Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

Class 8 NCERT Solutions- Chapter 8 Comparing Quantities – Exercise 8.3

  • Difficulty Level : Medium
  • Last Updated : 04 Mar, 2021

Question 1. Calculate the amount and compound interest on

(i) Rs 10,800 for 3 years at 12\mathbf{\frac{1}{2}} % per annum compounded annually.

(ii) Rs 18,000 for 2\mathbf{\frac{1}{2}} years at 10% per annum compounded annually.

Hey! Looking for some great resources suitable for young ones? You've come to the right place. Check out our self-paced courses designed for students of grades I-XII

Start with topics like Python, HTML, ML, and learn to make some games and apps all with the help of our expertly designed content! So students worry no more, because GeeksforGeeks School is now here!

 



(iii) Rs 62,500 for 1\mathbf{\frac{1}{2}} years at 8% per annum compounded half-yearly.

(iv) Rs 8,000 for 1 year at 9% per annum compounded half-yearly.

(v) Rs 10,000 for 1 year at 8% per annum compounded half-yearly. 

Solution: 

(i) Given values are,

P = Rs 10,800

R = 12 \frac{1}{2} % per annum = \frac{25}{2} %

T = 3 Years



As it is compounded annually then, n = 3 times.

We have,

A = P (1 + \mathbf{\frac{R}{100}} )n

A = 10,800 (1+ \frac{25}{2*100} )3 

A = 10,800 (1+ \frac{1}{8} )3

A = 10,800 (\frac{9}{8} )3

A = Rs 15,377.34

CI = A – P

CI = 15,377.34 – 10,800

CI = Rs 4,577.34



Hence, the amount = Rs 15,377.34 and 

Compound interest = Rs 4,577.34

(ii) Given values are,

P = Rs 18,000

R = 10 % per annum 

T = 2\frac{1}{2}  Years

As it is compounded annually then, n = 2\frac{1}{2}  times.

We have,

A = P (1 + \mathbf{\frac{R}{100}} )n

A = 18,000 (1+ \frac{10}{100} )



What we will do here is Firstly we know 2\frac{1}{2}  Years is 2 years and 6 months which can be calculated by first calculating the amount to 2 years using CI formula and then calculating the simple interest by using SI formula.

The amount for 2 years has to be calculated : 

A = 18,000 (1+ \frac{1}{10} )2

A = 18,000 (\frac{11}{10} )2

A = Rs 21,780

CI = A – P

CI = 21,780 – 18,000

CI = Rs 3,780

Now, The amount for \frac{1}{2}  year has to be calculated: 

New P is equal to the amount after 2 Years. Hence, 



P = Rs 21,780

R = 10 % per annum 

T = \frac{1}{2}  year

SI = \frac{PRT}{100}

SI = \frac{21,780 × 10 × \frac{1}{2}}{100}

SI = \frac{21,780 × 10 × 1}{200}

SI = Rs 1,089

Hence, the Total amount = A + SI

                                           = 21,780 + 1,809

                                           = Rs 22,869  

Total compound interest = CI + SI

                                          = 3,780 + 1,809

                                          = Rs 4,869   

(iii) Given values are,

P = Rs 62,500

R = 8 % per annum hence 4% Half Yearly

T = 1\frac{1}{2}  Years

As it is compounded Half yearly then, n = 3 times. (1\frac{1}{2}  Years contains 3 half years)

We have,

A = P (1 + \mathbf{\frac{R}{100}} )n



A = 62,500 (1+ \frac{4}{100} )3

A = 62,500 (1+ \frac{1}{25} )3

A = 62,500 (\frac{26}{25} )3

A = Rs 70,304

CI = A – P

CI = 70,304 – 62,500

CI = Rs 7,804

Hence, the amount = Rs 70,304 and

Compound interest = Rs 7,804

(iv) Given values are,

P = Rs 8,000

R = 9 % per annum hence, \frac{9}{2}  % Half Yearly

T = 1 Year

As it is compounded Half yearly then, n = 2 times. (1 Year contains 2 half years)

We have,

A = P (1 + \mathbf{\frac{R}{100}} )n

A = 8,000 (1+ \frac{9}{2*100} )2

A = 8,000 (1+ \frac{9}{200} )2

A = 8,000 (\frac{209}{200} )2

A = Rs 8,736.20



CI = A – P

CI = 8,736.20 – 8,000

CI = Rs 736.20

Hence, the amount = Rs 8,736.20 and

Compound interest = Rs 736.20

(v) Given values are,

P = Rs 10,000

R = 8 % per annum  hence, 4% Half Yearly

T = 1 Year

As it is compounded Half yearly then, n = 2 times. (1 Year contains 2 half years)

We have,

A = P (1+  \mathbf{\frac{R}{100}} )n

A = 10,000 (1+ (\frac{4}{100} ))2

A =10,000 (1+ (\frac{1}{25} ))2

A = 10,000 (\frac{26}{25} )2

A = Rs 10,816

CI = A – P

CI = 10,816- 10,000

CI = Rs 816

Hence, the amount = Rs 10,816 and

Compound interest = Rs 816

Question 2. Kamala borrowed Rs 26,400 from a Bank to buy a scooter at a rate of 15% p.a. compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan?

Solution:

Here, Given values are,

P = Rs 26,400

R = 15 % per annum 

T = 2 Years and 4 months, which is 2 \frac{1}{3}  years

As it is compounded annually then, n = 2\frac{1}{3}  times

We have,

A = P (1 + \mathbf{\frac{R}{100}} )n

A = 26,400 (1 + (\frac{15}{100} )2(1/3)



What we will do here is Firstly 2 years and 4 months which can be calculated by first calculating the amount to 2 years using CI formula and then calculating the simple interest by using SI formula.

The amount for 2 years has to be calculated:

A = 26,400 (1+ (\frac{15}{100} )2

A = 26,400 (1+ (\frac{3}{20} )2

 A = 26,400 (\frac{23}{20} )2

A = Rs 34,914

Now, The amount for (1/3) year (4 months) has to be calculated :

New P is equal to the amount after 2 Years. Hence,

P = Rs 34,914

R = 15 % per annum 

T = \frac{1}{3}  year

SI = \frac{P × R × T}{100}

SI = \frac{(34,914 × 15 × (1/3)}{100}

SI = \frac{(34,914 × 15 × 1)}{300}

SI = 1,745.70

Hence, the Total amount = A + SI

                                       = 34,914 + 1,745.70

                                       = Rs 36,659.70

Hence, the amount to be paid by Kamla = ₹ 36,659.70

Question 3. Fabina borrows Rs 12,500 at 12% per annum for 3 years at simple interest and Radha borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much?

Solution:



Let’s see each case 

Fabina Case: at simple interest

P = 12,500

R = 12% per annum  

T = 3 Years

SI = \frac{(P × R × T)}{100}

SI = \frac{(12,500 × 12 × 3)}{100}

SI = Rs 4,500

Radha Case: at compound interest

P = 12,500

R = 10% per annum  

T = 3 Years

As it is compounded annually then, n = 3 times

We have,

A = P (1 + \mathbf{\frac{R}{100}} )n

A = 12,500 (1 + (\frac{10}{100} ))3

A =12,500 (1 + \frac{1}{10} )3

A = 12,500 (\frac{11}{10} )3

A = Rs 16,637.5

CI = A – P

CI = 16,637.5 – 12,500

CI = 4,137.5

Clearly we can see that Fabina paid more interest, and she paid

4,500 – 4,137.5 = Rs 362.5 more than Radha

Question 4. I borrowed Rs 12,000 from Jamshed at 6% per annum simple interest for 2 years. Had I borrowed this sum at 6% per annum compound interest, what extra amount would I have to pay? 

Solution:

Lets see each case First

At simple interest

P = 12,000

R = 6% per annum  

T = 2 Years

SI = \frac{(P × R × T)}{100}

SI = \frac{(12,000 × 6 × 2)}{100}

SI = Rs 1,440

At compound interest

P = 12,000

R = 6% per annum  

T = 2 Years

As it is compounded annually then, n = 2 times

We have,

A = P (1 + \mathbf{\frac{R}{100}} )n



A = 12,000 (1+ (\frac{6}{100} ))2

A =12,000 (1+ (\frac{3}{50} ))2

A = 12,000 (\frac{53}{50} )2

A = Rs 13,483.2 

CI = A – P

CI = 13,483.2 – 12,000

CI = 1,483.2

Clearly we can see that,

1,483.2 – 1,440 = Rs 43.2

Hence, the extra amount to be paid = ₹ 43.20

Question 5. Vasudevan invested Rs 60,000 at an interest rate of 12% per annum compounded half-yearly. What amount would he get

(a) after 6 months?

(b) after 1 year?

Solution:

Let’s see each case 

(a) 

P = 60,000

R = 12% per annum (6% Half yearly)

T = 6 Months

As it is compounded Half Yearly then, n = 1 times (as 6 months is 1 half year)

We have,

A = P (1 + \mathbf{\frac{R}{100}} )n

A =60,000 (1+ (\frac{6}{100} ))1

A =60,000 (1+ (\frac{3}{50} ))1

A = 60,000 (\frac{53}{25} )1

A = Rs 63,600

He would get Rs 63,600 after 6 Months.

(b)

P = 60,000

R = 12% per annum (6% Half yearly)

T = 1 Year

As it is compounded Half Yearly then, n = 2 times (as 1 Year is 2 half year)

We have,

A = P (1 + \mathbf{\frac{R}{100}} )n

A = 60,000 (1+ (\frac{6}{100} ))2

A = 60,000 (1+ (\frac{3}{50} ))2

A = 60000 (\frac{53}{25} )2

A = Rs 67,416

He would get Rs 67,416 after 1 Year.

Question 6. Arif took a loan of Rs 80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he would be paying after 1\frac{1}{2}  years if the interest is

(a) compounded annually.

(b) compounded half-yearly. 



Solution:

Let’s see each case 

(a) Compounded Annually

P = 80,000

R = 10% per annum

T = 1\frac{1}{2}  Year

As it is compounded annually then, n = 1 \frac{1}{2}  times

We have,

A = P (1 + \mathbf{\frac{R}{100}} )n

A = 80,000 (1 + (\frac{10}{100} )

What we will do here is Firstly we know 1\frac{1}{2}  Years is 1 year and 6 months which can be calculated by first calculating the amount to 1 year using CI formula and then calculating the simple interest by using SI formula.

The amount for 1 years has to be calculated :

A = 80,000 (1+ (\frac{10}{100} ))1

A = 80,000 (1+ (\frac{1}{10} )1

A = 80,000 (\frac{11}{10} )1

A = Rs 88,000

Now, The amount for \frac{1}{2}  Year (6 months) has to be calculated :

New P is equal to the amount after 1 Year. Hence,

P = Rs 88,000

R = 10 % per annum

T =\frac{1}{2}  Year

SI = \frac{(P × R × T)}{ 100}

SI = \frac{(88,000 × 10 × \frac{1}{2})}{100}

SI = \frac{(88,000 × 10 × 1)}{200}

SI = 4,400

Hence, the Total amount = A + SI

                                        = 88,000 + 4,400

                                        = Rs 92,400

(b) Compounded Half-yearly

P = 80,000



R = 10% per annum (5 % Half Yearly)

T = 1\frac{1}{2}  Year

As it is compounded annually then, n = 3 times (as 1\frac{1}{2}  Year is 3 half year)

We have,

A = P (1 + \mathbf{\frac{R}{100}} )n

A = 80,000 (1+ (\frac{5}{100} )3

A = 80,000 (1+ (\frac{1}{20} )3

A = 80,000 (\frac{21}{20} )3

A = Rs 92,610

Hence, the Total amount = Rs 92,610

Question 7. Maria invested Rs 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find

(a) The amount credited against her name at the end of the second year.

(b) The interest for the 3rd year.

Solution:

Let’s see each case 

Here, 

P = 8,000

R = 5% Per annum

(a) The amount credited against Maria’s name at the end of the second year.

T = 2 Year

As it is compounded annually then, n = 2 times

We have,

A = P (1 + \mathbf{\frac{R}{100}} )n

A = 8,000 (1+ (\frac{5}{100} ))2

A = 8,000 (1+ (\frac{1}{20} ))2

A = 8,000 (\frac{21}{20} )2

A = Rs 8,820

Hence, the amount credited against Maria’s name at the end of the second year = Rs 8,820

(b) The interest for the 3rd year.

T = 3 Year

As it is compounded annually then, n = 3 times

We have,

A = P (1+ \mathbf{\frac{R}{100}} )n

A = 8,000 (1+ (\frac{5}{100} ))3

A = 8,000 (1+ (\frac{1}{20} ))3

A = 8,000 (\frac{21}{20} )3

A = Rs 9,261

The interest for the 3rd year = Amount after 3 years – Amount after 2 Years

                                                 = 9,261 – 8,820

                                                 = Rs 441

Another Solution for (b)



As we can calculate interest of 3rd year by having 2nd Year Amount as P.

P = 8,820

R = 5% per annum

T = 1 Year (2nd to 3rd year)

SI = \frac{(P × R × T)}{100}

SI = \frac{(8,820 × 5 × 1)}{100}

SI = Rs 441

The interest for the 3rd year = Rs 441

Question 8. Find the amount and the compound interest on Rs 10,000 for 1\frac{1}{2}  years at 10% per annum, compounded half-yearly. Would this interest be more than the interest he would get if it was compounded annually?

Solution:

Let’s see each cases

Compounded Annually

P = 10,000

R = 10% per annum

T = 1\frac{1}{2}  Year

As it is compounded annually then, n = 1 \frac{1}{2}  times

We have,

A = P (1 + \mathbf{\frac{R}{100}} )n

A = 10,000 (1 + (\frac{10}{100} )

What we will do here is Firstly we know 1½ Years is 1 year and 6 months which can be calculated by first calculating the amount to 1 year using CI formula and then calculating the simple interest by using SI formula.

The amount for 1 year has to be calculated:

A = 10,000 (1 + \frac{10}{100} )1

A = 10,000 (1+ \frac{1}{10} )1

A = 10,000 (\frac{11}{10} )1

A = Rs 11,000

CI = A – P

CI = 11,000-10,000

CI = 1,000

Now, The amount for \frac{1}{2}  Year (6 months) has to be calculated :

New P is equal to the amount after 1 Year. Hence,

P = Rs 11,000

R = 10 % per annum

T =\frac{1}{2}  Year

SI = \frac{(P × R × T)}{100}

SI = \frac{(11,000 × 10 × \frac{1}{2})}{100}

SI = \frac{11,000 × 10}{200}

SI = 550

Hence, the Total Interest (compounded annually)= CI + SI

                                       = 1,000 + 550

                                       = Rs 1,550

Compounded Half-yearly



P = 10,000

R = 10% per annum (5 % Half Yearly)

T = 1\frac{1}{2}  Year

As it is compounded annually then, n = 3 times (as 1\frac{1}{2}  Year is 3 half year)

We have,

A = P (1 + \mathbf{\frac{R}{100}} )n

A = 10,000 (1 + (\frac{5}{100} )3

A = 10,000 (1+ \frac{1}{20} )3

A = 10,000 (\frac{21}{20} )3

A = Rs 11,576.25

CI = A – P

CI = 11,576.25 – 10,000

CI = 1,576.25

Hence, the Total Interest (compounded Half Yearly) = Rs 11576.25

Difference between the two interests = 1,576.25 – 1,550 = Rs 26.25

Hence, the interest will be Rs 26.25 more when compounded half-yearly than the interest when compounded annually.

Question 9. Find the amount which Ram will get on Rs 4096, if he gave it for 18 months at 12 \frac{1}{2} % per annum, interest being compounded half-yearly

Solution:

Let’s see this case

P = Rs 4,096

R = 12 \frac{1}{2}  % per annum (\frac{25}{4}  % Half yearly)

T = 18 Months = 1\frac{1}{2}  Year

As it is compounded Half yearly then, n = 3 Times

We have,

A = P (1 + \mathbf{\frac{R}{100}} )n

A = 4,096 (1+ (\frac{\frac{25}{4}}{100} )3

A = 4,096 (1+ \frac{25}{400} )3

A = 4,096 (1+ (\frac{1}{16} )3

A = 4,096 (\frac{17}{16} )3

A = Rs 4,913

Ram will get the amount = Rs 4,913



Question 10. The population of a place increased to 54,000 in 2003 at a rate of 5% per annum

(a) find the population in 2001.

(b) what would be its population in 2005?

Solution:

Here,

P = 54,000 (in 2003)

R = 5% per annum

(a) Population in 2001

T = 2 Years (back)

n = 2

Population in 2003 = Population in 2001 (1 + \mathbf{\frac{R}{100}} )n

54,000 = P1 (1+(\frac{5}{100} ))2

54,000 = P1 (\frac{21}{20} )2

54,000 = P1 (\frac{441}{400} )

P1 = 54,000 (\frac{441}{400} )

P1 = 48,979.59

P1 = 48,980 (approx.).

Population in 2001 was 48,980 (approx.).

(b) Population in 2005

T = 2 Years

n = 2

We have,

A = P (1 + \mathbf{\frac{R}{100}} )n

A = 54,000 (1+ \frac{5}{100} )2

A = 54,000 (1+ (\frac{1}{20} )2

A = 54,000 (\frac{21}{20} )2

A = 59,535

Population in 2005 will be 59,535

Question 11. In a Laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5, 06,000.

Solution:

Here,

P = 5,06,000

R = 2.5% per hour

T = 2 hours

We have,

A = P (1 + \mathbf{\frac{R}{100}} )n

A = 5,06,000 (1+ \frac{2.5}{100} )2

A = 5,06,000 (1+ \frac{25}{1000} )2

A = 5,06,000 (1+ \frac{1}{40} )2

A = 5,06,000 (\frac{41}{40} )2

A = 5,31,616.25

A = 5,31,616 (approx.)

Bacteria at the end of 2 hours = 5,31,616 (approx.)

Question 12. A scooter was bought at Rs 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.

Solution:

Here,

P = 42,000

R = 8% per annum (depreciated)

T = 1 Year

We have,

A = P (1 + \mathbf{\frac{R}{100}} )n

A = 42,000 (1- \frac{8}{100} )1 (negative sign because the price is reduced)

A = 42,000 (1- (\frac{2}{25} )1

A = 42,000 (\frac{23}{25} )1

A = Rs 38,640

The value of scooter after one year will be = Rs 38,640




My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!