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Page 1
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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,Compound Interest - EXERCISE 14.4 | Mathematics (Maths) Class 8
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,Compound Interest - EXERCISE 14.4 | Mathematics (Maths) Class 8
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Additional Information about Compound Interest - EXERCISE 14.4 for Class 8 Preparation
Compound Interest - EXERCISE 14.4 Free PDF Download
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