths) Class 8 preparation. The notes and questions for Compound Interest - EXERCISE 14.4 have been prepared according to the Class 8 exam syllabus. Information about Compound Interest - EXERCISE 14.4 covers topics like and Compound Interest - EXERCISE 14.4 Example, for Class 8 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Compound Interest - EXERCISE 14.4.

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Full syllabus notes, lecture & questions for Compound Interest - EXERCISE 14.4 | Mathematics (Maths) Class 8 - Class 8 | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) Class 8 | Best notes, free PDF download

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	Mathematics (Maths) Class 8 79 videos\|408 docs\|31 tests

Mathematics (Maths) Class 8

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Compound Interest - EXERCISE 14.4 | Mathematics (Maths) Class 8

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